solucionario sadiku edición 3

Chapter 1, Solution 1 (a) q = 6.482x1017

x [-1.602x10-19 C] = -0.10384 C

(b) q = 1. 24x1018

x [-1.602x10-19 C] = -0.19865 C

(c) q = 2.46x1019

x [-1.602x10-19 C] = -3.941 C

(d) q = 1.628x1020

x [-1.602x10-19 C] = -26.08 C

Chapter 1, Solution 2

(a) i = dq/dt = 3 mA (b) i = dq/dt = (16t + 4) A (c) i = dq/dt = (-3e-t + 10e-2t) nA (d) i=dq/dt = 1200 120π πcos t pA (e) i =dq/dt = − +−e tt4 80 50 1000 50( cos sin ) Aµt


(a) C 1)(3t +=+= ∫ q(0)i(t)dt q(t)

(b) mC 5t)(t 2 +=++= ∫ q(v)dt s)(2tq(t)

(c) ( )q(t) 20 cos 10t / 6 q(0) (2sin(10 / 6) 1) Ctπ π µ= + + = + +∫

(d) C 40t) sin 0.12t(0.16cos40e 30t- +−=

−+

=+= ∫ t)cos 40-t40sin30(1600900

e10q(0)t40sin10eq(t)-30t

30t-


( ) mC 4.698=−=

−=== ∫ ∫

ππ

π

06.0cos165

tπ6cos6

5dt t π 6 5sinidtq10

0


µCmC )e1(21

e21-mC dteidtq

4

2

0

2t-2t-

490=−=

=== ∫ ∫


(a) At t = 1ms, mA 40===2

80dtdqi

(b) At t = 6ms, mA 0==dtdqi

(c) At t = 10ms, mA 20-===4

80dtdqi


<<<<<<

==8t6 25A,6t2 25A,-2t0 A,25

dtdqi

which is sketched below:


C 15 µ1102

110idtq =×+×

== ∫


(a) C 10=== ∫∫1

0dt 10idtq

(b) C 22.5=−+=

×+

×

−+×== ∫251015

152

1510110idtq3

0

(c) C 30=++== ∫ 101010idt5

0q


q ixt x x x= = =−8 10 15 10 1203 6 Cµ Chapter 1, Solution 11

q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J


For 0 < t < 6s, assuming q(0) = 0,

q t idt q tdt tt t

( ) ( ) .= + = + =∫ ∫0 3 0 150

2

0

At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s,

q t idt q dt tt t

( ) ( )= + = + = −∫ ∫6 18 54 18 56 6

4

66

At t=10, q(10) = 180 – 54 = 126 For 10<t<15s,

q t idt q dt tt t

( ) ( ) ( )= + = − + = − +∫ ∫10 12 126 12 24610 10

At t=15, q(15) = -12x15 + 246 = 66 For 15<t<20s,

q t dt qt

( ) ( )= + =∫ 0 1515

Thus,

q t

ttt

( )

.

,

=−

− +

1518 5412 24666

2 C, 0 < t < 6s C, 6 < t < 10s C, 10 < t < 15s

C 15 < t < 20s

The plot of the charge is shown below.

0 5 10 15 200

20

40

60

80

100

120

140

t

q(t)


kJ 2.486-=

−=

−=

==

==

∫∫∫

216sin4112008sin

821200

1)-2x cos 2xcos (since,1)dt -t8cos2(1200

dtt 4cos1200vidtw

2

0

2

0

2

2

0

22

0

tt


(a) ( ) ( )

( ) C 2.131=−+=

+=== ∫ ∫2e2110

2et10dte-110idtq0.5-

1

0

0.5t-1

0

0.5t-

(b) p(t) = v(t)i(t) p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935) = -8.188 W Chapter 1, Solution 15

(a)

( ) C 1.297=−−=

−=== ∫ ∫

1e5.1

e23dt3eidtq

2-

2

0

2t2

0

2t-

(b) We90

)(

t4−−==

−=−==

vip

e305e6dtdi5v 2t-2t

(c) J 22.5−=−−

=== ∫ ∫3

0

4t-3

0

4t- e490dt e-90pdtw


mJ 916.7

)(

(

(((

,

=

++

+−−++=

−+

++=

−+−+++==

<<<<<<

=<<<<

=

∫

∫ ∫∫∫∫

4

3

22

4

3

2

3

2

22

1

1

0

3

4

3

3

2

2

1

1

0

3t4t-t1625028

29122503

2250

3250

dtt)42502t-t4250

2250t

3250

25t)mJ-t)(1001040t)dt2510010t)dt2510(25t)dt10v(t)i(t)dtw

4t3 V10t -403t1 V 101t0 V10t

v(t)4t2 mA25t -1002t0 mA25t

i(t)

Chapter 1, Solution 17 Σ p = 0 → -205 + 60 + 45 + 30 + p3 = 0 p3 = 205 – 135 = 70 W Thus element 3 receives 70 W. Chapter 1, Solution 18

p1 = 30(-10) = -300 W p2 = 10(10) = 100 W p3 = 20(14) = 280 W p4 = 8(-4) = -32 W p5 = 12(-4) = -48 W


p I x x xs s= → − − − + = → =∑ 0 4 2 6 13 2 5 10 0 3 AI


Since Σ p = 0 -30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 V0 = 18 V


nA8.

(.

C/s100.8 C/s10611084

electron) / C1061photon

electron81

secphoton104

8-1911

1911

=×=×××=

×⋅

⋅

×=

∆∆

=

−

tqi


It should be noted that these are only typical answers.

(a) Light bulb 60 W, 100 W (b) Radio set 4 W (c) TV set 110 W (d) Refrigerator 700 W (e) PC 120 W (f) PC printer 18 W (g) Microwave oven 1000 W (h) Blender 350 W


(a) W12.5===1201500

vpi

(b) kWh 1.125. =×=⋅×××== kWh60451.5J60451051 3ptw

(c) Cost = 1.125 × 10 = 11.25 cents


p = vi = 110 x 8 = 880 W Chapter 1, Solution 25

cents 21.6 cents/kWh 930hr 64 kW 1.2 Cost =×××=


(a) mA 80.=

⋅=

10hhA80i

(b) p = vi = 6 × 0.08 = 0.48 W (c) w = pt = 0.48 × 10 Wh = 0.0048 kWh


∫ ∫ =××====

×==

kC 43.2 36004 3 T33dt idt q

36005 4 4h T Let (a)T

0

[ ]

kJ 475.2

..

.)((

=

××+×=

+=

+===

×

∫ ∫∫

36001625036004033600

250103

dt3600

t50103vidtpdtW b)

36004

0

2

0

T

0

tt

T

cents 1.188

(

=×=

==

cent 9kWh3600475.2 Cost

Ws)(J kWs, 475.2W c)


A 0.25===12030 (a)

VPi

$31.54(

=×==××==

262.8 $0.12CostkWh 262.8Wh 2436530ptW b)


cents 39.6

.

=×==+=

+

+++==

3.3 cents 12CostkWh 3.30.92.4

hr6030kW 1.8hr

6045)1540(20kW21ptw


Energy = (52.75 – 5.23)/0.11 = 432 kWh Chapter 1, Solution 31

Total energy consumed = 365(4 +8) W Cost = $0.12 x 365 x 12 = $526.60


cents 39.6

.

=×==+=

+

+++==

3.3 cents 12CostkWh 3.30.92.4

hr6030kW 1.8hr

6045)1540(20kW21ptw


C 6=××==→= ∫ 31032000idtqdtdqi


(b) Energy = ∑ = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 pt

= 10,000 kWh (c) Average power = 10,000/24 = 416.67 W


kWh 10.4

)((

=+=

×+×××+×+×== ∫28007200

24002120012200210006400W a) dttp

W/h 433.3( =h 24kW 10.4 b)


days 6,667,(

A 4

===

=⋅

=

day / 24hh000160

0.001A160Ah tb)

40hA160i (a)


( )J 901.2.

..−=×−==

−=×−×= −

12180WC 180106021105q 1920

qv


P = 10 hp = 7460 W W = pt = 7460 × 30 × 60 J = 13.43 × 106 J


A 16.667=×

==→=120

102vpivip

3


v = iR i = v/R = (16/5) mA = 3.2 mA


p = v2/R → R = v2/p = 14400/60 = 240 ohms


R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution 4

(a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA


n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Solution 6

n = 12; l = 8; b = n + l –1 = 19 Chapter 2, Solution 7

7 elements or 7 branches and 4 nodes, as indicated. 30 V 1 20Ω 2 3 -

2A 30Ω 60Ω 40Ω 4

++++ -

10Ω


dc

b

a

9 A

i3i2 12 A

12 A

i1

8 A At node a, 8 = 12 + i1 i1 = - 4A At node c, 9 = 8 + i2 i2 = 1A At node d, 9 = 12 + i3 i3 = -3A Chapter 2, Solution 9

Applying KCL, i1 + 1 = 10 + 2 i1 = 11A 1 + i2 = 2 + 3 i2 = 4A i2 = i3 + 3 i3 = 1A Chapter 2, Solution 10 At node 1, 4 + 3 = i1 i1 = 7A At node 3, 3 + i2 = -2 i2 = -5A

3

2-2A

3A

1

4A

i2i1


Applying KVL to each loop gives -8 + v1 + 12 = 0 v1 = 4v -12 - v2 + 6 = 0 v2 = -6v 10 - 6 - v3 = 0 v3 = 4v -v4 + 8 - 10 = 0 v4 = -2v Chapter 2, Solution 12 For loop 1, -20 -25 +10 + v1 = 0 v1 = 35v For loop 2, -10 +15 -v2 = 0 v2 = 5v For loop 3, -v1 +v2 +v3 = 0 v3 = 30v

+ 15v -

loop 3

loop 2

loop 1 + 20v -

+ 10v - – 25v + + v2 -

+ v1 -

+ v3 -


I2 1 I1

2A

7A I4 2 3 4

4A

3A I3

At node 2, 3 7 0 102 2+ + = → = −I I A

12

2 A

2 5 A

At node 1, I I I I A1 2 1 22 2+ = → = − =

At node 4, 2 4 2 44 4= + → = − = −I IAt node 3, 7 7 4 3 3+ = → = − =I I IHence, I A I A I A I1 2 3 412 10 5 2= = − = =, , , A−

V

11

8

Chapter 2, Solution 14 + + - 3V V1 I4 V2 - I3 - + 2V - + - + V3 - + + 4V I2 - I1 V4 + -

5V

For mesh 1, − + + = → =V V4 42 5 0 7 For mesh 2, + + + = → = − − = −4 0 4 73 4 3V V V V For mesh 3, − + − = → = + = −3 0 31 3 1 3V V V V V For mesh 4, − − − = → = − − =V V V V V1 2 2 12 0 2 6 Thus, V V V V V V V1 2 3 48 6 11= − = = − V7=, , ,


+ +

+ 12V 1 v2 - - 8V + - v1 - 3 + 2 - v3 10V

- + For loop 1,

8 12 0 42 2− + = → =v v V

V

V

For loop 2,

− − − = → = −v v3 38 10 0 18 For loop 3,

− + + = → = −v v v1 3 112 0 6 Thus, v V v V v1 2 36 4= − = = −, , V18 Chapter 2, Solution 16

+ v1 -

+ - + -

6V -+ loop 1

loop 2 12V

10V

+ v1 -

+ v2 -

Applying KVL around loop 1,

–6 + v1 + v1 – 10 – 12 = 0 v1 = 14V

Applying KVL around loop 2,

12 + 10 – v2 = 0 v2 = 22V

Chapter 2, Solution 17 + v1 -

+

- +

-+ 10V

12V

24V

loop 2

+ v3 -

v2

-loop 1

-+

It is evident that v3 = 10V Applying KVL to loop 2, v2 + v3 + 12 = 0 v2 = -22V Applying KVL to loop 1, -24 + v1 - v2 = 0 v1 = 2V Thus, v1 = 2V, v2 = -22V, v3 = 10V Chapter 2, Solution 18

Applying KVL, -30 -10 +8 + I(3+5) = 0

8I = 32 I = 4A -Vab + 5I + 8 = 0 Vab = 28V


Applying KVL around the loop, we obtain

-12 + 10 - (-8) + 3i = 0 i = -2A Power dissipated by the resistor: p 3Ω = i2R = 4(3) = 12W Power supplied by the sources: p12V = 12 (- -2) = 24W p10V = 10 (-2) = -20W

p8V = (- -2) = -16W Chapter 2, Solution 20

Applying KVL around the loop,

-36 + 4i0 + 5i0 = 0 i0 = 4A Chapter 2, Solution 21

10 Ω

+-45V -

+

+ v0 -

Apply KVL to obtain -45 + 10i - 3V0 + 5i = 0 But v0 = 10i, 3v0-45 + 15i - 30i = 0 i = -3A P3 = i2R = 9 x 5 = 45W 5 Ω

Chapter 2, Solution 22 Ω

+ v0 -

10A

2v0 Ω At the node, KCL re

0 v2104

v++

The current through i = 2V0 = -8. and the voltage acro

v = (6 + 4) i0

Hence, p2 vi = (-8.88 Chapter 2, Solution 8//12 = 4.8, 3//6 = 2The circuit is reduce 6A

6

quires that

0 = 0 v

the controlled

888A

ss it is

= 10 14

v0 −=

8)(-11.111) =

23

, (4 + 2)//(1.2d to that show

ix

+

2

4

0 = –4.444V

source is

111.1

98.75 W

+ 4.8) = 6//6 = 3 n below.

1Ω

vx -

Ω 3Ω

Applying current division,

i A A v ix x=+ +

= =2

2 1 36 2 1 2( ) , Vx =

The current through the 1.2- resistor is 0.5iΩ x = 1A. The voltage across the 12- resistor is 1 x 4.8 = 4.8 V. Hence the power is

Ω

p vR

W= = =2 24 8

12192. .


(a) I0 = 21 RR

Vs

+

α−=0V I0 ( )43 RR = 43

43

21

0

RRRR

RRV

+⋅

+−

α

( )( )4321

430

RRRRRR

Vs ++V −

=α

(b) If R1 = R2 = R3

= R4 = R,

1042

RR2V

V

S

0 =α

=⋅α

= α = 40


V0 = 5 x 10-3 x 10 x 103 = 50V

Using current division,

I20 =+

)5001.0(205

x=5 0.1 A

V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW


V0 = 5 x 10-3 x 10 x 103 = 50V


I20 =+

)5001.0(205

x=5 0.1 A

V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW Chapter 2, Solution 27


i1 = =+

)20(64

4 8 A

i2 = =+

)20(64

6 12 A


We first combine the two resistors in parallel =1015 6 Ω We now apply voltage division,

v1 = =+

)40(614

14 20 V

v2 = v3 = =+

)40(614

6 12 V

Hence, v1 = 28 V, v2 = 12 V, vs = 12 V


The series combination of 6 Ω and 3 Ω resistors is shorted. Hence i2 = 0 = v2

v1 = 12, i1 = =4

12 3 A

Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2 Chapter 2, Solution 30

i +v-

8 Ω

6 Ω

i1

9A 4 Ω

By current division, =+

= )9(126

12i 6 A

4 x 3 = ===−= 11 i4v,A369i 12 V p6 = 12R = 36 x 6 = 216 W Chapter 2, Solution 31

The 5 Ω resistor is in series with the combination of Ω=+ 5)64(10 . Hence by the voltage division principle,

=+

= )V20(55

5v 10 V

by ohm's law,

=+

=+

=64

1064

vi 1 A

pp = i2R = (1)2(4) = 4 W


We first combine resistors in parallel.

=3020 =50

30x20 12 Ω

=4010 =50

40x10 8 Ω

Using current division principle,

A12)20(2012ii,A8)20(

1288ii 4321 ==+=+

=+

== )8(5020i1 3.2 A

== )8(5030i2 4.8 A

== )12(5010i3 2.4A

== )12(5040i4 9.6 A

Chapter 2, Solution 33 Combining the conductance leads to the equivalent circuit below i

+v-

9A 1S

i

+ v -

4S

4S

1S

9A 2S

=SS 36 259

3x6= and 25 + 25 = 4 S


=+

= )9(

211

1i 6 A, v = 3(1) = 3 V

Chapter 2, Solution 34 By parallel and series combinations, the circuit is reduced to the one below:

-+

+ v1 -

8 Ωi1

=+ )132(10 Ω= 625

1510x

=+ )64(15 Ω= 625

1515x 28V 6 Ω Ω=+ 6)66(12

Thus i1 = =+ 6828 2 A and v1 = 6i1 = 12 V

We now work backward to get i2 and v2.

+ 6V -

1A

1A 6 Ω

-+ +

12V -

12 Ω

8 Ωi1 = 2A 28V 6 Ω 0.6A

+ 3.6V

-

4 Ω

+ 6V -

1A

1A

15 Ω

6 Ω

-+ +

12V -

12 Ω

8 Ω i1 = 2A 28V 6 Ω

Thus, v2 = ,123)63(1513

⋅=⋅ i2 = 24.013v2 =

p2 = i2R = (0.24)2 (2) = 0.1152 W i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W Chapter 2, Solution 35

i

20 Ω + V0 - i2

a b

5 Ω

30 Ω70 Ω

I0i1

+ V1 -

-+

50V

Combining the versions in parallel,

=3070 Ω= 21100

30x70 , =1520 =25

5x20 4 Ω

i = =+ 421

50 2 A

vi = 21i = 42 V, v0 = 4i = 8 V

i1 = =70v1 0.6 A, i2 = =

20v2 0.4 A

At node a, KCL must be satisfied i1 = i2 + I0 0.6 = 0.4 + I0 I0 = 0.2 A Hence v0 = 8 V and I0 = 0.2A Chapter 2, Solution 36

The 8-Ω resistor is shorted. No current flows through the 1-Ω resistor. Hence v0 is the voltage across the 6Ω resistor.

I0 = ==+ 4

41632

4 1 A

v0 = I0 ( ) == 0I263 2 V


Let I = current through the 16Ω resistor. If 4 V is the voltage drop across the R6 combination, then 20 - 4 = 16 V in the voltage drop across the 16Ω resistor.

Hence, I = =1616 1 A.

But I = 1R616

20=

+ 4 = =R6

R6R6+

R = 12 Ω


Let I0 = current through the 6Ω resistor. Since 6Ω and 3Ω resistors are in parallel. 6I0 = 2 x 3 R0 = 1 A The total current through the 4Ω resistor = 1 + 2 = 3 A. Hence vS = (2 + 4 + 32 ) (3 A) = 24 V

I = =v 10

S 2.4 A


(a) Req = =0R 0

(b) Req = =+ RRRR =+2R

2R R

(c) Req = ==++ R2R2)RR()RR( R

(d) Req = )R21R(R3)RRR(R +=+3

= =+ R

23R3

R23Rx3

R

(e) Req = R3R3R2R =

⋅

R3R2R

= R3 =+

=R

32R3

R32Rx3

R32 R

116


Req = =+=++ 23)362(43 5Ω

I = =5

10qRe=

10 2 A


Let R0 = combination of three 12Ω resistors in parallel

121

121

121

R1

o

++= Ro = 4

)R14(6030)RR10(6030R 0eq ++=+++=

R74

)R14(6030++

+=50 74 + R = 42 + 3R

or R = 16 Ω Chapter 2, Solution 42

(a) Rab = ==+=+25

20x5)128(5)30208(5 4 Ω

(b) Rab = =++=++=++++ 5.22255442)46(1058)35(42 6.5 Ω


(a) Rab = =+=+=+ 8450400

2520x54010205 12 Ω

(b) =302060 Ω==

++

−

10660

301

201

601 1

Rab = )1010( +80 =+

=100

2080 16 Ω


(a) Convert T to Y and obtain

R x x x1

20 20 20 10 10 2010

80010

80=+ +

= = Ω

R R2 380020

40= = =Ω

The circuit becomes that shown below. R1 a R3 R2 5Ω b R1//0 = 0, R3//5 = 40//5 = 4.444Ω R Rab = + = =2 0 4 444 40 4 444 4/ /( . ) / / . Ω

(b) 30//(20+50) = 30//70 = 21 Ω Convert the T to Y and obtain

R x x x1

20 10 10 40 40 2040

140040

35=+ +

= = Ω

R2140020

70= = Ω , R3140010

140= = Ω

The circuit is reduced to that shown below. 11Ω R2 30Ω

15Ω

Combining the resist

R1

R3

21Ω

21Ω

ors in parallel

R1//15 =35//15=10.5, 30//R2=30//70 = 21 leads to the circuit below. 11 10.5Ω Ω 21Ω 140Ω

21Ω 21Ω Coverting the T to Y leads to the circuit below. 11Ω 10.5Ω R4 R5 R6 21Ω R x x x R4 6

21 140 140 21 21 2121

632121

301=+ +

= = =Ω

R5

6321140

4515= = .

10.5//301 = 10.15, 301//21 = 19.63 R5//(10.15 +19.63) = 45.15//29.78 = 17.94 Rab = + =11 17 94 28 94. . Ω Chapter 2, Solution 45

(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8

Rab = + + =5 50 4 8 59 8. . Ω (b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus Rab = + + =5 12 8 15 32 5. . Ω


(a) Rab = =++ 2060407030 80

206040100

70x30 +++

=++ 154021= 76 Ω

(b) The 10-Ω, 50-Ω, 70-Ω, and 80-Ω resistors are shorted.

=3020 Ω= 1250

30x20

40 =60 24100

60x=

40

Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Ω Chapter 2, Solution 47

=205 Ω= 425

20x5

6 =3 Ω= 29

3x6

8 Ωa b

10 Ω

2 Ω 4 Ω Rab = 10 + 4 + 2 + 8 = 24 Ω


(a) Ra = 3010

100100100R

RRRRRR

3

133221 =++

=++

Ra = Rb = Rc = 30 Ω

(b) Ω==++

= 3.10330

310030

50x2050x3020x30R a

,15520

3100R b Ω== Ω== 6250

3100R c

Ra = 103.3 Ω, Rb = 155 Ω, Rc = 62 Ω


(a) R1 = Ω=+

=++

436

1212RRR

RR

cba

ca

R1 = R2 = R3 = 4 Ω

(b) Ω=++

= 18103060

30x60R1

Ω== 6100

10x60R 2

Ω== 3100

10x30R 3

R1 = 18Ω, R2 = 6Ω, R3 = 3Ω Chapter 2, Solution 50 Using = 3R∆R Y = 3R, we obtain the equivalent circuit shown below:

3R 30mA

R

R 3R

3R 3R

30mA 3R/2

=RR3 R43

R4RxR3

=

)R4/(3)R4/()RxR3(R =3

RR

23R3

R23Rx3

R23R3R

43R

43R3

=+==

+

P = I2R 800 x 10-3 = (30 x 10-3)2 R R = 889 Ω Chapter 2, Solution 51

(a) Ω= 153030 and Ω== 12)50/(20x302030

Rab = ==+ )39/(24x15)1212(15 9.31 Ω

b

15 Ω

a

20 Ω

30 Ω

b

a

20 Ω

30 Ω 30 Ω 30 Ω

12 Ω

12 Ω

(b) Converting the T-subnetwork into its equivalent network gives ∆ Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 Ω Rb'c' = 350/(10) = 35Ω, Ra'c' = 350/(20) = 17.5 Ω

Also Ω== 21)100/(70x307030 and 35/(15) = 35x15/(50) = 10.5

Rab = 25 + 5.315.1725)5.1021(5. +=+17 Rab = 36.25 Ω

b’

c’ c’ b

a

15 Ω 35 Ω 17.5 Ω

25 Ω 70 Ω a’

b

a

30 Ω

25 Ω

5 Ω

10 Ω

15 Ω

20 Ω

30 Ω

Chapter 2, Solution 52 (a) We first convert from T to ∆ .

R3 R2

b

a

100 Ω

100 Ω 100 Ω

100 Ω

100 Ω

100 Ω

100 Ω

b

a

200 Ω

200 Ω

100 Ω

100 Ω 100 Ω

100 Ω

100 Ω

100 Ω

100 Ω

100 Ω R1

Ω==++

= 800100

80000100

100x200200x200200x1001R

R2 = R3 = 80000/(200) = 400

But Ω== 80500

400x100400100

We connect the ∆ to Y.

Ra = Rc = Ω==++ 3

400960

000,648008080

800x80

Rc

Rb

b

aRa

100 Ω

100 Ω

100 Ω

100 Ω

100 Ω

b

a

800 Ω

80 Ω

80 Ω 100 Ω

100 Ω

100 Ω

100 Ω

100 Ω

Rb = Ω=320

96080x80

We convert T to . ∆

R3’

R2’

b

a

500/3 Ω

500/3 Ω

R1’

b

a

500/3 Ω

500/3 Ω

320/3 Ω

100 Ω

100 Ω

Ω==++

= 75.293)3/(320

)3/(000,94

3320

3320x100

3320x100100x100

'1R

33.313100

)3/(000,94RR 13

'2 ===

796.108)3/(1440

)3/(500x)3/(940)3/(500)30/(940 ==

Rab = ==36.511

6.217x75.293)796.108x2(75.293 125 Ω

(b) Converting the Ts to ∆ s, we have the equivalent circuit below.

100 Ω a

b

300 Ω

300 Ω 100 Ω

100 Ω

300 Ω

100 Ω 300 Ω

300 Ω

100 Ω

300 Ω

100 Ω

100 Ω

100 Ω

b

a 100 Ω

100 Ω

100 Ω

100 Ω

100 Ω

100 Ω ,75)400/(100x300100300 == 100)450/(150x300)7575( ==+300

Rab = 100 + )400/(300x100200100300 +=+100 Rab = 2.75 Ω

100 Ω

300 Ω

300 Ω

100 Ω

300 Ω

100 Ω

100 Ω

Chapter 2, Solution 53 (a) Converting one ∆ to T yields the equivalent circuit below:

20 Ω

b’

c’

a’

b 80 Ω

a 20 Ω

5 Ω

4 Ω

60 Ω

30 Ω

Ra'n = ,4501040

10x40Ω=

++ ,5

10050x10R n'b Ω== Ω== 20

10050x40R n'c

Rab = 20 + 80 + 20 + 6534120)560()430( +=++ Rab = 142.32 Ω

(a) We combine the resistor in series and in parallel.

Ω==+ 2090

60x30)3030(30

We convert the balanced ∆ s to Ts as shown below:

b

10 Ω

10 Ω 10 Ω

10 Ω

10 Ω

30 Ω30 Ω

30 Ω

30 Ω

30 Ω

30 Ω b

a

20 Ω

20 Ω 10 Ω

a

Rab = 10 + 40202010)102010()1010 +=++++( Rab = 33.33 Ω


(a) Rab = + + + = + =50 100 150 100 150 50 100 400 130/ /( ) / / Ω (b) Rab = + + + = + =60 100 150 100 150 60 100 400 140/ /( ) / / Ω

Chapter 2, Solution 55 We convert the T to . ∆

50 Ω

I0

-+

24 V

b

a

35 Ω

70 Ω

140 Ω

Req

-+

24 V 60 Ω

b

a

20 Ω

40 Ω

10 Ω

20 Ω

I0

60 Ω

70 Ω

Req

Rab = Ω==++

=++

3540

140040

20x1010x4040x20R

RRRRRR

3

133221

Rac = 1400/(10) = 140Ω, Rbc = 1400/(40) = 35Ω 357070 = and =160140 140x60/(200) = 42

Req = Ω=+ 0625.24)4235(35 I0 = 24/(Rab) = 0.9774A Chapter 2, Solution 56 We need to find Req and apply voltage division. We first tranform the Y network to ∆ .

c

35 Ω

16 Ω

30 Ω

37.5 Ω a b

30 Ω 45 Ω

Req

+100 V

-20 Ω

35 Ω

16 Ω

30 Ω

10 Ω

12 Ω

15 Ω

Req

+ 100 V

-

20 Ω

Rab = Ω==++ 5.37

12450

1215x1212x1010x15

Rac = 450/(10) = 45Ω, Rbc = 450/(15) = 30Ω Combining the resistors in parallel,

30||20 = (600/50) = 12 Ω,

37.5||30 = (37.5x30/67.5) = 16.667 Ω

35||45 = (35x45/80) = 19.688 Ω

Req = 19.688||(12 + 16.667) = 11.672Ω

By voltage division,

v = 10016672.11

672.11+

= 42.18 V


4 Ω

ec

f

a

bd

36 Ω

7 Ω

27 Ω

2 Ω

18 Ω

28 Ω 10 Ω

1 Ω 14 Ω

Rab = Ω==++ 18

12216

126x88x1212x6

Rac = 216/(8) = 27Ω, Rbc = 36 Ω

Rde = Ω=++ 7

856

84x88x22x4

Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 Ω Combining resistors in parallel,

,368.7382802810 Ω== Ω== 868.5

437x36736

Ω== 7.230

3x27327

4 Ω

0.5964 Ω 3.977 Ω1.829 Ω

4 Ω

7.568 Ω 14 Ω

2.7 Ω

14 Ω

18 Ω

5.868 Ω 7.568 Ω

Ω==++

= 829.1567.26

7.2x18867.57.218

7.2x18R an

Ω== 977.3567.26868.5x18R bn

Ω== 5904.0567.26

7.2x868.5R cn

)145964.0()368.7977.3(829.14R eq ++++=

=+ 5964.14346.11829.5= 12.21 Ω i = 20/(Req) = 1.64 A

Chapter 2, Solution 58 The resistor of the bulb is 120/(0.75) = 160Ω 2.25 A 1.5 A 40 Ω

0.75 A

80 Ω 160 Ω+ 90 V - +

120-+

VS Once the 160Ω and 80Ω resistors are in parallel, they have the same voltage 120V. Hence the current through the 40Ω resistor is 40(0.75 + 1.5) = 2.25 x 40 = 90 Thus vs = 90 + 120 = 210 V

Chapter 2, Solution 59 Total power p = 30 + 40 + 50 + 120 W = vi

or i = p/(v) = 120/(100) = 1.2 A


p = iv i = p/(v) i30W = 30/(100) = 0.3 A i40W = 40/(100) = 0.4 A i50W = 50/(100) = 0.5 A Chapter 2, Solution 61 There are three possibilities

(a) Use R1 and R2: R = Ω== 35.429080RR 21 p = i2R i = 1.2A + 5% = 1.2 ± 0.06 = 1.26, 1.14A p = 67.23W or 55.04W, cost = $1.50

(b) Use R1 and R3:

R = Ω== 44.4410080RR 31 p = I2R = 70.52W or 57.76W, cost = $1.35

(c) Use R2 and R3: R = Ω== 37.4710090RR 32 p = I2R = 75.2W or 61.56W, cost = $1.65

Note that cases (b) and (c) give p that exceed 70W that can be supplied. Hence case (a) is the right choice, i.e. R1 and R2


pA = 110x8 = 880 W, pB = 110x2 = 220 W Energy cost = $0.06 x 360 x10 x (880 + 220)/1000 = $237.60


Use eq. (2.61),

Rn = Ω=−

=− −

−

04.010x25

100x10x2RII 3

3

mm

mI

In = I - Im = 4.998 A p = (I = 9992.0)04.0()998.4R 22

n = ≅ 1 W Chapter 2, Solution 64

When Rx = 0, i R = A10x = Ω=1110

110

When Rx is maximum, ix = 1A Ω==+ 1101

110R xR

i.e., Rx = 110 - R = 99 Ω Thus, R = 11 Ω, Rx = 99 Ω Chapter 2, Solution 65

=Ω−=−= k1mA1050R

IV

R mfs

fsn 4 kΩ


20 kΩ/V = sensitivity = fsI1

i.e., Ifs = A50V/k20

µ=Ω1

The intended resistance Rm = Ω=Ω= k200)V/k20(10IV

fs

fs

(a) =Ω−µ

=−= k200A50V50R

iV

R mfs

fsn 800 kΩ

(b) p = =Ωµ= )k800()A50(RI 2n

2fs 2 mW


(a) By current division, i0 = 5/(5 + 5) (2 mA) = 1 mA V0 = (4 kΩ) i0 = 4 x 103 x 10-3 = 4 V

(b) .k4.2k6k4 Ω= By current division,

mA19.1)mA2(54.21

5i '0 =

++=

=Ω= )mA19.1)(k4.2(v'0 2.857 V

(c) % error = 0

'00

vvv −

x 100% = =100x4143.1 28.57%

(d) .k6.3k30k4 Ω=Ω By current division,

mA042.1)mA2(56.31

5i '0 =

++=

V75.3)mA042.1)(k6.3(v'0 =Ω

% error = ==−

4100x25.0%100x

vvv

0

'0 6.25%


(a) Ω= 602440

i = =+ 24164 0.1 A

(b) =++

=24116

4i ' 0.09756 A

(c) % error = =− %100x

1.009756.01.0 2.44%


With the voltmeter in place,

Sm2S1

m20 V

RRRRRR++

=V

where Rm = 100 kΩ without the voltmeter,

SS21

20 V

RRRR

++=V

(a) When R2 = 1 kΩ, Ω= k101100RR 2m

V0 = =

+

)40(

30101100101100

1.278 V (with)

V0 = =+

)40(301

1 1.29 V (without)

(b) When R2 = 10 kΩ, Ω== k091.91101000RR m2

V0 = =+

)40(30091.9

091.9 9.30 V (with)

V0 = =+

)40(3010

10 10 V (without)

(c) When R2 = 100 kΩ, Ω= k50RR m2

=+

= )40(3050

50V0 25 V (with)

V0 = =+

)40(30100

100 30.77 V (without)


(a) Using voltage division, v Va = +

=1212 8

25 15( )

v Vb = +=

1010 15

25 10( )

v v vab a b= − = V− =15 10 5

(b) + 8kΩ 15kΩ 25 V - a b

12kΩ 10kΩ o

v v V v v va b ab a b= = = V− = − = −0 10 0 10 1, , 0 Chapter 2, Solution 71

Vs RL

R1

+−

iL

Given that vs = 30 V, R1 = 20 Ω, IL = 1 A, find RL.

v i R R Rvi

Rs L L Ls

L= + → = − = − =( )1 1

301

20 10Ω


The system can be modeled as shown. +

9V - The n pa

9 12= x

Chapter 2,

Chapter 2,

12A

R R ••• R

rallel resistors R give a combined resistance of R/n. Thus,

129

12 159

20 → = = =Rn

n xR x

Solution 73

By the current division principle, the current through the ammeter will be one-half its previous value when

R = 20 + Rx 65 = 20 + Rx Rx = 45 Ω

Solution 74

With the switch in high position,

6 = (0.01 + R3 + 0.02) x 5 R3 = 1.17 Ω

At the medium position,

6 = (0.01 + R2 + R3 + 0.02) x 3 R2 + R3 = 1.97

or R2 = 1.97 - 1.17 = 0.8 Ω

At the low position, 6 = (0.01 + R1 + R2 + R3 + 0.02) x 1 R1 + R2 + R3 = 5.97 R1 = 5.97 - 1.97 = 4 Ω


MR 100 Ω

VS -+

12 Ω

(a) When Rx = 0, then

Im = Ifs = mRR

t+

R2 = mfs

2

RIE

− = Ω=− k9.1910010x1.0

23

(b) For half-scale deflection, Im = mA05.02Ifs =

Im = xm RRR

E++

Rx = =Ω−=+− − k2010x05.0

2)RR(IE

3mm

20 kΩ


For series connection, R = 2 x 0.4Ω = 0.8Ω

===8.0)120(

RV 22

p 18 kΩ (low)

For parallel connection, R = 1/2 x 0.4Ω = 0.2Ω

p = =2.0)120

=(

RV

22

72 kW (high)


(a) 5 Ω = 2020202010 =10

i.e., four 20 Ω resistors in parallel.

(b) 311.8 = 300 + 10 + 1.8 = 300 + 8.12020 + i.e., one 300Ω resistor in series with 1.8Ω resistor and a parallel combination of two 20Ω resistors.

(c) 40kΩ = 12kΩ + 28kΩ = k50k56k2424 + i.e., Two 24kΩ resistors in parallel connected in series with two 50kΩ resistors in parallel.

(d) 42.32kΩ = 42l + 320

= 24k + 28k = 320 = 24k = 20300k56k ++56

i.e., A series combination of 20Ω resistor, 300Ω resistor, 24kΩ resistor and a parallel combination of two 56kΩ resistors.


The equivalent circuit is shown below:

R

+ V0

--+

VS (1-α)R

V0 = S0S VR)1(VR)1(R

R)1α−=

α−+α−(

R)1(VV

S

0 α−=


Since p = v2/R, the resistance of the sharpener is R = v2/(p) = 62/(240 x 10-3) = 150Ω I = p/(v) = 240 mW/(6V) = 40 mA Since R and Rx are in series, I flows through both. IRx = Vx = 9 - 6 = 3 V Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75 Ω

Chapter 2, Solution 80 The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:

V -+

Case 1

R1 -+ V R2

Hence ,R

Vp2

= 2

1

1

2

RR

pp

= === )12(4

10pRR

12

12p 30 W

Case 2


Let R1 and R2 be in kΩ. 5RR 21eq +=R (1)

12

2

S

0

RR5R5

VV

+= (2)

From (1) and (2), 40R5

05. 1=0 2 = 2

22 R5

R5R5

+= or R2 = 3.33 kΩ

From (1), 40 = R1 + 2 R1 = 38 kΩ Thus R1 = 38 kΩ, R2 = 3.33 kΩ

Chapter 2, Solution 82 (a) 10 Ω

1 2

10 Ω

40 Ω 80 Ω R12

R12 = 80 + =+=+6

5080)4010(10 88.33 Ω

(b)

3 20 Ω 10 Ω

10 Ω

80 Ω

40 Ω

R13 1

R13 = 80 + =+=++ 501010020)4010(10 108.33 Ω

20 Ω 10 Ω

10 Ω

1

4

80 Ω

(c) R14 40 Ω

R14 = =++=++++ 2008020)104010(080 100 Ω Chapter 2, Solution 83 The voltage across the tube is 2 x 60 mV = 0.06 V, which is negligible

compared with 24 V. Ignoring this voltage amp, we can calculate the current through the devices.

I1 = mA5V9mW

V1

1 =45p

=

I2 = mA2024mW

V2

2 =480p

=

60 mA i2 = 20 mA

R1

iR2

iR1

i1 = 5 mA

R2 -+

24 V By applying KCL, we obtain and mA402060I

1R =−= mA35540I2R =−=

Hence, R1RI 1 = 24 - 9 = 15 V ==

mA40V15R1 375 Ω

V9RI 2R 2= ==

mA35V9

2R 257.14 Ω

Chapter 3, Solution 1.

8 Ω

v2

2 Ω

40 Ω v1

6 A 10 A

At node 1, 6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1) At node 2, v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2) Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V

P8Ω = ( )==

8143.9

8v 22

1 10.45 W

P4Ω = ( )

=−4vv 2

21 94.37 W

P2Ω = ( )=

==

2286.10

2v 21

2 52.9 W


At node 1,

2

vv6

5v

10v 2111 −

+=−− 60 = - 8v1 + 5v2 (1)

At node 2,

2

vv634

v 212 −++= 36 = - 2v1 + 3v2 (2)

Solving (1) and (2), v1 = 0 V, v2 = 12 V

Chapter 3, Solution 3 Applying KCL to the upper node,

10 = 60v

230v

20v

10v 0oo +++0 + v0 = 40 V

i1 = =10

0v 4 A , i2 = =

20v0 2 A, i3 = =

30v0 1.33 A, i4 = =

60v0 67 mA

Chapter 3, Solution 4 At node 1, 4 + 2 = v1/(5) + v1/(10) v1 = 20 At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A Chapter 3, Solution 5 Apply KCL to the top node.

k4

vk6v20

k2v30 000 =

−+

− v0 = 20 V

i1

10 Ω 10 Ω

2A

i4 i3

v1 v2

i2

4 A 5 Ω 5 Ω 5 A


i1 + i2 + i3 = 0 02

10v6v

412v 002 =

−++

−

or v0 = 8.727 V

Chapter 3, Solution 7 At node a,

babaaa VV

VVVV3610

10153010

−=→−

+=−

(1)

At node b,

babbba VV

VVVV72240

59

2012

10−=→=

−−+

−+

− (2)

Solving (1) and (2) leads to Va = -0.556 V, Vb = -3.444V Chapter 3, Solution 8

i2

5 Ω

2 Ω

i13 Ω v1 i3

1 Ω

+–

4V0

3V–+

+

V0

–

i1 + i2 + i3 = 0 05

v4v1

3v5v 0111 =

−+

−+

But 10 v52v = so that v1 + 5v1 - 15 + v1 - 0v

58

1 =

or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V


+ v1

–

i2

6 Ω i13 Ω v1 i3

8 Ω +–

+ v0 –

2v0 12V

–+

At the non-reference node,

6

v2v8v

3v 011112 −

+=− (1)

But -12 + v0 + v1 = 0 v0 = 12 - v1 (2) Substituting (2) into (1),

6

24v38v

3v 111 −

+=−12 v0 = 3.652 V

Chapter 3, Solution 10 At node 1,

8v4

1vv 112 +=

− 32 = -v1 + 8v2 - 8v0 (1)

1 Ω

4A 2i0

i0

v1 v0

8 Ω 2 Ω

v2 4 Ω

At node 0,

00 I2

2v

+=4 and 8vI 1

0 = 16 = 2v0 + v1 (2)

At node 2,

2I0 = 4

v1

v 212 +−v and

8v1

0 =I v2 = v1 (3)

From (1), (2) and (3), v0 = 24 V, but from (2) we get

io = 624242

22v4 o

−=−=−

= - 4 A


i3

6 Ω

vi1 i24 Ω 3 Ω

–+

10 V 5 A Note that i2 = -5A. At the non-reference node

6v5

4v10

=+− v = 18

i1 = =−4

v10 -2 A, i2 = -5 A


i3

40 Ω

v1 v210 Ω 20 Ω

–+ 5 A

50 Ω 24 V

At node 1, 40

0v20

vv10

v24 1211 −+

−=

− 96 = 7v1 - 2v2 (1)

At node 2, 50v

20vv 221 =

−+5 500 = -5v1 + 7v2 (2)

Solving (1) and (2) gives, v1 = 42.87 V, v2 = 102.05 V

==40vi 1

1 1.072 A, v2 = =50v2 2.041 A


At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts


1 Ω 2 Ω

4 Ω

v0 v1

5 A

8 Ω

20 V –+

40 V –+

At node 1, 1

v405

2vv 001 −

=+−

v1 + v0 = 70 (1)

At node 0, 8

20v4v

52

vv 0001 ++=+

− 4v1 - 7v0 = -20 (2)

Solving (1) and (2), v0 = 20 V


1 Ω 2 Ω

4 Ω

v0 v1

5 A

8 Ω

20 V –+

40 V –+

Nodes 1 and 2 form a supernode so that v1 = v2 + 10 (1) At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) 2 + 6v1 + 8v2 = 3v3 (2) At node 3, 2 + 4 = 3 (v3 - v2) v3 = v2 + 2 (3) Substituting (1) and (3) into (2),

2 + 6v2 + 60 + 8v2 = 3v2 + 6 v2 = 1156−

v1 = v2 + 10 = 1154

i0 = 6vi = 29.45 A

P65 = =

== 6

1154Gv

R

221

21v

144.6 W

P55 = =

−= 5

1156Gv

222 129.6 W

P35 = ( ) ==− 3)2(Gvv 22

3L 12 W

Chapter 3, Solution 16 2 S

+ v0 –

13 V –+

i0

1 S 4 S

8 Sv1 v2 v3

2 A At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 (1) But

v1 = v2 + 2v0 and v0 = v2. Hence

v1 = 3v2 (2) v3 = 13V (3)

Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V Chapter 3, Solution 17

60 V

i0

3i0

2 Ω

10 Ω

4 Ω

60 V –+

8 Ω

At node 1, 2

vv8v

4v60 2111 −

+=− 120 = 7v1 - 4v2 (1)

At node 2, 3i0 + 02

vv10

v60 212 =−

+−

But i0 = .4

v60 1−

Hence

( )

02

vv10

v604

v603 2121 =−

+−

+− 1020 = 5v1 - 12v2 (2)

Solving (1) and (2) gives v1 = 53.08 V. Hence i0 = =−4

v60 1 1.73 A


+ v3 –

+ v1 –

10 V

– + v1

v2

2 Ω 2 Ω

4 Ω 8 Ω

v3

5 A

(a) (b)

At node 2, in Fig. (a), 5 = 2

vv2

vv 3212 −+

− 10 = - v1 + 2v2 - v3 (1)

At the supernode, 8v

4v

2vv

2vv 313212 +=

−+

− 40 = 2v1 + v3 (2)

From Fig. (b), - v1 - 10 + v3 = 0 v3 = v1 + 10 (3) Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3


At node 1,

32112131 4716

48235 VVVVVVVV

−−=→+−

+−

+= (1)

At node 2,

32132221 270

428VVV

VVVVV−+−=→

−+=

− (2)

At node 3,

32132313 724360

42812

3 VVVVVVVV−+=−→=

−+

−+

−+ (3)

From (1) to (3),

BAVVVV

=→

−=

−−−−−

360

16

724271417

3

2

1

Using MATLAB,

V 267.12 V, 933.4 V, 10267.12933.410

3211 ===→

== − VVVBAV

Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence

040414 321

321 =++→=++ VVVVVV (1)

. V1 . V2 2Ω V3 4Ω 1Ω 4Ω

Between nodes 1 and 3,

12012 1331 −=→=++− VVVV (2) Similarly, between nodes 1 and 2, (3) iVV 221 +=But i . Combining this with (2) and (3) gives 4/3V=

.V (4) 2/6 12 V+= Solving (1), (2), and (4) leads to

V15 V,5.4 V,3 321 −==−= VVV Chapter 3, Solution 21

4 kΩ

2 kΩ

+ v0 –

3v0v1 v2v3

1 kΩ

+

3 mA

3v0

+ v2 –

+ v3 –

+

(b) (a) Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source. At node 1,

2000

vv4000

vv10x3 31213 −

+−

=− 12 = 3v1 - v2 - 2v3 (1)

At node 2,

1v

2vv

4vv 23121 =

−+

− 3v1 - 5v2 - 2v3 = 0 (2)

Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 v3 = - 2v2 (3) From (1) to (3), v1 = 1 V, v2 = 3 V


At node 1, 8

vv3

4v

2v 011012 −

++=−

24 = 7v1 - v2 (1)

At node 2, 3 + 1

v5v8

vv 2221 +=

−

But, v1 = 12 - v1 Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V 456 = 41v1 - 9v2 (2) Solving (1) and (2), v1 = - 10.91 V, v2 = - 100.36 V


At the supernode, 5 + 2 = 5v

10v 21 + 70 = v1 + 2v2 (1)

Considering Fig. (b), - v1 - 8 + v2 = 0 v2 = v1 + 8 (2) Solving (1) and (2), v1 = 18 V, v2 = 26 V

+ v1

–5 Ω 10 Ω

v1 v2

8 V

– +

5 A 2 A

+

v2

– (b)(a)


6mA 1 kΩ 2 kΩ 3 kΩ V1 V2 + io - 30V 15V - 4 kΩ 5 kΩ + At node 1,

212111 2796

246

130 VVVVVV

−=→−

++=− (1)

At node 2,

211222 311530

253)15(

6 VVVVVV+−=→

−+=

−−+ (2)

Solving (1) and (2) gives V1=16.24. Hence io = V1/4 = 4.06 mA Chapter 3, Solution 25 Using nodal analysis,

1 Ω

v0

2 Ω

4 Ω

2 Ω

10V –+

–+

40V –+

i0

20V

40v

2v10

2v40

1v20 0000 −

=−

+−

+−

v0 = 20V

i0 = 1

v20 0− = 0 A

Chapter 3, Solution 26 At node 1,

32121311 24745

5103

2015 VVVVVVVV

−−=−→−

+−

+=− (1)

At node 2,

554

532221 VVVIVV o −

=−

+− (2)

But 10

31 VVIo

−= . Hence, (2) becomes

321 31570 VVV +−= (3) At node 3,

32132331 52100

5510

103 VVV

VVVVV−+=−→=

−+

−−+

−+ (4)

Putting (1), (3), and (4) in matrix form produces

BAVVVV

=→

−

−=

−−

−−

10045

5213157247

3

2

1

Using MATLAB leads to

−−−

== −

96.1982.4835.9

1BAV

Thus, V 95.1 V, 982.4 V, 835.9 321 −=−=−= VVV Chapter 3, Solution 27 At node 1, 2 = 2v1 + v1 – v2 + (v1 – v3)4 + 3i0, i0 = 4v2. Hence, 2 = 7v1 + 11v2 – 4v3 (1) At node 2, v1 – v2 = 4v2 + v2 – v3 0 = – v1 + 6v2 – v3 (2) At node 3,

2v3 = 4 + v2 – v3 + 12v2 + 4(v1 – v3)

or – 4 = 4v1 + 13v2 – 7v3 (3) In matrix form,

−=

−−

−

402

vvv

71341614117

3

2

1

,1767134

1614117=

−−

−=∆ 110

71341604112

1 =−−

−−

=∆

,66744

101427

2 =−−

−=∆ 286

41340612117

3 =−

−=∆

v1 = ,V625.01761101 ==

∆∆

v2 = V375.0176662 ==

∆∆

v3 = .V625.11762863 ==

∆∆

v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V. Chapter 3, Solution 28 At node c,

dcbcbccd VVVVVVVV

211505410

−+−=→+−

=−

(1)

At node b,

cbabbcba VVVVVVVV

2445848

45+−=−→=

−+

−+ (2)

At node a,

dbabaada VVVVVVVV

427300845

16430

−−=→=−+

++−−

(3)

At node d,

dcacddda VVVVVVVV

72515010204

30−+=→

−+=

−− (4)

In matrix form, (1) to (4) become

BAV

VVVV

d

c

b

a

=→

−

=

−−−

−−−

15030450

72054027

024121150

We use MATLAB to invert A and obtain

−−

−

== −

17.29736.1

847.714.10

1BAV

Thus, V 17.29 V, 736.1 V, 847.7 V, 14.10 −=−==−= dcba VVVV Chapter 3, Solution 29 At node 1,

42121141 45025 VVVVVVVV −−=−→=−++−+ (1) At node 2,

32132221 4700)(42 VVVVVVVV −+−=→=−+=− (2) At node 3,

4324332 546)(46 VVVVVVV −+−=→−=−+ (3) At node 4,

43144143 5232 VVVVVVVV +−−=→=−+−+ (4) In matrix form, (1) to (4) become

BAV

VVVV

=→

−

=

−−−−

−−−−

2605

51011540

04711014

4

3

2

1

Using MATLAB,

−

== −

7076.0309.2209.17708.0

1BAV

i.e. V 7076.0 V, 309.2 V, 209.1 V, 7708.0 4321 ===−= VVVV


v1

10 Ω

20 Ω

80 Ω

40 Ω

1

v0

I0

2I0

2

v2

4v0

120 V

+––

+

– +

100 V At node 1,

20

vv410

v10040

vv 1o121 −+

−=

− (1)

But, vo = 120 + v2 v2 = vo – 120. Hence (1) becomes 7v1 – 9vo = 280 (2) At node 2,

Io + 2Io = 80

0v o −

80v

40v120v

3 oo1 =

−+

or 6v1 – 7vo = -720 (3)

from (2) and (3),

−

=

−−

720280

vv

7697

o

1

554497697

=+−=−−

=∆

844077209280

1 −=−−−

=∆ , 67207206

28072 −=

−=∆

v1 = ,16885

84401 −=−

=∆∆

vo = V13445

67202 −−

=∆∆

Io = -5.6 A

Chapter 3, Solution 31 1 Ω

i0

2v0 v3

1 Ω

2 Ω

4 Ω 10 V –+

v1 v2

+ v0 –

4 Ω

1 A At the supernode,

1 + 2v0 = 1

vv1

v4v 3121 −

++ (1)

But vo = v1 – v3. Hence (1) becomes, 4 = -3v1 + 4v2 +4v3 (2) At node 3,

2vo + 2

v10vv

4v 3

312 −

+−=

or 20 = 4v1 + v2 – 2v3 (3)

At the supernode, v2 = v1 + 4io. But io = 4

v 3 . Hence,

v2 = v1 + v3 (4) Solving (2) to (4) leads to, v1 = 4 V, v2 = 4 V, v3 = 0 V.


v3

v1 v2

10 V

loop 2

– +

20 V

+ –

12 V

loop 1

–+

+ v3 –

+ v1 –

10 kΩ

4 mA

5 kΩ

(b)(a) We have a supernode as shown in figure (a). It is evident that v2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain,

-v1 – 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V Thus, v1 = 2 V, v2 = 12 V, v3 = -8V. Chapter 3, Solution 33

(a) This is a non-planar circuit because there is no way of redrawing the circuit with no crossing branches.

(b) This is a planar circuit. It can be redrawn as shown below.

1 Ω

2 Ω

3 Ω

4 Ω

5 Ω 12 V

–+

Chapter 3, Solution 34 (a) This is a planar circuit because it can be redrawn as shown below,

7 Ω

10 V –+

1 Ω 2 Ω

3 Ω

4 Ω

6 Ω

5 Ω (b) This is a non-planar circuit. Chapter 3, Solution 35

5 kΩ i1

i2

+ v0 –2 kΩ

30 V –+

–+20 V

4 kΩ Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1,

-30 + 20 + 7i1 – 5i2 = 0 or 7i1 – 5i2 = 10 (1) For mesh 2, -20 + 9i2 – 5i1 = 0 or -5i1 + 9i2 = 20 (2) Solving (1) and (2), we obtain, i2 = 5. v0 = 4i2 = 20 volts.


i1 i2

2 Ω

4 Ω 10 V

+ –

12 V –+

I2

6 Ω I1

i3 Applying mesh analysis gives,

12 = 10I1 – 6I2

-10 = -6I1 + 8I2

or

−

−=

− 2

1

II

4335

56

,114335=

−−

=∆ ,94536

1 =−

−=∆ 7

5365

2 −=−−

=∆

,119I 1

1 =∆∆

= 11

7I 22

−=

∆∆

=

i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 – I2 = 10/11 = 1.4545 A.

vo = 6i2 = 6x1.4545 = 8.727 V.


5 Ω

4v0 2 Ω

1 Ω

3 Ω

3 V –+

+–

+ v0 –

i2

i1

Applying mesh analysis to loops 1 and 2, we get,

6i1 – 1i2 + 3 = 0 which leads to i2 = 6i1 + 3 (1) -1i1 + 6i2 – 3 + 4v0 = 0 (2) But, v0 = -2i1 (3)

Using (1), (2), and (3) we get i1 = -5/9. Therefore, we get v0 = -2i1 = -2(-5/9) = 1.111 volts Chapter 3, Solution 38

6 Ω

2v0 8 Ω

3 Ω

12 V –+ +

–

+ v0 –

i2

i1

We apply mesh analysis.

12 = 3 i1 + 8(i1 – i2) which leads to 12 = 11 i1 – 8 i2 (1) -2 v0 = 6 i2 + 8(i2 – i1) and v0 = 3 i1 or i1 = 7 i2 (2)

From (1) and (2), i1 = 84/69 and v0 = 3 i1 = 3x89/69 v0 = 3.652 volts Chapter 3, Solution 39 For mesh 1, 0610210 21 =−+− III x− But . Hence, 21 III x −=

212121 245610121210 IIIIII −=→−++−= (1) For mesh 2,

2112 43606812 IIII −=→=−+ (2) Solving (1) and (2) leads to -0.9A A, 8.0 21 == II


2 kΩ

i2

6 kΩ

4 kΩ

2 kΩ 6 kΩ

i3

i1

–+

4 kΩ

30V Assume all currents are in mA and apply mesh analysis for mesh 1.

30 = 12i1 – 6i2 – 4i3 15 = 6i1 – 3i2 – 2i3 (1) for mesh 2,

0 = - 6i1 + 14i2 – 2i3 0 = -3i1 + 7i2 – i3 (2) for mesh 2,

0 = -4i1 – 2i2 + 10i3 0 = -2i1 – i2 + 5i3 (3) Solving (1), (2), and (3), we obtain, io = i1 = 4.286 mA. Chapter 3, Solution 41 10 Ω

i3

ii2

2 Ω

1 Ω

8 V –+

6 V

+ –

i3

i2

i1

5 Ω 4 Ω

0

For loop 1, 6 = 12i1 – 2i2 3 = 6i1 – i2 (1) For loop 2, -8 = 7i2 – 2i1 – i3 (2) For loop 3,

-8 + 6 + 6i3 – i2 = 0 2 = 6i3 – i2 (3) We put (1), (2), and (3) in matrix form,

=

−−−

283

iii

610172016

3

2

1

,234610172016

−=−−−

=∆ 240620182036

2 −==∆

38210872316

3 −=−−−

=∆

At node 0, i + i2 = i3 or i = i3 – i2 = 234

2403823

−−−

=∆∆−∆ = 1.188 A

Chapter 3, Solution 42 For mesh 1, (1) 2121 3050120305012 IIII −=→=−+−For mesh 2, (2) 321312 40100308040301008 IIIIII −+−=→=−−+−For mesh 3, (3) 3223 50406040506 IIII +−=→=−+−Putting eqs. (1) to (3) in matrix form, we get

BAIIII

=→

=

−−−

−

68

12

50400401003003050

3

2

1

Using Matlab,

== −

44.040.048.0

1BAI

i.e. I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A Chapter 3, Solution 43

30 Ω

30 Ω

20 Ω

20 Ω

20 Ω

80 V –+

80 V –+

i3

i2

i1

a

+

Vab –

30 Ω b For loop 1, 80 = 70i1 – 20i2 – 30i3 8 = 7i1 – 2i2 – 3i3 (1)

For loop 2, 80 = 70i2 – 20i1 – 30i3 8 = -2i1 + 7i2 – 3i3 (2) For loop 3, 0 = -30i1 – 30i2 + 90i3 0 = i1 + i2 – 3i3 (3) Solving (1) to (3), we obtain i3 = 16/9

Io = i3 = 16/9 = 1.778 A Vab = 30i3 = 53.33 V. Chapter 3, Solution 44 6 V

1 Ω

4 Ω i3 i2

i1

3 A

+

6 V –+

2 Ω

5 Ω i2 i1 Loop 1 and 2 form a supermesh. For the supermesh, 6i1 + 4i2 - 5i3 + 12 = 0 (1) For loop 3, -i1 – 4i2 + 7i3 + 6 = 0 (2) Also, i2 = 3 + i1 (3) Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A

Chapter 3, Solution 45 4 Ω 8 Ω

1 Ω i1

2 Ω 6 Ω

3 Ω –+

i4i3

i2

30V For loop 1, 30 = 5i1 – 3i2 – 2i3 (1) For loop 2, 10i2 - 3i1 – 6i4 = 0 (2) For the supermesh, 6i3 + 14i4 – 2i1 – 6i2 = 0 (3) But i4 – i3 = 4 which leads to i4 = i3 + 4 (4) Solving (1) to (4) by elimination gives i = i1 = 8.561 A. Chapter 3, Solution 46

For loop 1, 12811081112 2121 =−→=−+− iiii (1)

For loop 2, 02148 21 =++− ovii

But v , 13io =

21121 706148 iiiii =→=++− (2) Substituting (2) into (1),

1739.012877 222 =→=− iii A and 217.17 21 == ii A

Chapter 3, Solution 47 First, transform the current sources as shown below.

- 6V + 2Ω I3 V1 8Ω V2 4Ω V3 4Ω 8Ω I1 2Ω I2 + + 20V 12V - - For mesh 1,

321321 47100821420 IIIIII −−=→=−−+− (1) For mesh 2,

321312 2760421412 IIIIII −+−=−→=−−+ (2) For mesh 3,

321123 7243084146 IIIIII +−−=→=−−+− (3) Putting (1) to (3) in matrix form, we obtain

BAIIII

=→

−=

−−−−−−

36

10

724271417

3

2

1

Using MATLAB,

8667.1,0333.0 ,5.28667.10333.02

3211 ===→

== − IIIBAI

But

1 120 20 4 10 V

4VI V−

= → = − =1I

2 1 22( ) 4.933 VV I I= − = Also,

32 3 2

12 12 8 12.267V8

VI V I−= → = + =

Chapter 3, Solution 48 We apply mesh analysis and let the mesh currents be in mA.

3kΩ I4 4kΩ 2kΩ 5kΩ Io 1kΩ I3 - I1 I2 6V + + 12 V + 10kΩ - 8V - For mesh 1,

421421 454045812 IIIIII −−=→=−−++− (1) For mesh 2,

43214312 2101380210138 IIIIIIII −−+−=→=−−−+− (2) For mesh 3,

432423 5151060510156 IIIIII −+−=→=−−+− (3) For mesh 4,

014524 4321 =+−−− IIII (4) Putting (1) to (4) in matrix form gives

BAI

IIII

=→

=

−−−−−−−−−−

0684

145245151002101314015

4

3

2

1

Using MATLAB,

== −

6791.7087.8217.7

1BAI

The current through the 10k resistor is IΩ o= I2 – I3 = 0.2957 mA


3 Ω

(a)

2i0

i20

16 V –+

1 Ω 2 Ω

i3

i2i1

i1

2 Ω

16V 2 Ω

i2

+ v0 –

+ v0 or–

i1

1 Ω

–+

2 Ω (b)

For the supermesh in figure (a), 3i1 + 2i2 – 3i3 + 16 = 0 (1) At node 0, i2 – i1 = 2i0 and i0 = -i1 which leads to i2 = -i1 (2) For loop 3, -i1 –2i2 + 6i3 = 0 which leads to 6i3 = -i1 (3) Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A i0 = -i1 = 10.667 A, from fig. (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V.


10 Ω

4 Ω 2 Ω i3

i2

i1

3i0

60 V –+

8 Ω i3 i2

For loop 1, 16i1 – 10i2 – 2i3 = 0 which leads to 8i1 – 5i2 – i3 = 0 (1) For the supermesh, -60 + 10i2 – 10i1 + 10i3 – 2i1 = 0 or -6i1 + 5i2 + 5i3 = 30 (2) Also, 3i0 = i3 – i2 and i0 = i1 which leads to 3i1 = i3 – i2 (3) Solving (1), (2), and (3), we obtain i1 = 1.731 and i0 = i1 = 1.731 A Chapter 3, Solution 51

5 A

+

v0

1 Ω

4 Ω

2 Ω

i3

i2

i1

20V –+40 V

–+

8 Ω

For loop 1, i1 = 5A (1) For loop 2, -40 + 7i2 – 2i1 – 4i3 = 0 which leads to 50 = 7i2 – 4i3 (2) For loop 3, -20 + 12i3 – 4i2 = 0 which leads to 5 = - i2 + 3 i3 (3) Solving with (2) and (3), i2 = 10 A, i3 = 5 A And, v0 = 4(i2 – i3) = 4(10 – 5) = 20 V. Chapter 3, Solution 52

i3

i2

2 Ω

4 Ω

8 Ω

i3

i2

i1

3A

+ v0 –

2V0 +–

–+

VS For mesh 1, 2(i1 – i2) + 4(i1 – i3) – 12 = 0 which leads to 3i1 – i2 – 2i3 = 6 (1) For the supermesh, 2(i2 – i1) + 8i2 + 2v0 + 4(i3 – i1) = 0 But v0 = 2(i1 – i2) which leads to -i1 + 3i2 + 2i3 = 0 (2) For the independent current source, i3 = 3 + i2 (3) Solving (1), (2), and (3), we obtain,

i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.


i3

i2

2 Ω

4 Ω

8 Ω

i3

i2

i1

3A

+ v0 –

2V0 +–

–+

VS For mesh 1, 2(i1 – i2) + 4(i1 – i3) – 12 = 0 which leads to 3i1 – i2 – 2i3 = 6 (1) For the supermesh, 2(i2 – i1) + 8i2 + 2v0 + 4(i3 – i1) = 0 But v0 = 2(i1 – i2) which leads to -i1 + 3i2 + 2i3 = 0 (2) For the independent current source, i3 = 3 + i2 (3) Solving (1), (2), and (3), we obtain,

i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.


Let the mesh currents be in mA. For mesh 1, 2121 22021012 IIII −=→=−++− (1)

For mesh 2, (2) 321312 3100310 IIIIII −+−=→=−−+−

For mesh 3, 3223 2120212 IIII +−=→=−+− (3)

Putting (1) to (3) in matrix form leads to

BAIIII

=→

=

−−−

−

12102

210131

012

3

2

1

Using MATLAB,

mA 25.10,mA 5.8 ,mA 25.525.105.825.5

3211 ===→

== − IIIBAI


dI1 i2

a 0

b c

1A

4A

i1

i3

I4

1A

I2

I3

8 V

12 Ω 4 Ω

2 Ω

6 Ω

10 V

+

+ –

4 A

I4

I3

I2

It is evident that I1 = 4 (1) For mesh 4, 12(I4 – I1) + 4(I4 – I3) – 8 = 0 (2) For the supermesh 6(I2 – I1) + 10 + 2I3 + 4(I3 – I4) = 0 or -3I1 + 3I2 + 3I3 – 2I4 = -5 (3) At node c, I2 = I3 + 1 (4) Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At node b, i1 = I2 – I1 = -1A At node a, i2 = 4 – I4 = 0A At node 0, i3 = I4 – I3 = 2A Chapter 3, Solution 56 + v1 –

2 Ω

2 Ω

2 Ω

2 Ω

2 Ω

i3

i2

i1 12 V

–+

+

v2 –

For loop 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3 (1) For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3 (2) For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 + 3i3 (3)

In matrix form (1), (2), and (3) become,

=

−−−−−−

006

iii

311131112

3

2

1

∆ = ,8311131112=

−−−−−−

∆2 = 24301131162=

−−−−

∆3 = 24011031612=

−−−

−, therefore i2 = i3 = 24/8 = 3A,

v1 = 2i2 = 6 volts, v = 2i3 = 6 volts Chapter 3, Solution 57

Assume R is in kilo-ohms. VVVVmAxkV 2872100100,72184 212 =−=−==Ω=

Current through R is

RR

RiViR

i RoR )18(3

3283

31, +

=→=+

=

This leads to R = 84/26 = 3.23 kΩ Chapter 3, Solution 58 30 Ω

10 Ω 30 Ω 10 Ω

i3

i2

i1

120 V

–+

30 Ω

For loop 1, 120 + 40i1 – 10i2 = 0, which leads to -12 = 4i1 – i2 (1) For loop 2, 50i2 – 10i1 – 10i3 = 0, which leads to -i1 + 5i2 – i3 = 0 (2) For loop 3, -120 – 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3 (3) Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A Chapter 3, Solution 59

i1

2I0

I0

10 Ω

20 Ω

40 Ω

+ v0 –

100V –+

120 V

– +

+–

4v0 i3

i2

80 Ω

i2 i3 For loop 1, -100 + 30i1 – 20i2 + 4v0 = 0, where v0 = 80i3 or 5 = 1.5i1 – i2 + 16i3 (1) For the supermesh, 60i2 – 20i1 – 120 + 80i3 – 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 – 12i3 (2) Also, 2I0 = i3 – i2 and I0 = i2, hence, 3i2 = i3 (3)

From (1), (2), and (3),

−−−

−

1301231

3223

=

06

10

iii

3

2

1

∆ = ,5130

12313223

=−−−

− ∆2 = ,28

1001261

32103−=

−−− ∆3 = 84

030631

1023−=−

−

I0 = i2 = ∆2/∆ = -28/5 = -5.6 A v0 = 8i3 = (-84/5)80 = -1344 volts

Chapter 3, Solution 60 0.5i0

i0

v1

1 Ω

4 Ω 8 Ω 10 V

10 V –+

v2 2 Ω At node 1, (v1/1) + (0.5v1/1) = (10 – v1)/4, which leads to v1 = 10/7 At node 2, (0.5v1/1) + ((10 – v2)/8) = v2/2 which leads to v2 = 22/7 P1Ω = (v1)2/1 = 2.041 watts, P2Ω = (v2)2/2 = 4.939 watts P4Ω = (10 – v1)2/4 = 18.38 watts, P8Ω = (10 – v2)2/8 = 5.88 watts Chapter 3, Solution 61

+ v0 –

20 Ω v2v1

+–30 Ω 5v0

40 Ω

10 Ω i0 is At node 1, is = (v1/30) + ((v1 – v2)/20) which leads to 60is = 5v1 – 3v2 (1) But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1 Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = -0.3


i1

4 kΩ A

–+

B8 kΩ

100V –+

i3i2

2 kΩ 40 V We have a supermesh. Let all R be in kΩ, i in mA, and v in volts. For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3 (1) At node A, i1 + 4 = i2 (2) At node B, i2 = 2i1 + i3 (3) Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA. Chapter 3, Solution 63

A

5 Ω

10 Ω

i2i1

+– 4ix

50 V –+

For the supermesh, -50 + 10i1 + 5i2 + 4ix = 0, but ix = i1. Hence, 50 = 14i1 + 5i2 (1) At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 – i2), hence, i1 + 2 = i2 (2) Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A vx = 2(i1 – i2) = -4 volts and ix = i2 – 2 = 4.105 amp


i0

i1

2 A

10 Ω

50 Ω A 10 Ω i2

+ −

0.2V0

4i0 +–

100V –+

i3

i2i1

40 Ω i1 i3B For mesh 2, 20i2 – 10i1 + 4i0 = 0 (1) But at node A, io = i1 – i2 so that (1) becomes i1 = (7/12)i2 (2) For the supermesh, -100 + 50i1 + 10(i1 – i2) – 4i0 + 40i3 = 0 or 50 = 28i1 – 3i2 + 20i3 (3) At node B, i3 + 0.2v0 = 2 + i1 (4) But, v0 = 10i2 so that (4) becomes i3 = 2 – (17/12)i2 (5) Solving (1) to (5), i2 = -0.674,

v0 = 10i2 = -6.74 volts, i0 = i1 - i2 = -(5/12)i2 = 0.281 amps Chapter 3, Solution 65

For mesh 1, 421 61212 III −−= (1) For mesh 2, 54321 81660 IIIII −−−+−= (2) For mesh 3, 532 1589 III −+−= (3) For mesh 4, 5421 256 IIII −+−−= (4) For mesh 5, 5432 8210 IIII +−−−= (5)

Casting (1) to (5) in matrix form gives

BAI

IIIII

=→

=

−−−−−−−−−−−−

−

10690

12

8211025011101580118166

010612

5

4

3

2

1

Using MATLAB leads to

== −

411.2864.2733.1824.1673.1

1BAI

Thus, A 411.2 A, 864.1 A, 733.1 A, 824.1 A, 673.1 54321 ===== IIIII

Chapter 3, Solution 66 Consider the circuit below.

2 kΩ 2 kΩ

+ + 20V I1 1 kΩ I2 10V - - 1 k 1 kΩ Ω Io

1 kΩ 2 kΩ 2 kΩ I3 I4 - 12V + We use mesh analysis. Let the mesh currents be in mA. For mesh 1, (1) 321420 III −−=For mesh 2, (2) 421 410 III −+−=−For mesh 3, (3) 431 412 III −+−=For mesh 4, (4) 432 412 III +−−=−

In matrix form, (1) to (4) become

BAI

IIII

=→

−

−=

−−−−−−

−−

121210

20

411014011041

0114

4

3

2

1

Using MATLAB,

−

−== −

5.275.375.15.5

1BAI

Thus, mA 75.33 −=−= II o Chapter 3, Solution 67

G11 = (1/1) + (1/4) = 1.25, G22 = (1/1) + (1/2) = 1.5 G12 = -1 = G21, i1 = 6 – 3 = 3, i2 = 5-6 = -1

Hence, we have,

−

=

−

−1

3vv

5.11125.1

2

1

∆

=

−

− −

25.1115.11

5.11125.1 1

, where ∆ = [(1.25)(1.5)-(-1)(-1)] = 0.875

=

−−

=

−

=

24

)4286.1(1)1429.1(3)1429.1(1)7143.1(3

13

4286.11429.11429.17143.1

vv

2

1

Clearly v1 = 4 volts and v2 = 2 volts

Chapter 3, Solution 68 By inspection, G11 = 1 + 3 + 5 = 8S, G22 = 1 + 2 = 3S, G33 = 2 + 5 = 7S

G12 = -1, G13 = -5, G21 = -1, G23 = -2, G31 = -5, G32 = -2

i1 = 4, i2 = 2, i3 = -1

We can either use matrix inverse as we did in Problem 3.51 or use Cramer’s Rule. Let us use Cramer’s rule for this problem.

First, we develop the matrix relationships.

−=

−−−−−−

124

vvv

725231518

3

2

1

85721232514

,34725231518

1 =−−

−−−

=∆=−−

−−−−

=∆

87125

231418

,109715221548

32 =−−−

−−

=∆=−−

−−−

=∆

v1 = ∆1/∆ = 85/34 = 3.5 volts, v2 = ∆2/∆ = 109/34 = 3.206 volts

v3 = ∆3/∆ = 87/34 = 2.56 volts


Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G11 = (1/2) + (1/4) + (1/1) = 1.75, G22 = (1/4) + (1/4) + (1/2) = 1, G33 = (1/1) + (1/4) = 1.25, G12 = -1/4 = -0.25, G13 = -1/1 = -1, G21 = -0.25, G23 = -1/4 = -0.25, G31 = -1, G32 = -0.25 i1 = 20, i2 = 5, and i3 = 10 – 5 = 5 The node-voltage equations are:

=

−−−−−−

55

20

vvv

25.125.0125.0125.0125.075.1

3

2

1


G11 = G1 + G2 + G4, G12 = -G2, G13 = 0, G22 = G2 + G3, G21 = -G2, G23 = -G3, G33 = G1 + G3 + G5, G31 = 0, G32 = -G3 i1 = -I1, i2 = I2, and i3 = I1

Then, the node-voltage equations are:

−=

++−−+−

−++

1

2

1

3

2

1

5313

3212

2421

III

vvv

GGGG0GGGG0GGGG


R11 = 4 + 2 = 6, R22 = 2 + 8 + 2 = 12, R33 = 2 + 5 = 7, R12 = -2, R13 = 0, R21 = -2, R23 = -2, R31 = 0, R32 = -2 v1 = 12, v2 = -8, and v3 = -20

Now we can write the matrix relationships for the mesh-current equations.

−−=

−−−

−

208

12

iii

7202122

026

3

2

1

Now we can solve for i2 using Cramer’s Rule.

4087200282

0126,452

7202122

026

2 −=−

−−−=∆=−

−−−

=∆

i2 = ∆2/∆ = -0.9026, p = (i2)2R = 6.52 watts Chapter 3, Solution 72

R11 = 5 + 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5, R12 = -2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0, R32 = -4, R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for all i not equal to j. v1 = 8, v2 = 4, v3 = -10, and v4 = -4

Hence the mesh-current equations are:

−−

=

−−−

−−−

41048

iiii

51001540

04620027

4

3

2

1


R11 = 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1 + 4 = 5, R44 = 1 + 1 = 2, R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1 v1 = 6, v2 = 4, v3 = 2, and v4 = -3

Hence,

−

=

−−−

−−−

3246

iiii

21001604

00830439

4

3

2

1


R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8, R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0, R24 = -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal to j.

The input voltage vector is =

−

−

4

3

2

1

VVV

V

−

−=

++−−−++−−++−

−−++

4

3

2

1

4

3

2

1

85385

88766

55424

64641

VVV

V

iiii

RRRRR0RRRR0RR0RRRR0RRRRR

Chapter 3, Solution 75 * Schematics Netlist * R_R4 $N_0002 $N_0001 30 R_R2 $N_0001 $N_0003 10 R_R1 $N_0005 $N_0004 30 R_R3 $N_0003 $N_0004 10 R_R5 $N_0006 $N_0004 30 V_V4 $N_0003 0 120V v_V3 $N_0005 $N_0001 0 v_V2 0 $N_0006 0 v_V1 0 $N_0002 0

i3

i2i1

Clearly, i1 = -3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees with the answers in Problem 3.44.

Chapter 3, Solution 76 * Schematics Netlist * I_I2 0 $N_0001 DC 4A R_R1 $N_0002 $N_0001 0.25 R_R3 $N_0003 $N_0001 1 R_R2 $N_0002 $N_0003 1 F_F1 $N_0002 $N_0001 VF_F1 3 VF_F1 $N_0003 $N_0004 0V R_R4 0 $N_0002 0.5 R_R6 0 $N_0001 0.5 I_I1 0 $N_0002 DC 2A R_R5 0 $N_0004 0.25

Clearly, v1 = 625 mVolts, v2 = 375 mVolts, and v3 = 1.625 volts, which agrees with the solution obtained in Problem 3.27.

Chapter 3, Solution 77 * Schematics Netlist * R_R2 0 $N_0001 4 I_I1 $N_0001 0 DC 3A I_I3 $N_0002 $N_0001 DC 6A R_R3 0 $N_0002 2 R_R1 $N_0001 $N_0002 1 I_I2 0 $N_0002 DC 5A

Clearly, v1 = 4 volts and v2 = 2 volts, which agrees with the answer obtained in Problem 3.51.

Chapter 3, Solution 78 The schematic is shown below. When the circuit is saved and simulated the node voltages are displaced on the pseudocomponents as shown. Thus,

,V15 V,5.4 V,3 321 −==−= VVV

.

Chapter 3, Solution 79 The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displaced. Thus,

V 88.26V V, 6944.0V V, 28.10V V, 278.5V dcba −===−=

Chapter 3, Solution 80 * Schematics Netlist * H_H1 $N_0002 $N_0003 VH_H1 6 VH_H1 0 $N_0001 0V I_I1 $N_0004 $N_0005 DC 8A V_V1 $N_0002 0 20V R_R4 0 $N_0003 4 R_R1 $N_0005 $N_0003 10 R_R2 $N_0003 $N_0002 12 R_R5 0 $N_0004 1 R_R3 $N_0004 $N_0001 2

Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20 volts, and v4 = -5.333 volts Chapter 3, Solution 81

Clearly, v1 = 26.67 volts, v2 = 6.667 volts, v3 = 173.33 volts, and v4 = -46.67 volts which agrees with the results of Example 3.4.

This is the netlist for this circuit. * Schematics Netlist * R_R1 0 $N_0001 2 R_R2 $N_0003 $N_0002 6 R_R3 0 $N_0002 4 R_R4 0 $N_0004 1 R_R5 $N_0001 $N_0004 3 I_I1 0 $N_0003 DC 10A V_V1 $N_0001 $N_0003 20V E_E1 $N_0002 $N_0004 $N_0001 $N_0004 3 Chapter 3, Solution 82

+ v0 –

4

3 kΩ

2 kΩ

4 kΩ 8 kΩ

6 kΩ

0

1 2 33v0

2i0

100V –+

4A +

This network corresponds to the Netlist.

Chapter 3, Solution 83 The circuit is shown below.

+ v0 –

4

3 kΩ

2 kΩ

4 kΩ 8 kΩ

6 kΩ

1 3

20 V –+ 30 Ω 2 A

0

50 Ω

0

1 2 3

3v0

2i0 2

100V –+

4A +

20 Ω 70 Ω When the circuit is saved and simulated, we obtain v2 = -12.5 volts Chapter 3, Solution 84

From the output loop, v0 = 50i0x20x103 = 106i0 (1) From the input loop, 3x10-3 + 4000i0 – v0/100 = 0 (2) From (1) and (2) we get, i0 = 0.5µA and v0 = 0.5 volt. Chapter 3, Solution 85

The amplifier acts as a source. Rs + Vs RL - For maximum power transfer, Ω== 9sL RR

Chapter 3, Solution 86 Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then,

[(0.03 – v1)/1k] + 400i = v1/2k (1) Assume that i is in mA. But, i = (0.03 – v1)/1 (2) Combining (1) and (2) yields, v1 = 29.963 mVolts and i = 37.4 nA, therefore, v0 = -5000x400x37.4x10-9 = -74.8 mvolts


v1 = 500(vs)/(500 + 2000) = vs/5 v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs, Therefore, v0/vs = -8

Chapter 3, Solution 88 Let v1 be the potential at the top end of the 100-ohm resistor. (vs – v1)/200 = v1/100 + (v1 – 10-3v0)/2000 (1) For the right loop, v0 = -40i0(10,000) = -40(v1 – 10-3)10,000/2000, or, v0 = -200v1 + 0.2v0 = -4x10-3v0 (2)

Substituting (2) into (1) gives, (vs + 0.004v1)/2 = -0.004v0 + (-0.004v1 – 0.001v0)/20

This leads to 0.125v0 = 10vs or (v0/vs) = 10/0.125 = -80

Chapter 3, Solution 89 vi = VBE + 40k IB (1) 5 = VCE + 2k IC (2) If IC = βIB = 75IB and VCE = 2 volts, then (2) becomes 5 = 2 +2k(75IB) which leads to IB = 20 µA.

Substituting this into (1) produces, vi = 0.7 + 0.8 = 1.5 volts. 2 kΩ IB 40 kΩ

+ VBE

–

-+

-+ vi 5 v

Chapter 3, Solution 90 1 kΩ

100 kΩ

IE

IB

+ VBE

–

+ VCE –

i2

i1 -+

18V -+

500 Ω + V0 –

vs For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 + 10,000IB + 500(1 + β)IB which leads to vs + 0.7 = 10,000IB + 500(151)IB = 85,500IB But, v0 = 500IE = 500x151IB = 4 which leads to IB = 5.298x10-5 Therefore, vs = 0.7 + 85,500IB = 5.23 volts

Chapter 3, Solution 91 We first determine the Thevenin equivalent for the input circuit.

RTh = 6||2 = 6x2/8 = 1.5 kΩ and VTh = 2(3)/(2+6) = 0.75 volts 5 kΩ

IC

1.5 kΩ

IE

IB

+ VBE

–

+ VCE –

i2

i1 -+

9 V -+

400 Ω + V0 –

0.75 V For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500IB + 400(1 + β)IB IB = 0.05/81,900 = 0.61 µA v0 = 400IE = 400(1 + β)IB = 49 mV For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = βIB and IE = (1 + β)IB VCE = 9 – 5kβIB – 400(1 + β)IB = 9 – 0.659 = 8.641 volts Chapter 3, Solution 92

IC

VC

10 kΩ

IE

IB

+ VBE

–

+ VCE –

12V -+

4 kΩ + V0 –

I1 5 kΩ

I1 = IB + IC = (1 + β)IB and IE = IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB + 5kI1 = 12 12 – 0.7 = 5k(1 + β)IB + 10kIB + 4k(1 + β)IB = 919kIB IB = 11.3/919k = 12.296 µA Also, 12 = 5kI1 + VC which leads to VC = 12 – 5k(101)IB = 5.791 volts Chapter 3, Solution 93

i1 i2i

4 Ω

4 Ω

2 Ω

2 Ω

1 Ω

8 Ω

3v0

+ v2 –

+ v1 –

+

–+

3v0 +

2 Ω i v2v1 i3

+ v0 –

24V

(a) (b) From (b), -v1 + 2i – 3v0 + v2 = 0 which leads to i = (v1 + 3v0 – v2)/2 At node 1 in (a), ((24 – v1)/4) = (v1/2) + ((v1 +3v0 – v2)/2) + ((v1 – v2)/1), where v0 = v2 or 24 = 9v1 which leads to v1 = 2.667 volts At node 2, ((v1 – v2)/1) + ((v1 + 3v0 – v2)/2) = (v2/8) + v2/4, v0 = v2 v2 = 4v1 = 10.66 volts Now we can solve for the currents, i1 = v1/2 = 1.333 A, i2 = 1.333 A, and i3 = 2.6667 A.


i io5 Ω

8 Ω

1 Ω

+ −

1 V 3 Ω

Ω=+ 4)35(8 , 51

411i =+

=

===101i

21io 0.1A


,3)24(6 Ω=+ A21i21 ==i

,41i

21i 1o == == oo i2v 0.5V

5 Ω 4 Ω

i2

8 Ω

i1 io

6 Ω

1 A 2 Ω If is = 1µA, then vo = 0.5µV Chapter 4, Solution 3.

R

+

vo

−

3R io

3R

3R

R

+ −

3R

+ − 1 V

1.5R Vs

(b)(a)

(a) We transform the Y sub-circuit to the equivalent ∆ .

,R43

R4R3R3R

2

== R23R

43R

43

=+

2v

v so = independent of R

io = vo/(R) When vs = 1V, vo = 0.5V, io = 0.5A (b) When vs = 10V, vo = 5V, io = 5A (c) When vs = 10V and R = 10Ω,

vo = 5V, io = 10/(10) = 500mA Chapter 4, Solution 4. If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω resistor is 2A.

+

v1

−

3A

Is 2 Ω 4 Ω

i1 3A 1A

Is

2A

6 Ω 4 Ω3 Ω

2 Ω 2 Ω

(a) (b)

Ω= 263 , vo = 3(4) = 12V, .A34

vo1 ==i

Hence Is = 3 + 3 = 6A If Is = 6A Io = 1 Is = 9A Io = 6/(9) = 0.6667A


If vo = 1V, V2131V1 =+

=

3

10v322V 1s =+

=

If vs = 3

10 vo = 1

Then vs = 15 vo = =15x103 4.5V

vo3 Ω2 Ω

+ − 6 Ω 6 Ω 6 Ω

v1

Vs


Let sT

ToT V

RRRV

RRRR

RRR132

3232 then ,//

+=

+==

133221

32

132

32

32

32

1 RRRRRRRR

RRRRRRRRR

RRR

VV

kT

T

s

o

++=

++

+=

+==

Chapter 4, Solution 7 We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below.

3Vx

5Ω 5Ω + + 4V 15Ω VTh - 6Ω - + Vx -

From the figure,

V3)4(515

15,0 =+

== Thx VV

To find RTh, consider the circuit below:

3Vx

5Ω 5Ω V1 V2 + 4V 15Ω 1A - 6Ω + Vx - At node 1,

122111 73258616,

5153

54 VVxVVVVVV

xx −=→==−

++=− (1)

At node 2,

9505

31 2121 −=→=

−++ VV

VVVx (2)

Solving (1) and (2) leads to V2 = 101.75 V

mW 11.2275.1014

94

,75.1011

2

max2 ===Ω==

xRV

pVRTh

ThTh

Chapter 4, Solution 8. Let i = i1 + i2, where i1 and iL are due to current and voltage sources respectively.

6 Ω

i1 + − 20V

i2

5 A 4 Ω 6 Ω

4 Ω (a) (b)

i1 = ,A3)5(46

6=

+ A2

4620

2 =+

=i

Thus i = i1 + i2 = 3 + 2 = 5A Chapter 4, Solution 9. Let i

2x1xx ii += where i is due to 15V source and i is due to 4A source,

1x 2x

12Ω

-4A

40Ω 10Ωix2

40Ω

i

10 Ω

ix1

12 Ω

+ −

15V

(a) (b)

For ix1, consider Fig. (a).

10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75

ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6 For ix2, consider Fig. (b).

12||40 = 480/52 = 120/13

ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92

ix = 0.6 – 1.92 = -1.32 A

p = vix = ix2R = (-1.32)210 = 17.43 watts

Chapter 4, Solution 10. Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources respectively.

+

vab2

−

10 Ω+ −

3vab2

2 A

10 Ω

+ −

+

vab1

−

+ −

3vab1

4V

(a) (b) For vab1, consider Fig. (a). Applying KVL gives,

- vab1 – 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V For vab2, consider Fig. (b). Applying KVL gives,

- vab2 – 3vab2 + 10x2 = 0, which leads to vab2 = 5

vab = 1 + 5 = 6 V

Chapter 4, Solution 11. Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to the 4-A source.

12V

4A

2Ω 2Ωix2

6Ω

4A

3Ω2Ω i2

3Ω

io

(a)

2 Ω

i1

6 Ω

+ −

(b) For i1, consider Fig. (a).

2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6

i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A For i2, consider Fig. (b), 6||3 = 2 ohm, i2 = 4/2 = 2 A

i = 1 + 2 = 3 A Chapter 4, Solution 12. Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V sources respectively. For vo1, consider the circuit below.

5 Ω

5 Ω

+ vo1 −

io

2A2A

3Ω

4 Ω

6Ω 12 Ω

5 Ω

+ vo1 −

6||3 = 2 ohms, 4||12 = 3 ohms. Hence,

io = 2/2 = 1, vo1 = 5io = 5 V

For vo2, consider the circuit below. 6 Ω 5 Ω 4 Ω 6 Ω 5 Ω

+ vo2 −

3 Ω3 Ω +

v1

−

+ − 12V

+ vo2 −

12 Ω

+ −

3 Ω

12V

3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5

vo2 = (5/8)v1 = (5/8)(16/5) = 2 V For vo3, consider the circuit shown below. 4 Ω 5 Ω 5 Ω 4 Ω

2Ω+ vo3 −

12 Ω+

v2

−

+ − 19V

+ −

19V 6Ω + vo3 −

12 Ω 3 Ω

7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975

v = (-5/7)v2 = -7.125

vo = 5 + 2 – 7.125 = -125 mV

,

Chapter 4, Solution 13 Let iiii 321o ++= where i1, i2, and i3 are the contributions to io due to 30-V, 15-V, and 6-mA sources respectively. For i1, consider the circuit below.

1 kΩ 2 kΩ 3 kΩ + i1 30V - 4 kΩ 5 kΩ

3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949 kohm. After combining the resistors except the 4-kohm resistor and transforming the voltage source, we obtain the circuit below. i1 30 mA 4 kΩ 0.7949 kΩ Using current division,

mA 4.973mA)30(7949.47949.0

1 ==i

For i2, consider the circuit below. 1 kΩ 2 kΩ 3 kΩ i2 - 15V 4 kΩ 5 kΩ +

After successive source transformation and resistance combinations, we obtain the circuit below: 2.42mA i2 4 kΩ 0.7949 kΩ


mA 4012.0mA)42.2(7949.47949.0

2 −=−=i

For i3, consider the circuit below.

6mA 1 kΩ 2 kΩ 3 kΩ i3 4 kΩ 5 kΩ

After successive source transformation and resistance combinations, we obtain the circuit below: 3.097mA i3 4 kΩ 0.7949 kΩ

mA 5134.0mA)097.3(7949.47949.0

3 −=−=i

Thus, mA 058.4321 =++= iiiio Chapter 4, Solution 14. Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A sources respectively. For vo1, consider the circuit below. 6 Ω

20V

+ −

+

vo1

−

4 Ω 2 Ω

3 Ω

6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V

For vo2, consider the circuit below. 6 Ω 6 Ω

1A

2 Ω 4 Ω

+

vo2

−

4V

− + 2 Ω 4 Ω

3 Ω

+

vo2

−

3 Ω

3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V For vo3, consider the circuit below. 6 Ω

3 Ω

− vo3 +

3 Ω

2A

2 Ω 4 Ω

3 Ω

+

vo3

−

2A

6||(4 + 2) = 3, vo3 = (-1)3 = -3 vo = 10 + 1 – 3 = 8 V

Chapter 4, Solution 15. Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For i1, consider the circuit below.

io

4Ω

3Ω

i1

1 Ω

+ −

20V 2 Ω

4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A For i3, consider the circuit below.

2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4

i3 = vo’/4 = -1 For i2, consider the circuit below.

3Ω i2

1 Ω (4/3)Ω

3Ωi2

1 Ω 2A

4Ω

+

vo’

−

4Ω

3Ω

i3

1 Ω2 Ω

− +

16V

2A 2 Ω

2||4 = 4/3, 3 + 4/3 = 13/3 Using the current division principle.

i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375

i = 2.5 + 0.375 - 1 = 1.875 A

p = i2R = (1.875)23 = 10.55 watts

Chapter 4, Solution 16. Let io = io1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below.

5Ω

io1

10 Ω

4 Ω

+ −

3 Ω 2 Ω

12V

10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A For io2, consider the circuit below.

io2

i1

4A

10Ω

2 Ω

4Ω

3 Ω

5 Ω

2 + 5 + 4||10 = 7 + 40/14 = 69/7 i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9

For io3, consider the circuit below.

i2

2 A

io3

10 Ω 5 Ω 4 Ω

3 Ω 2 Ω

3 + 2 + 4||10 = 5 + 20/7 = 55/7

i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9

io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA

Chapter 4, Solution 17. Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90-V, 6-A, and 40-V sources. For vx1, consider the circuit below.

30 Ω 10 Ω 20 Ω

12 Ω

+ − vx1

10 Ω

20 Ω 3 A

io

30 Ω

+ − vx1 60 Ω

+ −

90V

20||30 = 12 ohms, 60||30 = 20 ohms By using current division,

io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42 = 14.286 V For vx2, consider the circuit below.

io’

20 Ω30 Ω

io’

6A 60 Ω 30 Ω

10 Ω

+ − vx2 + vx2 −

6A 20 Ω 12 Ω

10 Ω

io’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io’ = -17.143 V

For vx3, consider the circuit below.

20 Ω

+ − vx3 7.5Ω

4A

io”

+ − 40V30 Ω 30 Ω

+ − vx3 60 Ω

10 Ω10 Ω 10 Ω

io” = [12/(12 + 30)]2 = 24/42, vx3 = -10io” = -5.714

vx = 14.286 – 17.143 – 5.714 = -8.571 V

Chapter 4, Solution 18. Let ix = i1 + i2, where i1 and i2 are due to the 10-V and 2-A sources respectively. To obtain i1, consider the circuit below. 2 Ω

+ −

10i1i1

2 Ω

+ − 10V

1 Ωi1

4 Ω5i1

1 Ω

+ −

4 Ω10V

-10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A For i2, consider the circuit below.

io+ −

10i2 2 Ω

+ − 2V

1 Ωio i2 1 Ω

4 Ω

2 Ω

10i2 2 A

+ − 4 Ω

-2 + 10i2 + 7io = 0, but i2 + 2 = io. Hence,

-2 + 10i2 +7i2 + 14 = 0, or i2 = (-12/17) A

vx = 1xix = 1(i1 + i2) = (10/17) – (12/17) = -2/17 = -117.6 mA Chapter 4, Solution 19. Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources respectively. ix v1 ix v2

+

v2

−

+

v1

−

8 Ω 2 Ω

4ix

6 A

− +

8Ω2 Ω

4ix

4 A

− +

(a) (b)

To find v1, consider the circuit in Fig. (a).

v1/8 = 4 + (-4ix – v1)/2 But, -ix = (-4ix – v1)/2 and we have -2ix = v1. Thus,

v1/8 = 4 + (2v1 – v1)/8, which leads to v1 = -32/3 To find v2, consider the circuit shown in Fig. (b).

v2/2 = 6 + (4ix – v2)/8 But ix = v2/2 and 2ix = v2. Therefore,

v2/2 = 6 + (2v2 – v2)/8 which leads to v2 = -16 Hence, vx = –(32/3) – 16 = -26.67 V Chapter 4, Solution 20. Transform the voltage sources and obtain the circuit in Fig. (a). Combining the 6-ohm and 3-ohm resistors produces a 2-ohm resistor (6||3 = 2). Combining the 2-A and 4-A sources gives a 6-A source. This leads to the circuit shown in Fig. (b).

i

6Ω 4A3Ω2Ω i

6A 2 Ω 2 Ω

2A (a) (b) From Fig. (b), i = 6/2 = 3 A Chapter 4, Solution 21. To get io, transform the current sources as shown in Fig. (a). 6 Ω

+

vo

−

2 A 3 Ωi

6 Ω2 A

+ − 6V

io 3 Ω

+ − 12V

(a) (b)

From Fig. (a), -12 + 9io + 6 = 0, therefore io = 666.7 mA To get vo, transform the voltage sources as shown in Fig. (b).

i = [6/(3 + 6)](2 + 2) = 8/3

vo = 3i = 8 V Chapter 4, Solution 22. We transform the two sources to get the circuit shown in Fig. (a). 5 Ω 5 Ω

10V

4Ω 10Ω 2A

i 10Ω1A

2A

(a)

10Ω4 Ω

− +

(b) We now transform only the voltage source to obtain the circuit in Fig. (b). 10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA

Chapter 4, Solution 23 If we transform the voltage source, we obtain the circuit below. 8Ω 10Ω 6Ω 3Ω 5A 3A 3//6 = 2-ohm. Convert the current sources to voltages sources as shown below. 10 8Ω Ω 2Ω + + 10V 30V -

- Applying KVL to the loop gives

A 10)2810(1030 =→=++++− II W82 === RIVIp

Chapter 4, Solution 24 Convert the current source to voltage source.

16Ω 1Ω

4Ω + 5Ω + 48 V 10Ω Vo

- + -

12 V - Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current Sources. We obtain the circuit below. 1Ω 2.4A 20 5Ω Ω 2.4A 10Ω Combine the resistors and current sources. 20//5 = (20x5)/25 = 4 , 2.4 + 2.4 = 4.8 A ΩConvert the current source to voltage source. We obtain the circuit below. 4Ω 1Ω + + 19.2V Vo 10Ω - - Using voltage division,

8.12)2.19(1410

10=

++=oV V

Chapter 4, Solution 25. Transforming only the current source gives the circuit below.

12V

30 V

5 Ω

9 Ω 18 V

+ − vo 4 Ω

2 Ω

i

+ −

− +

− +

+ −

30 V Applying KVL to the loop gives,

(4 + 9 + 5 + 2)i – 12 – 18 – 30 – 30 = 0

20i = 90 which leads to i = 4.5

vo = 2i = 9 V Chapter 4, Solution 26.

Transform the voltage sources to current sources. The result is shown in Fig. (a),

30||60 = 20 ohms, 30||20 = 12 ohms 10 Ω

+ + vx −

12 Ω10 Ω

− 96V i

20 Ω

+ − 60V

20Ω3A 2A

+ vx −

60Ω 30Ω6A

(a)

30Ω

(b)

Combining the resistors and transforming the current sources to voltage sources, we obtain the circuit in Fig. (b). Applying KVL to Fig. (b),

42i – 60 + 96 = 0, which leads to i = -36/42

vx = 10i = -8.571 V Chapter 4, Solution 27. Transforming the voltage sources to current sources gives the circuit in Fig. (a).

10||40 = 8 ohms Transforming the current sources to voltage sources yields the circuit in Fig. (b). Applying KVL to the loop,

-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4

vx 12i = -48 V

12 Ω

+ vx − 10Ω 20Ω40Ω5A 8A 2A

(a) 8 Ω 12 Ω 20 Ω

+ −

+ −

+ vx − 40V 200V i

(b)

Chapter 4, Solution 28. Transforming only the current sources leads to Fig. (a). Continuing with source transformations finally produces the circuit in Fig. (d).

3 Ωio 10 V

+ −

12 V

+ −

+ −

4 Ω 5 Ω2 Ω

10Ω

12 V (a) 4 Ω

5 Ω io

+ − 12V

4 Ω

+ − 11V io

io

+ − 12V 10Ω

4 Ω

2.2A10Ω

io

+ − 22 V

(b)

+ − 12V 10Ω

10 Ω

(c) (d) Applying KVL to the loop in fig. (d),

-12 + 9io + 11 = 0, produces io = 1/9 = 111.11 mA

Chapter 4, Solution 29. Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 = (4/3) k ohms

4 kΩ

It is clear that i = 3 mA which leads to vo = 1000i = 3 V If the use of source transformations was not required for this problem, the actual answer could have been determined by inspection right away since the only current that could have flowed through the 1 k ohm resistor is 3 mA. Chapter 4, Solution 30 Transform the dependent current source as shown below.

ix 24Ω 60Ω 10Ω + + 12V 30Ω 7ix - -

+

vo

−

1 kΩ +

vo

−

3 mA

2vo (4/3) kΩ

(b)

i

− +

3 mA

1.5vo

1 kΩ

2 kΩ

(a)

Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below.

ix 24Ω + 12V 30Ω 70Ω 0.1ix - Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the dependent current source as shown below.

ix 24Ω 21Ω + + 12V 2.1ix - - Applying KVL to the loop gives

mA 8.2541.47

1201.21245 ==→=+− xxx iii

Chapter 4, Solution 31. Transform the dependent source so that we have the circuit in Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in Fig. (b). 3 Ω

6 Ω

+ − vx + −

8 Ω vx/3 12V (a)

(24/7) Ω3 Ω

i +

+ − vx + −

(8/7)vx 12V

(b)

From Fig. (b),

vx = 3i, or i = vx/3. Applying KVL,

-12 + (3 + 24/7)i + (24/21)vx = 0

12 = [(21 + 24)/7]vx/3 + (8/7)vx, leads to vx = 84/23 = 3.625 V Chapter 4, Solution 32. As shown in Fig. (a), we transform the dependent current source to a voltage source,

15 Ω 10 Ω− +

50 Ω

+ −

5ix

40 Ω 60V

(a)

15 Ω

ix

50 Ω 50 Ω

(b)

+ −

ix 25 Ω

+ − 60V 2.5ix

15 Ω

−

60V 0.1ix

(c)

In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c), -60 + 40ix – 2.5ix = 0, or ix = 1.6 A Chapter 4, Solution 33.

(a) RTh = 10||40 = 400/50 = 8 ohms

VTh = (40/(40 + 10))20 = 16 V (b) RTh = 30||60 = 1800/90 = 20 ohms

2 + (30 – v1)/60 = v1/30, and v1 = VTh

120 + 30 – v1 = 2v1, or v1 = 50 V

VTh = 50 V

Chapter 4, Solution 34. To find RTh, consider the circuit in Fig. (a).

3 A

RTh = 20 + 10||40 = 20 + 400/50 = 28 ohms (b)

v2

+

VTh

20 Ω

40 Ω

10 Ω v1

+ − 40VRTh

20 Ω

40 Ω

10 Ω

(a)

To find VTh, consider the circuit in Fig. (b). At node 1, (40 – v1)/10 = 3 + [(v1 – v2)/20] + v1/40, 40 = 7v1 – 2v2 (1) At node 2, 3 + (v1- v2)/20 = 0, or v1 = v2 – 60 (2) Solving (1) and (2), v1 = 32 V, v2 = 92 V, and VTh = v2 = 92 V


RTh = Rab = 6||3 + 12||4 = 2 + 3 =5 ohms To find VTh, consider the circuit shown in Fig. (b).

RTh At node 1, 2 + (12 – v1)/6 = v1/3, or v1 = 8 At node 2, (19 – v2)/4 = 2 + v2/12, or v2 = 33/4 But, -v1 + VTh + v2 = 0, or VTh = v1 – v2 = 8 – 33/4 = -0.25

b

RTh = 5 Ω

10 Ω

+ − vo a

+ −

4 Ω

+ −

+

v2

−

+

v1

−

12Ω

v2+ VTh

3 Ω

2 A

v1

+ − 12V

6 Ω

(b)

a b

4 Ω12 Ω6 Ω 3 Ω

(a)

19V

VTh = (-1/4)V

vo = VTh/2 = -0.25/2 = -125 mV

Chapter 4, Solution 36. Remove the 30-V voltage source and the 20-ohm resistor.

a

b

10Ω +

VTh

+ − 50V

a

b

40Ω

RTh

10Ω

40Ω

(a) (b) From Fig. (a), RTh = 10||40 = 8 ohms From Fig. (b), VTh = (40/(10 + 40))50 = 40V

+ −

a

b

12 Ω

(c)

i

+ − 40V

8 Ω

30V The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL, 30 – 40 + (8 + 12)i = 0, which leads to i = 500mA

Chapter 4, Solution 37 RN is found from the circuit below.

20 Ω a 40Ω 12Ω b

Ω=+= 10)4020//(12NR IN is found from the circuit below.

2A

20 Ω a + 40Ω 120V 12Ω - IN b Applying source transformation to the current source yields the circuit below.

20Ω 40Ω + 80 V - + 120V IN - Applying KVL to the loop yields

A 6667.060/4006080120 ==→=++− NN II

Chapter 4, Solution 38 We find Thevenin equivalent at the terminals of the 10-ohm resistor. For RTh, consider the circuit below. 1Ω

4Ω 5Ω RTh 16 Ω

Ω=+=++= 541)164//(51ThR For VTh, consider the circuit below. 1Ω

V1 4Ω V2 5Ω + 3A 16Ω VTh

+ -

12 V - At node 1,

21211 4548

4163 VV

VVV−=→

−+= (1)

At node 2,

21221 95480

512

4VVVVV

+−=→=−

+− (2)

Solving (1) and (2) leads to 2.192 ==VVTh

Thus, the given circuit can be replaced as shown below. 5 Ω + + 19.2V Vo 10Ω - - Using voltage division,

8.12)2.19(510

10=

+=oV V


- 1 – 3 + 10io = 0, or io = 0.4

RTh = 1/io = 2.5 ohms To find VTh, consider the circuit shown in Fig. (b).

[(4 – v)/10] + 2 = 0, or v = 24 But, v = VTh + 3vab = 4VTh = 24, which leads to VTh = 6 V Chapter 4, Solution 40. To find RTh, consider the circuit in Fig. (a).

40V

a 10 Ω

(a)

+ −

1V50V

+

v1

−

+ −

3vab 10 Ω io

b

a

(b)

+ −

440 ΩV

+ −

3vab 20 Ω

2A

+ −

v

40Ω

RTh

a b 10 Ω 20 Ω

+

v2

−

+ VTh

8 A

+ −

(b)

10 Ω

+

vab = VTh

−

b

(a)

RTh = 10||40 + 20 = 28 ohms To get VTh, consider the circuit in Fig. (b). The two loops are independent. From loop 1,

v1 = (40/50)50 = 40 V For loop 2, -v2 + 20x8 + 40 = 0, or v2 = 200 But, VTh + v2 – v1 = 0, VTh = v1 = v2 = 40 – 200 = -160 volts This results in the following equivalent circuit. 28 Ω

vx = [12/(12 + 28)](-160) = -48 V Chapter 4, Solution 41 To find RTh, consider the circuit below 14Ω a 6Ω 5 Ω b

+ −

+

vx

−

12 Ω-160V

NTh RR =Ω=+= 4)614//(5 Applying source transformation to the 1-A current source, we obtain the circuit below.

6Ω - 14V + 14Ω VTh a + 6V 3A 5Ω - b At node a,

V 85

3146

614−=→+=

+−+

ThThTh VVV

A 24/)8( −=−==Th

ThN R

VI

Thus, A 2 V,8,4 −=−=Ω== NThNTh IVRR

Chapter 4, Solution 42. To find RTh, consider the circuit in Fig. (a). 20 Ω 30 Ω

10 Ω

20 Ω 10 Ω

10Ω 10 Ω

a

10 Ω10 Ω

30 Ω 30 Ωb

ba (a) (b) 20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b). 10||30 = 7.5 ohms. RTh = Rab = 30||(7.5 + 7.5) = 10 ohms. To find VTh, we transform the 20-V and the 5-V sources. We obtain the circuit shown in Fig. (c).

+ − 50V

10 Ω

10 Ω10 Ω

10 Ω − +

10 V

a 10 Ω+ b

+ − 30V

i2i1

(c) For loop 1, -30 + 50 + 30i1 – 10i2 = 0, or -2 = 3i1 – i2 (1) For loop 2, -50 – 10 + 30i2 – 10i1 = 0, or 6 = -i1 + 3i2 (2) Solving (1) and (2), i1 = 0, i2 = 2 A Applying KVL to the output loop, -vab – 10i1 + 30 – 10i2 = 0, vab = 10 V

VTh = vab = 10 volts Chapter 4, Solution 43. To find RTh, consider the circuit in Fig. (a).

RTh

a b

5 Ω+

vb

−

+

va

−

+ VTh

10 Ω

+ − 50V

10 Ω

5 Ω10Ω

a b

10Ω

(a)

2 A

(b)

RTh = 10||10 + 5 = 10 ohms

To find VTh, consider the circuit in Fig. (b).

vb = 2x5 = 10 V, va = 20/2 = 10 V But, -va + VTh + vb = 0, or VTh = va – vb = 0 volts Chapter 4, Solution 44. (a) For RTh, consider the circuit in Fig. (a).

RTh = 1 + 4||(3 + 2 + 5) = 3.857 ohms For VTh, consider the circuit in Fig. (b). Applying KVL gives,

10 – 24 + i(3 + 4 + 5 + 2), or i = 1

VTh = 4i = 4 V 3Ω 1Ω

+ a

+ −

VTh3Ω 1Ω a 4 Ω 24V

+ −

b RTh4 Ω 10V 2 Ω b 2 Ω

i 5 Ω 5 Ω

(b) (a) (b) For RTh, consider the circuit in Fig. (c).

3Ω 1Ω 3Ω 1Ω

+ − 24V

2 Ω

2A

c

b

5 Ω

+

VTh

vo4 Ω

2 Ω

5 Ω

4 Ω

RTh

c

b

(c) (d)

RTh = 5||(2 + 3 + 4) = 3.214 ohms To get VTh, consider the circuit in Fig. (d). At the node, KCL gives,

[(24 – vo)/9] + 2 = vo/5, or vo = 15

VTh = vo = 15 V Chapter 4, Solution 45. For RN, consider the circuit in Fig. (a).

6 Ω 6 Ω

RN6 Ω 4 Ω 4A 6 Ω 4 Ω

IN

(a) (b)

RN = (6 + 6)||4 = 3 ohms For IN, consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current is equally divided between the two 6-ohm resistors. Hence, IN = 4/2 = 2 A Chapter 4, Solution 46. (a) RN = RTh = 8 ohms. To find IN, consider the circuit in Fig. (a). 10 Ω 60 Ω

4 Ω

+ − 30V 2A IN

30 Ω

+ −

Isc

20V

(a) (b) IN = Isc = 20/10 = 2 A

(b) To get IN, consider the circuit in Fig. (b). IN = Isc = 2 + 30/60 = 2.5 A

Chapter 4, Solution 47 Since VTh = Vab = Vx, we apply KCL at the node a and obtain

V 19.1126/15026012

30==→+=

−ThTh

ThTh VVVV

To find RTh, consider the circuit below.

12Ω Vx a 2Vx 60Ω

1A

At node a, KCL gives

4762.0126/601260

21 ==→++= xxx

x VVV

V

5.24762.0/19.1,4762.01

===Ω==Th

ThN

xTh R

VI

VR

Thus, A 5.2,4762.0,19.1 =Ω=== NNThTh IRRVV

Chapter 4, Solution 48. To get RTh, consider the circuit in Fig. (a).

+

VTh

−

10Io

+ −

Io

2A4 Ω

2 Ω

Io

2 Ω

4 Ω

+ − +

V

−

10Io

1A

(a) (b) From Fig. (a), Io = 1, 6 – 10 – V = 0, or V = -4

RN = RTh = V/1 = -4 ohms

To get VTh, consider the circuit in Fig. (b),

Io = 2, VTh = -10Io + 4Io = -12 V

IN = VTh/RTh = 3A Chapter 4, Solution 49. RN = RTh = 28 ohms To find IN, consider the circuit below,

3A

vo

io

20 Ω

+ −

40 Ω

10 Ω

Isc = IN 40V At the node, (40 – vo)/10 = 3 + (vo/40) + (vo/20), or vo = 40/7 io = vo/20 = 2/7, but IN = Isc = io + 3 = 3.286 A Chapter 4, Solution 50. From Fig. (a), RN = 6 + 4 = 10 ohms

6 Ω 6 Ω

4 Ω 12V

+ −

4 Ω2A

Isc = IN

(b) (a) From Fig. (b), 2 + (12 – v)/6 = v/4, or v = 9.6 V

-IN = (12 – v)/6 = 0.4, which leads to IN = -0.4 A Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c).

i

4A5 Ω

0.4A 10 Ω

(c) i = [10/(10 + 5)] (4 – 0.4) = 2.4 A Chapter 4, Solution 51. (a) From the circuit in Fig. (a),

RN = 4||(2 + 6||3) = 4||4 = 2 ohms

2 Ω

+ − 120V

+

6A

VTh

4 Ω

3 Ω

6 Ω

RTh

4 Ω

3 Ω

6 Ω

2 Ω (a) (b) For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the circuit becomes that shown in Fig. (c).

i

2 Ω

+ − 12V

+ −

+ VTh

4 Ω2 Ω

40V (c) Applying KVL to the circuit in Fig. (c),

-40 + 8i + 12 = 0 which gives i = 7/2

VTh = 4i = 14 therefore IN = VTh/RN = 14/2 = 7 A

(b) To get RN, consider the circuit in Fig. (d).

RN = 2||(4 + 6||3) = 2||6 = 1.5 ohms

i

2 Ω

+ − 12V

+VTh

RN

4 Ω

3 Ω 2 Ω

6 Ω (d) (e) To get IN, the circuit in Fig. (c) applies except that it needs slight modification as in Fig. (e).

i = 7/2, VTh = 12 + 2i = 19, IN = VTh/RN = 19/1.5 = 12.667 A Chapter 4, Solution 52. For RTh, consider the circuit in Fig. (a). a

Io

b

2 kΩ3 kΩ 20Io

RTh (a)

+ + − Io

a

b

2 kΩ VTh 20Io

3 kΩ 6V (b) For Fig. (a), Io = 0, hence the current source is inactive and

RTh = 2 k ohms

For VTh, consider the circuit in Fig. (b).

Io = 6/3k = 2 mA

VTh = (-20Io)(2k) = -20x2x10-3x2x103 = -80 V Chapter 4, Solution 53. To get RTh, consider the circuit in Fig. (a). 0.25vo0.25vo

1/2 a 2 Ω

1A

+

vo

−

2 Ω

(b) b

+

vab

−

1/2

a

b

2 Ω

1A

+

vo

−

3 Ω

(a)

6 Ω From Fig. (b),

vo = 2x1 = 2V, -vab + 2x(1/2) +vo = 0

vab = 3V

RN = vab/1 = 3 ohms To get IN, consider the circuit in Fig. (c). 0.25vo

6 Ω a 2 Ω

+

vo

−

3 Ω

+ −

18V Isc = IN b (c)

[(18 – vo)/6] + 0.25vo = (vo/2) + (vo/3) or vo = 4V

But, (vo/2) = 0.25vo + IN, which leads to IN = 1 A


To find VTh =Vx, consider the left loop.

xoxo ViVi 2100030210003 +=→=++− (1) For the right loop, (2) oox iixV 20004050 −=−=Combining (1) and (2), mA13000400010003 −=→−=−= oooo iiii 222000 =→=−= Thox ViV To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independent source, as shown below.

1 k iΩ x .

io + + + 2Vx 40io Vx 50 Ω 1V

- - -

mA210002

,1 −=−== xox

ViV

-60mAA501mA80

5040 =+−=+= x

oxV

ii

Ω−=−== 67.16060.0/11

xTh iR

Chapter 4, Solution 55. To get RN, apply a 1 mA source at the terminals a and b as shown in Fig. (a).

80I+ −

vab/1000 +

vab

−

8 kΩ

a I

50 kΩ

1mA

b (a)

We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At node a,

(vab/50) + 80I = 1 (1) Also,

-8I = (vab/1000), or I = -vab/8000 (2) From (1) and (2), (vab/50) – (80vab/8000) = 1, or vab = 100

RN = vab/1 = 100 k ohms To get IN, consider the circuit in Fig. (b).

vab/1000

I

80I

a

50 kΩ

+

vab

−

+ −

+ −

8 kΩ

IN 2V b (b) Since the 50-k ohm resistor is shorted,

IN = -80I, vab = 0 Hence, 8i = 2 which leads to I = (1/4) mA

IN = -20 mA Chapter 4, Solution 56. We first need RN and IN.

16V

2A 4 Ω

2 Ω IN

1 Ω

− +

+ − 20V

4 Ω2 Ω RN

a

b

1 Ω

(a) (b)

To find RN, consider the circuit in Fig. (a).

RN = 1 + 2||4 = (7/3) ohms To get IN, short-circuit ab and find Isc from the circuit in Fig. (b). The current source can be transformed to a voltage source as shown in Fig. (c).

vo

i

RN

a

IN

1 Ω

− + 16V 2V

4 Ω

2 Ω IN − +

+ −

20V

3 Ω b (c) (d)

(20 – vo)/2 = [(vo + 2)/1] + [(vo + 16)/4], or vo = 16/7

IN = (vo + 2)/1 = 30/7 From the Norton equivalent circuit in Fig. (d),

i = RN/(RN + 3), IN = [(7/3)/((7/3) + 3)](30/7) = 30/16 = 1.875 A Chapter 4, Solution 57. To find RTh, remove the 50V source and insert a 1-V source at a – b, as shown in Fig. (a).

B A i

10 Ω6 Ω +

vx

− 0.5vx

a

b(a)

2 Ω

+ −

1V 3 Ω We apply nodal analysis. At node A,

i + 0.5vx = (1/10) + (1 – vx)/2, or i + vx = 0.6 (1) At node B,

(1 – vo)/2 = (vx/3) + (vx/6), and vx = 0.5 (2)

From (1) and (2), i = 0.1 and

RTh = 1/i = 10 ohms To get VTh, consider the circuit in Fig. (b).

10 Ω6 Ω+

vx

− 0.5vx

a

b(b)

2 Ω

+ −

v1 3 Ω v2 +

VTh

−

50V At node 1, (50 – v1)/3 = (v1/6) + (v1 – v2)/2, or 100 = 6v1 – 3v2 (3) At node 2, 0.5vx + (v1 – v2)/2 = v2/10, vx = v1, and v1 = 0.6v2 (4) From (3) and (4),

v2 = VTh = 166.67 V

IN = VTh/RTh = 16.667 A

RN = RTh = 10 ohms Chapter 4, Solution 58. This problem does not have a solution as it was originally stated. The reason for this is that the load resistor is in series with a current source which means that the only equivalent circuit that will work will be a Norton circuit where the value of RN = infinity. IN can be found by solving for Isc.

voib

+ −

β ib

R2

R1 Isc VS Writing the node equation at node vo,

ib + βib = vo/R2 = (1 + β)ib

But ib = (Vs – vo)/R1

vo = Vs – ibR1

Vs – ibR1 = (1 + β)R2ib, or ib = Vs/(R1 + (1 + β)R2)

Isc = IN = -βib = -βVs/(R1 + (1 + β)R2) Chapter 4, Solution 59. RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms To find VTh, consider the circuit below.

10 Ω

50 Ω

+ VTh

8A

i1 i2 20 Ω

40 Ω

i1 = i2 = 8/2 = 4, 10i1 + VTh – 20i2 = 0, or VTh = 20i2 –10i1 = 10i1 = 10x4 VTh = 40V, and IN = VTh/RTh = 40/22.5 = 1.7778 A Chapter 4, Solution 60. The circuit can be reduced by source transformations. 2A

5 Ω

10 Ω18 V

+ −

12 V

+ −

10 V

+ −

2A

b a

3A

2A

10 Ω

5 Ω

3A

10 V

+ − 3.333Ωa b 3.333Ω a b

Norton Equivalent Circuit Thevenin Equivalent Circuit Chapter 4, Solution 61. To find RTh, consider the circuit in Fig. (a). Let R = 2||18 = 1.8 ohms, RTh = 2R||R = (2/3)R = 1.2 ohms. To get VTh, we apply mesh analysis to the circuit in Fig. (d).

2 Ω a

2 Ω

6 Ω

6 Ω

2 Ω

6 Ω

b

(a)

1.8 Ω

1.8 Ω

a

b

1.8 Ω RTh

18 Ω

18 Ω 18 Ω2 Ω

a

b

(b)

2 Ω

2 Ω

(c)

+

+ − 12V

i3

i2i12 Ω

6 Ω

6 Ω

a

b

2 Ω

6 Ω

12V

− +

+ − 12V

2 Ω

VTh

(d) -12 – 12 + 14i1 – 6i2 – 6i3 = 0, and 7 i1 – 3 i2 – 3i3 = 12 (1) 12 + 12 + 14 i2 – 6 i1 – 6 i3 = 0, and -3 i1 + 7 i2 – 3 i3 = -12 (2) 14 i3 – 6 i1 – 6 i2 = 0, and -3 i1 – 3 i2 + 7 i3 = 0 (3) This leads to the following matrix form for (1), (2) and (3),

−=

−−−−−−

012

12

iii

733373337

3

2

1

100733373337=

−−−−−−

=∆ , 12070331233127

2 −=−

−−−−

=∆

i2 = ∆/∆2 = -120/100 = -1.2 A

VTh = 12 + 2i2 = 9.6 V, and IN = VTh/RTh = 8 A Chapter 4, Solution 62. Since there are no independent sources, VTh = 0 V To obtain RTh, consider the circuit below.

2

2vo

0.1io ix

v1io

40 Ω

10 Ω

+ −

+ − VS

+vo −1

20 Ω

At node 2,

ix + 0.1io = (1 – v1)/10, or 10ix + io = 1 – v1 (1) At node 1, (v1/20) + 0.1io = [(2vo – v1)/40] + [(1 – v1)/10] (2) But io = (v1/20) and vo = 1 – v1, then (2) becomes,

1.1v1/20 = [(2 – 3v1)/40] + [(1 – v1)/10]

2.2v1 = 2 – 3v1 + 4 – 4v1 = 6 – 7v1 or v1 = 6/9.2 (3) From (1) and (3),

10ix + v1/20 = 1 – v1

10ix = 1 – v1 – v1/20 = 1 – (21/20)v1 = 1 – (21/20)(6/9.2)

ix = 31.52 mA, RTh = 1/ix = 31.73 ohms.

Chapter 4, Solution 63. Because there are no independent sources, IN = Isc = 0 A RN can be found using the circuit below. 3 Ω

io10 Ω

+ −

+

vo

−

v1

0.5vo20 Ω

1V Applying KCL at node 1, 0.5vo + (1 – v1)/3 = v1/30, but vo = (20/30)v1 Hence, 0.5(2/3)(30)v1 + 10 – 10v1 =v1, or v1 = 10 and io = (1 – v1)/3 = -3 RN = 1/io = -1/3 = -333.3 m ohms Chapter 4, Solution 64.

With no independent sources, VTh = 0 V. To obtain RTh, consider the circuit shown below.

1 Ω

ix

io4 Ω

+ −

+ –

10ix

vo

2 Ω

1V

ix = [(1 – vo)/1] + [(10ix – vo)/4], or 2vo = 1 + 3ix (1)

But ix = vo/2. Hence,

2vo = 1 + 1.5vo, or vo = 2, io = (1 – vo)/1 = -1

Thus, RTh = 1/io = -1 ohm

Chapter 4, Solution 65 At the terminals of the unknown resistance, we replace the circuit by its Thevenin equivalent.

V 24)32(412

12,53212//42 =+

=Ω=+=+= ThTh VR

Thus, the circuit can be replaced by that shown below.

5Ω Io

+ + 24 V Vo

- - Applying KVL to the loop,

oooo I524V0VI524 −=→=++− Chapter 4, Solution 66. We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuit in Fig. (a).

3 Ω

5 Ω

b a

RTh

2 Ω

2 Ω 10V − +

i 20V

+ −

30V

3 Ω

+VTh

a b

− +

5 Ω

(a) (b)

RTh = 2||(3 + 5) = 2||8 = 1.6 ohms By performing source transformation on the given circuit, we obatin the circuit in (b).

We now use this to find VTh.

10i + 30 + 20 + 10 = 0, or i = -5

VTh + 10 + 2i = 0, or VTh = 2 V

p = VTh2/(4RTh) = (2)2/[4(1.6)] = 625 m watts

Chapter 4, Solution 67. We need to find the Thevenin equivalent at terminals a and b. From Fig. (a),

RTh = 4||6 + 8||12 = 2.4 + 4.8 = 7.2 ohms From Fig. (b),

10i1 – 30 = 0, or i1 = 3

+

−

4 Ω

12 Ω 8 Ω

6 Ω

i2

i1

+ −

+

−

30V

RTh

4 Ω

12 Ω 8 Ω

6 Ω

VTh

+

−

(b)(a)

20i2 + 30 = 0, or i2 = 1.5, VTh = 6i1 + 8i2 = 6x3 – 8x1.5 = 6 V For maximum power transfer,

p = VTh2/(4RTh) = (6)2/[4(7.2)] = 1.25 watts

Chapter 4, Solution 68. This is a challenging problem in that the load is already specified. This now becomes a "minimize losses" style problem. When a load is specified and internal losses can be adjusted, then the objective becomes, reduce RThev as much as possible, which will result in maximum power transfer to the load.

-+

-+

Removing the 10 ohm resistor and solving for the Thevenin Circuit results in:

RTh = (Rx20/(R+20)) and a Voc = VTh = 12x(20/(R +20)) + (-8) As R goes to zero, RTh goes to zero and VTh goes to 4 volts, which produces the maximum power delivered to the 10-ohm resistor.

P = vi = v2/R = 4x4/10 = 1.6 watts Notice that if R = 20 ohms which gives an RTh = 10 ohms, then VTh becomes -2 volts and the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts. It is also interesting to note that the internal losses for the first case are 122/20 = 7.2 watts and for the second case are = to 12 watts. This is a significant difference. Chapter 4, Solution 69. We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit below. 22 kΩ v1 Assume that all resistances are in k ohms and all currents are in mA.

1mA 3vo30 kΩ40 kΩ

+

vo

−

10 kΩ

10||40 = 8, and 8 + 22 = 30

1 + 3vo = (v1/30) + (v1/30) = (v1/15)

15 + 45vo = v1

But vo = (8/30)v1, hence,

15 + 45x(8v1/30) v1, which leads to v1 = 1.3636

RTh = v1/1 = -1.3636 k ohms To find VTh, consider the circuit below. 10 kΩ 22 kΩvo v1

(100 – vo)/10 = (vo/40) + (vo – v1)/22 (1)

3vo30 kΩ40 kΩ

+

vo

−

+ −

+

VTh

−

100V

[(vo – v1)/22] + 3vo = (v1/30) (2)

Solving (1) and (2),

v1 = VTh = -243.6 volts

p = VTh2/(4RTh) = (243.6)2/[4(-1363.6)] = -10.882 watts

Chapter 4, Solution 70 We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below.

3Vx

5Ω 5Ω + + 4V 15Ω VTh - 6Ω - + Vx -

From the figure,

V3)4(515

15,0 =+

== Thx VV

To find RTh, consider the circuit below:

3Vx

5Ω 5Ω V1 V2 + 4V 15Ω 1A - 6Ω + Vx - At node 1,

122111 73258616,

5153

54 VVxVVVVVV

xx −=→==−

++=− (1)

At node 2,

9505

31 2121 −=→=

−++ VV

VVVx (2)

Solving (1) and (2) leads to V2 = 101.75 V

mW 11.2275.1014

94

,75.1011

2

max2 ===Ω==

xRV

pVRTh

ThTh

Chapter 4, Solution 71. We need RTh and VTh at terminals a and b. To find RTh, we insert a 1-mA source at the terminals a and b as shown below. Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in volts. At node a,

a

b

− +

1mA 120vo40 kΩ1 kΩ

+

vo

−

10 kΩ

3 kΩ

1 = (va/40) + [(va + 120vo)/10], or 40 = 5va + 480vo (1) The loop on the left side has no voltage source. Hence, vo = 0. From (1), va = 8 V.

RTh = va/1 mA = 8 kohms To get VTh, consider the original circuit. For the left loop,

vo = (1/4)8 = 2 V For the right loop, vR = VTh = (40/50)(-120vo) = -192 The resistance at the required resistor is

R = RTh = 8 kohms

p = VTh2/(4RTh) = (-192)2/(4x8x103) = 1.152 watts

Chapter 4, Solution 72. (a) RTh and VTh are calculated using the circuits shown in Fig. (a) and (b) respectively. From Fig. (a), RTh = 2 + 4 + 6 = 12 ohms From Fig. (b), -VTh + 12 + 8 + 20 = 0, or VTh = 40 V

12V 4 Ω4 Ω 6 Ω 2 Ω 6 Ω − + +

VTh

−

+ −

RTh 2 Ω 8V 20V

+ − (b)(a)

(b) i = VTh/(RTh + R) = 40/(12 + 8) = 2A (c) For maximum power transfer, RL = RTh = 12 ohms (d) p = VTh

2/(4RTh) = (40)2/(4x12) = 33.33 watts. Chapter 4, Solution 73 Find the Thevenin’s equivalent circuit across the terminals of R.

10 Ω 25Ω RTh 20Ω 5Ω

Ω==+= 833.1030/3255//2520//10ThR

10 Ω 25Ω + + VTh - 60 V + + - Va Vb 20Ω 5Ω - -

10)60(305,40)60(

3020

==== ba VV

V 3010400 =−=−=→=++− baThbTha VVVVVV

W77.20833.104

304

22

max ===xR

Vp

Th

Th

Chapter 4, Solution 74. When RL is removed and Vs is short-circuited,

RTh = R1||R2 + R3||R4 = [R1 R2/( R1 + R2)] + [R3 R4/( R3 + R4)]

RL = RTh = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)/[( R1 + R2)( R3 + R4)] When RL is removed and we apply the voltage division principle,

Voc = VTh = vR2 – vR4

= ([R2/(R1 + R2)] – [R4/(R3 + R4)])Vs = [(R2R3) – (R1R4)]/[(R1 + R2)(R3 + R4)]Vs

pmax = VTh2/(4RTh)

= [(R2R3) – (R1R4)]2/[(R1 + R2)(R3 + R4)]2Vs

2[( R1 + R2)( R3 + R4)]/[4(a)]

where a = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)

pmax =

[(R2R3) – (R1R4)]2Vs2/[4(R1 + R2)(R3 + R4) (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)]


We need to first find RTh and VTh. R

+ −

+ − 2V

R

R

+ − 3V

+

VTh

−

vo

R R R RTh

1V (a) (b) Consider the circuit in Fig. (a).

(1/RTh) = (1/R) + (1/R) + (1/R) = 3/R

RTh = R/3 From the circuit in Fig. (b),

((1 – vo)/R) + ((2 – vo)/R) + ((3 – vo)/R) = 0

vo = 2 = VTh For maximum power transfer,

RL = RTh = R/3

Pmax = [(VTh)2/(4RTh)] = 3 mW

RTh = [(VTh)2/(4Pmax)] = 4/(4xPmax) = 1/Pmax = R/3

R = 3/(3x10-3) = 1 k ohms Chapter 4, Solution 76. Follow the steps in Example 4.14. The schematic and the output plots are shown below. From the plot, we obtain,

V = 92 V [i = 0, voltage axis intercept]

R = Slope = (120 – 92)/1 = 28 ohms

Chapter 4, Solution 77. (a) The schematic is shown below. We perform a dc sweep on a current source, I1, connected between terminals a and b. We label the top and bottom of source I1 as 2 and 1 respectively. We plot V(2) – V(1) as shown.

VTh = 4 V [zero intercept]

RTh = (7.8 – 4)/1 = 3.8 ohms

(b) Everything remains the same as in part (a) except that the current source, I1, is connected between terminals b and c as shown below. We perform a dc sweep on I1 and obtain the plot shown below. From the plot, we obtain,

V = 15 V [zero intercept]

R = (18.2 – 15)/1 = 3.2 ohms


he schematic is shown below. We perform a dc sweep on the current source, I1,

VTh = -80 V

Tconnected between terminals a and b. The plot is shown. From the plot we obtain,

[zero intercept]

RTh = (1920 – (-80))/1 = 2 k ohms

Chapter 4, Solution 79. After drawing and saving the schematic as shown below, we perform a dc sweep on I1 connected across a and b. The plot is shown. From the plot, we get,

V = 167 V [zero intercept]

R = (177 – 167)/1 = 10 ohms

Chapter 4, Solution 80. The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We perform dc sweep on I1. In the Trace/Add menu, type v(1) – v(2) which will result in the plot below. From the plot,


RTh = (40 – 17.5)/1 = 22.5 ohms [slope]

Chapter 4, Solution 81. The schematic is shown below. We perform a dc sweep on the current source, I2, connected between terminals a and b. The plot of the voltage across I2 is shown below. From the plot,


RTh = (10 – 6.4)/1 = 3.4 ohms.


VTh = Voc = 12 V, Isc = 20 A

RTh = Voc/Isc = 12/20 = 0.6 ohm. 0.6 Ω

i = 12/2.6 , p = i2R = (12/2.6)2(2) = 42.6 watts

i + − 2 Ω12V


VTh = Voc = 12 V, Isc = IN = 1.5 A

RTh = VTh/IN = 8 ohms, VTh = 12 V, RTh = 8 ohms

Chapter 4, Solution 84 Let the equivalent circuit of the battery terminated by a load be as shown below. RTh IL + + VTh - VL RL

-

For open circuit,

V 8.10, ===→∞= LocThL VVVR When RL = 4 ohm, VL=10.5,

7.24/8.10 ===L

LL R

VI

But

Ω=−

=−

=→+= 4444.07.2

8.1012

L

LThThThLLTh I

VVRRIVV

Chapter 4, Solution 85 (a) Consider the equivalent circuit terminated with R as shown below. RTh a + + VTh Vab R - - b

ThTh

ThTh

ab VR

VRRRV

+=→

+=

10106

or ThTh VR 10660 =+ (1)

where RTh is in k-ohm.

Similarly,

ThThThTh

VRVR

301236030

3012 =+→+

= (2)

Solving (1) and (2) leads to

Ω== kRV ThTh 30 V, 24

(b) V 6.9)24(3020

20=

+=abV

Chapter 4, Solution 86. We replace the box with the Thevenin equivalent. RTh

+

v

−

i

R

+ −

VTh

VTh = v + iRTh When i = 1.5, v = 3, which implies that VTh = 3 + 1.5RTh (1) When i = 1, v = 8, which implies that VTh = 8 + 1xRTh (2) From (1) and (2), RTh = 10 ohms and VTh = 18 V. (a) When R = 4, i = VTh/(R + RTh) = 18/(4 + 10) = 1.2857 A (b) For maximum power, R = RTH

Pmax = (VTh)2/4RTh = 182/(4x10) = 8.1 watts Chapter 4, Solution 87. (a) im = 9.975 mA im = 9.876 mA

+

vm

−

Rs Rm Rs Rs Rm

Is Is

(a) (b)

From Fig. (a),

vm = Rmim = 9.975 mA x 20 = 0.1995 V

Is = 9.975 mA + (0.1995/Rs) (1) From Fig. (b),

vm = Rmim = 20x9.876 = 0.19752 V

Is = 9.876 mA + (0.19752/2k) + (0.19752/Rs)

= 9.975 mA + (0.19752/Rs) (2) Solving (1) and (2) gives,

Rs = 8 k ohms, Is = 10 mA (b)

im’ = 9.876 mA

8k||4k = 2.667 k ohms

im’ = [2667/(2667 + 20)](10 mA) = 9.926 mA Chapter 4, Solution 88

To find RTh, consider the circuit below. RTh 5kΩ A B

30k 20kΩ Ω

Rs

(b)

RsIs Rm

10kΩ Ω=++= kRTh 445//201030

To find VTh , consider the circuit below.

5kΩ A B io +

30k 20kΩ Ω 4mA 60 V - 10kΩ

V 72,48)60(2520,120430 =−===== BAThBA VVVVxV

Chapter 4, Solution 89 It is easy to solve this problem using Pspice. (a) The schematic is shown below. We insert IPROBE to measure the desired ammeter reading. We insert a very small resistance in series IPROBE to avoid problem. After the circuit is saved and simulated, the current is displaced on IPROBE as A99.99 µ .

(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown below. We obtain exactly the same result as in part (a).


Rx = (R3/R1)R2 = (4/2)R2 = 42.6, R2 = 21.3

which is (21.3ohms/100ohms)% = 21.3% Chapter 4, Solution 91.

Rx = (R3/R1)R2 (a) Since 0 < R2 < 50 ohms, to make 0 < Rx < 10 ohms requires that when R2

= 50 ohms, Rx = 10 ohms.

10 = (R3/R1)50 or R3 = R1/5

so we select R1 = 100 ohms and R3 = 20 ohms (b) For 0 < Rx < 100 ohms

100 = (R3/R1)50, or R3 = 2R1

So we can select R1 = 100 ohms and R3 = 200 ohms

Chapter 4, Solution 92. For a balanced bridge, vab = 0. We can use mesh analysis to find vab. Consider the circuit in Fig. (a), where i1 and i2 are assumed to be in mA.

2 kΩ

5 kΩ

i2

i1

+ −

+ vab −

a b

3 kΩ 6 kΩ 220V 10 kΩ 0 (a)

220 = 2i1 + 8(i1 – i2) or 220 = 10i1 – 8i2 (1) 0 = 24i2 – 8i1 or i2 = (1/3)i1 (2) From (1) and (2),

i1 = 30 mA and i2 = 10 mA Applying KVL to loop 0ab0 gives

5(i2 – i1) + vab + 10i2 = 0 V Since vab = 0, the bridge is balanced. When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes unbalanced. (1) remains the same but (2) becomes

0 = 32i2 – 8i1, or i2 = (1/4)i1 (3) Solving (1) and (3),

i1 = 27.5 mA, i2 = 6.875 mA

vab = 5(i1 – i2) – 18i2 = -20.625 V

VTh = vab = -20.625 V To obtain RTh, we convert the delta connection in Fig. (b) to a wye connection shown in Fig. (c).

2 kΩ

a RTh b

18 kΩ

6 kΩR2

R1a RTh b

5 kΩ 18 kΩ R3

6 kΩ3 kΩ

(b) (c)

R1 = 3x5/(2 + 3 + 5) = 1.5 k ohms, R2 = 2x3/10 = 600 ohms,

R3 = 2x5/10 = 1 k ohm.

RTh = R1 + (R2 + 6)||(R3 + 18) = 1.5 + 6.6||9 = 6.398 k ohms

RL = RTh = 6.398 k ohms

Pmax = (VTh)2/(4RTh) = (20.625)2/(4x6.398) = 16.622 mWatts Chapter 4, Solution 93.

Rs Ro ix

βRoix+ −

ix

+ −

VS

-Vs + (Rs + Ro)ix + βRoix = 0

ix = Vs/(Rs + (1 + β)Ro)


(a) Vo/Vg = Rp/(Rg + Rs + Rp) (1)

Req = Rp||(Rg + Rs) = Rg

Rg = Rp(Rg + Rs)/(Rp + Rg + Rs)

RgRp + Rg2 + RgRs = RpRg + RpRs

RpRs = Rg(Rg + Rs) (2)

From (1), Rp/α = Rg + Rs + Rp Rg + Rs = Rp((1/α) – 1) = Rp(1 - α)/α (1a)

Combining (2) and (1a) gives, Rs = [(1 - α)/α]Req (3) = (1 – 0.125)(100)/0.125 = 700 ohms

From (3) and (1a),

Rp(1 - α)/α = Rg + [(1 - α)/α]Rg = Rg/α

Rp = Rg/(1 - α) = 100/(1 – 0.125) = 114.29 ohms

(b)

RTh

+ −

I RLVTh

VTh = Vs = 0.125Vg = 1.5 V

RTh = Rg = 100 ohms

I = VTh/(RTh + RL) = 1.5/150 = 10 mA


Let 1/sensitivity = 1/(20 k ohms/volt) = 50 µA For the 0 – 10 V scale,

Rm = Vfs/Ifs = 10/50 µA = 200 k ohms For the 0 – 50 V scale,

Rm = 50(20 k ohms/V) = 1 M ohm

+ − VTh

RTh

Rm

VTh = I(RTh + Rm)

(a) A 4V reading corresponds to

I = (4/10)Ifs = 0.4x50 µA = 20 µA

VTh = 20 µA RTh + 20 µA 250 k ohms

= 4 + 20 µA RTh (1)

(b) A 5V reading corresponds to

I = (5/50)Ifs = 0.1 x 50 µA = 5 µA

VTh = 5 µA x RTh + 5 µA x 1 M ohm

VTh = 5 + 5 µA RTh (2) From (1) and (2)

0 = -1 + 15 µA RTh which leads to RTh = 66.67 k ohms

From (1),

VTh = 4 + 20x10-6x(1/(15x10-6)) = 5.333 V

Chapter 4, Solution 96. (a) The resistance network can be redrawn as shown in Fig. (a),

+ − VTh

RTh

+

Vo

−

+

VTh

−

R

10 Ω 8 Ω 10 Ω

40Ω 10 Ω

i2i1

+ − 60 Ω

8 Ω

9V R

(a) (b)

RTh = 10 + 10 + 60||(8 + 8 + 10||40) = 20 + 60||24 = 37.14 ohms Using mesh analysis, -9 + 50i1 - 40i2 = 0 (1) 116i2 – 40i1 = 0 or i1 = 2.9i2 (2) From (1) and (2), i2 = 9/105

VTh = 60i2 = 5.143 V From Fig. (b),

Vo = [R/(R + RTh)]VTh = 1.8

R/(R + 37.14) = 1.8/5.143 which leads to R = 20 ohms (b) R = RTh = 37.14 ohms Imax = VTh/(2RTh) = 5.143/(2x37.14) = 69.23 mA Chapter 4, Solution 97.

4 kΩ

4 kΩ

+ −

+B

VTh

− E

12V

RTh = R1||R2 = 6||4 = 2.4 k ohms

VTh = [R2/(R1 + R2)]vs = [4/(6 + 4)](12) = 4.8 V Chapter 4, Solution 98. The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be connected to the wye connection as shown in Fig. (b),

b

a

RTh

R2

R3

30 Ω

R1

b

a RTh

20 Ω

60 Ω

30 Ω

14 Ω

(a) (b)

R1 = 20x60/(20 + 60 + 14) = 1200/94 = 12.97 ohms

R2 = 20x14/94 = 2.98 ohms

R3 = 60x14/94 = 8.94 ohms

RTh = R3 + R1||(R2 + 30) = 8.94 + 12.77||32.98 = 18.15 ohms

To find VTh, consider the circuit in Fig. (c).

16 V

I1

IT

IT

+ −

b

a +

VTh

20 Ω

60 Ω

30 Ω

14 Ω

(c)

IT = 16/(30 + 15.74) = 350 mA

I1 = [20/(20 + 60 + 14)]IT = 94.5 mA

VTh = 14I1 + 30IT = 11.824 V

I40 = VTh/(RTh + 40) = 11.824/(18.15 + 40) = 203.3 mA

P40 = I402R = 1.654 watts

Chapter 5, Solution 1. (a) Rin = 1.5 MΩ (b) Rout = 60 Ω (c) A = 8x104

Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 0.1V Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V Chapter 5, Solution 4.

v0 = Avd = A(v2 - v1)

v2 - v1 = V2010x24

Av

50 µ−=

−=

If v1 and v2 are in mV, then

v2 - v1 = -20 mV = 0.02 1 - v1 = -0.02

v1 = 1.02 mV Chapter 5, Solution 5.

+ v0

-

-

vd

+

R0

Rin

I

vi -+

Avd +-

-vi + Avd + (Ri - R0) I = 0 (1) But vd = RiI, -vi + (Ri + R0 + RiA) I = 0

vd = i0

ii

R)A1(RRv++

(2)

-Avd - R0I + v0 = 0

v0 = Avd + R0I = (R0 + RiA)I = i0

ii0

R)A1(Rv)ARR(

+++

45

54

i0

i0

i

0 10)101(100

10x10100R)A1(R

ARRvv

⋅++

+=

+++

=

≅ ( ) =⋅+

45

9

10101

10=

001,100000,100 0.9999990


- vd

+ + vo

-

R0

Rin

I

vi

+ -

Avd +-

(R0 + Ri)R + vi + Avd = 0 But vd = RiI, vi + (R0 + Ri + RiA)I = 0

I = i0

i

R)A1(Rv++

− (1)

-Avd - R0I + vo = 0 vo = Avd + R0I = (R0 + RiA)I Substituting for I in (1),

v0 =

++

+

i0

i0

R)A1(RARR

− vi

= ( )( ) 65

356

10x2x10x21501010x2x10x250

++⋅+

−−

≅ mV10x2x001,20010x2x000,200

6

6−

v0 = -0.999995 mV Chapter 5, Solution 7. 100 kΩ

1 210 kΩ

-+

+ Vd -

+ Vout

-

Rout = 100 Ω

Rin AVd +-

VS

At node 1, (VS – V1)/10 k = [V1/100 k] + [(V1 – V0)/100 k] 10 VS – 10 V1 = V1 + V1 – V0

which leads to V1 = (10VS + V0)/12 At node 2, (V1 – V0)/100 k = (V0 – AVd)/100

But Vd = V1 and A = 100,000, V1 – V0 = 1000 (V0 – 100,000V1)

0= 1001V0 – 100,000,001[(10VS + V0)/12]

0 = -83,333,334.17 VS - 8,332,333.42 V0

which gives us (V0/ VS) = -10 (for all practical purposes) If VS = 1 mV, then V0 = -10 mV Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = -100 nV

Chapter 5, Solution 8. (a) If va and vb are the voltages at the inverting and noninverting terminals of the op

amp.

va = vb = 0

1mA = k2v0 0−

v0 = -2V

(b)

10 kΩ

2V

+ -+ va -

10 kΩ

ia

+ vo -

+ vo -

va

vb

ia

2 kΩ

2V -+

1V -+

- +

(b) (a)

Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b), -va + 2 + v0 = 0 va = va - 2 = 1 - 2 = -1V Chapter 5, Solution 9. (a) Let va and vb be respectively the voltages at the inverting and noninverting terminals of the op amp va = vb = 4V At the inverting terminal,

1mA = k2

0v4 − v0 = 2V

Since va = vb = 3V, -vb + 1 + vo = 0 vo = vb - 1 = 2V

+ vb -

+ vo -

+ -

(b) 1V


Since no current enters the op amp, the voltage at the input of the op amp is vs. Hence

vs = vo 2v

101010 o=

+

s

o

vv

= 2


8 kΩ

vb = V2)3(510

10=

+

io

b

a

+ −

5 kΩ

2 kΩ

4 kΩ

+

vo

−

10 kΩ

−+

3 V

At node a,

8

vv2v3 oaa −

=−

12 = 5va – vo

But va = vb = 2V,

12 = 10 – vo vo = -2V

–io = mA142

822

4v0

8vv ooa =+

+=

−+

−

i o = -1mA

Chapter 5, Solution 12. 4 kΩ

b

a

+ −

2 kΩ

1 kΩ

+

vo

− 4 kΩ

−+

1.2V

At node b, vb = ooo v32v

32v

244

==+

At node a, 4

vv1

v2.1 oaa −=

−, but va = vb = ov

32

4.8 - 4 x ooo vv32v

32

−= vo = V0570.27

8.4x3=

va = vb = 76.9v

32

o =

is = 7

2.11

v2. a −=

1 −

p = vsis = 1.2 =

−7

2.1 -205.7 mW

Chapter 5, Solution 13. By voltage division,

i1 i2

90 kΩ

10 kΩ

b

a

+ −

100 kΩ

4 kΩ

50 kΩ

+−

io

+

vo

− 1 V

va = V9.0)1(100

=90

vb = 3

vv

150o

o =50

But va = vb 9.03

v0 = vo = 2.7V

io = i1 + i2 = =+k150

vk10

v oo 0.27mA + 0.018mA = 288 µA

Chapter 5, Solution 14. Transform the current source as shown below. At node 1,

10

vv20

vv5

v10 o1211 −+

−=

−

But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo 40 = 7v1 - 2vo (1)

20 kΩ

vo

10 kΩ

+ −

v1 −+

v2

5 kΩ

10 kΩ

+

vo

−

10V

At node 2, 0v,10

vv20

vv2

o221 =−

=− or v1 = -2vo (2)

From (1) and (2), 40 = -14vo - 2vo vo = -2.5V Chapter 5, Solution 15

(a) Let v1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives

3321

3

1

2

1 11Rv

RRv

Rvv

Rvi oo

s −

+=

−+= (1)

At the inverting terminal,

111

10 RivRvi ss −=→

−= (2)

Combining (1) and (2) leads to

++−=→−=

++

2

3131

33

1

2

11RRRRR

iv

Rv

RR

RRi

s

oos

(b) For this case,

Ω=Ω

++−= k 92- k

2540204020 x

iv

s

o

Chapter 5, Solution 16 10kΩ ix 5kΩ va iy - vb + vo + 2kΩ 0.5V - 8kΩ

Let currents be in mA and resistances be in kΩ . At node a,

oaoaa vvvvv

−=→−

=−

31105

5.0 (1)

But

aooba vvvvv8

1028

8=→

+== (2)

Substituting (2) into (1) gives

148

81031 =→−= aaa vvv

Thus,

A 28.14mA 70/15

5.0µ−=−=

−= a

xv

i

A 85.71mA 148

46.0)

810(6.0)(6.0

102µ==−=−=

−+

−= xvvvv

vvvvi aaao

aoboy


(a) G = =−=−=5

12RR

vv

1

2

i

o -2.4

(b) 5

80vv

i

o −= = -16

(c) =−=5

2000vv

i

o -400

Chapter 5, Solution 18. Converting the voltage source to current source and back to a voltage source, we have the circuit shown below:

3202010 = kΩ

1 MΩ

3v2

32050

1000v io ⋅

+−= =−=

17200

v1

ov -11.764

Chapter 5, Solution 19. We convert the current source and back to a voltage source.

3442 =

5 kΩ

vo

0V

(4/3) kΩ 10 kΩ

−+

4 kΩ

+ − (2/3)V

(20/3) kΩ

+

vo

−

−+

50 kΩ

+ − 2vi/3

=

−=32

k34x4

k10vo -1.25V

=−

+=k10

0vk5

vi oo

o -0.375mA

Chapter 5, Solution 20. 8 kΩ

+ − vs

+ −

4 kΩ

+

vo

−

−+

2 kΩ4 kΩ

a b

9 V At node a,

4

vv8

vv4v9 baoaa −

+−

=−

18 = 5va – vo - 2vb (1)

At node b,

2

vv4

vv obba −=

− va = 3vb - 2vo (2)

But vb = vs = 0; (2) becomes va = –2vo and (1) becomes

-18 = -10vo – vo vo = -18/(11) = -1.6364V


Eqs. (1) and (2) remain the same. When vb = vs = 3V, eq. (2) becomes

va = 3 x 3 - 2v0 = 9 - 2vo

Substituting this into (1), 18 = 5 (9-2vo) – vo – 6 leads to

vo = 21/(11) = 1.909V Chapter 5, Solution 22.

Av = -Rf/Ri = -15.

If Ri = 10kΩ, then Rf = 150 kΩ. Chapter 5, Solution 23

At the inverting terminal, v=0 so that KCL gives

121

000RR

vv

Rv

RRv f

s

o

f

os −= →−

+=−


v1 Rf

R1 R2 - vs + - + + R4 R3 vo v2 - We notice that v1 = v2. Applying KCL at node 1 gives

f

os

ff

os

Rv

Rvv

RRRRvv

Rvv

Rv

=−

++→=

−+

−+

21

21

1

2

1

1

1 1110)(

(1)

Applying KCL at node 2 gives

ss v

RRR

vRvv

Rv

43

31

4

1

3

1 0+

=→=−

+ (2)

Substituting (2) into (1) yields

sf

fo vRRR

RRR

RR

RRRv

−

+

−+=

243

3

2

43

1

3 1

i.e.

−

+

−+=

243

3

2

43

1

3 1RRR

RRR

RR

RRRk

ff


vo = 2 V

−

+

va

-va + 3 + vo = 0 Chapter 5, Solution 26

+

0.4V -

8.028

84.0 =+

== oob vvv

Hence,

05

5.05

===kk

vooi

+

which lea

+

vb 8kΩ

→

mA 1.

+

vo

ds to va = vo + 3 = 5 V.

- io + 5k Ω 2kΩ vo

-

V 5.08.0/4.0 ==ov

Chapter 5, Solution 27. (a) Let va be the voltage at the noninverting terminal. va = 2/(8+2) vi = 0.2vi

ia0 v2.10v20

1 =

+

1000v =

G = v0/(vi) = 10.2

(b) vi = v0/(G) = 15/(10.2) cos 120πt = 1.471 cos 120πt V Chapter 5, Solution 28.

−+

+ −

At node 1, k50vv

k10v0 o11 −

=−

But v1 = 0.4V, -5v1 = v1 – vo, leads to vo = 6v1 = 2.4V Alternatively, viewed as a noninverting amplifier, vo = (1 + (50/10)) (0.4V) = 2.4V io = vo/(20k) = 2.4/(20k) = 120 µA

Chapter 5, Solution 29 R1 va + vb - + + vi R2 R2 vo - R1 -

obia vRR

RvvRR

Rv21

1

21

2 ,+

=+

=

But oiba vRR

RvRR

Rv21

1

21

2

+=

+→=v

Or

1

2

RR

vv

i

o =

Chapter 5, Solution 30. The output of the voltage becomes vo = vi = 12 Ω= k122030 By voltage division,

V2.0)2.1(6012

12vx =+

=

===k202.0

k20v

i xx 10µA

===k20

04.0Rvp

2x 2µW

Chapter 5, Solution 31. After converting the current source to a voltage source, the circuit is as shown below: 12 kΩ

2

vo6 kΩ

+−

+ − 6 kΩ

3 kΩ 1 v1

vo 12 V At node 1,

12

vv6

vv3

v o1o1112 −+

−=

− 48 = 7v1 - 3vo (1)

At node 2,

xoo1 i6

0v6

vv=

−=

− v1 = 2vo (2)

From (1) and (2),

1148vo =

==k6

vi o

x 0.7272mA

Chapter 5, Solution 32. Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting amplifier.

=xv

+

10501 (4 mV) = 24 mV

Ω= k203060


vo = mV122

vv

2020o

o ==+20

ix = ( ) ==+ k40

mV24k2020

vx 600nA

p = =−

3

62o

10x6010x

=144

Rv

204nW

Chapter 5, Solution 33. After transforming the current source, the current is as shown below: 1 kΩ This is a noninverting amplifier.

3 kΩ

vi

va+−

+ − 2 kΩ

4 kΩ vo

4 V

iio v23v

211v =

+=

Since the current entering the op amp is 0, the source resistor has a OV potential drop. Hence vi = 4V.

V6)4(23vo ==

Power dissipated by the 3kΩ resistor is

==k3

36Rv2

o 12mW

=−

=−

=k1

64R

vvi oa

x -2mA


0R

vvR

vv

2

in1

1

in1 =−

+− (1)

but

o43

3a v

RRRv+

= (2)

Combining (1) and (2),

0vRRv

RRvv a

2

12

2

1a1 =−+−

22

11

2

1a v

RRv

RR1v +=

+

22

11

2

1

43

o3 vRRv

RR1

RRvR

+=

+

+

+

+

+= 2

2

11

2

13

43o v

RRv

RR1R

RRv

vO = )vRv()RR(R

RR221

213

43 ++

+


10RR1

vv

Ai

f

i

ov =+== Rf = 9Ri

If Ri = 10kΩ, Rf = 90kΩ

Chapter 5, Solution 36 V abTh V=

But abs VRR

R

21

1

+=v . Thus,

ssabTh vRRv

RRRVV )1(

1

2

1

21 +=+

==

To get RTh, apply a current source Io at terminals a-b as shown below.

v1 + v2 - a + R2

vo io R1 - b Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes through R1 and consequently R2 . Thus, vo=0 and

0==o

oTh i

vR


++−= 3

3

f2

2

f1

1

fo v

RR

vRR

vRR

v

−++−= )3(

3030)2(

2030)1(

1030

vo = -3V


+++−= 4

4

f3

3

f2

2

f1

1

fo v

RR

vRR

vRR

vRR

v

−++−+−= )100(

5050)50(

1050)20(

2050)10(

2550

= -120mV Chapter 5, Solution 39

This is a summing amplifier.

2233

22

11

5.29)1(5050

2050)2(

1050 vvv

RR

vRR

vRR

v fffo −−=

−++−=

++−=

Thus, V 35.295.16 22 =→−−=−= vvvo


R1 R2 va + R3 vb - + + v1 + - v2 Rf vo - + v3 R - - Applying KCL at node a,

)111(03213

3

2

2

1

1

3

3

2

2

1

1

RRRv

Rv

Rv

Rv

Rvv

Rvv

Rvv

aaaa ++=++→=

−+

−+

− (1)

But

of

ba vRRRv+

==v (2)

Substituting (2) into (1)gives

)111(3213

3

2

2

1

1

RRRRRRv

Rv

Rv

Rv

f

o +++

=++

or

)111/()(3213

3

2

2

1

1

RRRRv

Rv

Rv

RRR

v fo ++++

+=

Chapter 5, Solution 41. Rf/Ri = 1/(4) Ri = 4Rf = 40kΩ The averaging amplifier is as shown below: Chapter 5, Solution 42

v1 R2 = 40 kΩ

v2 R3 = 40 kΩ

v3 R4 = 40 kΩ

v4

10 kΩ

−+

R1 = 40 kΩ

vo

Ω== k 10R31R 1f

Chapter 5, Solution 43. In order for

+++= 4

4

f3

3

f2

2

f1

1

fo v

RR

vRR

vRR

vRR

v

to become

( )4321o vvvv41v +++−=

41

RR

i

f = ===4

124

RR if 3kΩ

Chapter 5, Solution 44. R4

At node b, 0R

vvR

vv

2

2b

1

1b =−

+−

21

2

2

1

1

b

R1

R1

Rv

Rv

v+

+= (1)

b

a

R1 v1

R2 v2

R3

vo

−+

At node a, 4

oa

3

a

Rvv

Rv0 −

=−

34

oa R/R1

vv

+= (2)

But va = vb. We set (1) and (2) equal.

21

1112

34

o

RRvRvR

R/R1v

++

=+

or

vo = ( )( ) ( )1112

213

43 vRvRRRR

RR+

++

Chapter 5, Solution 45. This can be achieved as follows:

( )

+−−= 21o v

2/RRv

3/RRv

( )

+−−= 2

2

f1

1

f vRR

vRR

i.e. Rf = R, R1 = R/3, and R2 = R/2 Thus we need an inverter to invert v1, and a summer, as shown below (R<100kΩ).

R/3

R/2 v2

R

−+

R v1

-v1

R

−+

vo Chapter 5, Solution 46.

33

f2

2

x1

1

f32

1o v

RR

)v(RR

vRR

v21)v(

31

3v

v +−+=+−+=−

i.e. R3 = 2Rf, R1 = R2 = 3Rf. To get -v2, we need an inverter with Rf = Ri. If Rf = 10kΩ, a solution is given below.

v1

30 kΩ

20 kΩv3

10 kΩ

−+

10 kΩ v2

-v2

10 kΩ

−+

30 kΩ

vo


If a is the inverting terminal at the op amp and b is the noninverting terminal, then,

V6vv,V6)8(13

3v bab ===+

= and at node a,4

vv2

v10 oaa −=

−

which leads to vo = –2 V and io = k4

)vv(k5

v oao −− = –0.4 – 2 mA = –2.4 mA

Chapter 5, Solution 48. Since the op amp draws no current from the bridge, the bridge may be treated separately as follows: v1

v2

i2

i1

+ −

For loop 1, (10 + 30) i1 = 5 i1 = 5/(40) = 0.125µA For loop 2, (40 + 60) i2 = -5 i2 = -0.05µA But, 10i + v1 - 5 = 0 v1 = 5 - 10i = 3.75mV 60i + v2 + 5 = 0 v2 = -5 - 60i = -2mV As a difference amplifier,

( ) [ ]mV)2(75.32080vv

RR

v 121

2o −−=−=

= 23mV

Chapter 5, Solution 49. R1 = R3 = 10kΩ, R2/(R1) = 2 i.e. R2 = 2R1 = 20kΩ = R4

Verify: 11

22

43

21

1

2o v

RR

vR/R1R/R1

RR

v −++

=

( )1212 vv2v2v5.01)5.01(2 −=−

++

=

Thus, R1 = R3 = 10kΩ, R2 = R4 = 20kΩ

Chapter 5, Solution 50. (a) We use a difference amplifier, as shown below:

( ) ( ,vv2vvRR

v 12121

2o −=−= ) i.e. R2/R1 = 2

R1 v2

R2

v1

R2

−+

R1

vo

If R1 = 10 kΩ then R2 = 20kΩ (b) We may apply the idea in Prob. 5.35.

210 v2v2v −=

( )

+−−= 21 v

2/RRv

2/RR

( )

+−−= 2

2

f1

1

f vRR

vRR

i.e. Rf = R, R1 = R/2 = R2

We need an inverter to invert v1 and a summer, as shown below. We may let R = 10kΩ.

R/2

R/2 v2

R

−+

R v1

-v1

R

−+

vo


We achieve this by cascading an inverting amplifier and two-input inverting summer as shown below: Verify: vo = -va - v1 But va = -v2. Hence

R

R v1

R

−+

R v2

va

R

−+

vo

vo = v2 - v1.

io2

112 vv

RRvv ==−

=i

o

vv

1

2

RR

(b)

At node A, 2/Rvv

Rvv

2/Rvv

1

aA

g

AB

1

A1 −=

−+

−

v1

R2

vb

va

vB

vA

R1/2 v2

R1/2 Rg

R2

+

vo

−

−+

R1/2 R1/2

−

vi

+

or ( ) aAABg

1A1 vvvv

R2Rvv −=−+− (1)

At node B, g

bB

1

AB

1

B2

Rvv

2/Rvv

2/Rvv −

+−

=−

or bBABg

1B2 vv)vv(

R2Rvv −=−−− (2)

Subtracting (1) from (2),

( ) abABABg

1AB12 vvvvvv

R2R2vvv +−−=−−+−−v

Since, va = vb,

( )2v

vvR2R

12

vv iAB

g

112 =−

+=

−

or

g

1

iAB

R2R1

12vvv

+⋅=− (3)

But for the difference amplifier,

( )AB1

2o vv

2/RR

−=v

or o2

1AB v

R2Rvv =− (4)

Equating (3) and (4),

g

1

io

2

1

R2R1

12vv

R2R

+⋅=

g

11

2

i

o

R2R1

1RR

vv

+⋅=

(c) At node a, 2/R

vvR

vv

2

Aa

1

a1 −=

−

A2

1a

2

1a1 v

RR2v

RR2v −=−v (1)

At node b, B2

1b

2

1b2 v

RR2v

RR2vv −=− (2)

Since va = vb, we subtract (1) from (2),

2v)vv(

RR2v i

AB2

112 =−

−=−v

or i1

2AB v

R2Rvv −

=− (3)

At node A,

2/Rvv

Rvv

2/Rvv oA

g

AB

2

Aa −=

−+

−

( ) oAABg

2Aa vvvv

R2Rvv −=−+− (4)

At node B, 2/R0v

Rvv

2/Rvv B

g

ABBb −=

−−

−

( ) BABg

2Bb vvv

R2Rv =−−−v (5)


( ) oBAABg

2AB vvvvv

RRv −−=−+−v

( ) og

2AB v

R2R

1vv2 −=

+− (6)


og

2i

1

2 vR2

R1vRR

−=

+

−

+=

g

2

1

2

i

o

R2R

1RR

vv

Chapter 5, Solution 54. (a) A0 = A1A2A3 = (-30)(-12.5)(0.8) = 300 (b) A = A1A2A3A4 = A0A4 = 300A4

But dB60ALog20 10 = 3ALog10 =

A = 103 = 1000 A4 = A/(300) = 3.333 Chapter 5, Solution 55. Let A1 = k, A2 = k, and A3 = k/(4) A = A1A2A3 = k3/(4) 42ALog20 10 = A = 101.2ALog10 = 2 ⋅1 = 125.89 k3 = 4A = 503.57 k = 956.57. 75033 = Thus A1 = A2 = 7.956, A3 = 1.989

Chapter 5, Solution 56. There is a cascading system of two inverting amplifiers.

sso v6v612

412v =

−−

=

mAv3k2

vi s

so ==

(a) When vs = 12V, io = 36mA (b) When vs = 10 cos 377t V, io = 30 cos 377t mA


The first stage is a difference amplifier. Since R1/R2 = R3/R4,

mA10)41(50

100)vv(RRv 12

1

2o =+=−=′

The second stage is a non-inverter.

)given(mV40mA1040R1v

40R1v oo =

+=′

+=

Which leads to, R = 120 kΩ

Chapter 5, Solution 58. By voltage division, the input to the voltage follower is:

V45.0)6.0(13

3v1 =+

=

Thus

15.3v7v5

10v210v 111o −=−=−

−=

=−

=k4v0

i oo 0.7875mA

Chapter 5, Solution 59. Let a be the node between the two op amps. va = vo

The first stage is a summer

oosa vv2010v

510v =−

−=

1.5vs = -2vs or

=−

=2v

5.1vs

o -1.333

Chapter 5, Solution 60. Transform the current source as shown below: 4 kΩ Assume all currents are in mA. The first stage is a summer

v1

3 Ω

5 kΩ

3 kΩ

+−

10 kΩ

−+

+ −

io

5is

2 kΩ

( ) osos1 v5.2i10v4

10i5510

−−=−−

=v (1)


oo1 v21v

333

=+

=v (2)

Alternatively, we notice that the second stage is a non-inverter.

11o v2v33

1=

+=v

From (1) and (2), 0 3voso v5.2i10v5. −−= o = 10is

3i10

i2 soo −=−=v ==

35

ii

s

o 1.667


Let v01 be the voltage at the left end of R5. The first stage is an inverter, while the second stage is a summer.

11

201 v

RRv −=

015

40 v

RR

v −= 23

4 vRR

−

v1 = 151RR42 v

RR2

3

4 vRR

−

Chapter 5, Solution 62. Let v1 = output of the first op amp v2 = output of the second op amp The first stage is a summer

i1

21 v

RR

v −= – of

2 vRR (1)

The second stage is a follower. By voltage division

143

42o v

RRR

vv+

== o4

431 v

RRR

v+

= (2)

From (1) and (2),

i1

2o

4

3 vRR

vRR

1 −=

+ o

f

2 vRR

−

i1

2o

f

2

4

3 vRR

vRR

RR

1 −=

++

4

2

4

31

2

i

o

RR

RR1

1RR

vv

++⋅−=

( )4321

42

RRRRRR++

−=

Chapter 5, Solution 63. The two op amps are summer. Let v1 be the output of the first op amp. For the first stage,

o3

2i

1

21 v

RRv

RRv −−= (1)

For the second stage,

io

41

5

4o v

RRv

RRv −−= (2)


i6

4o

3

2

5

4i

1

2

5

4o v

RR

vRR

RR

vRR

RR

v −

+

=

i6

4

51

42

53

42o v

RR

RRRR

RRRR

1v

−=

−

53

42

6

4

31

42

i

o

RRRR

1

RR

RRRR

vv

−

−=


G4 G G3 G1 1 G 2 - - + 0V + v 0V + + vs G2 vo

- -

At node 1, v1=0 so that KCL gives

GvvGvG os −=+ 41 (1) At node 2,

GvvGvG os −=+ 32 (2) From (1) and (2),

ososos vGGvGGvGvGvGvG )()( 43213241 −=−→+=+ or

43

21

GGGG

vv

s

o

−−

=


The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so that its output is

mV -18mV)6(1030' =−=ov

The third op amp is a noninverter so that

mV 6.21'4048

84040' −==→+

= oooo vvvv


)2(10100)4(

2040

20100)6(

25110vo −

−−

−=

=−+−= 204024 -4V Chapter 5, Solution 67.

vo = )2.0(2080)5.0(

2040−

−

8080

−

=−= 8.02.3 2.4V Chapter 5, Solution 68.

If Rq = ∞, the first stage is an inverter.

mV30)10(5

15Va −=−=

when Va is the output of the first op amp. The second stage is a noninverting amplifier.

=−+=

+= )30)(31(v

261v ao -120mV


In this case, the first stage is a summer

ooa v5.130v1015)10(

515v −−=−−=

For the second stage,

( )oaao v5.1304v4v261v −−==

+=

120v7 o −= =−=7

120ov -17.143mV


The output of amplifier A is

9)2(1030)10(

1030vA −=−−=

The output of amplifier B is

14)4(1020)3(

1020vB −=−−=

40 kΩ

V2)14(1060

60vb −=−+

=

vA

vB

a

b

60 kΩ

20 kΩ

−+

10 kΩ

vo

At node a, 40

vv20

vv oaaA −=

−

But va = vb = -2V, 2(-9+2) = -2-vo

Therefore, vo = 12V

Chapter 5, Solution 71 20kΩ 5k 100kΩ Ω

- 40kΩ +

+ v2 2V 80kΩ - - 10kΩ + + vo 20kΩ - - 10kΩ + v1 + - v3 + 3V 50kΩ - 30kΩ

8)30501(,8)2(

520,3 1321 =+=−=−== vvvv

V 10)1020(80

10040

10032 =+−−=

+−= vvvo


Since no current flows into the input terminals of ideal op amp, there is no voltage drop across the 20 kΩ resistor. As a voltage summer, the output of the first op amp is v01 = 0.4 The second stage is an inverter

012 v100150v −=

= =− )4.0(5.2 -1V Chapter 5, Solution 73.

The first stage is an inverter. The output is

V9)8.1(1050v01 −=−−=

The second stage is == 012 vv -9V

Chapter 5, Solution 74. Let v1 = output of the first op amp v2 = input of the second op amp.

The two sub-circuits are inverting amplifiers

V6)6.0(10100

1 −=−=v

V8)4.0(6.1

322 −=−=v

=+−

−=−

=k20

86k20vv 21

oi 100 µA

Chapter 5, Solution 75. The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are involved as shown to measure vo and i respectively. Once the circuit is saved, we click Analysis | Simulate. The values of v and i are displayed on the pseudo-components as:

i = 200 µA

(vo/vs) = -4/2 = -2 The results are slightly different than those obtained in Example 5.11.

Chapter 5, Solution 76. The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as

io = -374.78 µA

Chapter 5, Solution 77. The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as

io = -374.78 µA

Chapter 5, Solution 78. The circuit is constructed as shown below. We insert a VIEWPOINT to display vo. Upon simulating the circuit, we obtain,

vo = 667.75 mV

Chapter 5, Solution 79. The schematic is shown below. A pseudo-component VIEWPOINT is inserted to display vo. After saving and simulating the circuit, we obtain,

vo = -14.61 V

Chapter 5, Solution 80. The schematic is shown below. VIEWPOINT is inserted to display vo. After simulation, we obtain,

vo = 12 V

Chapter 5, Solution 81. The schematic is shown below. We insert one VIEWPOINT and one IPROBE to measure vo and io respectively. Upon saving and simulating the circuit, we obtain,

vo = 343.37 mV

io = 24.51 µA


The maximum voltage level corresponds to

11111 = 25 – 1 = 31 Hence, each bit is worth (7.75/31) = 250 mV

Chapter 5, Solution 83. The result depends on your design. Hence, let RG = 10 k ohms, R1 = 10 k ohms, R2 = 20 k ohms, R3 = 40 k ohms, R4 = 80 k ohms, R5 = 160 k ohms, R6 = 320 k ohms, then,

-vo = (Rf/R1)v1 + --------- + (Rf/R6)v6

= v1 + 0.5v2 + 0.25v3 + 0.125v4 + 0.0625v5 + 0.03125v6

(a) |vo| = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies, [v1 v2 v3 v4 v5 v6] = [100110]

(b) |vo| = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV (c) This corresponds to [1 1 1 1 1 1]. |vo| = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V

Chapter 5, Solution 84. For (a), the process of the proof is time consuming and the results are only approximate, but close enough for the applications where this device is used. (a) The easiest way to solve this problem is to use superposition and to solve

for each term letting all of the corresponding voltages be equal to zero. Also, starting with each current contribution (ik) equal to one amp and working backwards is easiest.

2R R R R

+ − ik

2R

v2 v4

+ − v3

+ −

+ −

2R 2R

v1

R

For the first case, let v2 = v3 = v4 = 0, and i1 = 1A. Therefore, v1 = 2R volts or i1 = v1/(2R). Second case, let v1 = v3 = v4 = 0, and i2 = 1A. Therefore, v2 = 85R/21 volts or i2 = 21v2/(85R). Clearly this is not (1/4th), so where is the difference? (21/85) = 0.247 which is a really good approximation for 0.25. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Now for the third case, let v1 = v2 = v4 = 0, and i3 = 1A. Therefore, v3 = 8.5R volts or i3 = v3/(8.5R). Clearly this is not (1/8th), so where is the difference? (1/8.5) = 0.11765 which is a really good approximation for 0.125. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Finally, for the fourth case, let v1 = v2 = v4 = 0, and i3 = 1A.

Therefore, v4 = 16.25R volts or i4 = v4/(16.25R). Clearly this is not (1/16th), so where is the difference? (1/16.25) = 0.06154 which is a really good approximation for 0.0625. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Please note that a goal of a lot of electronic design is to come up with practical circuits that are economical to design and build yet give the desired results.

(b) If Rf = 12 k ohms and R = 10 k ohms,

-vo = (12/20)[v1 + (v2/2) + (v3/4) + (v4/8)]

= 0.6[v1 + 0.5v2 + 0.25v3 + 0.125v4] For [v1 v2 v3 v4] = [1 0 11],

|vo| = 0.6[1 + 0.25 + 0.125] = 825 mV

For [v1 v2 v3 v4] = [0 1 0 1],

|vo| = 0.6[0.5 + 0.125] = 375 mV


Av = 1 + (2R/Rg) = 1 + 20,000/100 = 201 Chapter 5, Solution 86.

vo = A(v2 – v1) = 200(v2 – v1)

(a) vo = 200(0.386 – 0.402) = -3.2 V (b) vo = 200(1.011 – 1.002) = 1.8 V


The output, va, of the first op amp is,

va = (1 + (R2/R1))v1 (1) Also, vo = (-R4/R3)va + (1 + (R4/R3))v2 (2)

Substituting (1) into (2), vo = (-R4/R3) (1 + (R2/R1))v1 + (1 + (R4/R3))v2 Or, vo = (1 + (R4/R3))v2 – (R4/R3 + (R2R4/R1R3))v1 If R4 = R1 and R3 = R2, then, vo = (1 + (R4/R3))(v2 – v1) which is a subtractor with a gain of (1 + (R4/R3)). Chapter 5, Solution 88. We need to find VTh at terminals a – b, from this,

vo = (R2/R1)(1 + 2(R3/R4))VTh = (500/25)(1 + 2(10/2))VTh

= 220VTh Now we use Fig. (b) to find VTh in terms of vi.

30 kΩ

80 kΩ

20 kΩ

40 kΩ

30 kΩ

b

a

vi + −

b

a

vi

20 kΩ

40 kΩ 80 kΩ

(a) (b)

va = (3/5)vi, vb = (2/3)vi

VTh = vb – va (1/15)vi

(vo/vi) = Av = -220/15 = -14.667

Chapter 5, Solution 89. If we use an inverter, R = 2 k ohms,

(vo/vi) = -R2/R1 = -6

R = 6R = 12 k ohms Hence the op amp circuit is as shown below. 12 kΩ

2 kΩ−+

+ −

+

vo

−

vi Chapter 5, Solution 90. Transforming the current source to a voltage source produces the circuit below, At node b, vb = (2/(2 + 4))vo = vo/3

20 kΩ

io

b

a

2 kΩ

5 kΩ −+

+

vo

−

+ −

4 kΩ

5is At node a, (5is – va)/5 = (va – vo)/20 But va = vb = vo/3. 20is – (4/3)vo = (1/3)vo – vo, or is = vo/30 io = [(2/(2 + 4))/2]vo = vo/6 io/is = (vo/6)/(vo/30) = 5


io

i2

i1

is

−+

vo

R1

R2

io = i1 + i2 (1) But i1 = is (2) R1 and R2 have the same voltage, vo, across them. R1i1 = R2i2, which leads to i2 = (R1/R2)i1 (3) Substituting (2) and (3) into (1) gives, io = is(1 + R1/R2) io/is = 1 + (R1/R2) = 1 + 8/1 = 9 Chapter 5, Solution 92

The top op amp circuit is a non-inverter, while the lower one is an inverter. The output at the top op amp is

v1 = (1 + 60/30)vi = 3vi while the output of the lower op amp is v2 = -(50/20)vi = -2.5vi Hence, vo = v1 – v2 = 3vi + 2.5vi = 5.5vi

vo/vi = 5.5

Chapter 5, Solution 93. R3

+

vi

−

+

vL

−

vb

R4

RL

io

iL

−+

va R1

R2

+

vo

−

At node a, (vi – va)/R1 = (va – vo)/R3 vi – va = (R1/R2)(va – vo) vi + (R1/R3)vo = (1 + R1/R3)va (1) But va = vb = vL. Hence, (1) becomes vi = (1 + R1/R3)vL – (R1/R3)vo (2)

io = vo/(R4 + R2||RL), iL = (RL/(R2 + RL))io = (R2/(R2 + RL))(vo/( R4 + R2||RL)) Or, vo = iL[(R2 + RL)( R4 + R2||RL)/R2 (3) But, vL = iLRL (4) Substituting (3) and (4) into (2),

vi = (1 + R1/R3) iLRL – R1[(R2 + RL)/(R2R3)]( R4 + R2||RL)iL

= [((R3 + R1)/R3)RL – R1((R2 + RL)/(R2R3)(R4 + (R2RL/(R2 + RL))]iL = (1/A)iL

Thus,

A =

+

+

+−

+

L2

L24

32

L21L

3

1

RRRRR

RRRRRR

RR1

1


( =+−== −− t3t3 e6e25dtdvCi ) 10(1 - 3t)e-3t A

p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W


2211 )120)(40(

21Cv

21w ==

w2 = 221 )80)(40(

21

2=Cv1

( ) =−=−=∆ 22

21 8012020www 160 kW Chapter 6, Solution 3.

i = C =−

= −

516028010x40

dtdv 3 480 mA


)0(vidtC1v

t

o+= ∫

∫ +1tdt4sin621

=

= 1 - 0.75 cos 4t Chapter 6, Solution 5.

v = ∫ +t

o)0(vidt

C1

For 0 < t < 1, i = 4t,

∫−=t

o6 t410x201v dt + 0 = 100t2 kV

v(1) = 100 kV

For 1 < t < 2, i = 8 - 4t,

∫ +−= −

t

16 )1(vdt)t48(10x201v

= 100 (4t - t2 - 3) + 100 kV

Thus v (t) =

<<−−

<<

2t1,kV)2tt4(1001t0,kVt100

2

2


610x30dtdvCi −== x slope of the waveform.

For example, for 0 < t < 2,

310x210

dtdv

−=

i = mA15010x210x10x30

dtdv

36 == −

−C

Thus the current i is sketched below.

t (msec)

150

12 10 2

8

6

4

-150

i(t) (mA) Chapter 6, Solution 7.

∫∫ +=+= −−

t

o

33o 10dt10tx4

10x501)t(vidt

C1v

= =+1050t2 2

0.04k2 + 10 V


(a) tt BCeACedtdvC 600100 600100 −− −−==i (1)

BABCACi 656001002)0( −−=→−−== (2)

BAvv +=→= −+ 50)0()0( (3) Solving (2) and (3) leads to A=61, B=-11

(b) J 5250010421)0(

21 32 === − xxxCvEnergy

(c ) From (1),

A 4.264.241041160010461100 60010060031003 tttt eeexxxexxxi −−−−−− −−=−−= Chapter 6, Solution 9.

v(t) = ( ) ( )∫ −− +=+−t

o

tt Vet120dte1621

1

v(2) = 12(2 + e-2) = 25.62 V

p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t)

p(2) = 72(2-e-4) = 142.68 W Chapter 6, Solution 10

dtdvx

dtdvCi 3102 −==

<<<<<<

=s4t316t,-64

s 3t116,s10,16

µµµtt

v

<<<<

<<=

s4t3,16x10-s 3t10,

s10,1016

6

6

µµµtx

dtdv

<<<<

<<=

s4t3kA, 32-s 3t10,

s10,kA 32)(

µµµt

ti


v = ∫ +t

o)0(vidt

C1

For 0 < t < 1,

∫ == −−

t

o

36 t10dt10x40

10x41v kV

v(1) = 10 kV For 1 < t < 2,

kV10)1(vvdtC1v

t

1=+= ∫

For 2 < t < 3,

∫ +−= −−

t

2

36 )2(vdt)10x40(

10x41v

= -10t + 30kV Thus

v(t) =

<<+−<<<<⋅

3t2,kV30t102t1,kV101t0,kVt10


π−π== − 4sin)(4(60x10x3dtdvCi 3 t)

= - 0.7e π sin 4πt A P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W

W = ∫ ∫ t dt ππ−=t

o81

o8sin6.21pdt

= πππ 8

86. cos21 8/1

o = -5.4J


Under dc conditions, the circuit becomes that shown below:

i250 Ω

20 Ω

+ − 60V

+

v1

−

i1

30 Ω

10 Ω

+

v2

−

i2 = 0, i1 = 60/(30+10+20) = 1A v1 = 30i2 = 30V, v2 = 60-20i1 = 40V

Thus, v1 = 30V, v2 = 40V Chapter 6, Solution 14. (a) Ceq = 4C = 120 mF

(b) 304

C4

C1

eq

== Ceq = 7.5 mF

Chapter 6, Solution 15. In parallel, as in Fig. (a), v1 = v2 = 100

C2+

v2

−

C1 + − 100V

+

v2

−

C2

+ − v1

C1 +

v1

−

+ −

100V (b)(a)

w20 = == − 262 100x10x20x21Cv

21 0.1J

w30 = =− 26 100x10x30x21 0.15J

(b) When they are connected in series as in Fig. (b):

,60100x5030V

CCC

v21

21 ==

+= v2 = 40

w20 = =− 26 60x10x30x21 36 mJ

w30 = =− 26 4010x302

xx1 24 mJ


F 203080

8014 µ=→=+

+= CCCxCeq

Chapter 6, Solution 17. (a) 4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F i.e. Ceq = 3F

(b) Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F (c) 3F in series with 6F = (3 x 6)/9 = 6F

131

61

21

C1

eq

=++=

Ceq = 1F


For the capacitors in parallel = 15 + 5 + 40 = 60 µF 1

eqC

Hence 101

601

301

201

C1

eq

=++=

Ceq = 10 µF Chapter 6, Solution 19. We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60µ F. The 60 -µ F capacitor in series with another 60- µ F capacitor gives 30 µ F. 30 + 50 = 80µ F, 80 + 40 = 120 µ F The circuit is reduced to that shown below.

12 120

12 80

120-µ F capacitor in series with 80µ F gives (80x120)/200 = 48 48 + 12 = 60 60-µ F capacitor in series with 12µ F gives (60x12)/72 = 10µ F Chapter 6, Solution 20.

3 in series with 6 = 6x3/(9) = 2 2 in parallel with 2 = 4 4 in series with 4 = (4x4)/8 = 2 The circuit is reduced to that shown below:

20

1 6

2

8

6 in parallel with 2 = 8 8 in series with 8 = 4 4 in parallel with 1 = 5 5 in series with 20 = (5x20)/25 = 4 Thus Ceq = 4 mF

Chapter 6, Solution 21. 4µF in series with 12µF = (4x12)/16 = 3µF 3µF in parallel with 3µF = 6µF 6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF

Hence Ceq = 2.5µF Chapter 6, Solution 22. Combining the capacitors in parallel, we obtain the equivalent circuit shown below:

a b

40 µF 60 µF 30 µF

20 µF Combining the capacitors in series gives C , where 1

eq

101

301

201

601

C11eq

=++= C = 10µF 1eq

Thus

Ceq = 10 + 40 = 50 µF


(a) 3µF is in series with 6µF 3x6/(9) = 2µF v4µF = 1/2 x 120 = 60V v2µF = 60V

v6µF = =(3+

)6036

20V

v3µF = 60 - 20 = 40V

(b) Hence w = 1/2 Cv2 w4µF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ w2µF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ w6µF = 1/2 x 6 x 10-6 x 400 = 1.2mJ w3µF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ


20µF is series with 80µF = 20x80/(100) = 16µF

14µF is parallel with 16µF = 30µF (a) v30µF = 90V

v60µF = 30V v14µF = 60V

v20µF = =+

60x8020

80 48V

v80µF = 60 - 48 = 12V

(b) Since w = 2Cv21

w30µF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ w60µF = 1/2 x 60 x 10-6 x 900 = 27mJ w14µF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ w20µF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ w80µF = 1/2 x 80 x 10-6 x 144 = 5.76mJ


(a) For the capacitors in series,

Q1 = Q2 C1v1 = C2v2 1

2

2

1

CC

vv

=

vs = v1 + v2 = 21

2122

1

2 vC

CCvvCC +

=+ s21

12 v

CCC+

=v

Similarly, s21

21 v

CCCv+

=

(b) For capacitors in parallel

v1 = v2 = 2

2

1

1

CQ

CQ

=

Qs = Q1 + Q2 = 22

2122

2

1 QC

CCQQCC +

=+

or

Q2 = 21

2

CCC+

s21

11 Q

CCC

Q+

=

i = dtdQ s

21

11 i

CCC+

=i , s21

22 i

CCC+

=i


(a) Ceq = C1 + C2 + C3 = 35µF (b) Q1 = C1v = 5 x 150µC = 0.75mC

Q2 = C2v = 10 x 150µC = 1.5mC Q3 = C3v = 20 x 150 = 3mC

(c) w = J150x35x21

222

eq µ=vC1 = 393.8mJ


(a) 207

201

101

51

C1

C1

C1

C1

321eq

=++=++=

Ceq = =µF720 2.857µF

(b) Since the capacitors are in series,

Q1 = Q2 = Q3 = Q = Ceqv = =µV200x720 0.5714mV

(c) w = =µ= J200x720x

21

222

eq vC1 57.143mJ

Chapter 6, Solution 28. We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C.

Ca

Cb

Cc

50 µF 20 µF

301

401

301

301

101

401

101

C1

a

+

+

=

= 102

401

101

40=++

3

Ca = 5µF

302

101

12001

3001

4001

C1

6

=++

=

Cb = 15µF

154

401

12001

3001

4001

C1

c

=++

=

Cc = 3.75µF Cb in parallel with 50µF = 50 + 15 = 65µF Cc in series with 20µF = 23.75µF

65µF in series with 23.75µF = F39.1775.88

75.23x65µ=

17.39µF in parallel with Ca = 17.39 + 5 = 22.39µF Hence Ceq = 22.39µF Chapter 6, Solution 29. (a) C in series with C = C/(2) C/2 in parallel with C = 3C/2

2C3 in series with C =

5C3

2C5

2C3Cx=

5C3 in parallel with C = C + =

5C3 1.6 C

(b) 2C

Ceq 2C

C1

C21

C21

C1

eq

=+=

Ceq = C


vo = ∫ +t

o)0(iidt

C1

For 0 < t < 1, i = 60t mA,

kVt100tdt6010x3

10v 2t

o6

3

o =+= ∫−

−

vo(1) = 10kV For 1< t < 2, i = 120 - 60t mA,

vo = ∫ +−−

− t

1 o6

3

)1(vdt)t60120(10x3

10 t + = [40t – 10t2 kV10] 1

= 40t – 10t2 - 20

<<−−

<<=

2t1,kV20t10t401t0,kVt10

)t(v2

2

o


<<+−<<<<

=5t3,t10503t1,mA201t0,tmA20

)t(is

Ceq = 4 + 6 = 10µF

)0(vidtC1v

t

oeq

+= ∫

For 0 < t < 1,

∫−

−

=t

o6

3

t2010x10

10v dt + 0 = t2 kV

For 1 < t < 3,

∫ +−=+=t

1

3

kV1)1t(2)1(vdt201010v

kV1t2 −= For 3 < t < 5,

∫ +−=t

3

3

)3(vdt)5t(101010v

kV11t5tkV55t 2t3

2 +−=++−=

<<+−

<<−<<

=

5t3,kV11t5t3t1,kV1t21t0,kVt

)t(v2

2

dtdv10x6

dtdvCi 6

11−==

=

<<−<<<<

5t3,mA30123t1,mA121t0,tmA12

dtdv10x4

dtdvCi 6

21−==

<<−<<<<

=5t3,mA20t83t1,mA81t0,tmA8


(a) Ceq = (12x60)/72 = 10 µ F

13001250501250)0(301012

10 2

00

21

26

3

1 +−=+−=+= −−−−

−

∫ tt

ttt eevdtex

v

23025020250)0(301060

10 2

00

22

26

3

2 −=+=+= −−−−

−

∫ tt

ttt eevdtex

v

(b) At t=0.5s,

03.138230250,15.84013001250 12

11 −=−==+−= −− evev

J 235.4)15.840(101221 26

12 == − xxxw Fµ

J 1905.0)03.138(102021 26

20 =−= − xxxw Fµ

J 381.0)03.138(104021 26

40 =−= − xxxFµw


Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals. 3F + 2F = 5F 1/5 +1/5 = 2/5 or 2.5F The voltage will divide equally across the two 5F capacitors. Therefore, we get: VTh = 7.5 V, CTh = 2.5 F


i = 6e-t/2

2/t3 e21)6(10x10

dtdiL −−

==v

= -30e-t/2 mV v(3) = -300e-3/2 mV = -0.9487 mV p = vi = -180e-t mW

p(3) = -180e-3 mW = -0.8 mW


dtdiLv = ==

∆∆=

−

)2/(6.010x60

t/iVL

3

200 mH


V)t2sin)(2)(12(10x41

dtdiLv 3 −== −

= - 6 sin 2t mV p = vi = -72 sin 2t cos 2t mW But 2 sin A cos A = sin 2A

p = -36 sin 4t mW Chapter 6, Solution 37.

t100cos)100(4x10x12dtdiLv 3−==

= 4.8 cos 100t V p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t

w = t200sin6.9pdtt

o

200/11

o∫ ∫=

Jt200cos200

6.9 200/11o−=

= =−π− mJ)1(cos48 96 mJ Chapter 6, Solution 38.

( )dtte2e10x40dtdiLv t2t23 −−− −==

= 240 0t,mVe)t1( t2 >− −


)0(iidtL1i

dtdiLv t

0 +∫=→=

1dt)4t2t3(10x200

1i t0

23 +∫ ++= −

1)t4tt(5t

023 +++=

i(t) = 5t3 + 5t2 + 20t + 1 A Chapter 6, Solution 40

dtdix

dtdiLv 31020 −==

<<+<<

<<=

ms 4t310t,40-ms 3t110t,-20

ms 10,10 tti

<<<<<<

=ms 4t3,10x10ms 3t1,10x10-

ms 10,1010

3

3

3 tx

dtdi

<<<<<<

=ms 4t3V, 200ms 3t1V, 200-

ms 10,V 200 tv

which is sketched below. v(t) V 200 0 1 2 3 4 t(ms) -200


( )∫∫ +−

=+= −t

o

t2t

03.0dt2120

21)0(ivdt

L1i

= Aetet tto

t 745103021 22 .. −+=+

+ −−10

At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A

L21w = i2 = 35.72J


∫ ∫ −=+=t

o

t

o1dt)t(v

51)0(ivdt

L1i

For 0 < t < 1, ∫ −=−=t

01t21dt

510i A

For 1 < t < 2, i = 0 + i(1) = 1A

For 2 < t < 3, i = ∫ +=+ 1t2)2(idt1051 2

t

= 2t - 3 A For 3 < t < 4, i = 0 + i(3) = 3 A

For 4 < t < 5, i = ∫ +=+t

4

t4 3t2)4(idt10

51

= 2t - 5 A

Thus,

2 1 , 0 11 , 1 2

( ) 2 3 , 2 33 , 3 42 5, 4

t A tA t

i t t A tA tt t

− <

5

<< <

= − < < < < − < <


w = L )(Li21)t(Li

21idt

2t

−∞−=∫ ∞−

( ) 010x60x10x80x21 33 −= −−

= 144 µJ Chapter 6, Solution 44.

( )∫ ∫ −+=+=t

ot

t

oo 1dt)t2cos104(51tivdt

L1i

= 0.8t + sin 2t -1


i(t) = ∫ +t

o)0(i)t(v

L1

For 0 < t < 1, v = 5t

∫−=t

o3 t510x101i dt + 0

= 0.25t2 kA

For 1 < t < 2, v = -10 + 5t

∫ ++−= −

t

13 )1(idt)t510(10x101i

∫= ( +−t

1kA25.0dt)1t5.0

= 1 - t + 0.25t2 kA

<<+−

<<=

2t1,kAt25.0t11t0,kAt25.0

)t(i2

2

Chapter 6, Solution 46. Under dc conditions, the circuit is as shown below: 2 Ω

+

vC

−3 A

iL

4 Ω

By current division,

=+

= )3(24

4iL 2A, vc = 0V

L21w L = =

= 22

L )2(21

21i 1J

C21w c = == )v)(2(

21v2

c 0J

Chapter 6, Solution 47. Under dc conditions, the circuit is equivalent to that shown below: R

+ vC −

5 A

iL

2 Ω

,2R

10)5(2R

2iL +=

+=

2RR10Riv Lc +

==

2

262

cc )2R(R100x10x80Cv

21w

+== −

232

1L )2R(100x10x2Li

21w

+== −

If wc = wL,

2

3

2

26

)2R(100x10x2

)2Rx(R100x10x80

+=

−− 80 x 10-3R2 = 2

R = 5Ω

Chapter 6, Solution 48. Under dc conditions, the circuit is as shown below:

+

vC2

−

+ −

+

vC1

−

iL1

6 Ω

4 Ω

iL2 30V

=+

==64

30ii2L1L 3A

==

1L1C i6v 18V

=2Cv 0V


(a) ( ) =+=++= 36544165Leq 7H (b) ( ) ==+= 41266112Leq 3H (c) ( ) ==+= 446324Leq 2H


( )63124510Leq ++=

= 10 + 5||(3 + 2) = 10 + 2.5 = 12.5 mH


101

301

201

601

L1

=++= L = 10 mH

( )45

35x10102510Leq =+=

= 7.778 mH

Chapter 6, Solution 52. 3//2//6 = 1H, 4//12 = 3H After the parallel combinations, the circuit becomes that shown below. 3H a 1H 1 H b Lab = (3+1)//1 = (4x1)/5 = 0.8 H


[ ])48(6)128(58106Leq +++++= 416)44(816 +=++= Leq = 20 mH


( )126010)39(4Leq +++= 34)40(124 +=++= Leq = 7H Chapter 6, Solution 55.

(a) L//L = 0.5L, L + L = 2L

LLLLLxLLLLLeq 4.1

5.025.025.0//2 =

++=+=

(b) L//L = 0.5L, L//L + L//L = L Leq = L//L = 0.5L


3L

L31LLL ==

Hence the given circuit is equivalent to that shown below: L

L/3 L

L/3

=+

=

+=

L35L

L35Lx

L32LLLeq L

85


Let dtdiLeqv = (1)

221 vdtdi4vvv +=+= (2)

i = i1 + i2 i2 = i – i1 (3)

3

vdtdi

ordtdi

3v 2112 == (4)

and

0dtdi5

dtdi2v 2

2 =++−

dtdi5

dtdi2v 2

2 += (5)

Incorporating (3) and (4) into (5),

3

v5dtdi7

dtdi5

dtdi5

dtdi2v 21

2 −=−+=

dtdi7

351v2 =

+

dtdi

835v2 =

Substituting this into (2) gives

dtdi

835

dtdi4v +=

dtdi

867

=

Comparing this with (1),

==867Leq 8.375H


===dtdi3

dtdiLv 3 x slope of i(t).

Thus v is sketched below:

6

t (s)

7 5 1 4 3 2

-6

v(t) (V) 6 Chapter 6, Solution 59.

(a) ( )dtdiLLv 21s +=

21

s

LLv

dtdi

+=

,dtdiL11 =v

dtdiL22 =v

,vLL

Lv s

21

11 += s

21

2L v

LLL

v+

=

(b) dtdi

Ldtdi

Lvv 22

112i ===

21s iii +=

( )

21

21

21

21s

LLLLv

Lv

Lv

dtdi

dtdi

dtdi +

=+=+=

∫∫ =+

== dtdtdi

LLLL

L1vdt

L1i s

21

21

111 s

21

2 iLL

L+

=+

== ∫∫ dtdtdi

LLLL

L1vdt

L1i s

21

21

222 s

21

1 iLL

L+


8

155//3 ==eqL

( ) tteqo ee

dtd

dtdiLv 22 154

815 −− −===

∫∫ −−− +=+=−+=+=t

tt

ttt

ooo eeeidttvLIi

0

2

0

22

0

A 5.15.05.12)15(512)0()(


(a) is = i1 + i2 i )0(i)0(i)0( 21s += 6 i)0(i4 2+= 2(0) = 2mA (b) Using current division:

( ) ==+

= − t2s1 e64.0i

203020i 2.4e-2t mA

=−= 1s2 iii 3.6e-2t mA

(c) mH1250

20x302030 ==

( ) === −−− 3t231 10xe6

dtd10x10

dtdiLv -120e-2t µV

( ) === −−− 3t232 10xe6

dtd10x12

dtdiLv -144e-2t µV

(d) ( )6t43mH10 10xe3610x30x

21w −−−=

Je8.021t

t4 µ=

−=

= 24.36nJ

( ) 2/1t6t43

mH30 10xe76.510x30x21w =

−−−=

= 11.693nJ

( ) 2/1t6t43

mH20 10xe96.1210x20x21w =

−−−=

= 17.54 nJ


(a) mH 4080

60202560//2025 =+=+=xLeq

∫∫ +−−=+=+=→= −−−

− ttt

eqeq ieidte

xidttv

Li

dtdiLv

0

333

3

)0()1(1.0)0(121040

10)0()(1


iiiii41,

43

8060

21 ===

01333.0)0(01.0)0(75.0)0(43)0(1 −=→−=→= iiii

mA 67.2125e- A )08667.01.0(41 3t-3

2 +=+−= − tei

mA 33.367.2125)0(2 −=+−=i

(b) mA 6575e- A )08667.01.0(43 3t-3

1 +=+−= − tei

mA 67.2125e- -3t2 +=i

Chapter 6, Solution 63. We apply superposition principle and let

21 vvvo += where v1 and v2 are due to i1 and i2 respectively.

<<−<<

===63,2

30,22 11

1 tt

dtdi

dtdiLv

<<−<<<<

===64,4

42,020,4

2 222

ttt

dtdi

dtdiLv

v1 v2 2 4 0 3 6 t 0 2 4 6 t -2 -4 Adding v1 and v2 gives vo, which is shown below. vo(t) V 6 2 0 2 3 4 6 t (s) -2 -6 Chapter 6, Solution 64.

(a) When the switch is in position A, i=-6 =i(0) When the switch is in position B,

8/1/,34/12)( ====∞ RLi τ

A 93)]()0([)()( 8/ tt eeiiiti −− −=∞−+∞= ι (b) -12 + 4i(0) + v=0, i.e. v=12 – 4i(0) = 36 V (c) At steady state, the inductor becomes a short circuit so that v= 0 V


(a) === 22115 )4(x5x

21iL

21w 40 W

=−= 220 )2)(20(

21w 40 W

(b) w = w5 + w20 = 80 W

(c) ( )( )3t20011 10xe502005

dtdvLi −−−==

= -50e-200tA

( )3t20022 10xe50)200(20

dtdvLi −−−==

= -200e-200tA

( )3t20022 10xe50)200(20

dtdvLi −−−==

= -200e-200t A (d) i = i1 + i2 = -250e-200t A

Chapter 6, Solution 66. mH60243640601620Leq =+=++=

dtdiLv =

∫ +=t

o)0(ivdt

L1i

∫= 121 dt + 0 mA −

t

o3 t4sin10x60

=−= tot4cos50i 50(1 - cos 4t) mA

mH244060 =

mV)t4cos1)(50(dtd10x24

dtdiLv 3 −== −

= 4.8 sin 4t mV


∫−= viRC1vo dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3

∫−

= t50sin10210v

3

o dt

vo = 100 cos 50t mV Chapter 6, Solution 68.

∫−= viRC1vo dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5

vo = ∫ −=+−t

ot20dt10

51

The op amp will saturate at vo = 12± -12 = -2t t = 6s

Chapter 6, Solution 69. RC = 4 x 106 x 1 x 10-6 = 4

∫ ∫−=−= dtv41dtv

RC1v iio

For 0 < t < 1, vi = 20, ∫ =−=t

oo dt2041v -5t mV

For 1 < t < 2, vi = 10, ∫ −−−=+−=t

1o 5)1t(5.2)1(vdt1041v

= -2.5t - 2.5mV

For 2 < t < 4, vi = - 20, ∫ −−=++=t

2o 5.7)2t(5)2(vdt2041v

= 5t - 17.5 mV

For 4 < t < 5m, vi = -10, 5.2)4t(5.2)4(vdt1041v

t

4o +−=+= ∫

= 2.5t - 7.5 mV

For 5 < t < 6, vi = 20, ∫ +−−=+−=t

5o 5)5t(5)5(vdt2041v

= - 5t + 30 mV

Thus vo(t) is as shown below:

2 5

6

5

751 432

5

0 Chapter 6, Solution 70.

One possibility is as follows:

50RC

=1

Let R = 100 kΩ, F2.010x100x50

13 µ==C

Chapter 6, Solution 71. By combining a summer with an integrator, we have the circuit below:

∫ ∫ ∫−−−= dtvCR

1dtvCR

1dtvCR

1v 22

22

11

o

−+

For the given problem, C = 2µF, R1C = 1 R1 = 1/(C) = 1006/(2) = 500 kΩ R2C = 1/(4) R2 = 1/(4C) = 500kΩ/(4) = 125 kΩ R3C = 1/(10) R3 = 1/(10C) = 50 kΩ


The output of the first op amp is

∫−= i1 vRC1v dt = ∫ −=−

t

o63 2t100idt

10x2x10x101

−

= - 50t

∫−= io vRC1v dt = ∫ −−

t

o63 dt)t50(10x5.0x10x20

1−

= 2500t2 At t = 1.5ms, == −62

o 10x)5.1(2500v 5.625 mV Chapter 6, Solution 73.

Consider the op amp as shown below: Let va = vb = v

At node a, R

vvR

v0 o−=

− 2v - vo = 0 (1)

v +

vo

−

b + − vi C

R

v R

a

R

−+

R

At node b, dtdvC

Rvv

Rvv oi +

−=

−

dtdvRCvv2v oi +−= (2)


dt

dv2

RCvvv oooi +−=

or

∫= io vRC2v dt

showing that the circuit is a noninverting integrator. Chapter 6, Solution 74.

RC = 0.01 x 20 x 10-3 sec

secmdtdv2.0

dtdvRCv i

o −=−=

<<−<<<<−

=4t3,V23t1,V21t0,V2

vo

Thus vo(t) is as sketched below:

3

t (ms)

2 1

-2

vo(t) (V) 2


,dt

dvRCv i0 −= 5.210x10x10x250RC 63 == −

=−= )t12(dtd5.2vo -30 mV


,dt

dvRCv io −= RC = 50 x 103 x 10 x 10-6 = 0.5

<<<<−

==5t5,55t0,10

dtdv5.0v i

o

The input is sketched in Fig. (a), while the output is sketched in Fig. (b).

t (ms)

5

5

0 10

(b)

15

-10

t (ms)

vi(t) (V)

5

5

0 10

(a)

15

vo(t) (V) Chapter 6, Solution 77. i = iR + iC

( )oF

0i v0dtdC

Rv0

R0v

−+−

=−

110x10CR 66F == −

Hence

+−=

dtdv

vv ooi

Thus vi is obtained from vo as shown below: –dvo(t)/dt – vo(t) (V)

4

-4

t (ms)

1

4

0 2 3

vi(t) (V)

3

8

2 1

t (ms)

-8

4 -4

4

-4

t (ms)

1

4

0 2 3


ooo

2

vdtdv2

t2sin10dtvd

−−=

Thus, by combining integrators with a summer, we obtain the appropriate analog computer as shown below:


We can write the equation as

)(4)( tytfdtdy

−=

which is implemented by the circuit below. 1 V t=0 C R R R R/4 R dy/dt - - -

+ -y + + R dy/dt

f(t)

R

t = 0

-dvo/dt

dvo/dt

d2vo/dt2

d2vo/dt2

-sin2t

2vo

C C

R

R/10

R

R

R

R/2

R

R

− +

−+

− +

−+

− +

+ − sin2t

−+

vo

R


From the given circuit,

dt

dvk200k1000v

k5000k1000)t(f

dtvd o

o2o

2

ΩΩ

−ΩΩ

−=

or

)t(fv2dt

dv5

dtvd

oo

2o

2

=++


We can write the equation as

)(252

2

tfvdtvd

−−=

which is implemented by the circuit below. C C R R - R R/5 d2v/dt2 + - -dv/dt + v - + d2v/dt2 R/2 f(t)

Chapter 6, Solution 82 The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown below. 10R R R R - + - vo + R C=1/(2R) R - + + vs - Chapter 6, Solution 83.

Since two 10µF capacitors in series gives 5µF, rated at 600V, it requires 8 groups in parallel with each group consisting of two capacitors in series, as shown below:

+

600

−

Answer: 8 groups in parallel with each group made up of 2 capacitors in series.


tqI∆∆

=∆ ∆I x ∆t = ∆q

∆q = 0.6 x 4 x 10-6

= 2.4µC

62.4 10 150

(36 20)q xC nv

−∆= = =∆ −

F

Chapter 6, Solution 85. It is evident that differentiating i will give a waveform similar to v. Hence,

dtdiLv =

<<−

<<=

2t1,t481t0,t4

i

<<−

<<==

2t1,L41t0,L4

dtdiLv

But,

<<−

<<=

2t1,mV51t0,mV5

v

Thus, 4L = 5 x 10-3 L = 1.25 mH in a 1.25 mH inductor Chapter 6, Solution 86. (a) For the series-connected capacitor

Cs = 8C

C1....

CC

=+++

111

For the parallel-connected strings,

=µ=== F3

1000x108C10

C10C sseq 1250µF

(b) vT = 8 x 100V = 800V

( ) 262Teq )800(10x1250

21vC

21w −==

= 400J


Applying KVL to Fig. 7.1.

0RidtiC1 t

-=+∫

∞

Taking the derivative of each term,

0dtdi

RCi

=+

or RCdt

idi

−=

Integrating,

RCt-

I)t(i

ln0

=

RCt-0eI)t(i =

RCt-0eRI)t(Ri)t(v ==

or RCt-0eV)t(v =


CR th=τ where is the Thevenin equivalent at the capacitor terminals. thR

Ω=+= 601280||120R th =××=τ -3105.060 ms30


(a) ms 10102105,510//10 63 ===Ω== −xxxCRkR ThTh τ (b) 6s3.020,208)255//(20 ===Ω=++= xCRR ThTh τ


eqeqCR=τ

where 21

21eq CC

CC+

=C , 21

21eq RR

RRR

+=

=τ)CC)(RR(

CCRR

2121

2121

++


τ= 4)-(t-e)4(v)t(v where 24)4(v = , 2)1.0)(20(RC ===τ

24)-(t-e24)t(v = == 26-e24)10(v V195.1


Ve4)t(v25210x2x10x40RC,ev)t(v

V4)24(210

2)0(vv

t5.12

36/to

o

−

−τ−

=

===τ=

=+

==


τ= t-e)0(v)t(v , CR th=τwhere is the Thevenin resistance across the capacitor. To determine , we insert a 1-V voltage source in place of the capacitor as shown below.

thR thR

8 Ω i2 i

i1

10 Ω0.5 V +

v = 1

−

+ −

+ − 1 V

1.0101

i1 == , 161

85.01

i2 =−

=

8013

161

1.0iii 21 =+=+=

1380

i1

R th ==

138

1.01380

CR th =×==τ

=)t(v V20 813t-e


(a) 41

RC ==τ

dtdv

Ci- =

=→= Ce-4))(10(Ce0.2- -4t-4t mF5

==C41

R Ω50

(b) ===τ41

RC s25.0

(c) =×== )100)(105(21

CV21

)0(w 3-20C mJ250

(d) ( )τ−=×= 02t-20

20R e1CV

21

CV21

21

w

21

ee15.0 00 8t-8t- =→−=

or 2e 08t =

== )2(ln81

t 0 ms6.86


τ= t-e)0(v)t(v , CR eq=τ

Ω=++=++= 82423||68||82R eq

2)8)(25.0(CR eq ===τ

=)t(v Ve20 2t-


10 Ω

10 mF +

v

−

i

15 Ωio

iT

4 Ω

A215

)3)(10(ii10i15 oo ==→=

i.e. if i , then A3)0( = A2)0(io = A5)0(i)0(i)0(i oT =+=

V502030)0(i4)0(i10)0(v T =+=+= across the capacitor terminals.

Ω=+=+= 106415||104R th 1.0)1010)(10(CR -3

th =×==τ -10tt- e50e)0(v)t(v == τ

)e500-)(1010(dtdv

Ci 10t-3-C ×==

=Ci Ae5- -10t By applying the current division principle,

==+

= CC i-0.6)i-(1510

15)t(i Ae3 -10t


Applying KCL to the RL circuit,

0Rv

dtvL1

=+∫

Differentiating both sides,

0vLR

dtdv

0dtdv

R1

Lv

=+→=+ LRt-eAv =

If the initial current is , then 0I

ARI)0(v 0 ==

τ= t-0 eRIv ,

RL

=τ

∫∞

=t

-dt)t(v

L1

i

t-

t-0 eL

RI-i ∞

ττ

= τ= t-

0 eRI-i τ= t-

0 eI)t(i Chapter 7, Solution 12. When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).

3 Ω

i(0-)

+ −

12 V 2 H 4 Ω

(a) (b)

A43

12)0(i ==−

Since the current through an inductor cannot change abruptly, A4)0(i)0(i)0(i === +−

When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b).

5.042

RL

===τ

Hence, == τt-e)0(i)t(i Ae4 -2t


thRL

=τ

where is the Thevenin resistance at the terminals of the inductor. thR

Ω=+=+= 37162120||8030||70R th

=×

=τ37102 -3

s08.81 µ


Converting the wye-subnetwork to delta gives 16Ω R2 80mH R1 R3 30Ω

Ω==Ω==Ω==++

= 17010

1700,3450

1700,8520/170020

105050202010321 RRxxxR

30//170 = (30x170)/200 = 25.5 Ω , 34//16=(34x16)/50 =10.88Ω

sxRLxRTh

Th m 14.3476.251080,476.25

38.12138.3685)88.105.25//(85

3

===Ω==+=−

τ


(a) sRLRTh

Th 25.020/5,2040//1012 ===Ω=+= τ

(b) ms 5.040/)1020(,408160//40 3 ===Ω=+= −xRLRTh

Th τ


eq

eq

RL

=τ

(a) LLeq = and 31

31312

31

312eq RR

RR)RR(RRR

RRRR

+++

=+

+=

=τ31312

31

RR)RR(R)RR(L

+++

(b) where 21

21eq LL

LL+

=L and 21

21213

21

213eq RR

RR)RR(RRR

RRRR

+++

=+

+=

=τ)RR)RR(R()LL(

)RR(LL

2121321

2121

++++


τ= t-e)0(i)t(i , 161

441

RL

eq===τ

-16te2)t(i =

16t-16t-

o e2)16-)(41(e6dtdi

Li3)t(v +=+=

=)t(vo Ve2- -16t

Chapter 7, Solution 18. If , the circuit can be redrawn as shown below. 0)t(v =

+

vo(t)

− i(t)

Req0.4 H

56

3||2R eq == , 31

65

52

RL

=×==τ -3tt- ee)0(i)t(i == τ

=== 3t-o e-3)(

52-

dtdi

-L)t(v Ve2.1 -3t


i1 i2

i2 i1

i/2 10 Ω 40 Ω

− +

1 V i

To find we replace the inductor by a 1-V voltage source as shown above. thR

0i401i10 21 =+− But 2iii 2 += and 1ii =i.e. i2i2i 21 ==

301

i0i201i10 =→=+−

Ω== 30i1

R th

s2.0306

RL

th===τ

=)t(i Ae2 -5t


(a). L50R501

RL

=→==τ

dtdi

Lv- =

=→= Le-50))(30(Le150- -50t-50t H1.0 == L50R Ω5

(b). ===501

RL

τ ms20

(c). === 22 )30)(1.0(21

)0(iL21

w J45

(d). Let p be the fraction

( )τ−=⋅ 02t-00 e1IL

21

pIL21

3296.0e1e1p -0.450(2)(10)- =−=−= i.e. =p %33


The circuit can be replaced by its Thevenin equivalent shown below.

Rth

+ −

Vth 2 H

V40)60(4080

80Vth =

+=

R3

80R80||40R th +=+=

R38040

RV

)(i)0(iIth

th

+==∞==

1380R

40)2(

21

IL21

w2

2 =

+==

340

R1380R

40=→=

+

=R Ω33.13


τ= t-e)0(i)t(i , eqR

L=τ

Ω=+= 5120||5R eq , 52

=τ

=)t(i Ae10 -2.5t Using current division, the current through the 20 ohm resistor is

2.5t-o e-2

5i-

-i)(205

5i ==

+=

== oi20)t(v Ve04- -2.5t


Since the 2 Ω resistor, 1/3 H inductor, and the (3+1) Ω resistor are in parallel, they always have the same voltage.

-1.5)0(i5.113

222

i- =→=+

+=

The Thevenin resistance at the inductor’s terminals is thR

34

)13(||2R th =+= , 41

3431

RL

th===τ

0t,e-1.5e)0(i)t(i -4tt- >== τ

4t-oL e/3)-1.5(-4)(1

dtdi

Lvv ===

=ov 0t,Ve2 -4t >

=+

= Lx v13

1v 0t,Ve5.0 -4t >


(a) =)t(v u(t)5-

(b) [ ] [ ])5t(u)3t(u10)3t(u)t(u-10)t(i −−−+−−=

= )5t(u10)3t(u20)t(u10- −−−+

(c) [ ] [ ])3t(u)2t(u)2t(u)1t(u)1t()t(x −−−+−−−−= [ ])4t(u)3t(u)t4( −−−−+

= )4t(u)4t()3t(u)3t()2t(u)2t()1t(u)1t( −−+−−−−−−−− = )4t(r)3t(r)2t(r)1t(r −+−−−−−

(d) [ ])1t(u)t(u5)t-(u2)t(y −−−= = )1t(u5)t(u5)t-(u2 −+−


v(t) = [u(t) + r(t – 1) – r(t – 2) – 2u(t – 2)] V Chapter 7, Solution 26.

(a) [ ])t(u)1t(u)t(u)1t(u)t(v1 −−+−+=

=)t(v1 )1t(u)t(u2)1t(u −+−+

(b) [ ])4t(u)2t(u)t4()t(v2 −−−−= )4t(u)4t()2t(u)4t(-)t(v2 −−+−−=

=)t(v2 4)r(t2)r(t2)u(t2 −+−−−

(c) [ ] [ ]6)u(t4)u(t44)u(t2)u(t2)t(v3 −−−+−−−= =)t(v3 6)u(t44)u(t22)u(t2 −−−+−

(d) [ ] )2t(ut1)u(t-t)2t(u)1t(u-t)t(v4 −+−=−−−=

)2t(u)22t()1t(u)11t-()t(v4 −+−+−−+= =)t(v4 2)u(t22)r(t1)u(t1)r(t- −+−+−−−


v(t) is sketched below.

1

2

4 3 2 0 1 t -1

v(t)

Chapter 7, Solution 28. i(t) is sketched below.

1

4 3 2 0 1 t

-1

i(t)


x(t) (a) 3.679 0 1 t (b) y(t) 27.18

0 t

(c) )1(6536.0)1(4cos)1(4cos)( −−=−=−= tttttz δδδ , which is sketched below. z(t) 0 1 t -0.653 ( )tδ Chapter 7, Solution 30.

(a) ==−δ =∫ 1t210

02 t4dt)1t(t4 4

(b) =π=π=−δπ =

∞

∞∫ cos)t2cos(dt)5.0t()t2cos( 5.0t-1-


(a) [ ] ===−δ =

∞

∞∫ 16-

2t4t-

-4t- eedt)2t(e 22 -910112×

(b) [ ] ( ) =++=π++=δπ+δ+δ =

∞

∞∫ 115)t2cos(e5dt)t(t2cos)t(e)t(5 0t

t--

t- 7 Chapter 7, Solution 32.

(a) 11)(111

−=== ∫∫ tdduttt

λλλλ

(b) 5.42

)1(0)1( 41

4

1

21

0

4

0

=−=−+=− ∫∫ ttdttdtdttr∫

(c ) 16)6()2()6( 22

5

1

2 =−=−− =∫ ttdttt δ


)0(idt)t(vL1

)t(it

0+= ∫

0dt)2t(201010

10)t(i

t

03-

-3

+−δ×

= ∫

=)t(i A)2t(u2 −


(a) [ ]

)1t()1t(01)1t()1t()1t(u

)1t(u)1t()1t(u )1t(udtd

−δ=+δ•+•−δ=+δ−

++−δ=+−

(b) [ ]

)6t(u)2t(01)6t(u)2t()6t(r

)2t(u)6t(u)2t(u )6t(rdtd

−=−δ•+•−=−δ−

+−−=−−

(c)

[ ]

)3t(5366.0)3t(u t4cos4)3t(3x4sin)3t(u t4cos4

)3t(t4sin)3t(u t4cos4)3t(u t4sindtd

−δ−−=−δ+−=

−δ+−=−


(a) 35t-eA)t(v = , -2A)0(v == =)t(v Ve2- 35t-

(b) 32teA)t(v = , 5A)0(v == =)t(v Ve5 32t


(a) 0t,eBA)t(v -t >+=

1A = , B10)0(v +== or B -1= =)t(v 0t,Ve1 -t >−

(b) 0t,eBA)t(v 2t >+=

-3A = , B-3-6)0(v +== or -3B = =)t(v ( ) 0t,Ve13- 2t >+

Chapter 7, Solution 37. Let v = vh + vp, vp =10.

4/041 t

hh Aevvhv −•

=→=+

tAev 25.010 −+= 8102)0( −=→+== AAv tev 25.0810 −−=

(a) s4=τ (b) 10)( =∞v V (c ) te 25.0810 −−=v


Let i = ip +ih

)(03 3 tuAeiii thhh

−•

=→=+

Let 32)(2)(3,0),( =→===

•

ktutkuitku ppi

)(32 tui p =

)()32( 3 tuAei t += −

If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus

)()1(32 3 tuei t−−=


(a) Before t = 0,

=+

= )20(14

1)t(v V4

After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

8)2)(4(RC ===τ , 4)0(v = , 20)(v =∞ 8t-e)208(20)t(v −+=

=)t(v Ve1220 8t-− (b) Before t = 0, 21 vvv += , where is due to the 12-V source and is

due to the 2-A source. 1v 2v

V12v1 = To get v , transform the current source as shown in Fig. (a). 2

V-8v2 = Thus,

=−= 812v V4 After t = 0, the circuit becomes that shown in Fig. (b).

2 F 2 F 4 Ω

12 V

+ −

+ − v2 8 V

+ −

3 Ω 3 Ω

(a) (b)

[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 12)(v =∞ , 4)0(v = , 6)3)(2(RC ===τ

6t-e)124(12)t(v −+= =)t(v Ve812 6t-−


(a) Before t = 0, =v V12 .

After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 4)(v =∞ , 12)0(v = , 6)3)(2(RC ===τ

6t-e)412(4)t(v −+= =)t(v Ve84 6t-+

(b) Before t = 0, =v V12 .

After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v After transforming the current source, the circuit is shown below.

t = 0

4 Ω

2 Ω

+ −

12 V 5 F

12)0(v = , 12)(v =∞ , 10)5)(2(RC ===τ

=v V12 Chapter 7, Solution 41.

0)0(v = , 10)12(1630

)(v ==∞

536

)30)(6()1)(30||6(CR eq ===

[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

5t-e)100(10)t(v −+=

=)t(v V)e1(10 -0.2t−


(a) [ ] τ∞−+∞= t-

oooo e)(v)0(v)(v)t(v

0)0(vo = , 8)12(24

4)(vo =

+=∞

eqeqCR=τ , 34

4||2R eq ==

4)3(34

==τ 4t-

o e88)t(v −= =)t(vo V)e1(8 -0.25t−

(b) For this case, 0)(vo =∞ so that

τ= t-oo e)0(v)t(v

8)12(24

4)0(vo =

+= , 12)3)(4(RC ===τ

=)t(vo Ve8 12t- Chapter 7, Solution 43. Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source.

40v

2i5.0 o−= , 80v

i o=

Hence, 645

320v

40v

280v

21

ooo ==→−=

==80v

i o A8.0

After t = 0, the circuit is as shown in Fig. (b).

τ= t-CC e)0(v)t(v , CR th=τ

To find , we replace the capacitor with a 1-V voltage source as shown in Fig. (c). thR

0.5i vC

+ − 0.5i

i

1 V 80 Ω

(c)

801

80v

i C == , 80

5.0i5.0io ==

Ω=== 1605.0

80i1

Ro

th , 480CR th ==τ

V64)0(vC = 480t-

C e64)t(v =

480t-CC e64

4801

-3dt

dv-Ci-i5.0

===

=)t(i Ae8.0 480t- Chapter 7, Solution 44.

Ω== 23||6R eq , 4RC ==τ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

Using voltage division,

V10)30(63

3)0(v =

+= , V4)12(

633

)(v =+

=∞

Thus, 4t-4t- e64e)410(4)t(v +=−+=

=

== 4t-e41-

)6)(2(dtdv

C)t(i Ae3- -0.25t


For t < 0, 0)0(v0)t(u5vs =→==

For t > 0, , 5vs = 45

)5(124

4)(v =

+=∞

1012||47R eq =+= , 5)21)(10(CR eq ===τ

[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

=)t(v V)e1(25.1 5t-−

5t-e

51-

45-

21

dtdv

C)t(i

==

=)t(i Ae125.0 5t-


30566)(,0)0(,225.0)62( ===∞==+== xivvsxCR sThτ

)1(30)]()0([)()( 2// tt eevvvtv −− −=∞−+∞= τ V Chapter 7, Solution 47.

For t < 0, , 0)t(u = 0)1t(u =− , 0)0(v = For 0 < t < 1, 1)1.0)(82(RC =+==τ

0)0(v = , 24)3)(8()(v ==∞ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

( )t-e124)t(v −= For t > 1, ( ) 17.15e124)1(v 1- =−=

30)(v024-)v(6- =∞→=∞+ 1)--(te)3017.15(30)t(v −+=

1)--(te83.1430)t(v −= Thus,

=)t(v( )

>−<<−1t,Ve83.1430

1t0,Ve124-1)(t-

t-


For t < 0, , 1-t)(u = V10)0(v = For t > 0, , 0-t)(u = 0)(v =∞

301020R th =+= , 3)1.0)(30(CR th ===τ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

=)t(v Ve10 3t-

3t-e1031-

)1.0(dtdv

C)t(i

==

=)t(i Ae31- 3t-


For 0 < t < 1, , 0)0(v = 8)4)(2()(v ==∞

1064R eq =+= , 5)5.0)(10(CR eq ===τ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

( ) Ve18)t(v 5t-−= For t > 1, ( ) 45.1e18)1(v -0.2 =−= , 0)(v =∞

[ ] τ−∞−+∞= )1t(-e)(v)1(v)(v)t(v Ve45.1)t(v 5)1t(- −=

Thus,

=)t(v ( )

><<−

− 1t,Ve45.11t0,Ve18

5)1t(-

5t-


For the capacitor voltage, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

0)0(v = For t < 0, we transform the current source to a voltage source as shown in Fig. (a).

1 kΩ1 kΩ

+

v

−

+ −

30 V 2 kΩ

(a)

V15)30(112

2)(v =

++=∞

Ω=+= k12||)11(R th

41

1041

10CR 3-3th =××==τ

( ) 0t,e115)t(v -4t >−=

We now obtain i from v(t). Consider Fig. (b). x

ix

1/4 mF

v 1 kΩ

1 kΩ

iT

30 mA 2 kΩ

(b)

Tx imA30i −=

But dtdv

CRv

i3

T +=

( ) Ae-15)(-4)(1041

mAe15.7)t(i 4t-3-4t-T ×+−=

( ) mAe15.7)t(i -4tT +=

Thus,

mAe5.75.730)t(i -4tx −−=

=)t(i x ( ) 0t,mAe35.7 -4t >− Chapter 7, Solution 51.

Consider the circuit below.

fter the switch is closed, applying KVL gives

R-di=

L

R

i

t = 0

+ −

+

v

−

VS

A

dtLRiVS +=

di

−=

RV

iR-dtdi

L S or

dtLRVi S−

tegrating both sides, In

tLR-

Riln I0

=

−

V )t(iS

τ=

−− t-

RVIRVi

lnS0

S

τ=−− t-

S0

S eRVIRVi

or

τ

−+= t-S0

S eRV

IRV

)t(i

which is the same as Eq. (7.60).

hapter 7, Solution 52.

C

A21020

)0(i == , A2)(i =∞

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i

=)t A2 (i


(a) Before t = 0,

C

=+

=23

25i A5

After t = 0, τ= t-e)0(i)t(i L

224

R===τ , 5)0(i =

=)t(i Ae5 2t-

(b) Before t = 0, the inductor acts as a short circuit so that the 2 Ω and 4 Ω resistors are short-circuited.

=)t(i A6 After t = 0, we have an RL circuit.

= e)0(i)t(i , 23

R==τ τt- L

=)t(i Ae6 3t2-

Chapter 7, Solution 54. (a) Before t = 0, i is obtained by current division or

=+

= )2(44

4)t(i A1

After t = 0, [ ] τ∞−+∞= t-e)(i)0(i)(i)t(i

eqRL

=τ , Ω=+= 712||44R eq

21

75.3==τ

1)0(i = , 76

)2(34

3)2(

12||4412||4

)(i =+

=+

=∞

t2-e76

176

)t(i

−+=

=)t(i ( ) Ae671 2t-−

(b) Before t = 0, =+

=32

10)t(i A2

After t = 0, 5.42||63R eq =+=

94

5.42

RL

eq===τ

2)0(i = To find , consider the circuit below, at t = when the inductor becomes a short circuit,

)(i ∞

v

2 Ω

24 V

6 Ω

+ −

+ −

i

2 H 10 V

3 Ω

9v3v

6v24

2v10

=→=−

+−

A33v

)(i ==∞ 4t9-e)32(3)t(i −+=

=)t(i Ae3 4t9-−

Chapter 7, Solution 55. For t < 0, consider the circuit shown in Fig. (a).

+

v

−

4io io

+ −

io 0.5 H

+ −

0.5 H

+

v

−

+ −

i 3 Ω 8 Ω

2 Ω 2 Ω24 V 20 V

(a) (b)

24i0i424i3 ooo =→=−+

== oi4)t(v V96 A482v

i ==

For t > 0, consider the circuit in Fig. (b).

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i

48)0(i = , A228

20)(i =

+=∞

Ω=+= 1082R th , 201

105.0

RL

th===τ

-20t-20t e462e)248(2)t(i +=−+= == )t(i2)t(v Ve924 -20t+


Ω=+= 105||206R eq , 05.0RL==τ

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i i(0) is found by applying nodal analysis to the following circuit.

0.5 H

+ −

12 Ω

5 Ω

20 Ω

6 Ω

+

v

−

vx i

20 V

2 A

12v6

v20v

12v

5v20

2 xxxxx =→++=

−+

A26

v)0(i x ==

45||20 =

6.1)4(64

4)(i =

+=∞

-20t0.05t- e4.06.1e)6.12(6.1)t(i +=−+=

Since ,

20t-e-20)()4.0(21

dtdi

L)t(v ==

=)t(v Ve4- -20t Chapter 7, Solution 57.

At , the circuit has reached steady state so that the inductors act like short circuits.

−= 0t

+ −

6 Ω i

5 Ω

i1 i2

20 Ω30 V

31030

20||5630

i ==+

= , 4.2)3(2520

i1 == , 6.0i2 =

A4.2)0(i1 = , A6.0)0(i2 =

For t > 0, the switch is closed so that the energies in L and flow through the closed switch and become dissipated in the 5 Ω and 20 Ω resistors.

1 2L

1t-11 e)0(i)t(i τ= ,

21

55.2

RL

1

11 ===τ

=)t(i1 Ae4.2 -2t

2t-22 e)0(i)t(i τ= ,

51

204

RL

2

22 ===τ

=)t(i2 Ae6.0 -5t


For t < 0, 0)t(vo =

For t > 0, , 10)0(i = 531

20)(i =

+=∞

Ω=+= 431R th , 161

441

RL

th===τ

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i =)t(i ( ) Ae15 -16t+

( ) 16t-16t-o e-16)(5)(

41

e115dtdi

Li3)t(v ++=+=

=)t(vo Ve515 -16t− Chapter 7, Solution 59.

Let I be the current through the inductor. For t < 0, , 0vs = 0)0(i =

For t > 0, 63||64Req =+= , 25.065.1

RL

eq===τ

1)3(42

2)(i =

+=∞

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i -4te1)t(i −=

)-4)(-e)(5.1(dtdi

L)t(v 4t-o ==

=)t(vo Ve6 -4t


Let I be the inductor current. For t < 0, 0)0(i0)t(u =→=

For t > 0, Ω== 420||5Req , 248

RL

eq===τ

4)(i =∞ [ ] τ∞−+∞= t-e)(i)0(i)(i)t(i

( )2t-e14)t(i −=

2t-e21-

)4-)(8(dtdi

L)t(v

==

=)t(v Ve16 -0.5t Chapter 7, Solution 61.

The current source is transformed as shown below.

4 Ω

20u(-t) + 40u(t)

+ −

0.5 H

81

421

RL

===τ , 5)0(i = , 10)(i =∞

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i =)t(i Ae510 -8t−

8t-e)8-)(5-(

21

dtdi

L)t(v

==

=)t(v Ve20 -8t Chapter 7, Solution 62.

16||3

2RL

eq===τ

For 0 < t < 1, so that 0)1t(u =−

0)0(i = , 61

)(i =∞

( )t-e161

)t(i −=

For t > 1, ( ) 1054.0e161

)1(i 1- =−=

21

61

31

)(i =+=∞ 1)--(te)5.01054.0(5.0)t(i −+=

1)--(te3946.05.0)t(i −= Thus,

=)t(i ( )

>−

<<−

1tAe3946.05.0

1t0Ae161

-1)(t-

t-


For t < 0, , 1)t-(u = 25

10)0(i ==

For t > 0, , 0-t)(u = 0)(i =∞

Ω== 420||5R th , 81

45.0

RL

th===τ

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i =)t(i Ae2 -8t

8t-e)2)(8-(

21

dtdi

L)t(v

==

=)t(v Ve8- -8t Chapter 7, Solution 64.

Let i be the inductor current. For t < 0, the inductor acts like a short circuit and the 3 Ω resistor is short-circuited so that the equivalent circuit is shown in Fig. (a).

v 6 Ω

(b)

i

3 Ω

2 Ω

+ − 10 Ω

(a)

i

3 Ω

6 Ω

+ −

io

10 Ω

A667.16

10)0(ii ===

For t > 0, Ω=+= 46||32R th , 144

RL

th===τ

To find , consider the circuit in Fig. (b). )(i ∞

610

v2v

3v

6v10

=→+=−

65

2v

)(ii ==∞=

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i

( ) Ae165

e65

610

65

)t(i t-t- −=

−+=

ov is the voltage across the 4 H inductor and the 2 Ω resistor

t-t-t-o e

610

610

e-1)(65

)4(e6

106

10dtdi

Li2)t(v −=

++=+=

=)t(vo ( ) Ve1667.1 -t− Chapter 7, Solution 65.

Since [ ])1t(u)t(u10vs −−= , this is the same as saying that a 10 V source is turned on at t = 0 and a -10 V source is turned on later at t = 1. This is shown in the figure below.

For 0 < t < 1, , 0)0(i = 25

10)(i ==∞

vs

-10

10

1

t

420||5R th == , 21

42

RL

th===τ

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i

( ) Ae12)t(i -2t−= ( ) 729.1e12)1(i -2 =−=

For t > 1, since 0v0)(i =∞ s =

τ−= )1t(-e)1(i)t(i Ae729.1)t(i )1t(-2 −=

Thus,

=)t(i( )

><<−

− 1tAe729.11t0Ae12

)1t(2-

2t-


Following Practice Problem 7.14, τ= t-

T eV)t(v

-4)0(vVT == , 501

)102)(1010(CR 6-3f =××==τ

-50te-4)t(v = 0t,e4)t(v-)t(v -50t

o >==

=×

== 50t-3

o

oo e

10104

R)t(v

)t(i 0t,mAe4.0 -50t >


The op amp is a voltage follower so that vvo = as shown below.

R

vo

vo −+

+

vo

−

C

v1 R

R

At node 1,

o1o111o v

32

vR

vvR

0vR

vv=→

−+

−=

−

At the noninverting terminal,

0R

vvdt

dvC 1oo =

−+

ooo1oo v

31v

32vvv

dtdv

RC =−=−=−

RC3v

dtdv oo −=

3RCt-To eV)t(v =

V5)0(vV oT == , 100

3)101)(1010)(3(RC3 6-3 =××==τ

=)t(vo Ve5 3100t-

Chapter 7, Solution 68. This is a very interesting problem and has both an important ideal solution as well as an important practical solution. Let us look at the ideal solution first. Just before the switch closes, the value of the voltage across the capacitor is zero which means that the voltage at both terminals input of the op amp are each zero. As soon as the switch closes, the output tries to go to a voltage such that the input to the op amp both go to 4 volts. The ideal op amp puts out whatever current is necessary to reach this condition. An infinite (impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp). So vo will be equal to 8 volts for all t > 0. What happens in a real circuit? Essentially, the output of the amplifier portion of the op amp goes to whatever its maximum value can be. Then this maximum voltage appears across the output resistance of the op amp and the capacitor that is in series with it. This results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts.

vC(t) = Vop amp max(1 – e-t/(RoutC)) volts, for all values of vC less than 8 V,

= 8 V when t is large enough so that the 8 V is reached. Chapter 7, Solution 69.

Let v be the capacitor voltage. x

For t < 0, 0)0(vx =

For t > 0, the 20 kΩ and 100 kΩ resistors are in series since no current enters the op amp terminals. As ∞→t , the capacitor acts like an open circuit so that

1348

)4(1010020

10020)(vx =

+++

=∞

Ω=+= k12010020R th , 3000)1025)(10120(CR 3-3th =××==τ

[ ] τ∞−+∞= t-xxxx e)(v)0(v)(v)t(v

( )3000t-x e1

1348

)t(v −=

== )t(v120100

)t(v xo ( ) Ve11340 3000t-−


Let v = capacitor voltage. For t < 0, the switch is open and 0)0(v = . For t > 0, the switch is closed and the circuit becomes as shown below.

2

1

C R

+ −v + −

vo

+−

vS

s21 vvv == (1)

dtdv

CR

v0 s =−

(2)

where (3) vvvvvv soos −=→−= From (1),

0RCv

dtdv s ==

∫ =+=RC

vt-)0(vdtv

RC1-

v ss

Since v is constant,

1.0)105)(1020(RC -63 =××=

mVt-200mV0.1

t20-v ==

From (3),

t20020vvv so +=−= =ov mV)t101(20 +


Let v = voltage across the capacitor. Let v = voltage across the 8 kΩ resistor. o

For t < 2, so that 0v = 0)2(v = . For t > 2, we have the circuit shown below.

+

vo

−

10 kΩ

+ −

20 kΩ

+

v

−

100 mF

10 kΩ−+ io

4 V 8 kΩ

Since no current enters the op amp, the input circuit forms an RC circuit.

1000)10100)(1010(RC 3-3 =××==τ [ ] τ−∞−+∞= )2t(-e)(v)2(v)(v)t(v

( )1000)2t(-e14)t(v −−= As an inverter,

( )1e2vk20k10-

v 1000)2t(-o −== −

==8v

i oo ( ) A1e25.0 1000)2t(- −−


The op amp acts as an emitter follower so that the Thevenin equivalent circuit is shown below.

C

+ −v + −

io

3u(t) R

Hence,

[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v V2-)0(v = , V3)(v =∞ , 1.0)1010)(1010(RC -63 =××==τ

-10t-10t e53e3)--2(3)t(v −=+=

10t-6-o e)10-)(5-)(1010(

dtdv

Ci ×==

=oi 0t,mAe5.0 -10t > Chapter 7, Solution 73.


Rf

+

vo

−

+ −v

+ −

R1 −+

C v2v1 v3

v1

At node 2,

dtdv

CR

vv

1

21 =−

(1)

At node 3,

f

o3

Rvv

dtdv

C−

= (2)

But and 0v3 = 232 vvvv =−= . Hence, (1) becomes

dtdv

CR

vv

1

1 =−

dtdv

CRvv 11 =−

or CR

vCR

vdtdv

1

1

1=+

which is similar to Eq. (7.42). Hence,

( )

>−+<

= τ 0tevvv0tv

)t(v t-1T1

T

where and 1)0(vvT == 4v1 = 2.0)1020)(1010(CR 6-3

1 =××==τ

>−<

=0te340t1

)t(v 5t-

From (2),

)e15)(1020)(1020(dtdv

CR-v 5t-6-3fo ××==

0t,e-6v -5to >= =ov V)t(ue6- -5t

Chapter 7, Solution 74. Let v = capacitor voltage.

Rf

+

vo

−

+ −v

+ −

R1 −+

C v2v1 v3

v1

For t < 0, 0)0(v =For t > 0, . Consider the circuit below. A10is µ=

Rv

dtdv

Cis += (1)

[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v (2) It is evident from the circuit that

1.0)1050)(102(RC 36 =××==τ −

is

+

vo

−

C Rf

−+

R

is

At steady state, the capacitor acts like an open circuit so that i passes through R. Hence,

s

V5.0)1050)(1010(Ri)(v 36s =××==∞ −

Then,

( ) Ve15.0)t(v -10t−= (3)

But fsof

os Ri-v

Rv0

i =→−

= (4)

Combining (1), (3), and (4), we obtain

dtdv

CRvRR-

v ff

o −=

dtdv

)102)(1010(v51-

v 6-3o ××−=

( )-10t-2-10to e10-)5.0)(102(e0.1-0.1v ×−+=

1.0e2.0v -10to −= =ov ( ) V1e21.0 -10t −

Chapter 7, Solution 75. Let v = voltage at the noninverting terminal. 1

Let 2v = voltage at the inverting terminal.

For t > 0, 4vvv s21 ===

o1

s iR

v0=

−, Ω= k20R1

Ri-v oo = (1)

Also, dtdv

CRv

i2

o += , Ω= k10R 2 , F2C µ=

i.e. dtdv

CRv

Rv-

21

s += (2)

This is a step response.

[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v , 1)0(v =

where 501

)102)(1010(CR 6-32 =××==τ

At steady state, the capacitor acts like an open circuit so that i passes through

. Hence, as o

2R ∞→t

2o

1

s

R)(v

iRv- ∞

==

i.e. -2)4(2010-

vRR-

)(v s1

2 ===∞

-50te2)(1-2)t(v ++=

-50te3-2)t(v += But os vvv −=

or t50-so e324vvv −+=−=

=ov Ve36 t50-−

mA-0.220k

4-Rv-

i1

so ===

or =+=dtdv

CRv

i2

o mA0.2-

Chapter 7, Solution 76. The schematic is shown below. For the pulse, we use IPWL and enter the corresponding values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and simulating the circuit, we select Trace/Add and display –V(C1:2). The plot of V(t) is shown below.

Chapter 7, Solution 77. The schematic is shown below. We click Marker and insert Mark Voltage Differential at the terminals of the capacitor to display V after simulation. The plot of V is shown below. Note from the plot that V(0) = 12 V and V(∞) = -24 V which are correct.

Chapter 7, Solution 78. (a) When the switch is in position (a), the schematic is shown below. We insert

IPROBE to display i. After simulation, we obtain,

i(0) = 7.714 A from the display of IPROBE.

(b) When the switch is in position (b), the schematic is as shown below. For inductor I1, we let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below. Note that i(∞) = 12A, which is correct.

Chapter 7, Solution 79. When the switch is in position 1, io(0) = 12/3 = 4A. When the switch is in position 2,

3/80,3/84//)53(A, 5.035

4)( ===+=−=+

−=∞LR

Ri ThTho τ

A 5.45.0)]()0([)()( 80/3/ tt

oooo eeiiiti −− +−=∞−+∞= τ Chapter 7, Solution 80.

(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are short-

circuited so that 0)0()0()0( 21 === ovii but the current through the 4-H inductor is iL(0) =30/10 = 3A. (b) When the switch is in position B,

5.04/2,26//3 ===Ω==LR

R ThTh τ

A 330)]()0([)()( 25.0// ttt

LLLL eeeiiiti −−− =+=∞−+∞= τ

(c) A 0)(93)(,A 2

51030)( 21 =∞−=∞=+

=∞ Liii

V 0)()( =∞→= oL

o vdtdiLtv

Chapter 7, Solution 81. The schematic is shown below. We use VPWL for the pulse and specify the attributes as shown. In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final Step = 3 S. By inserting a current marker at one termial of LI, we automatically obtain the plot of i after simulation as shown below.


C

=××

=τ

=→=τ 6-

-3

10100103

CRRC Ω30


sxxxRCvv 51010151034,0)0(,120)( 66 =====∞ −τ

)1(1206.85)]()0([)()( 510// tt eevvvtv −− −=→∞−+∞= τ Solving for t gives st 16.637488.3ln510 == speed = 4000m/637.16s = 6.278m/s


Let Io be the final value of the current. Then 50/18/16.0/),1()( / ===−= − LReIti t

o ττ

. ms 33.184.0

1ln501)1(6.0 50 ==→−= − teII t

oo


(a) s24)106)(104(RC -66 =××==τ Since [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v

[ ] τ∞−=∞− 1t-1 e)(v)0(v)(v)t(v (1)

[ ] τ∞−=∞− 2t-2 e)(v)0(v)(v)t(v (2)

Dividing (1) by (2),

τ−=∞−∞− )tt(

2

1 12e)(v)t(v)(v)t(v

∞−∞−

τ=−=)(v)t(v)(v)t(v

lnttt2

1120

==

−−

= )2(ln241203012075

ln24t0 s63.16

(b) Since , the light flashes repeatedly every tt0 <==τ RC s24


[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 12)(v =∞ , 0)0(v = ( )τ−= t-e112)t(v

( )τ−== 0t-0 e1128)t(v

31

ee1128

00 t-t- =→−= ττ

)3(lnt0 τ= For , Ω= k100R

s2.0)102)(10100(RC -63 =××==τ s2197.0)3(ln2.0t0 ==

For , Ω= M1R

s2)102)(101(RC -66 =××==τ s197.2)3(ln2t0 ==

Thus,

s197.2ts2197.0 0 <<


Let i be the inductor current.

For t < 0, A2.1100120

)0(i ==−

For t > 0, we have an RL circuit

1.0400100

50RL

=+

==τ , 0)(i =∞

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i -10te2.1)t(i =

At t = 100 ms = 0.1 s,

== -1e2.1)1.0(i A441.0 which is the same as the current through the resistor.


(a) s60)10200)(10300(RC -123 µ=××==τAs a differentiator,

ms6.0s60010T =µ=τ> i.e. =minT ms6.0

(b) s60RC µ==τAs an integrator,

s61.0T µ=τ< i.e. =maxT s6 µ


Since s1T1.0 µ=<τ

s1RL

µ<

)101)(10200(10RL -63-6 ××=×<

mH200L <


We determine the Thevenin equivalent circuit for the capacitor C . s

ips

sth v

RRR

v+

= , psth R||RR =

Rth

+ −

Vth Cs

The Thevenin equivalent is an RC circuit. Since

ps

sith RR

R101

v101

v+

=→=

===96

R91

R ps ΩM32

Also,

s15CR sth µ==τ

where Ω=+

== M6.0326)32(6

R||RR spth

=××

=τ

= 6

6-

ths 100.6

1015R

C pF25


mA2405012

)0(io == , 0)(i =∞

[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i τ= t-e240)t(i

R2

RL==τ

τ== 0t-0 e24010)t(i

)24(lnt24e 0t0 τ=→=τ

R2

573.1)24(ln

5)24(ln

t 0 ====τ

==573.12

R Ω271.1


<<×

<<×⋅×==

DR6-

R3-9-

ttt10510-

tt0102

10

104dtdv

Ci

µ+<<<<µ

=s5ms2tms2mA8-

ms2t0A20)t(i

which is sketched below.

20 µA

2 ms

5 µs

-8 mA

t

i(t)

(not to scale)

Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a).

+

vL

−

6 Ω

10 H

+

v

−(a)

6 Ω

+ −

6 Ω

VS

10 µF

(b)

i(0-) = 12/6 = 2A, v(0-) = 12V

At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V (b) For t > 0, we have the equivalent circuit shown in Figure (b).

vL = Ldi/dt or di/dt = vL/L Applying KVL at t = 0+, we obtain,

vL(0+) – v(0+) + 10i(0+) = 0

vL(0+) – 12 + 20 = 0, or vL(0+) = -8 Hence, di(0+)/dt = -8/2 = -4 A/s Similarly, iC = Cdv/dt, or dv/dt = iC/C

iC(0+) = -i(0+) = -2

dv(0+)/dt = -2/0.4 = -5 V/s (c) As t approaches infinity, the circuit reaches steady state.

i(∞) = 0 A, v(∞) = 0 V

Chapter 8, Solution 2. (a) At t = 0-, the equivalent circuit is shown in Figure (a). 25 kΩ 20 kΩ

iR + −

+

v

−

iL

60 kΩ

80V

(a) 25 kΩ 20 kΩ

iR + −

iL 80V

(b)

60||20 = 15 kohms, iR(0-) = 80/(25 + 15) = 2mA. By the current division principle,

iL(0-) = 60(2mA)/(60 + 20) = 1.5 mA

vC(0-) = 0 At t = 0+,

vC(0+) = vC(0-) = 0

iL(0+) = iL(0-) = 1.5 mA

80 = iR(0+)(25 + 20) + vC(0-)

iR(0+) = 80/45k = 1.778 mA But, iR = iC + iL

1.778 = iC(0+) + 1.5 or iC(0+) = 0.278 mA

(b) vL(0+) = vC(0+) = 0

But, vL = LdiL/dt and diL(0+)/dt = vL(0+)/L = 0

diL(0+)/dt = 0

Again, 80 = 45iR + vC

0 = 45diR/dt + dvC/dt

But, dvC(0+)/dt = iC(0+)/C = 0.278 mohms/1 µF = 278 V/s

Hence, diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45

diR(0+)/dt = -6.1778 A/s

Also, iR = iC + iL

diR(0+)/dt = diC(0+)/dt + diL(0+)/dt

-6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s (c) As t approaches infinity, we have the equivalent circuit in Figure

(b).

iR(∞) = iL(∞) = 80/45k = 1.778 mA

iC(∞) = Cdv(∞)/dt = 0. Chapter 8, Solution 3. At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). iL(0-) = 0, and vR(0-) = 0. But, -vR(0-) + vC(0-) + 10 = 0, or vC(0-) = -10V. (a) At t = 0+, since the inductor current and capacitor voltage cannot change abruptly,

the inductor current must still be equal to 0A, the capacitor has a voltage equal to –10V. Since it is in series with the +10V source, together they represent a direct short at t = 0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore, vR(0+) = 0 V.

(b) At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0, thus, diL/dt = 0A/s,

iC(0+) = 2 A, this means that dvC(0+)/dt = 2/C = 8 V/s. Now for the value of dvR(0+)/dt. Since vR = vC + 10, then dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s.

40 Ω 40 Ω

+ − 10V

+

vC

−

10 Ω

2A

iL +

vR

−

+

vR

−

+ − 10V

+

vC

− 10 Ω

(b) (a) (c) As t approaches infinity, we end up with the equivalent circuit shown in Figure (b).

iL(∞) = 10(2)/(40 + 10) = 400 mA

vC(∞) = 2[10||40] –10 = 16 – 10 = 6V

vR(∞) = 2[10||40] = 16 V Chapter 8, Solution 4. (a) At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a).

i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V. Hence, i(0+) = i(0-) = 5A v(0+) = v(0-) = 25V

3 Ω

5 Ω

i +

v

−

+ −

40V

(a)

0.25 H3 Ω

iRiC

+ −

+ vL −i

5 Ω0.1F

4 A

40V (b) (b) iC = Cdv/dt or dv(0+)/dt = iC(0+)/C For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i and v cannot change abruptly,

iR = v/5 = 25/5 = 5A, i(0+) + 4 =iC(0+) + iR(0+)

5 + 4 = iC(0+) + 5 which leads to iC(0+) = 4

dv(0+)/dt = 4/0.1 = 40 V/s Chapter 8, Solution 5. (a) For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0).

iL(0-) = 0 and vC(0-) = 0.

For t = 0+, 4u(t) = 4. Consider the circuit below.

iL

iC + vL −

1 H

+

v

−

4 Ω 0.25F +

vC

−

Ai

6 Ω

4A

Since the 4-ohm resistor is in parallel with the capacitor,

i(0+) = vC(0+)/4 = 0/4 = 0 A Also, since the 6-ohm resistor is in series with the inductor, v(0+) = 6iL(0+) = 0V.

(b) di(0+)/dt = d(vR(0+)/R)/dt = (1/R)dvR(0+)/dt = (1/R)dvC(0+)/dt

= (1/4)4/0.25 A/s = 4 A/s

v = 6iL or dv/dt = 6diL/dt and dv(0+)/dt = 6diL(0+)/dt = 6vL(0+)/L = 0

Therefore dv(0+)/dt = 0 V/s

(c) As t approaches infinity, the circuit is in steady-state.

i(∞) = 6(4)/10 = 2.4 A

v(∞) = 6(4 – 2.4) = 9.6 V Chapter 8, Solution 6. (a) Let i = the inductor current. For t < 0, u(t) = 0 so that

i(0) = 0 and v(0) = 0.

For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0. vR(0+) = Ri(0+) = 0 V

Also, since v(0+) = vR(0+) + vL(0+) = 0 = 0 + vL(0+) or vL(0+) = 0 V. (1)

(b) Since i(0+) = 0, iC(0+) = VS/RS

But, iC = Cdv/dt which leads to dv(0+)/dt = VS/(CRS) (2)

From (1), dv(0+)/dt = dvR(0+)/dt + dvL(0+)/dt (3) vR = iR or dvR/dt = Rdi/dt (4)

But, vL = Ldi/dt, vL(0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0 (5)

From (4) and (5), dvR(0+)/dt = 0 V/s

From (2) and (3), dvL(0+)/dt = dv(0+)/dt = Vs/(CRs) (c) As t approaches infinity, the capacitor acts like an open circuit, while the inductor acts like a short circuit.

vR(∞) = [R/(R + Rs)]Vs

vL(∞) = 0 V


s2 + 4s + 4 = 0, thus s1,2 = 2

4x444 2 −±− = -2, repeated roots.

v(t) = [(A + Bt)e-2t], v(0) = 1 = A

dv/dt = [Be-2t] + [-2(A + Bt)e-2t]

dv(0)/dt = -1 = B – 2A = B – 2 or B = 1.

Therefore, v(t) = [(1 + t)e-2t] V


s2 + 6s + 9 = 0, thus s1,2 = 2

3666 2 −±− = -3, repeated roots.

i(t) = [(A + Bt)e-3t], i(0) = 0 = A

di/dt = [Be-3t] + [-3(Bt)e-3t]

di(0)/dt = 4 = B.

Therefore, i(t) = [4te-3t] A


s2 + 10s + 25 = 0, thus s1,2 = 2

101010 −±− = -5, repeated roots.

i(t) = [(A + Bt)e-5t], i(0) = 10 = A

di/dt = [Be-5t] + [-5(A + Bt)e-5t]

di(0)/dt = 0 = B – 5A = B – 50 or B = 50.

Therefore, i(t) = [(10 + 50t)e-5t] A


s2 + 5s + 4 = 0, thus s1,2 = 2

16255 −±− = -4, -1.

v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A

dv/dt = (-4Ae-4t - Be-t)

dv(0)/dt = 10 = – 4A – B = –3A or A = –10/3 and B = 10/3.

Therefore, v(t) = (–(10/3)e-4t + (10/3)e-t) V Chapter 8, Solution 11.

s2 + 2s + 1 = 0, thus s1,2 = 2

442 −±− = -1, repeated roots.

v(t) = [(A + Bt)e-t], v(0) = 10 = A

dv/dt = [Be-t] + [-(A + Bt)e-t]

dv(0)/dt = 0 = B – A = B – 10 or B = 10.

Therefore, v(t) = [(10 + 10t)e-t] V

Chapter 8, Solution 12. (a) Overdamped when C > 4L/(R2) = 4x0.6/400 = 6x10-3, or C > 6 mF (b) Critically damped when C = 6 mF (c) Underdamped when C < 6mF

Chapter 8, Solution 13. Let R||60 = Ro. For a series RLC circuit,

ωo = LC1 =

4x01.01 = 5

For critical damping, ωo = α = Ro/(2L) = 5

or Ro = 10L = 40 = 60R/(60 + R)

which leads to R = 120 ohms

Chapter 8, Solution 14. This is a series, source-free circuit. 60||30 = 20 ohms

α = R/(2L) = 20/(2x2) = 5 and ωo = LC1 =

04.01 = 5

ωo = α leads to critical damping

i(t) = [(A + Bt)e-5t], i(0) = 2 = A

v = Ldi/dt = 2[Be-5t] + [-5(A + Bt)e-5t]

v(0) = 6 = 2B – 10A = 2B – 20 or B = 13.


Chapter 8, Solution 15. This is a series, source-free circuit. 60||30 = 20 ohms

α = R/(2L) = 20/(2x2) = 5 and ωo = LC1 =

04.01 = 5


i(t) = [(A + Bt)e-5t], i(0) = 2 = A

v = Ldi/dt = 2[Be-5t] + [-5(A + Bt)e-5t]

v(0) = 6 = 2B – 10A = 2B – 20 or B = 13.


Chapter 8, Solution 16. At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V For t > 0, we have a source-free RLC circuit.

α = R/(2L) = (40 + 60)/5 = 20 and ωo = LC1 =

5.2x1013−

= 20


i(t) = [(A + Bt)e-20t], i(0) = 0 = A

di/dt = [Be-20t] + [-20(Bt)e-20t],

but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24]

Hence, B = -9.6 or i(t) = [-9.6te-20t] A


.iswhich,20

412

10L2

R

10

251

411

LC1

240)600(4)VRI(L1

dt)0(di

6015x4V)0(v,0I)0(i

o

o

00

00

ω>===α

===ω

−=+−=+−=

=====

( )t268t32.3721

2121

t32.372

t68.21

2o

2

ee928.6)t(i

A928.6AtoleadsThis

240A32.37A68.2dt

)0(di,AA0)0(i

eAeA)t(i

32.37,68.23102030020s

−−

−−

−=

−=−=

−=−−=+==

+=

−−=±−=±−=ω−α±α−=

getwe,60dt)t(iC1)t(v,Since t

0 +∫=

v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V

Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit.

5.02

1,2125.0

11=====

RCxLCo αω

936.125.04case dunderdampe 22d =−=−=→< αωωωα oo

Io(0) = i(0) = initial inductor current = 20/5 = 4A Vo(0) = v(0) = initial capacitor voltage = 0 V

)936.1sin936.1cos()sincos()( 215.0

21 tAtAetAtAetv tdd

t +=+= −− αα ωω v(0) =0 = A1

)936.1cos936.1936.1sin936.1()936.1sin936.1cos)(5.0( 215.0

215.0 tAtAetAtAe

dtdv tt +−++−= −− αα

066.2936.15.041

)40()()0(221 −=→+−=−=

+−=

+−= AAA

RCRIV

dtdv oo

Thus,

tetv t 936.1sin066.2)( 5.0−−= Chapter 8, Solution 19. For t < 0, the equivalent circuit is shown in Figure (a). 10 Ω i

+ −

+

v

−

i L C

120V (a) (b)

i(0) = 120/10 = 12, v(0) = 0

+

v

−

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α.

ωo = LC1 =

41 = 0.5 = ωd

i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A

v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t],

which leads to -v(0)/L = 0 = B

Hence, i(t) = 12cos0.5t A and v = 0.5

However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V

Chapter 8, Solution 20. For t < 0, the equivalent circuit is as shown below.

2 Ω

+ − 12

− +vC

i

v(0) = -12V and i(0) = 12/2 = 6A For t > 0, we have a series RLC circuit.

α = R/(2L) = 2/(2x0.5) = 2

ωo = 1/ 2241x5.0/1LC ==

Since α is less than ωo, we have an under-damped response.

24822od =−=α−ω=ω

i(t) = (Acos2t + Bsin2t)e-2t

i(0) = 6 = A

di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αt

di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0

Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A Chapter 8, Solution 21. By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25) = 24 ohms.

12 Ω

+ − +

v

−

t = 0 i

24 Ω

6 Ω 3 H

24V (1/27)F At t = 0-, i(0) = 0, v(0) = 24x24/36 = 16V For t > 0, we have a series RLC circuit. R = 30 ohms, L = 3 H, C = (1/27) F

α = R/(2L) = 30/6 = 5

27/1x3/1LC/1o ==ω = 3, clearly α > ωo (overdamped response)

s1,2 = 222o

2 355 −±−=ω−α±α− = -9, -1

v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B (1)

i = Cdv/dt = C[-Ae-t - 9Be-9t]

i(0) = 0 = C[-A – 9B] or A = -9B (2) From (1) and (2), B = -2 and A = 18.

Hence, v(t) = (18e-t – 2e-9t) V

Chapter 8, Solution 22. α = 20 = 1/(2RC) or RC = 1/40 (1)

22od 50 α−ω==ω which leads to 2500 + 400 = ωo

2 = 1/(LC) Thus, LC 1/2900 (2) In a parallel circuit, vC = vL = vR But, iC = CdvC/dt or iC/C = dvC/dt

= -80e-20tcos50t – 200e-20tsin50t + 200e-20tsin50t – 500e-20tcos50t = -580e-20tcos50t

iC(0)/C = -580 which leads to C = -6.5x10-3/(-580) = 11.21 µF

R = 1/(40C) = 106/(2900x11.21) = 2.23 kohms

L = 1/(2900x11.21) = 30.76 H

Chapter 8, Solution 23. Let Co = C + 0.01. For a parallel RLC circuit,

α = 1/(2RCo), ωo = 1/ oLC

α = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF

ωo = 1/ 5.0x5.0 = 6.32 > α (underdamped)

Co = C + 10 mF = 50 mF or 40 mF Chapter 8, Solution 24. For t < 0, u(-t) 1, namely, the switch is on.

v(0) = 0, i(0) = 25/5 = 5A For t > 0, the voltage source is off and we have a source-free parallel RLC circuit.

α = 1/(2RC) = 1/(2x5x10-3) = 100

ωo = 1/ 310x1.0/1LC −= = 100

ωo = α (critically damped)

v(t) = [(A1 + A2t)e-100t]

v(0) = 0 = A1

dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000

But, dv/dt = [(A2 + (-100)A2t)e-100t]

Therefore, dv(0)/dt = -5000 = A2 – 0 v(t) = -5000te-100t V

Chapter 8, Solution 25. In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0.

(1/4)F

+ −

8 Ω

2 Ω

t=0, note this is a make before break

switch so the inductor current is

not interrupted.

1 H io(t) 30V

Figure 8.78 For Problem 8.25. At t = 0-, vo(0) = (8/(2 + 8)(30) = 24 For t > 0, we have a source-free parallel RLC circuit.

α = 1/(2RC) = ¼

ωo = 1/ 241x1/1LC ==

Since α is less than ωo, we have an under-damped response.

9843.1)16/1(422od =−=α−ω=ω

vo(t) = (A1cosωdt + A2sinωdt)e-αt

+

vo(t)

−

vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0.

dvo/dt = -α(A1cosωdt + A2sinωdt)e-αt + (-ωdA1sinωdt + ωdA2cosωdt)e-αt

at t = 0, we get dvo(0)/dt = 0 = -αA1 + ωdA2

Thus, A2 = (α/ωd)A1 = (1/4)(24)/1.9843 = 3.024

vo(t) = (24cosωdt + 3.024sinωdt)e-t/4 volts


s2 + 2s + 5 = 0, which leads to s1,2 = 2

2042 −±− = -1±j4

i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2

i(0) = 2 = = 2 + A1, or A1 = 0

di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1

i(t) = 2 + sin4te-t A


s2 + 4s + 8 = 0 leads to s = 2j22

32164±−=

−±−

v(t) = Vs + (A1cos2t + A2sin2t)e-2t

8Vs = 24 means that Vs = 3

v(0) = 0 = 3 + A1 leads to A1 = -3

dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t

0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3

v(t) = [3 – 3(cos2t + sin2t)e-2t] volts


The characteristic equation is s2 + 6s + 8 with roots

2,42

323662,1 −−=

−±−=s

Hence,

tts BeAeIti 42)( −− ++=

5.1128 =→= ss II

BAi ++=→= 5.100)0( (1)

tt BeAe

dtdi 42 42 −− −−=

BABAdtdi 210422)0(

++=→−−== (2)

Solving (1) and (2) leads to A=-2 and B=0.5.

tt eeti 42 5.025.1)( −− +−= A Chapter 8, Solution 29.

(a) s2 + 4 = 0 which leads to s1,2 = ±j2 (an undamped circuit)

v(t) = Vs + Acos2t + Bsin2t

4Vs = 12 or Vs = 3

v(0) = 0 = 3 + A or A = -3

dv/dt = -2Asin2t + 2Bcos2t

dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V

(b) s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4

i(t) = (Is + Ae-t + Be-4t)

4Is = 8 or Is = 2

i(0) = -1 = 2 + A + B, or A + B = -3 (1)

di/dt = -Ae-t - 4Be-4t

di(0)/dt = 0 = -A – 4B, or B = -A/4 (2)

From (1) and (2) we get A = -4 and B = 1

i(t) = (2 – 4e-t + e-4t) A

(c) s2 + 2s + 1 = 0, s1,2 = -1, -1

v(t) = [Vs + (A + Bt)e-t], Vs = 3.

v(0) = 5 = 3 + A or A = 2

dv/dt = [-(A + Bt)e-t] + [Be-t]

dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3

v(t) = [3 + (2 + 3t)e-t] V Chapter 8, Solution 30.

22

222

1 800,500 oo ss ωααωαα −−−=−=−+−=−=

LRss

26502130021 ==→−=−=+ αα

Hence,

mH 8.1536502

2002

===x

RLα

LCss oo

145.6232300 2221 ==→−==− ωωα

F25.16)45.632(

12 µ==L

C

Chapter 8, Solution 31. For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent circuit is shown in Figure (b). By KVL,

v(0+) = v(0-) = 40, i(0+) = i(0-) = 1 By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+) = 0 which leads to vL(0+) = 40x1 + 40 = 80

vL(0+) = 80 V, vC(0+) = 40 V 40 Ω 10 Ω i1

0.5H

+

v

−50V

+ −

+

vL

−

40 Ω 10 Ω

i +

v

−

50V

+ −

(a) (b) Chapter 8, Solution 32. For t = 0-, the equivalent circuit is shown below. 2 A

i

6 Ω

+ −v

i(0-) = 0, v(0-) = -2x6 = -12V For t > 0, we have a series RLC circuit with a step input.

α = R/(2L) = 6/2 = 3, ωo = 1/ 04.0/1LC =

s = 4j32593 ±−=−±−

Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]

where Vf = final capacitor voltage = 50 V

v(t) = 50 + [(Acos4t + Bsin4t)e-3t]

v(0) = -12 = 50 + A which gives A = -62

i(0) = 0 = Cdv(0)/dt

dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]

0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5

v(t) = 50 + [(-62cos4t – 46.5sin4t)e-3t] V Chapter 8, Solution 33. We may transform the current sources to voltage sources. For t = 0-, the equivalent circuit is shown in Figure (a).

1 H

i

+ − 30V

+

v

−

4F

i

+ − 5 Ω

10 Ω

+

v

−

10 Ω 30V (a) (b)

i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V

For t > 0, we have a series RLC circuit.

α = R/(2L) = 5/2 = 2.5

4/1LC/1o ==ω = 0.25, clearly α > ωo (overdamped response)

s1,2 = 25.025.65.22o

2 −±−=ω−α±α− = -4.95, -0.05

v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20.

v(0) = 10 = 20 + A1 + A2 (1)

i(0) = Cdv(0)/dt or dv(0)/dt = 2/4 = 1/2

Hence, ½ = -4.95A1 – 0.05A2 (2) From (1) and (2), A1 = 0, A2 = -10.

v(t) = 20 – 10e-0.05t V Chapter 8, Solution 34. Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short circuit.

i(0) = 0, v(0) = 20 V For t > 0, the LC circuit is disconnected from the voltage source as shown below.

+ − Vx

(1/16)F

(¼) H

i This is a lossless, source-free, series RLC circuit.

α = R/(2L) = 0, ωo = 1/ LC = 1/41

161

+ = 8, s = ±j8

Since α is less than ωo, we have an underdamped response. Therefore,

i(t) = A1cos8t + A2sin8t where i(0) = 0 = A1

di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80 However, di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10 Now we have i(t) = -10sin8t A

Chapter 8, Solution 35. At t = 0-, iL(0) = 0, v(0) = vC(0) = 8 V For t > 0, we have a series RLC circuit with a step input.

α = R/(2L) = 2/2 = 1, ωo = 1/ LC = 1/ 5/1 = 5

s1,2 = 2j12

o2 ±−=ω−α±α−

v(t) = Vs + [(Acos2t + Bsin2t)e-t], Vs = 12.

v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0.

But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t]

0 = dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2

v(t) = 12 – (4cos2t + 2sin2t)e-t V. Chapter 8, Solution 36. For t = 0-, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V. For t > 0, we have the series RLC circuit shown below.

20 V 2 Ω0.2 F

i 10 Ω

+ −

+ −

5 H 10 Ω +

v

−

15V

α = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8

ωo = 1/ LC = 1/ 2.0x5 = 1

s1,2 = 6.0j8.02o

2 ±−=ω−α±α−

v(t) = Vs + [(Acos0.6t + Bsin0.6t)e-0.8t]

Vs = 15 + 20 = 35V and v(0) = 20 = 35 + A or A = -15

i(0) = Cdv(0)/dt = 0

But dv/dt = [-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t]

0 = dv(0)/dt = -0.8A + 0.6B which leads to B = 0.8x(-15)/0.6 = -20

v(t) = 35 – [(15cos0.6t + 20sin0.6t)e-0.8t] V i = Cdv/dt = 0.2[0.8(15cos0.6t + 20sin0.6t)e-0.8t] + [0.6(15sin0.6t – 20cos0.6t)e-0.8t]

i(t) = [(5sin0.6t)e-0.8t] A

Chapter 8, Solution 37. For t = 0-, the equivalent circuit is shown below.

6 Ω

+ − 10V

+ −

i2

i1

+

v(0)

−

6 Ω6 Ω

30V

18i2 – 6i1 = 0 or i1 = 3i2 (1) -30 + 6(i1 – i2) + 10 = 0 or i1 – i2 = 10/3 (2)

From (1) and (2). i1 = 5, i2 = 5/3

i(0) = i1 = 5A

-10 – 6i2 + v(0) = 0

v(0) = 10 + 6x5/3 = 20

For t > 0, we have a series RLC circuit.

R = 6||12 = 4

ωo = 1/ LC = 1/ )8/1)(2/1( = 4

α = R/(2L) = (4)/(2x(1/2)) = 4

α = ωo, therefore the circuit is critically damped

v(t) = Vs +[(A + Bt)e-4t], and Vs = 10

v(0) = 20 = 10 + A, or A = 10

i = Cdv/dt = -4C[(A + Bt)e-4t] + C[(B)e-4t]

i(0) = 5 = C(-4A + B) which leads to 40 = -40 + B or B = 80

i(t) = [-(1/2)(10 + 80t)e-4t] + [(10)e-4t]

i(t) = [(5 – 40t)e-4t] A

Chapter 8, Solution 38. At t = 0-, the equivalent circuit is as shown.

2 A

10 Ω

i

i1

5 Ω

+

v

− 10 Ω

i(0) = 2A, i1(0) = 10(2)/(10 + 15) = 0.8 A

v(0) = 5i1(0) = 4V

For t > 0, we have a source-free series RLC circuit.

R = 5||(10 + 10) = 4 ohms

ωo = 1/ LC = 1/ )4/3)(3/1( = 2

α = R/(2L) = (4)/(2x(3/4)) = 8/3

s1,2 = =ω−α±α− 2o

2 -4.431, -0.903

i(t) = [Ae-4.431t + Be-0.903t]

i(0) = A + B = 2 (1)

di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333

Hence, -5.333 = -4.431A – 0.903B (2)

From (1) and (2), A = 1 and B = 1.

i(t) = [e-4.431t + e-0.903t] A

Chapter 8, Solution 39. For t = 0-, the equivalent circuit is shown in Figure (a). Where 60u(-t) = 60 and 30u(t) = 0.

+ − 30V 20 Ω

+ −

+ v −

20 Ω

30 Ω 0.5F 0.25H 30 Ω 60V (a) (b)

v(0) = (20/50)(60) = 24 and i(0) = 0

For t > 0, the circuit is shown in Figure (b).

R = 20||30 = 12 ohms

ωo = 1/ LC = 1/ )4/1)(2/1( = 8

α = R/(2L) = (12)/(0.5) = 24

Since α > ωo, we have an overdamped response.

s1,2 = =ω−α±α− 2o

2 -47.833, -0.167 Thus, v(t) = Vs + [Ae-47.833t + Be-0.167t], Vs = 30 v(0) = 24 = 30 + A + B or -6 = A + B (1)

i(0) = Cdv(0)/dt = 0

But, dv(0)/dt = -47.833A – 0.167B = 0 B = -286.43A (2)

From (1) and (2), A = 0.021 and B = -6.021

v(t) = 30 + [0.021e-47.833t – 6.021e-0.167t] V Chapter 8, Solution 40. At t = 0-, vC(0) = 0 and iL(0) = i(0) = (6/(6 + 2))4 = 3A For t > 0, we have a series RLC circuit with a step input as shown below.

+ −v

6 Ω+ −

+ − 12V

24V

i 14 Ω0.02 F 2 H

ωo = 1/ LC = 1/ 02.0x2 = 5

α = R/(2L) = (6 + 14)/(2x2) = 5

Since α = ωo, we have a critically damped response.

v(t) = Vs + [(A + Bt)e-5t], Vs = 24 – 12 = 12V

v(0) = 0 = 12 + A or A = -12

i = Cdv/dt = C[Be-5t] + [-5(A + Bt)e-5t]

i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90

Thus, i(t) = 0.02[90e-5t] + [-5(-12 + 90t)e-5t]

i(t) = (3 – 9t)e-5t A

Chapter 8, Solution 41. At t = 0-, the switch is open. i(0) = 0, and

v(0) = 5x100/(20 + 5 + 5) = 50/3 For t > 0, we have a series RLC circuit shown in Figure (a). After source transformation, it becomes that shown in Figure (b).

1 H

+ − 20V

+

v

−

i 4 Ω

20 Ω 10 µF

10 H

5A

(a)

5 Ω

0.04F (b)

ωo = 1/ LC = 1/ 25/1x1 = 5

α = R/(2L) = (4)/(2x1) = 2

s1,2 = =ω−α±α− 2o

2 -2 ± j4.583 Thus, v(t) = Vs + [(Acosωdt + Bsinωdt)e-2t],

where ωd = 4.583 and Vs = 20

v(0) = 50/3 = 20 + A or A = -10/3

i(t) = Cdv/dt = C(-2) [(Acosωdt + Bsinωdt)e-2t] + Cωd[(-Asinωdt + Bcosωdt)e-2t]

i(0) = 0 = -2A + ωdB

B = 2A/ωd = -20/(3x4.583) = -1.455

i(t) = C[(0cosωdt + (-2B - ωdA)sinωdt)]e-2t

= (1/25)[(2.91 + 15.2767) sinωdt)]e-2t

i(t) = 0.7275sin(4.583t)e-2t A

Chapter 8, Solution 42. For t = 0-, we have the equivalent circuit as shown in Figure (a).

i(0) = i(0) = 0, and v(0) = 4 – 12 = -8V

+

v

−

6 Ω

− + 12V

1 H i 4V + − − +

+

v(0)

−

12V

1 Ω

5 Ω

0.04F

(a) (b) For t > 0, the circuit becomes that shown in Figure (b) after source transformation.

ωo = 1/ LC = 1/ 25/1x1 = 5

α = R/(2L) = (6)/(2) = 3

s1,2 = =ω−α±α− 2o

2 -3 ± j4 Thus, v(t) = Vs + [(Acos4t + Bsin4t)e-3t], Vs = -12

v(0) = -8 = -12 + A or A = 4

i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]

i(0) = -3A + 4B or B = 3

v(t) = -12 + [(4cos4t + 3sin4t)e-3t] A


For t>0, we have a source-free series RLC circuit.

Ω===→= 85.08222

xxLRLR αα

836649003022 =−=→=−= ood ωαωω

mF 392.25.0836

1112 ===→=

xLC

LC oo ω

ω


4

910

1010011,500

121000

2======

−xLCxLR

oωα

→α>ωo underdamped. Chapter 8, Solution 45.

ωo = 1/ LC = 1/ 5.0x1 = 2

α = R/(2L) = (1)/(2x2x0.5) = 0.5

Since α < ωo, we have an underdamped response.

s1,2 = =ω−α±α− 2o

2 -0.5 ± j1.323 Thus, i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t], Is = 4

i(0) = 1 = 4 + A or A = -3

v = vC = vL = Ldi(0)/dt = 0

di/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t]

di(0)/dt = 0 = 1.323B – 0.5A or B = 0.5(-3)/1.323 = -1.134 Thus, i(t) = 4 – [(3cos1.323t + 1.134sin1.323t)e-0.5t] A

Chapter 8, Solution 46. For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0. For t > 0, we have a parallel RLC circuit with a step input, as shown below.

5µF8mH

+

v

−

i 2 kΩ

6mA

α = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50

ωo = 1/ LC = 1/ 63 10x5x10x8 − = 5,000

Since α < ωo, we have an underdamped response.

s1,2 = ≅ω−α±α− 2o

2 -50 ± j5,000 Thus, i(t) = Is + [(Acos5,000t + Bsin5,000t)e-50t], Is = 6mA

i(0) = 0 = 6 + A or A = -6mA

v(0) = 0 = Ldi(0)/dt

di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t]

di(0)/dt = 0 = 5,000B – 50A or B = 0.01(-6) = -0.06mA Thus, i(t) = 6 – [(6cos5,000t + 0.06sin5,000t)e-50t] mA Chapter 8, Solution 47. At t = 0-, we obtain, iL(0) = 3x5/(10 + 5) = 1A

and vo(0) = 0.

For t > 0, the 20-ohm resistor is short-circuited and we have a parallel RLC circuit with a step input.

α = 1/(2RC) = (1)/(2x5x0.01) = 10

ωo = 1/ LC = 1/ 01.0x1 = 10


s1,2 = -10

Thus, i(t) = Is + [(A + Bt)e-10t], Is = 3

i(0) = 1 = 3 + A or A = -2

vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t]

vo(0) = 0 = B – 10A or B = -20

Thus, vo(t) = (200te-10t) V

Chapter 8, Solution 48. For t = 0-, we obtain i(0) = -6/(1 + 2) = -2 and v(0) = 2x1 = 2.

For t > 0, the voltage is short-circuited and we have a source-free parallel RLC circuit.

α = 1/(2RC) = (1)/(2x1x0.25) = 2

ωo = 1/ LC = 1/ 25.0x1 = 2


s1,2 = -2

Thus, i(t) = [(A + Bt)e-2t], i(0) = -2 = A

v = Ldi/dt = [Be-2t] + [-2(-2 + Bt)e-2t]

vo(0) = 2 = B + 4 or B = -2

Thus, i(t) = [(-2 - 2t)e-2t] A

and v(t) = [(2 + 4t)e-2t] V

Chapter 8, Solution 49. For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0.

For t > 0, we have a parallel RLC circuit with a step input.

α = 1/(2RC) = (1)/(2x5x0.05) = 2

ωo = 1/ LC = 1/ 05.0x5 = 2


s1,2 = -2

Thus, i(t) = Is + [(A + Bt)e-2t], Is = 3

i(0) = 6 = 3 + A or A = 3

v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t]

v(0)/L = 0 = di(0)/dt = B – 2x3 or B = 6

Thus, i(t) = 3 + [(3 + 6t)e-2t] A Chapter 8, Solution 50. For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A. For t > 0, we have a parallel RLC circuit.

i

40 Ω 6A 3A

10 mF+

v

−

10 Ω

10 H

Is = 3 + 6 = 9A and R = 10||40 = 8 ohms

α = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25

ωo = 1/ LC = 1/ 01.0x4 = 5

Since α > ωo, we have a overdamped response.

s1,2 = =ω−α±α− 2o

2 -10, -2.5

Thus, i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9

i(0) = 3 = 9 + A + B or A + B = -6

di/dt = [-10Ae-10t] + [-2.5Be-2.5t],

v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A

Thus, A = 2 and B = -8

Clearly, i(t) = 9 + [2e-10t] + [-8e-2.5t] A Chapter 8, Solution 51. Let i = inductor current and v = capacitor voltage.

At t = 0, v(0) = 0 and i(0) = io.

For t > 0, we have a parallel, source-free LC circuit (R = ∞).

α = 1/(2RC) = 0 and ωo = 1/ LC which leads to s1,2 = ± jωo

v = Acosωot + Bsinωot, v(0) = 0 A

iC = Cdv/dt = -i

dv/dt = ωoBsinωot = -i/C

dv(0)/dt = ωoB = -io/C therefore B = io/(ωoC)

v(t) = -(io/(ωoC))sinωot V where ωo = LC Chapter 8, Solution 52.

RC21300 ==α (1)

LCood1575.264300400400 2222 ==−=→=−= ωαωω (2)

From (2),

F71.2851050)575.264(

132 µ== −xx

C

From (1),

Ω=== 833.5)3500(30021

21

xCR

α


C1 R2

+ −v1

i2i1

+ −

C2

+

vo

−

R1

vS

i2 = C2dvo/dt (1) i1 = C1dv1/dt (2) 0 = R2i2 + R1(i2 – i1) +vo (3) Substituting (1) and (2) into (3) we get, 0 = R2C2dvo/dt + R1(C2dvo/dt – C1dv1/dt) (4) Applying KVL to the outer loop produces,

vs = v1 + i2R2 + vo = v1 + R2C2dvo/dt + vo, which leads to v1 = vs – vo – R2C2dvo/dt (5) Substituting (5) into (4) leads to,

0 = R1C2dvo/dt + R1C2dvo/dt – R1C1(dvs/dt – dvo/dt – R2C2d2vo/dt2)

Hence, (R1C1R2C2)(d2vo/dt2) + (R1C1 + R2C2 +R1C2)(dvo/dt) = R1C1(dvs/dt) Chapter 8, Solution 54. Let i be the inductor current.

dtdvvi 5.0

4+=− (1)

dtdii += 2v (2)


035.221

41

2 2

2

2

2

=++→+++=− vdtdv

dtvd

dtvd

dtdv

dtdvvv

199.125.1035.22 jsss ±−=→=++

tBetAev tt 199.1sin199.1cos 25.125.1 −− +=

v(0) = 2=A. Let w=1.199

)cossin()sincos(25.1 25.125.125.125.1 wtBewtAewwtBewtAedtdv tttt −−−− +−++−=

085.2199.1

225.125.10)0(==→+−==

XBBwAdtdv

V 199.1sin085.2199.1cos2 25.125.1 tetev tt −− +=

Chapter 8, Solution 55. At the top node, writing a KCL equation produces,

i/4 +i = C1dv/dt, C1 = 0.1

5i/4 = C1dv/dt = 0.1dv/dt i = 0.08dv/dt (1)

But, v = )idt)C/1(i2( 2 ∫+− , C2 = 0.5 or -dv/dt = 2di/dt + 2i (2)

Substituting (1) into (2) gives,

-dv/dt = 0.16d2v/dt2 + 0.16dv/dt

0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0

Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s1,2 = 0, -7.25

v(t) = A + Be-7.25t (3) v(0) = 4 = A + B (4)

From (1), i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25

But, dv/dt = -7.25Be-7.25t, which leads to,

dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 – B = 4 + 3.448 = 7.448

Thus, v(t) = 7.45 – 3.45e-7.25t V Chapter 8, Solution 56. For t < 0, i(0) = 0 and v(0) = 0. For t > 0, the circuit is as shown below. 4 Ω

io

i 6 Ω

+ −

0.04F

i 20 0.25H Applying KVL to the larger loop,

-20 +6io +0.25dio/dt + 25 ∫ + dt)ii( o = 0 Taking the derivative,

6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0 (1) For the smaller loop, 4 + 25 ∫ + dt)ii( o = 0 Taking the derivative, 25(i + io) = 0 or i = -io (2) From (1) and (2) 6dio/dt + 0.25d2io/dt2 = 0

This leads to, 0.25s2 + 6s = 0 or s1,2 = 0, -24

io(t) = (A + Be-24t) and io(0) = 0 = A + B or B = -A

As t approaches infinity, io(∞) = 20/10 = 2 = A, therefore B = -2

Thus, io(t) = (2 - 2e-24t) = -i(t) or i(t) = (-2 + 2e-24t) A Chapter 8, Solution 57. (a) Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is closed and the circuit has reached steady-state.

v(0-) = 16V and i(0-) = 16/8 = 2A At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2. We now have a source-free RLC circuit.

R 8 + 12 = 20 ohms, L = 1H, C = 4mF.

α = R/(2L) = (20)/(2x1) = 10

ωo = 1/ LC = 1/ )36/1(x1 = 6

Since α > ωo, we have a overdamped response.

s1,2 = =ω−α±α− 2o

2 -18, -2

Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0. (b) i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B (1)

To get di(0)/dt, consider the circuit below at t = 0+.

+

vL

−

12 Ω

(1/36)F+

v

−

i 8 Ω

1 H

-v(0) + 20i(0) + vL(0) = 0, which leads to,

-16 + 20x2 + vL(0) = 0 or vL(0) = -24

Ldi(0)/dt = vL(0) which gives di(0)/dt = vL(0)/L = -24/1 = -24 A/s

Hence -24 = -2A – 18B or 12 = A + 9B (2) From (1) and (2), B = 1.25 and A = 0.75

i(t) = [0.75e-2t + 1.25e-18t] = -ix(t) or ix(t) = [-0.75e-2t - 1.25e-18t] A

v(t) = 8i(t) = [6e-2t + 10e-18t] A Chapter 8, Solution 58.

(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4

V/s 85.0

)04()]0()0([)0(−=

+−=

+−=

RCRiv

dtdv

(b) For , the circuit is a source-free RLC parallel circuit. 0≥t

2125.0

11,115.02

12

1======

xLCxxRC oωα

732.11422 =−=−= αωω od

Thus,

)732.1sin732.1cos()( 21 tAtAetv t += − v(0) = 4 = A1

tAetAetAetAedtdv tttt 732.1cos732.1732.1sin732.1sin732.1732.1cos 2211

−−−− +−−−=

309.2732.18)0(221 −=→+−=−= AAA

dtdv

)732.1sin309.2732.1cos4()( ttetv t −= − V


Let i = inductor current and v = capacitor voltage v(0) = 0, i(0) = 40/(4+16) = 2A For t>0, the circuit becomes a source-free series RLC with

2,216/14

11,242

162

==→====== oo xLCxLR ωαωα

tt BteAeti 22)( −− += i(0) = 2 = A

ttt BteBeAedtdi 222 22 −−− −+−=

4),032(412)]0()0([12)0(

−=+−=+−→+−=+−= BBAvRiL

BAdtdi

tt teeti 22 42)( −− −= t

ttt

ttt

tt

teedttedtevidtC

v0

2

00

22

0

2

0

)12(4

64166432)0(1−−−−=−=+= −−−− ∫∫∫

ttev 232 −= V


At t = 0-, 4u(t) = 0 so that i1(0) = 0 = i2(0) (1)

Applying nodal analysis,

4 = 0.5di1/dt + i1 + i2 (2) Also, i2 = [1di1/dt – 1di2/dt]/3 or 3i2 = di1/dt – di2/dt (3) Taking the derivative of (2), 0 = d2i1/dt2 + 2di1/dt + 2di2/dt (4) From (2) and (3), di2/dt = di1/dt – 3i2 = di1/dt – 3(4 – i1 – 0.5di1/dt)

= di1/dt – 12 + 3i1 + 1.5di1/dt Substituting this into (4),

d2i1/dt2 + 7di1/dt + 6i1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6)

Thus, i1(t) = Is + [Ae-t + Be-6t], 6Is = 24 or Is = 4

i1(t) = 4 + [Ae-t + Be-6t] and i1(0) = 4 + [A + B] (5)

i2 = 4 – i1 – 0.5di1/dt = i1(t) = 4 + -4 - [Ae-t + Be-6t] – [-Ae-t - 6Be-6t]

= [-0.5Ae-t + 2Be-6t] and i2(0) = 0 = -0.5A + 2B (6) From (5) and (6), A = -3.2 and B = -0.8

i1(t) = 4 + [-3.2e-t – 0.8e-6t] A

i2(t) = [1.6e-t – 1.6e-6t] A Chapter 8, Solution 61. For t > 0, we obtain the natural response by considering the circuit below.

0.25F+

vC

−

a iL

4 Ω

1 H 6 Ω At node a, vC/4 + 0.25dvC/dt + iL = 0 (1) But, vC = 1diL/dt + 6iL (2) Combining (1) and (2),

(1/4)diL/dt + (6/4)iL + 0.25d2iL/dt2 + (6/4)diL/dt + iL = 0

d2iL/dt2 + 7diL/dt + 10iL = 0

s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s1,2 = -2, -5

Thus, iL(t) = iL(∞) + [Ae-2t + Be-5t],

where iL(∞) represents the final inductor current = 4(4)/(4 + 6) = 1.6

iL(t) = 1.6 + [Ae-2t + Be-5t] and iL(0) = 1.6 + [A+B] or -1.6 = A+B (3)

diL/dt = [-2Ae-2t - 5Be-5t]

and diL(0)/dt = 0 = -2A – 5B or A = -2.5B (4)

From (3) and (4), A = -8/3 and B = 16/15

iL(t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t]

v(t) = 6iL(t) = 9.6 + [-16e-2t + 6.4e-5t] V

vC = 1diL/dt + 6iL = [ (16/3)e-2t - (16/3)e-5t] + 9.6 + [-16e-2t + 6.4e-5t]

vC = 9.6 + [-(32/3)e-2t + 1.0667e-5t]

i(t) = vC/4 = 2.4 + [-2.667e-2t + 0.2667e-5t] A Chapter 8, Solution 62. This is a parallel RLC circuit as evident when the voltage source is turned off.

α = 1/(2RC) = (1)/(2x3x(1/18)) = 3

ωo = 1/ LC = 1/ 18/1x2 = 3


s1,2 = -3

Let v(t) = capacitor voltage

Thus, v(t) = Vs + [(A + Bt)e-3t] where Vs = 0

But -10 + vR + v = 0 or vR = 10 – v

Therefore vR = 10 – [(A + Bt)e-3t] where A and B are determined from initial conditions.

Chapter 8, Solution 63. - R R v1 + vo vs v2 C C At node 1,

dtdvC

Rvvs 11 =

− (1)

At node 2,

dtdv

CRvv oo =

−2 (2)

As a voltage follower, vvv == 21 . Hence (2) becomes

dtdv

RCvv oo += (3)

and (1) becomes

dtdvRCvvs += (4)


2

222

dtvdCR

dtdvRC

dtdvRCvv ooo

os +++=

or

sooo vvdtdvRC

dtvdCR =++ 22

222

Chapter 8, Solution 64. C2

vs R1 2 1

v1

R2

−+

C1

vo

At node 1, (vs – v1)/R1 = C1 d(v1 – 0)/dt or vs = v1 + R1C1dv1/dt (1) At node 2, C1dv1/dt = (0 – vo)/R2 + C2d(0 – vo)/dt or –R2C1dv1/dt = vo + C2dvo/dt (2) From (1) and (2), (vs – v1)/R1 = C1 dv1/dt = -(1/R2)(vo + C2dvo/dt) or v1 = vs + (R1/R2)(vo + C2dvo/dt) (3) Substituting (3) into (1) produces,

vs = vs + (R1/R2)(vo + C2dvo/dt) + R1C1dvs + (R1/R2)(vo + C2dvo/dt)/dt

= vs + (R1/R2)(vo)+ (R1C2/R2) dvo/dt) + R1C1dvs/dt + (R1R1C1/R2)dvo/dt + (R1

2 C1C2/R2)[d2vo/dt2]

Simplifying we get,

d2vo/dt2 + [(1/ R1C1) + (1/ C2)]dvo/dt + [1/(R1C1C2)](vo) = - [R2/(R1C2)]dvs/dt

Chapter 8, Solution 65. At the input of the first op amp,

(vo – 0)/R = Cd(v1 – 0) (1) At the input of the second op amp, (-v1 – 0)/R = Cdv2/dt (2) Let us now examine our constraints. Since the input terminals are essentially at ground, then we have the following,

vo = -v2 or v2 = -vo (3) Combining (1), (2), and (3), eliminating v1 and v2 we get,

0v100dt

vdv

CR1

dtvd

o2o

2

o222o

2

=−=

−

Which leads to s2 – 100 = 0

Clearly this produces roots of –10 and +10. And, we obtain,

vo(t) = (Ae+10t + Be-10t)V

At t = 0, vo(0+) = – v2(0+) = 0 = A + B, thus B = –A

This leads to vo(t) = (Ae+10t – Ae-10t)V. Now we can use v1(0+) = 2V.

From (2), v1 = –RCdv2/dt = 0.1dvo/dt = 0.1(10Ae+10t + 10Ae-10t)

v1(0+) = 2 = 0.1(20A) = 2A or A = 1

Thus, vo(t) = (e+10t – e-10t)V It should be noted that this circuit is unstable (clearly one of the poles lies in the right-half-plane). Chapter 8, Solution 66. C2

R4

R2

+–

C1

vS 1

2

R3

R1

vo Note that the voltage across C1 is v2 = [R3/(R3 + R4)]vo This is the only difference between this problem and Example 8.11, i.e. v = kv, where k = [R3/(R3 + R4)].

At node 1,

(vs – v1)/R1 = C2[d(v1 – vo)/dt] + (v1 – v2)/R2

vs/R1 = (v1/R1) + C2[d(v1)/dt] – C2[d(vo)/dt] + (v1 – kvo)/R2 (1) At node 2,

(v1 – kvo)/R2 = C1[d(kvo)/dt] or v1 = kvo + kR2C1[d(vo)/dt] (2) Substituting (2) into (1), vs/R1 = (kvo/R1) + (kR2C1/R1)[d(vo)/dt] + kC2[d(vo)/dt] + kR2C1C2[d2(vo)/dt2] – (kvo/R2) + kC1[d(vo)/dt] – (kvo/R2) + C2[d(vo)/dt] We now rearrange the terms. [d2(vo)/dt2] + [(1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1)][d(vo)/dt] + [vo/(R1R2C1C2)] = vs/(kR1R2C1C2) If R1 = R2 10 kohms, C1 = C2 = 100 µF, R3 = 20 kohms, and R4 = 60 kohms,

k = [R3/(R3 + R4)] = 1/3

R1R2C1C2 = 104 x104 x10-4 x10-4 = 1

(1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1) = 1 + 1 + 1 – 3 = 3 – 3 = 0 Hence, [d2(vo)/dt2] + vo = 3vs = 6, t > 0, and s2 + 1 = 0, or s1,2 = ±j

vo(t) = Vs + [Acost + B sint], Vs = 6

vo(0) = 0 = 6 + A or A = –6

dvo/dt = –Asint + Bcost, but dvo(0)/dt = 0 = B

Hence, vo(t) = 6(1 – cost)u(t) volts.

Chapter 8, Solution 67. At node 1,

dt

)0v(dCdt

)vv(dC

Rvv 1

2o1

11

1in −+

−=

− (1)

At node 2, 2

o12 R

v0dt

)0v(dC−

=− , or

22

o1

RCv

dtdv −

= (2)

From (1) and (2),

2

o1

o11

o

22

111in R

vR

dtdv

CRdt

dvRCCRvv −−−=−

2

o1

o11

o

22

11in1 R

vR

dtdv

CRdt

dvRCCRvv +++= (3)

C1

From (2) and (3),

0Vvin

C2

R2

−+v1

1 2 R1

vo

dtdv

RR

dtvd

CRdt

dvRCCR

dtdv

dtdv

RCv o

2

12

o2

11o

22

11in1

22

o +++==−

dtdv

CR1

RRCCv

dtdv

C1

C1

R1

dtvd in

111221

oo

2122

o2

−=+

++

But C1C2R1R2 = 10-4 x10-4 x104 x104 = 1

210x10

2CR

2C1

C1

R1

4412212

===

+ −

dtdvv

dtdv

2dt

vd ino

o2

o2

−=++

Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = –1, –1

Therefore, vo(t) = [(A + Bt)e-t] + Vf

As t approaches infinity, the capacitor acts like an open circuit so that

Vf = vo(∞) = 0

vin = 10u(t) mV and the fact that the initial voltages across each capacitor is 0

means that vo(0) = 0 which leads to A = 0.

vo(t) = [Bte-t]

dtdvo = [(B – Bt)e-t] (4)

From (2), 0RC

)0(vdt

)0(dv

22

oo =+

−=+

From (1) at t = 0+,

dt)0(dv

CR

01 o1

1

+−=

− which leads to 1RC1

dt)0(dv

11

o −=−=+

Substituting this into (4) gives B = –1

Thus, v(t) = –te-tu(t) V

Chapter 8, Solution 68. The schematic is as shown below. The unit step is modeled by VPWL as shown. We insert a voltage marker to display V after simulation. We set Print Step = 25 ms and final step = 6s in the transient box. The output plot is shown below.

Chapter 8, Solution 69. The schematic is shown below. The initial values are set as attributes of L1 and C1. We set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker is inserted at the terminal of L1 to automatically display i(t) after simulation. The result is shown below.

Chapter 8, Solution 70. The schematic is shown below.

After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown below.

Chapter 8, Solution 71. The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot of v(t) after simulation. The result is shown below.

Chapter 8, Solution 72. When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the inductor. When the switch is in position 2, the schematic of the circuit is shown below.

When the circuit is simulated, we obtain i(t) as shown below.

Chapter 8, Solution 73. (a) For t < 0, we have the schematic below. When this is saved and simulated, we

obtain the initial inductor current and capacitor voltage as

iL(0) = 3 A and vc(0) = 24 V.

(b) For t > 0, we have the schematic shown below. To display i(t) and v(t), we insert current and voltage markers as shown. The initial inductor current and capacitor voltage are also incorporated. In the Transient box, we set Print Step = 25 ms and the Final Time to 4s. After simulation, we automatically have io(t) and vo(t) displayed as shown below.


10Ω 5Ω + 20 V 2F 4H - Hence the dual circuit is shown below. 2H 4F 0.2Ω 20A 0.1 Ω

Chapter 8, Solution 75. The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b).

0.5 H 2 F

12A

24A

0.25 Ω

0.1 Ω

10 µF

12A + − 24V

10 H

10 Ω

0.25 Ω

24A

+ − 12V

0.5 F

10 H

(a)

4 Ω

0.1 Ω

(b)

Chapter 8, Solution 76. The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in Figure (b).

120 A

2 V

+ −

2 A

30 Ω

1/3 Ω

10 Ω

0.1 Ω

120 V

– +

60 V

+ − 60 A

4F

1 F 4 H 1 H

20 Ω

0.05 Ω (a) 0.05 Ω

60 A 1 H

120 A

1/4 F

0.1 Ω

+ − 2V

1/30 Ω (b)

Chapter 8, Solution 77. The dual is constructed in Figure (a) and redrawn in Figure (b).

+ − 5 V

1/4 F

1 H

2 Ω

1 Ω

1/3 Ω12 A

1 Ω

12 A

1/3 Ω

3 Ω1/2 Ω

5 V

– + 5 A

+ − 12V1/4 F

1 F 1/4 H

1 H

(a)

1 Ω

2 Ω

(b)

Chapter 8, Solution 78. The voltage across the igniter is vR = vC since the circuit is a parallel RLC type.

vC(0) = 12, and iL(0) = 0.

α = 1/(2RC) = 1/(2x3x1/30) = 5

ωo 30/1x10x60/1LC/1 3−== = 22.36

α < ωo produces an underdamped response.

2o

22,1s ω−α±α−= = –5 ± j21.794

vC(t) = e-5t(Acos21.794t + Bsin21.794t) (1)

vC(0) = 12 = A dvC/dt = –5[(Acos21.794t + Bsin21.794t)e-5t] + 21.794[(–Asin21.794t + Bcos21.794t)e-5t] (2)

dvC(0)/dt = –5A + 21.794B

But, dvC(0)/dt = –[vC(0) + RiL(0)]/(RC) = –(12 + 0)/(1/10) = –120

Hence, –120 = –5A + 21.794B, leads to B (5x12 – 120)/21.794 = –2.753

At the peak value, dvC(to)/dt = 0, i.e.,

0 = A + Btan21.794to + (A21.794/5)tan21.794to – 21.794B/5

(B + A21.794/5)tan21.794to = (21.794B/5) – A

tan21.794to = [(21.794B/5) – A]/(B + A21.794/5) = –24/49.55 = –0.484

Therefore, 21.7945to = |–0.451|

to = |–0.451|/21.794 = 20.68 ms


For critical damping of a parallel RLC circuit,

LCRCo1

21

=→=ωα

Hence,

F434144425.0

4 2 µ===xR

LC


t1 = 1/|s1| = 0.1x10-3 leads to s1 = –1000/0.1 = –10,000

t2 = 1/|s2| = 0.5x10-3 leads to s1 = –2,000

2o

21s ω−α−α−=

2o

22s ω−α+α−=

s1 + s2 = –2α = –12,000, therefore α = 6,000 = R/(2L)

L = R/12,000 = 60,000/12,000 = 5H

2o

22s ω−α+α−= = –2,000

2o

2 ω−α−α = 2,000

2o

2000,6 ω−α− = 2,000

2o

2 ω−α = 4,000

α2 – = 16x102oω

6

2oω = α2 – 16x106 = 36x106 – 16x106

ωo = 103 LC/120 =

C = 1/(20x106x5) = 10 nF


t = 1/α = 0.25 leads to α = 4

But, α 1/(2RC) or, C = 1/(2αR) = 1/(2x4x200) = 625 µF

22od α−ω=ω

232322

d2o 010x42(16)10x42( π≅+π=α+ω=ω = 1/(LC)

This results in L = 1/(64π2x106x625x10-6) = 2.533 µH

Chapter 8, Solution 82. For t = 0-, v(0) = 0. For t > 0, the circuit is as shown below.

+

vo

−

C1

+

v

−

a

R2

R1 C2 At node a,

(vo – v/R1 = (v/R2) + C2dv/dt

vo = v(1 + R1/R2) + R1C2 dv/dt

60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt

60 = 3v + 25dv/dt

v(t) = Vs + [Ae-3t/25]

where 3Vs = 60 yields Vs = 20

v(0) = 0 = 20 + A or A = –20

v(t) = 20(1 – e-3t/25)V

Chapter 8, Solution 83. i = iD + Cdv/dt (1) –vs + iR + Ldi/dt + v = 0 (2) Substituting (1) into (2),

vs = RiD + RCdv/dt + Ldi/dt + LCd2v/dt2 + v = 0

LCd2v/dt2 + RCdv/dt + RiD + Ldi/dt = vs

d2v/dt2 + (R/L)dv/dt + (R/LC)iD + (1/C)di/dt = vs/LC


(a) angular frequency ω = 103 rad/s

(b) frequency f = πω2

= 159.2 Hz

(c) period T = f

= 1

6.283 ms

(d) Since sin(A) = cos(A – 90°),

vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°) vs in cosine form is vs = 12 cos(103t – 66°) V

(e) vs(2.5 ms) = 12 )24)105.2)(10sin(( 3-3 °+×

= 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°) = 2.65 V


(a) amplitude = 8 A (b) ω = 500π = 1570.8 rad/s

(c) f = πω2

= 250 Hz

(d) Is = 8∠-25° A

Is(2 ms) = )25)102)(500cos((8 3- °−×π= 8 cos(π − 25°) = 8 cos(155°) = -7.25 A


(a) 4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°) (b) -2 sin(6t) = 2 cos(6t + 90°) (c) -10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)


(a) v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°) (b) i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)

Chapter 9, Solution 5. v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°) v2 = 60 cos(ωt − 10°)

This indicates that the phase angle between the two signals is 20° and that v1 lags v2.


(a) v(t) = 10 cos(4t – 60°) i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°) Thus, i(t) leads v(t) by 20°.

(b) v1(t) = 4 cos(377t + 10°)

v2(t) = -20 cos(377t) = 20 cos(377t + 180°) Thus, v2(t) leads v1(t) by 170°.

(c) x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°)

X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04° x(t) = 13.928 cos(2t – 21.04°) y(t) = 15 cos(2t – 11.8°) phase difference = -11.8° + 21.04° = 9.24° Thus, y(t) leads x(t) by 9.24°.

Chapter 9, Solution 7. If f(φ) = cosφ + j sinφ,

)(fj)sinj(cosjcosj-sinddf

φ=φ+φ=φ+φ=φ

φ= djfdf

Integrating both sides ln f = jφ + ln A f = Aejφ = cosφ + j sinφ f(0) = A = 1 i.e. f(φ) = ejφ = cosφ + j sinφ


(a) 4j3

4515−

°∠+ j2 =

°∠°∠

53.13-54515

+ j2

= 3∠98.13° + j2 = -0.4245 + j2.97 + j2 = -0.4243 + j4.97

(b) (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57°

j4)-j)(3(220-8

+°∠

+j125-

10+

= °∠

°∠26.57-11.1820-8

+14425

)10)(12j5-(+

−

= 0.7156∠6.57° − 0.2958 − j0.71

= 0.7109 + j0.08188 − 0.2958 − j0.71

= 0.4151 − j0.6281 (c) 10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38°

= 109.25 – j31.07 Chapter 9, Solution 9.

(a) 2 +8j54j3

−+

= 2 +6425

)8j5)(4j3(+

++

= 2 +89

3220j24j15 −++

= 1.809 + j0.4944

(b) 4∠-10° +°∠

−632j1

= 4∠-10° +°∠

°∠6363.43-236.2

= 4∠-10° + 0.7453∠-69.43° = 3.939 – j0.6946 + 0.2619 – j0.6978 = 4.201 – j1.392

(c) °∠−°∠°∠+°∠

50480920-6108

= 064.3j571.2863.8j5628.1052.2j638.53892.1j879.7

−−+−++

= 799.5j0083.1

6629.0j517.13+−

− =

°∠°∠

86.99886.581.2-533.13

= 2.299∠-102.67° = -0.5043 – j2.243


(a) z 9282.64z and ,566.8z ,86 321 jjj −−=−=−=

93.1966.10321 jzzz −=++

(b) 499.7999.93

21 jzzz

+=


(a) = (-3 + j4)(12 + j5) 21zz = -36 – j15 + j48 – 20 = -56 + j33

(b) ∗2

1

zz

= 5j124j3-

−+

= 25144

)5j12)(4j3(-+

++ = -0.3314 + j0.1953

(c) = (-3 + j4) + (12 + j5) = 9 + j9 21 zz +

21 zz − = (-3 + j4) – (12 + j5) = -15 – j

21

21

zzzz

−+

= )j15(-)j1(9

++

= 22 115j)-15)(j1(9-

−+

= 226

)14j16(9- +

= -0.6372 – j0.5575


(a) = (-3 + j4)(12 + j5) 21zz = -36 – j15 + j48 – 20 = -56 + j33

(b) ∗2

1

zz

= 5j124j3-

−+

= 25144

)5j12)(4j3(-+

++ = -0.3314 + j0.1953

(c) = (-3 + j4) + (12 + j5) = 9 + j9 21 zz +

21 zz − = (-3 + j4) – (12 + j5) = -15 – j

21

21

zzzz

−+

= )j15(-)j1(9

++

= 22 115j)-15)(j1(9-

−+

= 226

)14j16(9- +

= -0.6372 – j0.5575 Chapter 9, Solution 13.

(a) 1520.02749.1)2534.08425.0()4054.04324.0 jjj( +−=−−++−

(b) 0833.215024

3050−=

∠−∠

o

o

(c) (2+j3)(8-j5) –(-4) = 35 +j14


(a) 5116.05751.01115

143 jjj

+−=+−

−

(b) 55.11922.17.213406.246

9.694424186)5983.1096.16)(8467(

)8060)(8056.13882.231116.62( jjjj

jjj−−=

+−

=++

−−++

(c) ( ) 89.2004.256)120260(42 2 jjj −−=−+−


(a) j1-5-3j26j10

+−+

= -10 – j6 + j10 – 6 + 10 – j15

= -6 – j11

(b) °∠°∠°∠°−∠

45301610-4-3020

= 60∠15° + 64∠-10°

= 57.96 + j15.529 + 63.03 – j11.114 = 120.99 – j4.415

(c)

j1j0jj1

j1j1j1j

0jj1

−−−

+−

−−

= 1 )j1(j)j1(j0101 22 ++−+−−++

= 1 )j1j1(1 ++−− = 1 – 2 = -1


(a) -10 cos(4t + 75°) = 10 cos(4t + 75° − 180°) = 10 cos(4t − 105°)

The phasor form is 10∠-105° (b) 5 sin(20t – 10°) = 5 cos(20t – 10° – 90°)

= 5 cos(20t – 100°) The phasor form is 5∠-100°

(c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°)

The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87° Chapter 9, Solution 17.

(a) Let A = 8∠-30° + 6∠0° = 12.928 – j4 = 13.533∠-17.19°

a(t) = 13.533 cos(5t + 342.81°)

(b) We know that -sinα = cos(α + 90°). Let B = 20∠45° + 30∠(20° + 90°)

= 14.142 + j14.142 – 10.261 + j28.19 = 3.881 + j42.33 = 42.51∠84.76°

b(t) = 42.51 cos(120πt + 84.76°) (c) Let C = 4∠-90° + 3∠(-10° – 90°)

= -j4 – 0.5209 – j2.954 = 6.974∠265.72°

c(t) = 6.974 cos(8t + 265.72°) Chapter 9, Solution 18.

(a) = )t(v1 60 cos(t + 15°) (b) = 6 + j8 = 10∠53.13° 2V

)t(v2 = 10 cos(40t + 53.13°) (c) = )t(i1 2.8 cos(377t – π/3) (d) = -0.5 – j1.2 = 1.3∠247.4° 2I

)t(i 2 = 1.3 cos(103t + 247.4°) Chapter 9, Solution 19.

(a) 3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5 = -1.376 + j3.021 = 3.32∠114.49°

Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°)

(b) 4∠-90° + 3∠-45° = -j40 + 21.21 – j21.21

= 21.21 – j61.21 = 64.78∠-70.89°

Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°) (c) Using sinα = cos(α − 90°),

20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699 = 6.7101 – j6.641 = 9.44∠-44.7°

Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°) = 9.44 cos(400t – 44.7°)

Chapter 9, Solution 20. (a) oooo jj 399.4966.82139.383.32464.340590604 −∠=−−−−=∠−−−∠=V Hence,

)399.4377cos(966.8 otv −= (b) 5,90208010 =−∠+∠= ωω ooo jI , i.e. ooI 04.1651.49204010 ∠=∠+=

)04.165cos(51.49 oti +=


(a) oooo jF 86.343236.8758.48296.690304155 ∠=+=−−∠−∠=

)86.3430cos(324.8)( ottf +=

(b) G ooo j 49.62565.59358.4571.2504908 −∠=−=∠+−∠=

)49.62cos(565.5)( ottg −=

(c) ( ) 40,9050101=−∠+∠= ω

ωoo

jH

i.e. ooo jH 6.1162795.0125.025.0180125.09025.0 −∠=−−=−∠+−∠=

)6.11640cos(2795.0)( otth −= Chapter 9, Solution 22.

Let f(t) = ∫∞−

−+t

dttvdtdvtv )(24)(10

oVjVVjVF 3020,5,2410 −∠==−+= ωω

ω

ojjVjVjVF 97.921.440)1032.17)(6.1910(4.02010 −∠=−−=−+=

)97.925cos(1.440)( ottf −=


(a) v(t) = 40 cos(ωt – 60°) (b) V = -30∠10° + 50∠60°

= -4.54 + j38.09 = 38.36∠96.8° v(t) = 38.36 cos(ωt + 96.8°)

(c) I = j6∠-10° = 6∠(90° − 10°) = 6∠80°

i(t) = 6 cos(ωt + 80°)

(d) I = j2

+ 10∠-45° = -j2 + 7.071 – j7.071

= 11.5∠-52.06° i(t) = 11.5 cos(ωt – 52.06°)


(a)

1,010j

=ω°∠=ω

+V

V

10)j1( =−V

°∠=+=−

= 45071.75j5j1

10V

Therefore, v(t) = 7.071 cos(t + 45°) (b)

4),9010(20j4

5j =ω°−°∠=ω

++ωV

VV

°∠=

++ 80-20

4j4

54jV

°∠=+

°∠= 96.110-43.3

3j580-20

V

Therefore, v(t) = 3.43 cos(4t – 110.96°)


(a) 2,45-432j =ω°∠=+ω II

°∠=+ 45-4)4j3(I

°∠=°∠°∠

=+

°∠= 98.13-8.0

13.53545-4

j4345-4

I

Therefore, i(t) = 0.8 cos(2t – 98.13°) (b)

5,2256jj

10 =ω°∠=+ω+ω

III

°∠=++ 225)65j2j-( I

°∠=°∠

°∠=

+°∠

= 56.4-745.056.26708.6

2253j6

225I

Therefore, i(t) = 0.745 cos(5t – 4.56°) Chapter 9, Solution 26.

2,01j

2j =ω°∠=ω

++ωI

II

12j1

22j =

++I

°∠=+

= 87.36-4.05.1j2

1I

Therefore, i(t) = 0.4 cos(2t – 36.87°) Chapter 9, Solution 27.

377,10-110j

10050j =ω°∠=ω

++ωV

VV

°∠=

−+ 10-110377100j

50377jV

°∠=°∠ 10-110)45.826.380(V °∠= 45.92-289.0V

Therefore, v(t) = 0.289 cos(377t – 92.45°).


===8

)t377cos(110R

)t(v)t(i s 13.75 cos(377t) A.


5.0j-)102)(10(j

1Cj

16-6 =

×=

ω=Z

°∠=°∠°∠== 65-2)90-5.0)(254(IZV

Therefore v(t) = 2 sin(106t – 65°) V.


Z 2j)104)(500(jLj 3- =×=ω=

°∠=°∠°∠

== 155-30902

65-60ZV

I

Therefore, i(t) = 30 cos(500t – 155°) A. Chapter 9, Solution 31.

i(t) = 10 sin(ωt + 30°) = 10 cos(ωt + 30° − 90°) = 10 cos(ωt − 60°) Thus, I = 10∠-60°

v(t) = -65 cos(ωt + 120°) = 65 cos(ωt + 120° − 180°) = 65 cos(ωt − 60°)

Thus, V = 65∠-60°

Ω=°∠°∠

== 5.660-1060-65

IV

Z

Since V and I are in phase, the element is a resistor with R = 6.5 Ω.

Chapter 9, Solution 32. V = 180∠10°, I = 12∠-30°, ω = 2

Ω+=°∠=°∠°∠

== 642.9j49.11041530-1201180

IV

Z

One element is a resistor with R = 11.49 Ω. The other element is an inductor with ωL = 9.642 or L = 4.821 H.

Chapter 9, Solution 33. 2

L2R vv110 +=

2R

2L v110v −=

=−= 22L 85110v 69.82 V


if 0vo = LC1

C1

L =ω→ω

=ω

=××

=ω−− )102)(105(

133

100 rad/s


°∠= 05sV2j)1)(2(jLj ==ω

2j-)25.0)(2(j

1Cj

1==

ω

=°∠°∠=°∠=+−

= )05)(901(0522j

2j2j22j

so VV 5∠90°

Thus, 5 cos(2t + 90°) = =)t(vo -5 sin(2t) V


Let Z be the input impedance at the source.

2010100200mH 100 3 jxxjLj ==→ −ω

5002001010

11F10 6 jxxjCj

−==→ −ωµ

1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84

ojZ 104.6225584.23962.2242 −∠=−=

mA 896.361.26104.62255

1060 oo

o

I −∠=−∠−∠

=

)896.3200cos(1.266 oti −=


5j)1)(5(jLj ==ω

j-)2.0)(5(j

1Cj

1==

ω

Let Z , j-1 = 5j210j

5j2)5j)(2(

5j||22 +=

+==Z

Then, s21

2x I

ZZZ

I+

= , where °∠= 0s 2I

°∠=+

=

++

+= 3212.2

8j520j

)2(

5j210j

j-

5j210j

xI

Therefore, =)t(i x 2.12 sin(5t + 32°) A


(a) 2j-)6/1)(3(j

1Cj

1F

61

==ω

→

°∠=°∠−

= 43.18-472.4)4510(2j4

2j-I

Hence, i(t) = 4.472 cos(3t – 18.43°) A

°∠=°∠== 43.18-89.17)43.18-472.4)(4(4IV Hence, v(t) = 17.89 cos(3t – 18.43°) V

(b) 3j-)12/1)(4(j

1Cj

1F

121

==ω

→

12j)3)(4(jLjH3 ==ω→

°∠=−

°∠== 87.3610

j34050

ZV

I

Hence, i(t) = 10 cos(4t + 36.87°) A

°∠=°∠+

= 69.336.41)050(j128

12jV

Hence, v(t) = 41.6 cos(4t + 33.69°) V Chapter 9, Solution 39.

10j810j5j

)10j-)(5j(8)10j-(||5j8 +=

−+=+=Z

°∠=°∠

=+

°∠== 34.51-124.3

34.51403.620

j108040

ZV

I

°∠==−

= 34.51-248.6210j5j

10j-1 III

°∠=== 66.1283.124-5j-5j

2 III

Therefore, =)t(i1 6.248 cos(120πt – 51.34°) A

=)t(i2 3.124 cos(120πt + 128.66°) A


(a) For ω , 1=j)1)(1(jLjH1 ==ω→

20j-)05.0)(1(j

1Cj

1F05.0 ==

ω→

802.0j98.120j2

40j-j)20j-(||2j +=

−+=+=Z

°∠=°∠

°∠=

+°∠

== 05.22-872.105.22136.2

04j0.8021.9804

o ZV

I

Hence, i =)t(o 1.872 cos(t – 22.05°) A

(b) For ω , 5=5j)1)(5(jLjH1 ==ω→

4j-)05.0)(5(j

1Cj

1F05.0 ==

ω→

2.4j6.12j1

4j-5j)4j-(||25j +=

−+=+=Z

°∠=°∠

°∠=

+°∠

== 14.69-89.014.69494.4

04j41.6

04o Z

VI

Hence, i =)t(o 0.89 cos(5t – 69.14°) A

(c) For ω , 10=10j)1)(10(jLjH1 ==ω→

2j-)05.0)(10(j

1Cj

1F05.0 ==

ω→

9j12j2

4j-10j)2j-(||210j +=

−+=+=Z

°∠=°∠

°∠=

+°∠

== 66.38-4417.066.839.055

049j1

04o Z

VI

Hence, i =)t(o 0.4417 cos(10t – 83.66°) A


, 1=ωj)1)(1(jLjH1 ==ω→

j-)1)(1(j

1Cj

1F1 ==

ω→

j21

1j-1)j-(||)j1(1 −=

++=++=Z

j210s

−==

ZV

I , II )j1(c +=

°∠=−

−=−=+= 18.43-325.6

j2)10)(j1(

)j1()j1)(j-( IIV

Thus, v(t) = 6.325 cos(t – 18.43°) V


ω 200=

100j-)1050)(200(j

1Cj

1F50 6- =

×=

ω→µ

20j)1.0)(200(jLjH1.0 ==ω→

20j40j2-1

j100-j10050

)(50)(-j100-j100||50 −==

−=

°∠=°∠=°∠−++

= 9014.17)060(7020j

)060(20j403020j

20joV

Thus, =)t(vo 17.14 sin(200t + 90°) V or =)t(vo 17.14 cos(200t) V


ω 2=2j)1)(2(jLjH1 ==ω→

5.0j-)1)(2(j

1Cj

1F1 ==

ω→

°∠=°∠+

=+−

−= 69.33328.304

5.1j15.1j

15.0j2j5.0j2j

o II

Thus, =)t(io 3.328 cos(2t + 33.69°) A


ω 200=2j)1010)(200(jLjmH10 -3 =×=ω→

j-)105)(200(j

1Cj

1mF5 3- =

×=

ω→

4.0j55.010

j35.0j25.0

j31

2j1

41

−=+

+−=−

++=Y

865.0j1892.14.0j55.0

11+=

−==

YZ

°∠=+°∠

=+

°∠= 7.956-96.0

865.0j1892.606

506Z

I

Thus, i(t) = 0.96 cos(200t – 7.956°) A

Chapter 9, Solution 45. We obtain I by applying the principle of current division twice. o

I I2 I2 Io

Z1 -j2 ΩZ2 2 Ω

(a) (b)

2j-1 =Z , 3j1j2-2

j4-j42||-j2)(4j2 +=+=+=Z

j1j10-

)05(3j12j-

2j-

21

12 +

=°∠++

=+

= IZZ

ZI

=+

=

+

==

1110-

j1j10-

j-1j-

j2-2j2-

2o II -5 A


°∠=→°+= 405)40t10cos(5i ss I

j-)1.0)(10(j

1Cj

1F1.0 ==

ω→

2j)2.0)(10(jLjH2.0 ==ω→

Let 6.1j8.02j4

8j2j||41 +=

+==Z , j32 −=Z

)405(6.0j8.3

j1.60.8s

21

1o °∠

++

=+

= IZZ

ZI

°∠=°∠

°∠°∠= 46.94325.2

97.8847.3)405)(43.63789.1(

oI

Thus, =)t(io 2.325 cos(10t + 94.46°) A


First, we convert the circuit into the frequency domain.

-j10

Ix

+ −

2 Ω j4 5∠0˚ 20 Ω

°∠=°−∠

=−+

=

++−+−

+= 63.524607.0

63.52854.105

626.8j588.425

4j2010j)4j20(10j2

5Ix

is(t) = 0.4607cos(2000t +52.63˚) A


Converting the circuit to the frequency domain, we get: 10 Ω 30 ΩV1

+ − j20

Ix -j20 20∠-40˚

We can solve this using nodal analysis.

A)4.9t100sin(4338.0i

4.94338.020j30

29.24643.15I

29.24643.1503462.0j12307.0

402V

402)01538.0j02307.005.0j1.0(V

020j300V

20j0V

104020V

x

x

1

1

111

°+=

°∠=−

°−∠=

°−∠=−

°∠=

°−∠=++−

=−−

+−

+°−∠−


4j1

)j1)(2j(2)j1(||2j2T =

+−

+=−+=Z

1 ΩI Ix

j2 Ω -j Ω

IIIj1

2jj12j

2jx +

=−+

= , where 21

05.x =°∠0=I

4jj1

2jj1

x

+=

+= II

°∠=−=+

=+

== 45-414.1j1j

j1)4(

4jj1

Ts ZIV

=)t(vs 1.414 sin(200t – 45°) V Chapter 9, Solution 50. Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3) = -j10Ω.

j10 Ix

-j10 +

vx

−

20 Ω

5∠40˚

Using the current dividing rule:

V)50t100cos(50v5050I20V

505.2405.2j40510j2010j

10jI

x

xx

x

°−=°−∠==

°−∠=°∠−=°∠++−

−=


5j-)1.0)(2(j

1Cj

1F1.0 ==

ω→

j)5.0)(2(jLjH5.0 ==ω→ The current I through the 2-Ω resistor is

4j32j5j11 s

s −=

++−=

III , where °∠= 010I

°∠=−= 13.53-50)4j3)(10(sI Therefore,

=)t(is 50 cos(2t – 53.13°) A Chapter 9, Solution 52.

5.2j5.2j1

5j5j5

25j5j||5 +=

+=

+=

101 =Z , 5.2j5.25.2j5.25j-2 −=++=Z

I2

Z1 Z2IS

sss21

12 j5

45.2j5.12

10III

ZZZ

I−

=−

=+

=

)5.2j5.2( += 2o IV

ss j5)j1(10

)j1)(5.2(j5

4308 II

−+

=+

−

=°∠

=+

−°∠=

)j1(10)j5)(308(

sI 2.884∠-26.31° A


Convert the delta to wye subnetwork as shown below. Z1 Z2 Io 2 Ω Z3 + 10Ω

60 V 8o30−∠ Ω - Z

,3077.24615.0210

46,7692.01532.021042

21 jjxjZj

jxjZ +−=

−=−=

−−

=

2308.01538.1210

123 j

jZ +=

−=

6062.0726.4)3077.25385.9//()2308.01538.9()10//()8( 23 jjjZZ +=++=++

163.0878.66062.0726.42 1 jjZZ −=+++=

A 64.28721.83575.188.63060

Z3060I o

o

ooo −∠=

−∠

−∠=

−∠=


Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. Vs

−

V1

+

+

V2

−

+ −

Z 2 Z

)j1(2)j1(o1 −=−= IV )j1(42 12 −== VV

)j1(621s −=+= VVV =sV 8.485∠-45° V


-j4 Ω

I I1

+

Vo

−

I2+ −

Z 12 Ω

-j20 V j8 Ω

-j0.58j4

8jo

1 ===V

I

j8j4-

)8j((-j0.5)j4-

)8j(12 +=

+=

+=

ZZZII

5.0j8

j8

-j0.521 +=++=+=ZZ

III

)8j(12j20- 1 ++= ZII

)8j(2j-

2j

812j20- ++

+= ZZ

−=21

j23

j26-4- Z

°∠=°∠°∠

=−

= 279.6864.1618.43-5811.1

25.26131.26

21

j23

j26-4-Z

Z = 2.798 – j16.403 Ω


30H3 jLj =→ ω

30/13F jCj

−=→ω

15/11.5F jCj

−=→ω

06681.0

1530

1530

)15///(30 jjj

jxjjj −=

−

−

=−

Ω−=−+−−−

=−−

= m 333606681.02033.0

)06681.02(033.0)06681.02//(30

jjjjjjjZ


2H2 jLj =→ ω

jCj

−=→ω11F

2.1j6.2j22j)j2(2j1)j2//(2j1Z +=

−+−

+=−+=

S 1463.0j3171.0Z

1Y −==


(a) 2j-)1010)(50(j

1Cj

1mF10 3- =

×=

ω→

5.0j)1010)(50(jLjmH10 -3 =×=ω→

)2j1(||15.0jin −+=Z

2j22j1

5.0jin −−

+=Z

)j3(25.05.0jin −+=Z =inZ 0.75 + j0.25 Ω

(b) 20j)4.0)(50(jLjH4.0 ==ω→

10j)2.0)(50(jLjH2.0 ==ω→

20j-)101)(50(j

1Cj

1mF1 3- =

×=

ω→

For the parallel elements,

20j-1

10j1

2011

p++=

Z

10j10p +=Z Then,

=inZ 10 + j20 + = pZ 20 + j30 Ω Chapter 9, Solution 59.

)4j2(||)2j1(6eq +−+=Z

)4j2()2j1()4j2)(2j1(

6eq ++−+−

+=Z

5385.1j308.26eq −+=Z

=eqZ 8.308 – j1.5385 Ω


Ω+=−++=+−++= 878.91.51122.5097.261525)1030//()5020()1525( jjjjjjZ


All of the impedances are in parallel.

3j11

5j1

2j11

j111

eq +++

++

−=

Z

4.0j8.0)3.0j1.0()2.0j-()4.0j2.0()5.0j5.0(1

eq

−=−++−++=Z

=−

=4.0j8.0

1eqZ 1 + j0.5 Ω


2 20j)102)(1010(jLjmH -33 =××=ω→

100j-)101)(1010(j

1Cj

1F1 6-3 =

××=

ω→µ

50 Ω j20 Ω

+

+ −V +

Vin

−

1∠0° A 2V

-j100 Ω

50)50)(01( =°∠=V

)50)(2()100j20j50)(01(in +−+°∠=V 80j15010080j50in −=+−=V

=°∠

=01in

in

VZ 150 – j80 Ω

Chapter 9, Solution 63. First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta.

5.22j1020

450j200z,333.13j3015j

450j200z

45j2010

300j150j200z

32

1

+=+

=−=+

=

+=++

=

–j12 Ω –j16 Ω8 Ω

z2

z3

–j16 Ω

z1

10 Ω

ZT 10 Ω Now all we need to do is to combine impedances.

Ω−=−+−+−=

−=−

−=−

−−=−

93.6j69.34)821.3j7.21938.8j721.8(z12j8Z

821.3j70.21)16j10(z

938.8j721.833.29j40

)16j10)(333.13j30()16j10(z

1T

3

2


A7.104527.14767.1j3866.0Z

9030I

5j192j6

)8j6(10j4Z

T

T

°∠=+−=°∠

=

Ω−=−+−

+=


)4j3(||)6j4(2T +−+=Z

2j7)4j3)(6j4(

2T −+−

+=Z

=TZ 6.83 + j1.094 Ω = 6.917∠9.1° Ω

=°∠

°∠==

1.9917.610120

TZV

I 17.35∠0.9° A


)j12(145170

5j60)10j40)(5j20(

)10j40(||)5j20(T −=+

+−=+−=Z

=TZ 14.069 – j1.172 Ω = 14.118∠-4.76°

°∠=°∠

°∠== 76.9425.4

76.4-118.149060

TZV

I

I

I1

20 Ω

+ −Vab

I2

j10 Ω

IIIj122j8

5j6010j40

1 ++

=++

=

IIIj12j4

5j605j20

2 +−

=+−

=

21ab 10j20- IIV +=

IIVj12

40j10j12

)40j(160-ab +

++

++

=

IIV145

j)(150)-12(j12

150-ab

+=

+=

)76.9725.4)(24.175457.12(ab °∠°∠=V

=abV 52.94∠273° V


(a) 20j)1020)(10(jLjmH20 -33 =×=ω→

80j-)105.12)(10(j

1Cj

1F5.12 6-3 =

×=

ω→µ

)80j60(||20j60in −+=Z

60j60)80j60)(20j(

60in −−

+=Z

°∠=+= 22.20494.6733.23j33.63inZ

==in

in

1Z

Y 0.0148∠-20.22° S

(b) 10j)1010)(10(jLjmH10 -33 =×=ω→

50j-)1020)(10(j

1Cj

1F20 6-3 =

×=

ω→µ

2060||30 =

)10j40(||2050j-in ++=Z

10j60)10j40)(20(

50j-in ++

+=Z

°∠=−= 56.74-75.5092.48j5.13inZ

==in

in

1Z

Y 0.0197∠74.56° S = 5.24 + j18.99 mS


4j-

1j3

12j5

1eq +

++

−=Y

)25.0j()1.0j3.0()069.0j1724.0(eq +−++=Y

=eqY 0.4724 + j0.219 S


)2j1(41

2j-1

411

o+=+=

Y

6.1j8.05

)2j1)(4(2j1

4o −=

−=

+=Y

6.0j8.0jo −=+Y

)6.0j8.0()333.0j()1(6.0j8.0

13j-

1111

o

+++=−

++=′Y

°∠=+=′ 41.27028.2933.0j8.11

oY

2271.0j4378.041.27-4932.0o −=°∠=′Y

773.4j4378.05jo +=+′Y

97.22773.4j4378.0

5.0773.4j4378.0

1211

eq

−+=

++=

Y

2078.0j5191.01

eq−=

Y

=−

=3126.0

2078.0j5191.0eqY 1.661 + j0.6647 S


Make a delta-to-wye transformation as shown in the figure below.

cb

n

a

Zan

Zcn Zbn

2 Ω

Zeq

8 Ω

-j5 Ω

9j75j15

)10j15)(10(15j1010j5)15j10)(10j-(

an −=+−

=++−

+=Z

5.3j5.45j15

)15j10)(5(bn +=

++

=Z

3j1-5j15

)10j-)(5(cn −=

+=Z

)5j8(||)2( cnbnaneq −+++= ZZZZ

)8j7(||)5.3j5.6(9j7eq −++−=Z

5.4j5.13)8j7)(5.3j5.6(

9j7eq −−+

+−=Z

2.0j511.59j7eq −+−=Z

=−= 2.9j51.12eqZ 15.53∠-36.33° Ω


We apply a wye-to-delta transformation.

j4 Ω

Zeq

-j2 ΩZbc

Zab

Zac

a b

c

1 Ω

j12j

2j22j

4j2j2ab −=

+=

+−=Z

j12

2j2ac +=

+=Z

j2-2j-2j2

bc +=+

=Z

8.0j6.13j1

)j1)(4j()j1(||4j||4j ab −=

+−

=−=Z

2.0j6.0j2

)j1)(1()j1(||1||1 ac +=

++

=+=Z

6.0j2.2||1||4j acab −=+ ZZ

6.0j2.21

2j2-1

2j-11

eq −+

++=

Z

1154.0j4231.025.0j25.05.0j ++−−= °∠=+= 66.644043.03654.0j173.0

=eqZ 2.473∠-64.66° Ω = 1.058 – j2.235 Ω


Transform the delta connections to wye connections as shown below. a

R3

R2R1

-j18 Ω -j9 Ω

j2 Ω

j2 Ωj2 Ω

b

6j-18j-||9j- = ,

Ω=++

= 8102020)20)(20(

R1 , Ω== 450

)10)(20(R 2 , Ω== 4

50)10)(20(

3R

44)j6(j2||)82j(j2ab ++−++=Z

j4)(4||)2j8(j24ab −+++=Z

j2-12)4jj2)(4(8

j24ab

−+++=Z

4054.1j567.3j24ab −++=Z

=abZ 7.567 + j0.5946 Ω

Chapter 9, Solution 73. Transform the delta connection to a wye connection as in Fig. (a) and then transform the wye connection to a delta connection as in Fig. (b).

a

R3

R2R1

-j18 Ω -j9 Ω

j2 Ω

j2 Ωj2 Ω

b

8.4j-10j48

6j8j8j)6j-)(8j(

1 ==−+

=Z

-j4.812 == ZZ

4.6jj1064-

10j)8j)(8j(

3 ===Z

=++++++ ))(2())(4()4)(2( 313221 ZZZZZZ

6.9j4.46)4.6j)(8.4j2()4.6j)(8.4j4()8.4j4)(8.4j2( +=−+−+−−

25.7j5.14.6j

6.9j4.46a −=

+=Z

688.6j574.38.4j4

6.9j4.46b +=

−+

=Z

945.8j727.18.4j2

6.9j4.46c +=

−+

=Z

3716.3j07407688.12j574.3

)88.61583.7)(906(||6j b +=

+°∠°∠

=Z

602.2j186.025.11j5.1

j7.25)-j4)(1.5(||4j- a −=

−−

=Z

1693.5j5634.0945.20j727.1

)07.7911.9)(9012(||12j c +=

+°∠°∠

=Z

)||12j||4j-(||)||6j( cabeq ZZZZ +=

)5673.2j7494.0(||)3716.3j7407.0(eq ++=Z =eqZ 1.508∠75.42° Ω = 0.3796 + j1.46 Ω


One such RL circuit is shown below.

Z

+

Vi = 1∠0° j20 Ω

20 Ω 20 ΩV

j20 Ω+

Vo

−

We now want to show that this circuit will produce a 90° phase shift.

)3j1(42j120j20-

40j20)20j20)(20j(

)20j20(||20j +=++

=++

=+=Z

)j1(31

3j63j1

)01(12j2412j4

20 i +=++

=°∠++

=+

= VZ

ZV

°∠==

+

+

=+

= 903333.03j

)j1(31

j1j

20j2020j

o VV

This shows that the output leads the input by 90°.


Since , we need a phase shift circuit that will cause the output to lead the input by 90°.

)90tsin()tcos( °+ω=ωThis is achieved by the RL circuit shown

below, as explained in the previous problem.

10 Ω 10 Ω

+

Vi

−

j10 Ω j10 Ω+

Vo

−

This can also be obtained by an RC circuit.


Let Z = R – jX, where fC2

1C

1Xπ

=ω

=

394.9566116R|Z|XXR|Z| 222222 ===−=→+=

F81.27394.95x60x2

1fX21C µ=

π=

π=


(a) ic

co jXR

jX-VV

−=

where 979.3)1020)(102)(2(

1C1

X 9-6c =××π

=ω

=

))53.979(tan-90(979.35

979.3j3.979-5

j3.979- 1-22

i

o +°∠+

==VV

)51.38-90(83.1525

979.3

i

o °−°∠+

=VV

°∠= 51.49-6227.0i

o

VV

Therefore, the phase shift is 51.49° lagging

(b) )RX(tan-90-45 c-1+°=°=θ

C1

XR)RX(tan45 cc1-

ω==→=°

RC1

f2 =π=ω

=×π

=π

=)1020)(5)(2(

1RC21

f 9- 1.5915 MHz

Chapter 9, Solution 78. 8+j6 R Z -jX

[ ] 5)6(8

)]6(8[)6(8// =−++−+

=−+=XjRXjRXjRZ

i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X Equating real and imaginary parts:

8R = 5R + 40 which leads to R=13.33Ω 6R-XR =30-5 which leads to X=4.125Ω.


(a) Consider the circuit as shown.

40 Ω

Z1

V1V2

j30 Ω+

Vi

−

j60 Ω+

Vo

−

j10 Ω

30 Ω20 Ω

Z2

21j390j30

)60j30)(30j()60j30(||30j1 +=

++

=+=Z

°∠=+=++

=+= 21.80028.9896.8j535.131j43

)21j43)(10j()40(||10j 12 ZZ

Let V . °∠= 01i

896.8j535.21)01)(21.80028.9(

20 i2

22 +

°∠°∠=

+= V

ZZ

V

°∠= 77.573875.02V

°∠°∠°∠

=++

=+

=03.2685.47

)77.573875.0)(87.81213.21(21j43

21j340 22

1

11 VV

ZZ

V

°∠= 61.1131718.01V

111o )j2(52

2j12j

60j3060j

VVVV +=+

=+

=

)6.1131718.0)(56.268944.0(o °∠°∠=V °∠= 2.1401536.0oV

Therefore, the phase shift is 140.2°

(b) The phase shift is leading. (c) If , then V120i =V

°∠=°∠= 2.14043.18)2.1401536.0)(120(oV V and the magnitude is 18.43 V.

Chapter 9, Solution 80. Ω=×π=ω→ 4.75j)10200)(60)(2(jLjmH200 3-

)0120(4.75j50R

4.75j4.75j50R

4.75jio °∠

++=

++= VV

(a) When Ω= 100R ,

°∠°∠°∠

=°∠+

=69.2688.167

)0120)(904.75()0120(

4.75j1504.75j

oV

=oV 53.89∠63.31° V

(b) When Ω= 0R ,

°∠°∠°∠

=°∠+

=45.5647.90

)0120)(904.75()0120(

4.75j504.75j

oV

=oV 100∠33.55° V

(c) To produce a phase shift of 45°, the phase of = 90° + 0° − α = 45°. oVHence, α = phase of (R + 50 + j75.4) = 45°. For α to be 45°, R + 50 = 75.4 Therefore, R = 25.4 Ω


Let Z , 11 R=2

22 Cj1

Rω

+=Z , 33 R=Z , and x

xx Cj1

Rω

+=Z .

21

3x Z

ZZ

Z =

ω+=

ω+

22

1

3

xx Cj

1R

RR

Cj1

R

=== )600(400

1200R

RR

R 21

3x 1.8 kΩ

=×

==→

= )103.0(

1200400

CRR

CC1

RR

C1 6-

23

1x

21

3

x 0.1 µF


=×

== )1040(2000100

CRR

C 6-s

2

1x 2 µF


=×

== )10250(1200500

LRR

L 3-s

1

2x 104.17 mH


Let s

11 Cj1

||Rω

=Z , 22 R=Z , 33 R=Z , and . xxx LjR ω+=Z

1CRjR

Cj1

R

CjR

s1

1

s1

s

1

1 +ω=

ω+

ω=Z

Since 21

3x Z

ZZ

Z = ,

)CRj1(R

RRR

1CRjRRLjR s1

1

32

1

s132xx ω+=

+ω=ω+

Equating the real and imaginary components,

1

32x R

RRR =

)CR(R

RRL s1

1

32x ω=ω implies that

s32x CRRL =

Given that , Ω= k40R1 Ω= k6.1R 2 , Ω= k4R 3 , and F45.0Cs µ=

=Ω=Ω== k16.0k40

)4)(6.1(R

RRR

1

32x 160 Ω

=== )45.0)(4)(6.1(CRRL s32x 2.88 H Chapter 9, Solution 85.

Let , 11 R=Z2

22 Cj1

Rω

+=Z , 33 R=Z , and 4

44 Cj1

||Rω

=Z .

jCRRj-

1CRjR

44

4

44

44 −ω

=+ω

=Z

Since 324121

34 ZZZZZ

ZZ

Z =→= ,

ω−=

−ω 223

44

14

Cj

RRjCR

RRj-

2

3232

424

24414

CjR

RR1CR

)jCR(RRj-ω

−=+ω

+ω

Equating the real and imaginary components,

3224

24

241 RR

1CRRR

=+ω

(1)

2

324

24

24

241

CR

1CRCRR

ω=

+ωω

(2) Dividing (1) by (2),

2244

CRCR

1ω=

ω

4422

2

CRCR1

=ω

4422 CRCR1

f2 =π=ω

4242 CCRR21

fπ

=


84j-1

95j1

2401

++=Y

0119.0j01053.0j101667.4 3- +−×=Y

°∠=

+==

2.183861.41000

37.1j1667.410001

YZ

Z = 228∠-18.2° Ω


)102)(102)(2(j-

50Cj

150 6-31 ××π

+=ω

+=Z

79.39j501 −=Z

)1010)(102)(2(j80Lj80 -33

2 ××π+=ω+=Z

66.125j802 +=Z

1003 =Z

321

1111ZZZZ

++=

66.125j801

79.39j501

10011

++

−+=

Z

)663.5j605.3745.9j24.1210(101 3- −+++=Z

3-10)082.4j85.25( ×+= °∠×= 97.81017.26 -3

Z = 38.21∠-8.97° Ω


(a) 20j12030j20j- −++=Z

Z = 120 – j10 Ω

(b) If the frequency were halved, Cf2

1C1

π=

ωLf2L

would cause the capacitive

impedance to double, while π=ω would cause the inductive impedance to halve. Thus,

40j12015j40j- −++=Z Z = 120 – j65 Ω


ω

+ω=Cj

1R||LjinZ

ω−ω+

ω+=

ω+ω+

ω

+ω=

C1

LjR

RLjCL

Cj1

LjR

Cj1

RLj

inZ

22

in

C1

LR

C1

LjRRLjCL

ω−ω+

ω−ω−

ω+=Z

To have a resistive impedance, 0)Im( in =Z . Hence,

0C1

LCL

RL 2 =

ω−ω

−ω

C1

LCR 2

ω−ω=ω

1LCCR 2222 −ω=ω

C1CR

L 2

222

ω+ω

=

(1) Ignoring the +1 in the numerator in (1),

=×== )1050()200(CRL 9-22 2 mH Chapter 9, Solution 90.

Let , °∠= 0145sV L377jL)60)(2(jLjX =π=ω=

jXR800145

jXR80s

++°∠

=++

=V

I

jXR80)145)(80(

801 ++== IV

jXR80)145)(80(

50++

= (1)

jXR80)0145)(jXR(

)jXR(o ++°∠+

=+= IV

jXR80)145)(jXR(

110++

+= (2)

From (1) and (2),

jXR80

11050

+=

=+5

11)80(jXR

30976XR 22 =+ (3)

From (1),

23250

)145)(80(jXR80 ==++

53824XRR1606400 22 =+++

47424XRR160 22 =++ (4)


=→= R16448R160 102.8 Ω From (3),

204081056830976X2 =−=

=→== LL37786.142X 0.3789 H


Lj||RCj

1in ω+

ω=Z

LjRLRj

Cj-

in ω+ω

+ω

=Z

222

222

LRLRjRL

Cj-

ω+ω+ω

+ω

=

To have a resistive impedance, 0)Im( in =Z . Hence,

0LR

LRC1-

222

2

=ω+

ω+

ω

222

2

LRLR

C1

ω+ω

=ω

22

222

LRLR

Cω

ω+=

where 7102f2 ×π=π=ω

)109)(1020)(104()10400)(104(109

C 46142

121424

×××π××π+×

= −

−

nF72

169C 2

2

ππ+

=

C = 235 pF


(a) Ω∠=∠

∠== −

oo

o

o xYZZ 5.134.471

104845075100

6

(b) ooo xxZY 5.612121.0104845075100 6 ∠=∠∠== −γ


Ls 2 ZZZZ ++= )9.186.05.0(j)2.238.01( +++++=Z

20j25+=Z

°∠°∠

==66.3802.32

0115SL Z

VI

=LI 3.592∠-38.66° A

Chapter 10, Solution 1. ω

1=

°∠→°− 45-10)45tcos(10 °∠→°+ 60-5)30tsin(5

jLjH1 =ω→

j-Cj

1F1 =

ω→

The circuit becomes as shown below.

Vo3 Ω

+ − 10∠-45° V +

− 2 Io

j Ω

5∠-60° V


j-j)60-5(

3)45-10( ooo VVV

=−°∠

+−°∠

oj60-1545-10j V=°∠+°∠ °∠=°∠+°∠= 9.24715.73150-1545-10oV

Therefore, =)t(vo 15.73 cos(t + 247.9°) V


ω 10=°∠→π− 45-4)4t10cos(4

°∠→π+ 150-20)3t10sin(20 10jLjH1 =ω→

5j-2.0j

1Cj

1F02.0 ==

ω→

The circuit becomes that shown below.

Io

Vo10 Ω

j10 Ω -j5 Ω20∠-150° V + − 4∠-45° A


5j-10j45-4

10)150-20( ooo VVV

+=°∠+−°∠

o)j1(1.045-4150-20 V+=°∠+°∠

°∠=+

°∠+°∠== 98.150816.2

)j1(j45-4150-2

10jo

o

VI

Therefore, =)t(io 2.816 cos(10t + 150.98°) A Chapter 10, Solution 3.

ω 4=°∠→ 02)t4cos(2

-j1690-16)t4sin(16 =°∠→ 8jLjH2 =ω→

3j-)121)(4(j

1Cj

1F121 ==

ω→

The circuit is shown below.

Vo

+ − 2∠0° A-j16 V

4 Ω -j3 Ω 6 Ω

1 Ω

j8 Ω


8j612

3j416j- ooo

++=+

−− VVV

o8j61

3j41

123j4

16j-V

+

+−

+=+−

°∠=°∠°∠

=+−

= 02.35-835.388.12207.115.33-682.4

04.0j22.156.2j92.3

oV

Therefore, =)t(vo 3.835 cos(4t – 35.02°) V


16 4,10-16)10t4sin( =ω°∠→°−

4jLjH1 =ω→

j-)41)(4(j

1Cj

1F25.0 ==

ω→

j4 Ω

1 Ω

Ix -j Ω

16∠-10° V + −

V1

0.5 Ix

+

Vo

−

j121

4j)10-16( 1

x1

−=+

−°∠ VI

V

But

4j)10-16( 1

x

VI

−°∠=

So, j18j

))10-16((3 11

−=

−°∠ VV

4j1-10-48

1 +°∠

=V


°∠=+°∠

=−

= 04.69-232.8)4j1j)(--(1

10-48j1

11o VV

Therefore, =)t(vo 8.232 sin(4t – 69.04°) V

Chapter 10, Solution 5. Let the voltage across the capacitor and the inductor be Vx and we get:

03j

V2j

V4

3010I5.0V xxxx =+−

+°∠−−

xx

xxx V5.0j2j

VIbut3030I5.1V)4j6j3( =−

=°∠=−−+

Combining these equations we get:

A38.97615.425.1j3

30305.0jI

25.1j33030Vor3030V)75.0j2j3(

x

xx

°∠=+

°∠=

+°∠

=°∠=−+

Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get:

010j20

V3

20V4V oxo =

++−

− where ox V

10j2020V+

=

Combining these we get:

30j60V)35.0j1(010j20

V3

10j20V4

20V

oooo +=−+→=+

+−+

−

=+−

=+−+

=5.0j2

)3(20Vor5.0j2

30j60V xo 29.11∠–166˚ V.

Chapter 10, Solution 7. At the main node,

++

+

=−−+−

→+−

+∠=+

−−∠

501

30j

20j401V

3j196.520j40

058.31j91.11550V

30jV306

20j40V15120 o

o

V 15408.1240233.0j04.0

7805.4j1885.3V o−∠=+−−

=


,200=ω

20j1.0x200jLjmH100 ==ω→

100j10x50x200j

1Cj

1F50 6 −==ω

→µ−

The frequency-domain version of the circuit is shown below.

0.1 Vo 40Ω V1 Io V2 + -j100Ω

6 o15∠ 20Ω Vo j20Ω -

At node 1,

40VV

100jV

20VV1.0156 2111

1o −

+−

+=+∠

or 21 025.0)01.0025.0(5529.17955.5 VVjj −+−=+ (1)

At node 2,

212

121 V)2j1(V30

20jV

V1.040

VV−+=→+=

− (2)

From (1) and (2),

BAVor0

)5529.1j7955.5(VV

)2j1(3025.0)01.0j025.0(

2

1 =

+=

−

−+−

Using MATLAB,

V = inv(A)*B leads to V 09.1613.110,23.12763.70 21 jVj +−=−−=

o21

o 17.82276.740

VVI −∠=

−=

Thus, A )17.82t200cos(276.7)t(i o

o −= Chapter 10, Solution 9.

10 33 10,010)t10cos( =ω°∠→

10jLjmH10 =ω→

20j-)1050)(10(j

1Cj

1F50 6-3 =

×=

ω→µ

Consider the circuit shown below.

V120 Ω V2

-j20 Ω

20 Ω

Io

30 Ω

+

Vo

−

4 Io+ − 10∠0° V

j10 Ω At node 1,

20j-202010 2111 VVVV −

+=−

21 j)j2(10 VV −+= (1)

At node 2,

10j3020)4(

20j-2121

++=

− VVVV, where

201

o

VI = has been substituted.

21 )8.0j6.0()j4-( VV +=+

21 j4-8.0j6.0

VV++

= (2)

Substituting (2) into (1)

22 jj4-

)8.0j6.0)(j2(10 VV −

+++

=

or 2.26j6.0

1702 −=V

°∠=−

⋅+

=+

= 26.70154.62.26j6.0

170j3

310j30

302o VV

Therefore, =)t(vo 6.154 cos(103 t + 70.26°) V Chapter 10, Solution 10.

2000,100j10x50x2000jLj mH 50 3 =ω==ω→ −

250j10x2x2000j

1Cj

1F2 6 −==ω

→µ−

Consider the frequency-domain equivalent circuit below. V1 -j250 V2 36<0o 2kΩ j100 0.1V1 4kΩ

At node 1,

212111 V004.0jV)006.0j0005.0(36

250jVV

100jV

2000V

36 −−=→−−

++= (1)

At node 2,

212

121 V)004.0j00025.0(V)004.0j1.0(0

4000V

V1.0250j

VV++−=→+=

−−

(2)

Solving (1) and (2) gives

o2o 43.931.89515.893j6.535VV ∠=+−==

vo (t) = 8.951 sin(2000t +93.43o) kV


cos( 2,01)t2 =ω°∠→

°∠→°+ 60-8)30t2sin(8

2jLjH1 =ω→ j-)21)(2(j

1Cj

1F2 ==

ω→1

4jLjH2 =ω→ 2j-)41)(2(j

1Cj

1F4 ==

ω→1


-j Ω

-j Ω

2 Io

2 Io

2 Io 2 Io 2 I

2 2

2 Io

2

At node 1,

2jj-2)60-8( 2111 VVVV −

+=−°∠

21 j)j1(60-8 VV ++=°∠ (1)

At node 2,

02j4j)60-8(

2j1 221 =

−−°∠

+−

+VVV

12 5.0j60-4 VV ++°∠= (2)

Substituting (2) into (1), 1)5.1j1(30460-81 V+=°∠−°∠+

5.1j130460-81

1 +°∠−°∠+

=V

°∠=−

°∠−°∠+== 46.55-024.5

j5.130460-81

j-1

o

VI

Therefore, =)t(io 5.024 cos(2t – 46.55°) Chapter 10, Solution 12.

20 1000,020)t1000sin( =ω°∠→

10jLjmH10 =ω→

20j-)1050)(10(j

1Cj

1F50 6-3 =

×=

ω→µ

The frequency-domain equivalent circuit is shown below.

2 Io

-j20 Ω20 Ω

V2V1

20∠0° A

10 Ω

Io

j10 Ω

At node 1,

1020220 211

o

VVVI

−++= ,

where

10j2

o

VI =

102010j2

20 2112 VVVV −++=

21 )4j2(3400 VV +−= (1)

At node 2,

10j20j-1010j2 22212 VVVVV

+=−

+

21 )2j3-(2j VV +=

or (2) 21 )5.1j1( VV +=Substituting (2) into (1),

222 )5.0j1()4j2()5.4j3(400 VVV +=+−+=

5.0j1400

2 +=V

°∠=+

== 6.116-74.35)5.0j1(j

4010j

2o

VI

Therefore, =)t(io 35.74 sin(1000t – 116.6°) A Chapter 10, Solution 13. Nodal analysis is the best approach to use on this problem. We can make our work easier by doing a source transformation on the right hand side of the circuit. –j2 Ω 18 Ω j6 Ω

+ − 50∠0º V

+ −

+

Vx

−

3 Ω

40∠30º V

06j1850V

3V

2j3040V xxx =

+−

++−

°∠−

which leads to Vx = 29.36∠62.88˚ A.


At node 1,

°∠=−

+−

+−

30204j10

02j-

0 1211 VVVV

100j2.1735.2j)5.2j1(- 21 +=−+ VV (1)

At node 2,

°∠=−

++ 30204j5j-2j

1222 VVVV

100j2.1735.2j5.5j- 12 +=+ VV (2)

Equations (1) and (2) can be cast into matrix form as

°∠°∠

=

+3020030200-

5.5j-5.2j5.2j5.2j1

2

1

VV

°∠=−=+

=∆ 38.15-74.205.5j205.5j-5.2j5.2j5.2j1

°∠=°∠=°∠°∠

=∆ 120600)30200(3j5.5j-302005.2j30200-

1

°∠=+°∠=°∠°∠+

=∆ 7.1081020)5j1)(30200(302005.2j30200-5.2j1

2

°∠=∆∆

= 38.13593.2811V

°∠=∆∆

= 08.12418.4922V

)5.31381.15)(905.0(2j-1 °∠°∠==

VI

=I 7.906∠43.49° A


At node 1,

5j-10202j 21211 VVVVV −

+−

+=

21 )4j2()4j3(40j VV +−+= At node 2,

10jj1

5j-1022121 VVVVV

=++−

+−

21 )j1()2j1(-)j1(10 VV +++=+ Thus,

++++

=

+ 2

1

j1j2)(1-j2)(12-4j3

)j1(1040j

VV

°∠=−=++++

=∆ 11.31-099.5j5j1j2)(1-j2)(12-4j3

°∠=+−=+++

=∆ 96.12062.116100j60j1j)(110j2)(12-40j

1

°∠=+=++

+=∆ 29.129142.13j110-90

j)(110j2)(1-40j4j3

2

=∆∆

= 11V 22.87∠132.27° V

=∆∆

= 22V 27.87∠140.6° V



At node 1,

IoV2V1

j4 Ω

+ − 100∠20° V

2 Ω

-j2 Ω

1 Ω

3 Ω

234j20100 2111 VVVV −

+=−°∠

21 2j)10j3(

320100 V

V−+=°∠ (1)

At node 2,

2j-2120100 2212 VVVV

=−

+−°∠

21 )5.0j5.1(5.0-20100 VV ++=°∠ (2)

From (1) and (2),

+

+=

°∠°∠

2

1

2j-310j1)j3(5.05.0-

2010020100

VV

5.4j1667.02j-310j1

5.0j5.15.0-−=

++

=∆

2.286j45.55-2j-20100

5.0j5.1201001 −=

°∠+°∠

=∆

5.364j95.26-20100310j1201005.0-

2 −=°∠+°∠

=∆

°∠=∆∆

= 08.13-74.6411V

°∠=∆∆

= 35.6-17.8122V

9j3333.031.78j5.28-

222121

o −+

=∆∆−∆

=−

=VV

I

=oI 9.25∠-162.12°



j6 Ω

2 Ω +

Vo

−

-j2 Ω

4 Ω j5 Ω

4∠45° A -j Ω+

Vx

−

2 Vx

8 ΩV1 V2

At node 1,

6j82454 211

+−

+=°∠VVV

21 )3j4()3j29(45200 VV −−−=°∠ (1)

At node 2,

2j5j4j-2

6j822

x21

−++=+

+− VV

VVV

, where VV =

1x

21 )41j12()3j104( VV +=−

21 3j10441j12

VV−+

= (2)

Substituting (2) into (1),

22 )3j4(3j104)41j12(

)3j29(45200 VV −−−+

−=°∠

2)17.8921.14(45200 V°∠=°∠

°∠°∠

=17.8921.14

452002V

222o 258j6-

3j42j-

2j5j42j-

VVVV−

=+

=−+

=

°∠°∠

⋅°∠

=17.8921.14

4520025

13.23310oV

=oV 5.63∠189° V


We have a supernode as shown in the circuit below.

j2 Ω

-j4 Ω+

Vo

−

V2V3

V1 4 Ω

2 Ω 0.2 Vo

Notice that . 1o VV =At the supernode,

2j24j-4311223 VVVVVV −

++=−

321 )2j1-()j1()2j2(0 VVV ++++−= (1)

At node 3,

42j2.0 2331

1

VVVVV

−=

−+

0)2j1-()2j8.0( 321 =+++− VVV (2)


21 j2.10 VV += (3) But at the supernode,

21 012 VV +°∠= or (4) 1212 −= VVSubstituting (4) into (3),

)12(j2.10 11 −+= VV

o1 j2.112j

VV =+

=

°∠°∠

=81.39562.1

9012oV

=oV 7.682∠50.19° V


The circuit is converted to its frequency-domain equivalent circuit as shown below.

R

Cj1ω

+

Vo

−

jωL+ − Vm∠0°

Let LC1

Lj

Cj1

Lj

CL

Cj1

||Lj 2ω−ω

=

ω+ω

=ω

ω=Z

m2m

2

2

mo VLj)LC1(R

LjV

LC1Lj

R

LC1Lj

VR ω+ω−

ω=

ω−ω

+

ω−ω

=+

=Z

ZV

ω−ω

−°∠ω+ω−

ω=

)LC1(RL

tan90L)LC1(R

VL2

1-22222

moV

If , then

φ∠= AoV

=A22222

m

L)LC1(RVL

ω+ω−

ω

and =φ)LC1(R

Ltan90 2

1-

ω−ω

−°


(a) RCjLC1

1

Cj1

LjR

Cj1

2i

o

ω+ω−=

ω+ω+

ω=

VV

At , 0=ω ==11

i

o

VV

1

As ω , ∞→ =i

o

VV

0

At LC1

=ω , =⋅

=

LC1

jRC

1

i

o

VV

CL

Rj-

(b) RCjLC1

LC

Cj1

LjR

Lj2

2

i

o

ω+ω−ω−

=

ω+ω+

ω=

VV

At , 0=ω =i

o

VV

0

As ω , ∞→ ==11

i

o

VV

1

At LC1

=ω , =⋅

−=

LC1

jRC

1

i

o

VV

CL

Rj


Consider the circuit in the frequency domain as shown below.

R1

Cj1ω

R2 +

Vo

−jωL

+ −

Vs

Let Cj

1||)LjR( 2 ω

ω+=Z

LCRj1LjR

Cj1

LjR

)LjR(Cj

1

22

2

2

2

ω−ω+ω+

=

ω+ω+

ω+ω

=Z

CRjLC1LjR

R

CRjLC1LjR

R2

22

1

22

2

1s

o

ω+ω−ω+

+

ω+ω−ω+

=+

=Z

ZVV

=s

o

VV

)CRRL(jLCRRRLjR

2112

21

2

+ω+ω−+ω+


0CVj

Cj1Lj

VR

VV s =ω+

ω+ω

+−

s2 VRCVj1LC

RCVjV =ω++ω−

ω+

s2

232VV

LC1RLCjRCjRCjLC1

=

ω−

ω−ω+ω+ω−

)LC2(RCjLC1V)LC1(V 22

s2

ω−ω+ω−

ω−=


For mesh 1,

22

121

s Cj1

Cj1

Cj1

IIVω

−

ω+

ω= (1)

For mesh 2,

22

12 Cj

1LjR

Cj1

0 II

ω+ω++

ω−

= (2)

Putting (1) and (2) into matrix form,

ω+ω+

ω−

ω−

ω+

ω=

2

1

22

221s

Cj1LjR

Cj1

Cj1

Cj1

Cj1

0 IIV

212

221 CC1

Cj1

LjRCj1

Cj1

ω+

ω+ω+

ω+

ω=∆

ω+ω+=∆

2s1 Cj

1LjRV and

2

s2 Cjω=∆

V

=∆∆

= 11I

212

221

2s

CC1

Cj1

LjRCj1

Cj1

Cj1

LjRV

ω+

ω+ω+

ω+

ω

ω+ω+

=∆∆

= 22I

212

221

2

s

CC1

Cj1

LjRCj1

Cj1

CjV

ω+

ω+ω+

ω+

ω

ω


2=ω

°∠→ 010)t2cos(10

-j690-6)t2sin(6 =°∠→ 4jLjH2 =ω→

2j-)41)(2(j

1Cj

1F25.0 ==

ω→


4 Ω j4 Ω

For loop 1,

I2+ − 10∠0° V +

− -j2 ΩI1

Io

6∠-90° V

02j)2j4(10- 21 =+−+ II

21 j)j2(5 II +−= (1) For loop 2,

0)6j-()2j4j(2j 21 =+−+ II 321 =+ II (2)

In matrix form (1) and (2) become

=

−35

11jj2

2

1

II

)j1(2 −=∆ , 3j51 −=∆ , 3j12 −=∆

°∠=+=−

=∆∆−∆

=−= 45414.1j1)j1(2

42121o III

Therefore, =)t(io 1.414 cos(2t + 45°) A Chapter 10, Solution 26.

We apply mesh analysis to the circuit shown below. For mesh 1,

0204010- 21 =−+ II

21 241 II −= (1) For the supermesh,

0)10j30(20)20j20( 312 =++−− III 0)j3()2j2(2- 321 =++−+ III (2)

At node A, 21o III −= (3)

At node B, o32 4III += (4)

Substituting (3) into (4) 2132 44 IIII −+=

123 45 III −= (5) Substituting (5) into (2) gives

21 )3j17()4j14(-0 II +++= (6) From (1) and (6),

++

=

2

1

3j17)4j14(-2-4

01

II

4j40 +=∆

3j17j3170

2-11 +=

+=∆ , 4j14

0j4)(14-14

2 +=+

=∆

4j408j245

45 12123 +

+=

∆∆−∆

=−= III

°∠=++

== 25.70154.6j10

)4j1(1530 3o IV

Therefore, =)t(vo 6.154 cos(103 t + 70.25°) V Chapter 10, Solution 27.

For mesh 1,

020j)20j10j(3040- 21 =+−+°∠ II

21 2jj-304 II +=°∠ (1) For mesh 2,

020j)20j40(050 12 =+−+°∠ II

21 )2j4(2j-5 II −−= (2) From (1) and (2),

−

=

°∠

2

1

)2j4(-2j-2jj-

5304

II

°∠=+=∆ 56.116472.4j42-

°∠=−−°∠=∆ 8.21101.2110j)2j4)(304(-1

°∠=°∠+=∆ 27.15444.412085j-2

=∆∆

= 11I 4.698∠95.24° A

=∆∆

= 22I 0.9928∠37.71° A


25.0j4x1j

1Cj

1F1,4jLjH1 −==ω

→=ω→

The frequency-domain version of the circuit is shown below, where

o2

o1 3020V,010V −∠=∠= .

1 j4 j4 1 -j0.25 + + V1 I1 1 I2 V2 - -

o2

o1 3020V,010V −∠=∠=

Applying mesh analysis,

21 I)25.0j1(I)75.3j2(10 −−+= (1)

(2) 21

o I)75.3j2(I)025.0j1(3020 ++−−=−∠− From (1) and (2), we obtain

++−+−+

=

+− 2

1II

75.3j225.0j125.0j175.3j2

10j32.1710

Solving this leads to

o2

o1 1536505.4111.2j1438.4I,69.356747.19769.0j3602.1I ∠=+−=−∠=−=

Hence,

A )15346cos(651.4i A, )69.35t4cos(675.1i o2

o1 +=−=


For mesh 1,

02030)j2()5j5( 21 =°∠−+−+ II

21 )j2()5j5(2030 II +−+=°∠ (1) For mesh 2,

0)j2()6j3j5( 12 =+−−+ II

21 )3j5()j2(-0 II −++= (2) From (1) and (2),

+

++=

°∠

2

1

j3-5j)(2-j)(2-5j5

02030

II

°∠=+=∆ 21.948.376j37

°∠=°∠°∠=∆ 96.10-175)96.30-831.5)(2030(1 °∠=°∠°∠=∆ 56.4608.67)56.26356.2)(2030(2

=∆∆

= 11I 4.67∠-20.17° A

=∆∆

= 22I 1.79∠37.35° A

Chapter 10, Solution 30. Consider the circuit shown below.

For mesh 1,

I3

I1

I2

Io

j4 Ω

+ − 10∠0° V

2 Ω

3 Ω

1 Ω

-j2 Ω

321 34j)4j3(20100 III −−+=°∠ (1) For mesh 2,

321 2j)4j3(4j-0 III −++= (2) For mesh 3,

321 )2j5(23-0 III −+−= (3) Put (1), (2), and (3) into matrix form.

°∠=

++

00

20100

j2-52-3-j2-j43j4-3-j4-4j3

3

2

1

III

30j106j2-52-3-

j2-j43j4-3-j4-4j3

+=++

=∆

)26j8)(20100(j2-503-

j2-0j4-3-201004j3

2 +°∠=°∠+

=∆

)20j9)(20100(02-3-0j43j4-

20100j4-4j3

3 +°∠=+°∠+

=∆

30j106)6j1)(20100(23

23o +−°∠

=∆∆−∆

=−= III

=oI 5.521∠-76.34° A


Consider the network shown below.

80 Ω j60 Ω 20 ΩIo

-j40 ΩI1 I3-j40 Ω+ − 100∠120° V +

− I2 60∠-30° V

For loop 1,

040j)40j80(20100- 21 =+−+°∠ II

21 4j)j2(42010 II +−=°∠ (1) For loop 2,

040j)80j60j(40j 321 =+−+ III

321 220 III +−= (2)

For loop 3, 040j)40j20(30-60 23 =+−+°∠ II

32 )2j1(24j30-6- II −+=°∠ (3) From (2),

123 22 III −= Substituting this equation into (3),

21 )2j1()2j1(2-30-6- II ++−=°∠ (4) From (1) and (4),

+−

−=

°∠°∠

2

1

2j1)2j1(2-4j)j2(4

30-6-12010

II

°∠=+=++

−=∆ 3274.3720j32

2j14j2-4j-4j8

°∠=+=°∠+°∠−

=∆ 44.9325.8211.82j928.4-30-6-4j2-

120104j82

=∆∆

== 22o II 2.179∠61.44° A



j4 Ω

3 Vo -j2 Ω

Io

I1 I2 + −

+

Vo

−

2 Ω4∠-30° V

For mesh 1,

03)30-4(2)4j2( o1 =+°∠−+ VI where )30-4(2 1o IV −°∠= Hence,

0)30-4(630-8)4j2( 11 =−°∠+°∠−+ II

1)j1(30-4 I−=°∠ or °∠= 15221I

)30-4)(2(2j-

32j-

31

oo I

VI −°∠==

)152230-4(3jo °∠−°∠=I =oI 8.485∠15° A

==32j- o

o

IV 5.657∠-75° V



5 A

I1 + −

2 Ω

I

-j2 Ω 2 I I2

I4

I3

j Ω

-j20 V 4 Ω

For mesh 1,

02j)2j2(20j 21 =+−+ II 10j-j)j1( 21 =+− II (1)

For the supermesh,

0j42j)2jj( 4312 =−++− IIII (2) Also,

)(22 2123 IIIII −==−

213 2 III −= (3) For mesh 4,

54 =I (4) Substituting (3) and (4) into (2),

5j)j4-()2j8( 21 =+−+ II (5) Putting (1) and (5) in matrix form,

=

−+

−5j10j-

j42j8jj1

2

1

II

5j3- −=∆ , 40j5-1 +=∆ , 85j15-2 +=∆

=−−

=∆∆−∆

=−=5j3-

45j102121 III 7.906∠43.49° A



-j2 Ω

j4 Ω

8 Ω

10 Ω

3 A I2

I3

I1+ − 40∠90° V

5 Ω

Io

20 Ω

j15 Ω

For mesh 1,

0)4j10()2j8()2j18(40j- 321 =+−−−++ III (1) For the supermesh,

0)2j18()19j30()2j13( 132 =+−++− III (2) Also,

332 −= II (3) Adding (1) and (2) and incorporating (3),

0)15j20()3(540j- 33 =++−+ II

°∠=++

= 48.38465.13j58j3

3I

== 3o II 1.465∠38.48° A



4 Ω j2 Ω

1 Ω

I1

I3

I2+ −

-j3 Ω8 Ω

-j4 A

10 Ω

20 V

-j5 Ω

For the supermesh,

0)3j9()8j11(820- 321 =−−−++ III (1) Also,

4j21 += II (2) For mesh 3,

0)3j1(8)j13( 213 =−−−− III (3) Substituting (2) into (1),

32j20)3j9()8j19( 32 −=−−− II (4) Substituting (2) into (3),

32j)j13()3j9(- 32 =−+− II (5) From (4) and (5),

−=

−−−−

32j32j20

j13)3j9(-)3j9(-8j19

3

2

II

69j167 −=∆ , 148j3242 −=∆

°∠°∠

=−−

=∆∆

=45.22-69.18055.24-2.356

69j167148j3242

2I

=2I 1.971∠-2.1° A



j4 Ω -j3 Ω

I1 I2

I3

+

Vo

−

2 Ω2 Ω

+ − 4∠90° A 2 Ω 12∠0° V

2∠0° A

Clearly, 4j9041 =°∠=I and 2-3 =I

For mesh 2, 01222)3j4( 312 =+−−− III

01248j)3j4( 2 =++−− I

64.0j52.3-3j48j16-

2 −=−+

=I

Thus, 28.9j04.7)64.4j52.3)(2()(2 21o +=+=−= IIV

=oV 11.648∠52.82° V Chapter 10, Solution 37. I1 + Ix 120 V Z o90−∠ - I2 Z=80-j35Ω Iz - Iy

o30120 −∠ V Z + I3 For mesh x,

120jZIZI zx −=− (1) For mesh y,

60j92.10330120ZIZI ozy +−=∠−=− (2)

For mesh z,

0ZI3ZIZI zyx =+−− (3) Putting (1) to (3) together leads to the following matrix equation:

BAI0

60j92.103120j

III

)105j240()35j80()35j80()35j80()35j80(0)35j80(0)35j80(

z

y

x=→

+−

−=

−+−+−+−−+−−

Using MATLAB, we obtain

+

+==

j0.15251.3657-j0.954- 2.1806-j1.4115 1.9165-

B*inv(A)I

A 6.1433802.24115.1j9165.1II ox1 ∠=+−==

A 37.963802.23655.2j2641.0III oxy2 −∠=−−=−=

A 63.233802.2954.0j1806.2II oy3 ∠=+=−=



Io

I1

j2 Ω

1 Ω

2 Ω-j4 Ω

10∠90° V

1 Ω

I2

I3 I44∠0° A

+ − 2∠0° A

A

Clearly, 21 =I (1)

For mesh 2, 090104j2)4j2( 412 =°∠++−− III (2)

Substitute (1) into (2) to get 5j22j)2j1( 42 −=+− II

For the supermesh, 04j)4j1(2j)2j1( 2413 =+−+−+ IIII

4j)4j1()2j1(4j 432 =−+++ III (3) At node A,

443 −= II (4) Substituting (4) into (3) gives

)3j1(2)j1(2j 42 +=−+ II (5) From (2) and (5),

+−

=

−

−6j25j2

j12j2j2j1

4

2

II

3j3−=∆ , 11j91 −=∆

)j10-(31

3j3)11j9(--

- 12o +=

−−

=∆∆

== II

=oI 3.35∠174.3° A


For mesh 1,

o321 6412I15jI8I)15j28( ∠=+−− (1)

For mesh 2, 0I16jI)9j8(I8 321 =−−+− (2)

For mesh 3, 0I)j10(I16jI15j 321 =++− (3) In matrix form, (1) to (3) can be cast as

BAIor006412

III

)j10(16j15j16j)9j8(8

15j8)15j28( o

3

2

1=

∠=

+−−−−

−−

Using MATLAB, I = inv(A)*B

A 6.1093814.03593.0j128.0I o1 ∠=+−=

A 4.1243443.02841.0j1946.0I o2 ∠=+−=

A 42.601455.01265.0j0718.0I o3 −∠=−=

A 5.481005.00752.0j0666.0III o21x ∠=+=−=


Let i , where i is due to the dc source and is due to the ac source. For

, consider the circuit in Fig. (a). 2O1OO ii += 1O 2Oi

1Oi

4 Ω 2 Ω

iO1 + −

8 V

(a)Clearly,

A428i 1O == For , consider the circuit in Fig. (b). 2Oi

4 Ω 2 Ω

10∠0° V j4 Ω

IO2

+ −

(b)If we transform the voltage source, we have the circuit in Fig. (c), where Ω= 342||4 .

2 Ω2.5∠0° A 4 Ω

IO2

j4 Ω

(c) By the current division principle,

)05.2(4j34

342O °∠

+=I

°∠=−= 56.71-79.075.0j25.02OI

Thus, A)56.71t4cos(79.0i 2O °−= Therefore,

=+= 2O1OO iii 4 + 0.79 cos(4t – 71.56°) A

Chapter 10, Solution 41. Let vx = v1 + v2. For v1 we let the DC source equal zero. 5 Ω 1 Ω

j100V)j55j1(tosimplifieswhich01

Vj

V5

20V1

111 =+−=+−

+−

V1 = 2.56∠–39.8˚ or v1 = 2.56sin(500t – 39.8˚) V

Setting the AC signal to zero produces:

+ – 20∠0˚ –j

+

V1

−

1 Ω

+ – 6 V

+

V2

−

5 Ω

The 1-ohm resistor in series with the 5-ohm resistor creating a simple voltage divider yielding: v2 = (5/6)6 = 5 V.

vx = 2.56sin(500t – 39.8˚) + 5 V.

Chapter 10, Solution 42. Let ix = i1 + i2, where i1 and i2 which are generated by is and vs respectively. For i1 we let is = 6sin2t A becomes Is = 6∠0˚, where ω =2.

°−∠=−=−−

=−++

−= 63.41983.431.3j724.3

2j52j1126

4j22j34j2I1

i1= 4.983sin(2t – 41.63˚) A –j4 2 Ω

j2

3 Ω

i1 is For i2, we transform vs = 12cos(4t – 30˚) into the frequency domain and get Vs = 12∠–30˚.

Thus, °∠=++−°−∠

= 2.8385.54j32j2

30122I or i2 = 5.385cos(4t + 8.2˚) A

–j2 2 Ω

+ − Vs

j4

3 Ω

i2

ix = [5.385cos(4t + 8.2˚) + 4.983sin(2t – 41.63˚)] A.


Let i , where i is due to the dc source and is due to the ac source. For , consider the circuit in Fig. (a).

2O1OO ii += 1O 2Oi

1Oi

4 Ω 2 Ω

iO1 + −

8 V

(a)Clearly,

A428i 1O == For , consider the circuit in Fig. (b). 2Oi

4 Ω 2 Ω

10∠0° V j4 Ω

IO2

+ −

(b)If we transform the voltage source, we have the circuit in Fig. (c), where Ω= 342||4 .

2 Ω2.5∠0° A 4 Ω

IO2

j4 Ω

(c) By the current division principle,

)05.2(4j34

342O °∠

+=I

°∠=−= 56.71-79.075.0j25.02OI Thus, A)56.71t4cos(79.0i 2O °−= Therefore,

=+= 2O1OO iii 4 + 0.79 cos (89)(4t – 71.56°) A

Chapter 10, Solution 44. Let v , where v21x vv += 1 and v2 are due to the current source and voltage source respectively.

For v1 , , 6=ω 30jLjH 5 =ω→ The frequency-domain circuit is shown below. 20Ω j30 + 16Ω V1 Is -

Let o5.1631.12497.3j8.1130j36

)30j20(16)30j20//(16 ∠=+=+

Z +=+=

V )5.26t6cos(7.147v5.267.147)5.1631.12)(1012(ZIV o1

ooos1 +=→∠=∠∠==

For v2 , , 2=ω 10jLjH 5 =ω→ The frequency-domain circuit is shown below. 20Ω j10 + 16Ω V2 + Vs

- - -


V )52.15t2sin(41.21v52.1541.2110j36

)050(16V10j2016

16V o2

oo

s2 −=→−∠=+∠

=++

=

Thus,

V )52.15t2sin(41.21)5.26t6cos(7.147v oox −++=


Let I , where I is due to the voltage source and is due to the current source. For I , consider the circuit in Fig. (a).

21o II +=

1

1 2I

10 Ω IT

+ − 20∠-150° V -j5 Ωj10 Ω

I1

(a)

10j-5j-||10j =

j1150-2

10j10150-20

T −°∠

=−

°∠=I


°∠+=−

°∠⋅=

−= 150-)j1(-

j1150-2

5j5j-

5j10j5j-

T1 II

For , consider the circuit in Fig. (b). 2I

-j5 Ωj10 Ω

I2

4∠-45° A 10 Ω

(b)

j210j-

5j-||10−

=


°∠+=°∠+−−

= 45-)j1(2-)45-4(10j)j2(10j-

)j2(10j-2I

°∠−°∠=+= 022105-2-21o III °∠=+= 98.150816.2366.1j462.2-oI

Therefore, =oi 2.816 cos(10t + 150.98°) A Chapter 10, Solution 46.

Let v , where , , and are respectively due to the 10-V dc source, the ac current source, and the ac voltage source. For consider the circuit in Fig. (a).

321o vvv ++= 1v 2v 3v

1v

2 H 6 Ω

1/12 F + −

+

v1

−

10 V

(a) The capacitor is open to dc, while the inductor is a short circuit. Hence,

V10v1 = For , consider the circuit in Fig. (b). 2v

2=ω 4jLjH2 =ω→

6j-)12/1)(2(j

1Cj

1F

121

==ω

→

4∠0° A+

V2

−

-j6 Ω 6 Ω j4 Ω

(b)


2222

4j

6j

61

4j6j-64 V

VVV

−+=++=

°∠=−

= 56.2645.215.0j1

242V

Hence, V)56.26t2sin(45.21v2 °+= For , consider the circuit in Fig. (c). 3v

3=ω 6jLjH2 =ω→

4j-)12/1)(3(j

1Cj

1F

121

==ω

→

6 Ω j6 Ω

+

V3

−

-j4 Ω+ − 12∠0° V

(c)


6j4j-612 333 VVV

+=−

°∠=+

= 56.26-73.105.0j1

123V

Hence, V)56.26t3cos(73.10v3 °−= Therefore, =ov 10 + 21.45 sin(2t + 26.56°) + 10.73 cos(3t – 26.56°) V Chapter 10, Solution 47.

Let i , where i , i , and are respectively due to the 24-V dc source, the ac voltage source, and the ac current source. For , consider the circuit in Fig. (a).

321o iii ++= 1 2 3i

1i

− +

2 Ω

1 Ω 1/6 F 2 H 24 V

i1

4 Ω

(a) Since the capacitor is an open circuit to dc,

A424

24i1 =

+=

For , consider the circuit in Fig. (b). 2i

1=ω 2jLjH2 =ω→

6j-Cj

1F

61

=ω

→

1 Ω j2 Ω-j6 Ω

2 Ω I2I1+ − 10∠-30° V

I2

4 Ω

(b)For mesh 1,

02)6j3(30-10- 21 =−−+°∠ II

21 2)j21(330-10 II −−=°∠ (1)

For mesh 2,

21 )2j6(2-0 II ++=

21 )j3( II += (2)


215j1330-10 I−=°∠ °∠= 1.19504.02I

Hence, A)1.19tsin(504.0i2 °+= For , consider the circuit in Fig. (c). 3i

3=ω 6jLjH2 =ω→

2j-)6/1)(3(j

1Cj

1F

61

==ω

→

1 Ω j6 Ω-j2 Ω

2∠0° A2 Ω

I3

4 Ω

(c)

2j3)2j1(2

)2j1(||2−−

=−


3j13)2j1(2

2j3)2j1(2

6j4

)02(2j3

)2j1(2

3 +−

=

−−

++

°∠⋅−−

=I

°∠= 43.76-3352.03I Hence A)43.76t3cos(3352.0i3 °−= Therefore, =oi 4 + 0.504 sin(t + 19.1°) + 0.3352 cos(3t – 76.43°) A Chapter 10, Solution 48.

Let i , where i is due to the ac voltage source, is due to the dc voltage source, and is due to the ac current source. For , consider the circuit in Fig. (a).

3O2O1OO iii ++=

3Oi1O 2Oi

1Oi

2000=ω °∠→ 050)t2000cos(50

80j)1040)(2000(jLjmH40 3- =×=ω→

25j-)1020)(2000(j

1Cj

1F20 6- =

×=

ω→µ

I IO1

50∠0° V 80 Ω+ −

-j25 Ω

100 Ω

j80 Ω 60 Ω(a)

3160)10060(||80 =+

33j3230

25j80j316050

+=

−+=I


°∠°∠

==+

=9.4546

1801031-

16080I80-

1O II

°∠= 1.134217.01OI Hence, A)1.134t2000cos(217.0i 1O °+= For , consider the circuit in Fig. (b). 2Oi

+ −

100 Ω

24 V

iO2

80 Ω

60 Ω

(b)

A1.01006080

24i 2O =

++=

For , consider the circuit in Fig. (c). 3Oi

4000=ω °∠→ 02)t4000cos(2

160j)1040)(4000(jLjmH40 3- =×=ω→

5.12j-)1020)(4000(j

1Cj

1F20 6- =

×=

ω→µ

-j12.5 Ω

80 Ω

60 Ω

I2

I1

I3

2∠0° A

j160 Ω

IO3

100 Ω

(c) For mesh 1,

21 =I (1) For mesh 2,

080160j)5.12j160j80( 312 =−−−+ III Simplifying and substituting (1) into this equation yields

32j8)75.14j8( 32 =−+ II (2) For mesh 3,

08060240 213 =−− III Simplifying and substituting (1) into this equation yields

5.13 32 −= II (3) Substituting (3) into (2) yields

125.54j12)25.44j16( 3 +=+ I

°∠=++

= 38.71782.125.44j16

125.54j123I

°∠== 38.71782.1-- 33O II

Hence, A)38.7t4000sin(1782.1-i 3O °+= Therefore, =Oi 0.1 + 0.217 cos(2000t + 134.1°) – 1.1782 sin(4000t + 7.38°) A Chapter 10, Solution 49.

200,308)30t200sin(8 =ω°∠→°+ j)105)(200(jLjmH5 3- =×=ω→

5j-)101)(200(j

1Cj

1mF1 3- =

×=

ω→

After transforming the current source, the circuit becomes that shown in the figure below.

5 Ω 3 Ω I

+ − 40∠30° V

j Ω

-j5 Ω

°∠=−

°∠=

−++°∠

= 56.56472.44j8

30405jj35

3040I

=i 4.472 sin(200t + 56.56°) A


55 10,050)t10cos(50 =ω°∠→ 40j)104.0)(10(jLjmH4.0 3-5 =×=ω→

50j-)102.0)(10(j

1Cj

1F2.0 6-5 =

×=

ω→µ

After transforming the voltage source, we get the circuit in Fig. (a).

j40 Ω

20 Ω+

Vo

−

-j50 Ω2.5∠0° A 80 Ω

(a)

Let 5j2

100j-50j-||20

−==Z

and 5j2

250j-)05.2(s −

=°∠= ZV

With these, the current source is transformed to obtain the circuit in Fig.(b).

Z j40 Ω

+

Vo

−

+ − 80 ΩVs

(b)


5j2250j-

40j805j2

100j-80

40j8080

so −⋅

++−

=++

= VZ

V

°∠=−

= 6.40-15.3642j36

)250j-(8oV

Therefore, =ov 36.15 cos(105 t – 40.6°) V


The original circuit with mesh currents and a node voltage labeled is shown below.

Io

40 Ωj10 Ω -j20 Ω4∠-60° V 1.25∠0° A

The following circuit is obtained by transforming the voltage sources.

Io

4∠-60° V -j20 Ωj10 Ω 40 Ω 1.25∠0° A

Use nodal analysis to find . xV

x401

20j-1

10j1

025.160-4 V

++=°∠+°∠

x)05.0j025.0(464.3j25.3 V−=−

°∠=+=−−

= 61.1697.8429.24j42.8105.0j025.0

464.3j25.3xV

Thus, from the original circuit,

10j)29.24j42.81()20j64.34(

10j3040 x

1

+−+=

−°∠=

VI

=+=−

= 678.4j429.0-10j

29.4j78.46-1I 4.698∠95.24° A

4029.24j42.31

40050x

2

+=

°∠−=

VI

=°∠=+= 7.379928.06072.0j7855.02I 0.9928∠37.7° A Chapter 10, Solution 52.

We transform the voltage source to a current source.

12j64j2060

s −=+

°∠=I

The new circuit is shown in Fig. (a).

-j2 Ω

6 Ω

2 ΩIs = 6 – j12 A

-j3 Ωj4 Ω

4 Ω

Ix

5∠90° A

(a)

Let 8.1j4.24j8

)4j2(6)4j2(||6s +=

++

=+=Z

)j2(1818j36)8.1j4.2)(12j6(sss −=−=+−== ZIV With these, we transform the current source on the left hand side of the circuit to a voltage source. We obtain the circuit in Fig. (b).

Zs -j2 Ω

4 Ω

-j3 Ω

+ −

Vs

Ix

j5 A

(b) Let )j12(2.02.0j4.22jso −=−=−= ZZ

207.6j517.15)j12(2.0

)j2(18

o

so −=

−−

==ZV

I

With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit in Fig. (c).

Zo

-j3 Ω

4 Ω

Ix

Io j5 A

(c)


)207.1j517.15(2.3j4.62.0j4.2

)5j(3j4 o

o

ox −

−−

=+−+

= IZ

ZI

=+= 5625.1j5xI 5.238∠17.35° A Chapter 10, Solution 53.

We transform the voltage source to a current source to obtain the circuit in Fig. (a).

4 Ω j2 Ω+

Vo

−

2 Ω

j4 Ω-j3 Ω

5∠0° A -j2 Ω

(a)

Let 6.1j8.02j4

8j2j||4s +=

+==Z

j4)6.1j8.0)(5()05( ss +=+=°∠= ZV 8 With these, the current source is transformed so that the circuit becomes that shown in Fig. (b).

Zs -j3 Ω j4 Ω

+

Vo

−

+ −

Vs 2 Ω -j2 Ω

(b)Let 4.1j8.03jsx −=−= ZZ

6154.4j0769.34.1j8.0

8j4

s

sx +−=

−+

==ZVI

With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).

j4 Ω

+

Vo

−

Zx -j2 Ω2 ΩIx

(c)

Let 5714.0j8571.04.1j8.28.2j6.1

||2 xy −=−−

== ZZ

7143.5j)5714.0j8571.0()6154.4j0769.3(yxy =−⋅+−== ZIV With these, we transform the current source to obtain the circuit in Fig. (d).

Zy j4 Ω

+ − -j2 Ω

+

Vo

−

Vy

(d)


=−+−

=−+

=2j4j5714.0j8571.0

)7143.5j(2j-2j4j

2j-y

yo V

ZV (3.529 – j5.883) V


059.2224.133050

)30(50)30//(50 jjjxj −=

−−

=−

We convert the current source to voltage source and obtain the circuit below. 13.24 – j22.059Ω

40Ω j20Ω

+ + - 115.91 –j31.06V I V - + -

134.95-j74.912 V

Applying KVL gives -115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0

or 8055.17817.4059.224.53

97.10586.250 jjjI +−=

−+−

=

But I)20j40(VV0VI)20j40(V s +−=→=+++−

V 15406.124)8055.1j7817.4)(20j40(05.31j91.115V o−∠=+−+−−=

which agrees with the result in Prob. 10.7. Chapter 10, Solution 55.

(a) To find , consider the circuit in Fig. (a). thZ

j20 Ω 10 Ω

Zth-j10 Ω

(a)

10j20j)10j-)(20j(

10)10j-(||20j10thN −+=+== ZZ

=−= 20j10 22.36∠-63.43° Ω To find , consider the circuit in Fig. (b). thV

j20 Ω 10 Ω

+

Vth

−

+ − 50∠30° V -j10 Ω

(b)

=°∠−

= )3050(10j20j

10j-thV -50∠30° V

=°∠

°∠==

43.63-36.223050-

th

thN Z

VI 2.236∠273.4° A

(b) To find , consider the circuit in Fig. (c). thZ

-j5 Ω

j10 ΩZth

8 Ω

(c)

=−+−

=−==5j810j)5j8)(10j(

)5j8(||10jthN ZZ 10∠26° Ω

To obtain , consider the circuit in Fig. (d). thV

-j5 Ω

Io

4∠0° A

+

Vth

−

j10 Ω8 Ω

(d) By current division,

5j832

)04(5j10j8

8o +

=°∠−+

=I

=+

==5j8

320j10j oth IV 33.92∠58° V

=°∠°∠

==2610

5892.33

th

thN Z

VI 3.392∠32° A



j4 Ω

-j2 ΩZth

6 Ω

(a)

4j62j4j

)2j-)(4j(6)2j-(||4j6thN −=

−+=+== ZZ

= 7.211∠-33.69° Ω By placing short circuit at terminals a-b, we obtain,

=NI 2∠0° A

=°∠°∠== )02()69.33-211.7(ththth IZV 14.422∠-33.69° V

(b) To find , consider the circuit in Fig. (b). thZ

j10 Ω

-j5 Ω60 ΩZth

30 Ω

(b)

2060||30 =

5j20)10j20)(5j-(

)10j20(||5j-thN ++

=+== ZZ

= 5.423∠-77.47° Ω

To find and , we transform the voltage source and combine the 30 Ω and 60 Ω resistors. The result is shown in Fig. (c).

thV NI

a

4∠45° A -j5 Ω

j10 Ω

20 ΩIN

b (c)

)454)(j2(52

)454(10j20

20N °∠−=°∠

+=I

= 3.578∠18.43° A

)43.18578.3()47.77-423.5(Nthth °∠°∠== IZV = 19.4∠-59° V


To find , consider the circuit in Fig. (a). thZ

5 Ω -j10 Ω 2 Ω

j20 ΩZth

(a)

10j5)10j5)(20j(

2)10j5(||20j2thN +−

+=−+== ZZ

=− 12j18= 21.633∠-33.7° Ω To find , consider the circuit in Fig. (b). thV

5 Ω -j10 Ω 2 Ω

+

Vth

−

+ − j20 Ω60∠120° V

(b)

)12060(2j1

4j)12060(

20j10j520j

th °∠+

=°∠+−

=V

= 107.3∠146.56° V

=°∠°∠

==7.33-633.21

56.1463.107

th

thN Z

VI 4.961∠-179.7° A


Consider the circuit in Fig. (a) to find . thZ

Zth

-j6 Ω

j10 Ω8 Ω

(a)

)j2(54j8

)6j8)(10j()6j8(||10jth +=

+−

=−=Z

= 11.18∠26.56° Ω Consider the circuit in Fig. (b) to find . thV

+

Vth

Io

5∠45° A

-j6 Ω

j10 Ω8 Ω

(b)

)455(2j43j4

)455(10j6j8

6j8o °∠

+−

=°∠+−

−=I

=+

°∠−==

)j2)(2()455)(3j4)(10j(

10j oth IV 55.9∠71.56° V


The frequency-domain equivalent circuit is shown in Fig. (a). Our goal is to find and

across the terminals of the capacitor as shown in Figs. (b) and (c). thV

thZ

3 Ω j Ω 3 Ω j Ω

Zth

b

a+

Vo

−

-j Ω+ − 5∠-60° A10∠-45° V +

−

(b) (a)

3 Ω j Ω Zth

+

Vth

−

10∠-45° V + −

+ −

+

Vo

−b

a

(d)

+ −

Vth5∠-60° A -j Ω

(c) From Fig. (b),

)3j1(103

j33j

j||3th +=+

==Z

From Fig.(c),

0j

60-53

45-10 thth =−°∠

+−°∠ VV

3j1301545-10

th −°∠−°∠

=V

From Fig. (d),

°∠−°∠=−

= 301545-10j

j-th

tho V

ZV

=oV 15.73∠247.9° V Therefore, =ov 15.73 cos(t + 247.9°) V Chapter 10, Solution 60.


10 Ω -j4 Ω

b

a

j5 Ω 4 Ω

(a)

Zth

)4j24j-(||4)5j||104j-(||4th ++=+=Z

== 2||4thZ 1.333 Ω To find , consider the circuit in Fig. (b). thV

+

Vth

−

j5 Ω 4 Ω

V2V1

+ − 4∠0° A20∠0° V

10 Ω -j4 Ω

(b) At node 1,

4j-5j1020 2111 VVVV −

+=−

205.2j)5.0j1( 21 =−+ VV (1)

At node 2,

44j-4 221 VVV

=−

+

16j)j1( 21 +−= VV (2)

Substituting (2) into (1) leads to

2)3j5.1(16j28 V−=−

333.5j83j5.1

16j282 +=

−−

=V

Therefore,

== 2th VV 9.615∠33.69° V

(b) To find , consider the circuit in Fig. (c). thZ

c dZth

j5 Ω 4 Ω

10 Ω -j4 Ω

(c)

+

+=+=j2

10j4||4j-)5j||104(||4j-thZ

=+=+= )4j6(64j-

)4j6(||4j-thZ 2.667 – j4 Ω

To find ,we will make use of the result in part (a). thV

)2j3()38(333.5j82 +=+=V )j5()38(16j16j)j1( 21 −+=+−= VV

=+=−= 8j31621th VVV 9.614∠56.31° V

Chapter 10, Solution 61. First, we need to find and across the 1 Ω resistor. thV thZ

4 Ω -j3 Ω j8 Ω 6 Ω

Zth

(a) From Fig. (a),

8.0j4.45j10

)8j6)(3j4()8j6(||)3j4(th −=

++−

=+−=Z

=thZ 4.472∠-10.3° Ω

4 Ω -j3 Ω j8 Ω 6 Ω

+

Vth

−

+ − 2 A -j16 V

(b) From Fig. (b),

8j62

3j416j- thth

+=+

−− VV

°∠=+−

= 45.43-93.204.0j22.056.2j92.3

thV

°∠°∠

=+

=43.8-46.5

45.43-93.201 th

tho Z

VV

°∠= 02.35-835.3oV Therefore, =ov 3.835 cos(4t – 35.02°) V


First, we transform the circuit to the frequency domain. 1,012)tcos(12 =ω°∠→

2jLjH2 =ω→

4j-Cj

1F

41

=ω

→

8j-Cj

1F

81

=ω

→


3 Io

21

Io 4 Ω

-j8 Ω

Vx

-j4 Ω + −

1 V

j2 Ω Ix

(a)At node 1,

2j1

34j-4

xo

xx VI

VV −=++ , where

4- x

o

VI =

Thus, 2j

14

24j-

xxx VVV −=−

8.0j4.0x +=V At node 2,

2j1

8j-1

3 xox

VII

−+=+

83

j)5.0j75.0( xx −+= VI

425.0j1.0-x +=I

Ω°∠=−== 24.103-29.2229.2j5246.0-1

xth I

Z

To find , consider the circuit in Fig. (b). thV

3 Io

V2

12∠0° V

Io 4 Ω

-j8 Ω

1 2 +

Vth

−

V1

-j4 Ω+ −

j2 Ω

(b)At node 1,

2j4j-3

412 211

o1 VVV

IV −

++=−

, where 4

12 1o

V−=I

21 2j)j2(24 VV −+= (1)

At node 2,

8j-3

2j2

o21 V

IVV

=+−

21 3j)4j6(72 VV −+= (2)

From (1) and (2),

++

=

2

1

3j-4j62j-j2

7224

VV

6j5- +=∆ , 24j-2 =∆

°∠=∆∆

== 8.219-073.322th VV

Thus,

229.2j4754.1)8.219-073.3)(2(

22

thth

o −°∠

=+

= VZ

V

°∠=°∠°∠

= 3.163-3.25.56-673.28.219-146.6

oV

Therefore, =ov 2.3 cos(t – 163.3°) V


Transform the circuit to the frequency domain.

200,304)30t200cos(4 =ω°∠→°+ Ω==ω→ k2j)10)(200(jLjH10

Ω=×

=ω

→µ kj-)105)(200(j

1Cj

1F5 6-

NZ is found using the circuit in Fig. (a).

-j kΩ

2 kΩZN

j2 kΩ

(a)

Ω=++=+= k1j1j-2j||2j-NZ We find using the circuit in Fig. (b). NI

-j kΩ

j2 kΩ 2 kΩ4∠30° A IN

(b)

j12||2j += By the current division principle,

°∠=°∠−+

+= 75657.5)304(

jj1j1

NI

Therefore,

=Ni 5.657 cos(200t + 75°) A =NZ 1 kΩ


is obtained from the circuit in Fig. (a). NZ

ZN60 Ω 40 Ω

-j30 Ωj80 Ω

(a)

50j100)50j)(100(

50j||100)30j80j(||)4060(N +==−+=Z

=+= 40j20NZ 44.72∠63.43° Ω

To find , consider the circuit in Fig. (b). NI

IN

I2

I1

Is3∠60° A

j80 Ω

60 Ω 40 Ω

-j30 Ω

(b)

°∠= 603sI For mesh 1,

060100 s1 =− II °∠= 608.11I

For mesh 2,

080j)30j80j( s2 =−− II °∠= 608.42I

=−= 21N III 3∠60° A


2,05)t2cos(5 =ω°∠→ 8j)4)(2(jLjH4 ==ω→

2j-)4/1)(2(j

1Cj

1F

41

==ω

→

j-)2/1)(2(j

1Cj

1F

21

==ω

→

To find , consider the circuit in Fig. (a). NZ

2 Ω

ZN

-j2 Ω -j Ω

(a)

)10j2(131

3j2)2j2(j-

)2j2(||j-N −=−−

=−=Z

To find , consider the circuit in Fig. (b). NI

2 Ω+ −

IN-j2 Ω

5∠0° V

-j Ω

(b)

5jj-05

N =°∠

=I

The Norton equivalent of the circuit is shown in Fig. (c).

Io

ZN IN j8 Ω

(c)


94j210j50

8j)10j2)(131()5j)(10j2)(131(

8j NN

No +

+=

+−−

=+

= IZ

ZI

°∠=−= 47.77-05425294.0j1176.0oI Therefore, =oi 0.542 cos(2t – 77.47°) A


ω 10=

5j)5.0)(10(jLjH5.0 ==ω→

10j-)1010)(10(j

1Cj

1mF10 3- =

×=

ω→

Vx

j5 Ω 2 Vo +

Vo

−

-j10 Ω

1 A 10 Ω

(a)


10j105j21 xx

o −+=+

VVV , where

10j10 x

o −V

10=V

2j2110j10-

5j10j1019

1 xxx

++

=→=−

+ VVV

=°∠°∠

===44.5095.21

135142.141

xthN

VZZ 0.67∠129.56° Ω

To find and , consider the circuit in Fig. (b). thV NI

2 Vo I

− ++

Vth

−

+

Vo

−

-j10 Ω

j5 Ω10 Ω

12∠0° V

-j2 A

(b)

012)2(5j)2j-)(10()5j10j10( o =−+−+− VI where )2j-)(10(o IV −= Thus,

20j188-)105j10( −=− I

105j10-20j188

++

=I

200105j)40j21(5j)2(5j oth −=+=+= IIVIV

076.2j802.11-200105j10-

)20j188(105jth +=−

++

=V

=thV 11.97∠170° V

=°∠°∠

==56.12967.0

17097.11

th

thN Z

VI 17.86∠40.44° A


Ω+=++

+−−

=++−== 079.1j243.116j20

)6j8(125j23

)5j13(10)6j8//(12)5j13//(10ZZ ThN

Ω+=∠++

=+=∠−

= 37.454j93.25)4560(6j20)6j8(V,44.21j78.13)4560(

5j2310V o

bo

a

A 09.9734.38ZV

I V, 599.11.433VVV o

Th

ThN

obaTh −∠==−∠=−=


10j1x10jLj H1 ==ω→

2j

201x10j

1Cj

1 F201

−==ω

→

We obtain VTh using the circuit below.

Io 4 Ω a + + + -j2 j10 Vo 6<0o Vo/3 - 4Io - - b

5.2j2j10j)2j(10j)2j//(10j −=

−−

=−

ooo I10j)5.2j(xI4V −=−= (1)

0V31I46 oo =++− (2)

Combining (1) and (2) gives

oooTho 19.5052.11

3/10j460jI10jVV,

3/10j46I −∠=

−−

=−==−

=

)19.50t10sin(52.11v o

Th −= To find RTh, we insert a 1-A source at terminals a-b, as shown below. Io 4 Ω a + + -j2 j10 Vo Vo/3 - 4Io -

1<0o

12V

I0V31I4 o

ooo −=→=+

10jV

2jV

I41 ooo +

−=+

Combining the two equations leads to

4766.1j2293.14.0j333.0

1Vo −=+

=

Ω−== 477.12293.11

VZ o

Th


This is an inverting op amp so that

=ω

==Cj1

R--

i

f

s

o

ZZ

VV

-jωRC

When and ms V=V RC1=ω ,

°∠==⋅⋅⋅= 90-VVj-VRCRC1

j- mmmoV

Therefore,

=°−ω= )90tsin(V)t(v mo - Vm cos(ωt) Chapter 10, Solution 70.

This may also be regarded as an inverting amplifier.

44 104,02)t104cos(2 ×=ω°∠→×

Ω=××

=ω

→ k5.2j-)1010)(104(j

1Cj

1nF10 9-4

i

f

s

o -ZZ

VV

=

where and Ω= k50iZ Ω−

== kj40

100j-)k5.2j-(||k100fZ .

Thus, j40

2j-

s

o

−=

VV

If , °∠= 02sV

°∠=°∠

°∠=

−= 57.88-1.0

43.1-01.4090-4

j404j-

oV

Therefore,

=)t(vo 0.1 cos(4x104 t – 88.57°) V


oo 308)30t2cos(8 ∠→+

0. Ω−==ω

→µ−

k1j10x5.0x2j

1Cj

1F5 6

At the inverting terminal,

)j6.0(308)j1.0(Vk2308

k10308V

k1j308V

ooo

oo

o +∠=+→∠

=∠−

+−

∠−

o

o 747.4283.9j1.0

)j6.0)(4j9282.6(V ∠=+

++=

vo(t) = 9.283cos(2t + 4.75o) V


44 10,04)t10cos(4 =ω°∠→

Ω==ω

→ k100j-)10)(10(j

1Cj

1nF1 9-4

Consider the circuit as shown below.

4∠0° V

Vo Vo

-j100 kΩ

50 kΩ

+ −

+− Io

100 kΩ

At the noninverting node,

5.0j14

100j-504

ooo

+=→=

−V

VV

A56.26-78.35mA)5.0j1)(100(

4k100

oo µ°∠=

+==

VI

Therefore,

=)t(io 35.78 cos(104 t – 26.56°) µA


As a voltage follower, o2 VV =

Ω=××

=ω

→= k-j20)1010)(105(j

1Cj1

nF10C 9-31

1

Ω=××

=ω

→= k-j10)1020)(105(j

1Cj1

nF20C 9-32

2

Consider the circuit in the frequency domain as shown below.

-j20 kΩ

Zin

Io

V1

V2Is

-j10 kΩ

20 kΩ10 kΩ

+ −

+− Vo

VS

At node 1,

2020j-10o1o11s VVVVVV −

+−

=−

o1s )j1()j3(2 VVV +−+= (1) At node 2,

10j-0

20oo1 −

=− VVV

o1 )2j1( VV += (2) Substituting (2) into (1) gives

os 6j2 VV = or so 31

-j VV =

so1 31

j32

)2j1( VVV

−=+=

s1s

s k10)j1)(31(

k10V

VVI

−=

−=

k30j1

s

s −=

VI

k)j1(15j1

k30

s

sin +=

−==

IV

Z

=inZ 21.21∠45° kΩ Chapter 10, Solution 74.

11i Cj

1R

ω+=Z ,

22f Cj

1R

ω+=Z

=

ω+

ω+

===

11

22

i

f

s

ov

Cj1

R

Cj1

R-ZZ

VV

A

ω+ω+

11

22

2

1

CRj1CRj1

CC

At , 0=ω =vA2

1

CC

As ω , ∞→ =vA1

2

RR

At 11CR

1=ω , =vA

+

+

j1CRCRj1

CC 1122

2

1

At 22CR

1=ω , =vA

++

22112

1

CRCRj1j1

CC


3102×=ω

Ω=××

=ω

→== k-j500)101)(102(j

1Cj1

nF1CC 9-31

21


100 kΩ

-j500 kΩ

+

Vo

−

-j500 kΩ

20 kΩ

V1

V2

100 kΩ20 kΩ

+ −

+−

VS

At node 1,

500j-100500j-211o1s VVVVVV −

+−

=−

2o1s 5j)5j2( VVVV −−+= (1) At node 2,

100500j-221 VVV

=−

21 )5j1( VV −= (2) But

2RRR o

o43

32

VVV =

+= (3)

From (2) and (3),

o1 )5j1(21

VV −⋅= (4)

Substituting (3) and (4) into (1),

ooos 21

5j)5j1)(5j2(21

VVVV −−−+⋅=

os )25j26(21

VV −⋅=

=−

=25j26

2

s

o

VV

0.0554∠43.88°


Let the voltage between the -jkΩ capacitor and the 10kΩ resistor be V1.

o1o

o1o11o

V6.0jV)6.0j1(302

k20VV

k10VV

k4jV302

+−=∠

→−

+−

=−

−∠ (1)

Also,

o1oo1 V)5j1(Vk2j

Vk10VV

+=→−

=−

(2)

Solving (2) into (1) yields

V 34.813123.03088.0j047.0V oo −∠=−=



1

2

+ −

VS

+

Vo

−

C1

C2 R2 R1

V1

V1

R3

−+

At node 1,

11

1s CjR

VVV

ω=−

111s )CRj1( VV ω+= (1) At node 2,

)(CjRR

0o12

2

o1

3

1 VVVVV

−ω+−

=−

ω+−= 32

2

31o1 RCj

RR

)( VVV

13223

o RCj)RR(1

1 VV

ω++= (2)

From (1) and (2),

ω++

ω+=

3223

2

11

so RRCjR

R1

CRj1V

V

=s

o

VV

)RRCjR()CRj1(RRCjRR

322311

32232

ω+ω+ω++

Chapter 10, Solution 78. 400,02)t400sin(2 =ω°∠→

Ω=×

=ω

→µ k5j-)105.0)(400(j

1Cj

1F5.0 6-

Ω=×

=ω

→µ k10j-)1025.0)(400(j

1Cj

1F25.0 6-

Consider the circuit as shown below.

20 kΩ

At node 1,

10 kΩ2∠0° V

10 kΩ -j5 kΩ

20 kΩ

V1 V2

-j10 kΩ40 kΩ

+ −

+− Vo

205j-10j-102 o12111 VVVVVV −

+−

+=−

o21 4j)6j3(4 VVV −−+= (1) At node 2,

105j221 VVV

=−−

21 )5.0j1( VV −= (2) But

oo2 31

402020

VVV =+

= (3)

From (2) and (3),

o1 )5.0j1(31

VV −⋅= (4)

Substituting (3) and (4) into (1) gives

oooo 61

j134

j)5.0j1(31

)6j3(4 VVVV

−=−−−⋅⋅+=

°∠=−

= 46.9945.3j6

24oV

Therefore, =)t(vo 3.945 sin(400t + 9.46°) V


1000,05)t1000cos(5 =ω°∠→

Ω=×

=ω

→µ k10j-)101.0)(1000(j

1Cj

1F1.0 6-

Ω=×

=ω

→µ k5j-)102.0)(1000(j

1Cj

1F2.0 6-


20 kΩ

+ − Vs = 5∠0° V

V1

-j5 kΩ

-j10 kΩ 40 kΩ

+Vo−

− +

10 kΩ −+

Since each stage is an inverter, we apply ii

fo

-V

ZZ

V = to each stage.

1o 15j-40-

VV =

(1) and

s1 10)10j-(||20-

VV =

(2) From (1) and (2),

°∠

−

= 0510j20

)10-j)(20(-10

8j-oV

°∠=+= 56.2678.35)j2(16oV

Therefore, =)t(vo 35.78 cos(1000t + 26.56°) V Chapter 10, Solution 80. 4

1000,60-4)60t1000cos( =ω°∠→°−

Ω=×

=ω

→µ k10j-)101.0)(1000(j

1Cj

1F1.0 6-

Ω=×

=ω

→µ k5j-)102.0)(1000(j

1Cj

1F2.0 6-

The two stages are inverters so that

+°∠⋅=

10j5-

5020

)60-4(10j-

20oo VV

o52

2j-)60-4()2j(

2j- V⋅+°∠⋅⋅=

°∠=+ 60-4)5j1( oV

°∠=+

°∠= 31.71-922.3

5j160-4

oV

Therefore, =)t(vo 3.922 cos(1000t – 71.31°) V

Chapter 10, Solution 81. The schematic is shown below. The pseudocomponent IPRINT is inserted to print the value of Io in the output. We click Analysis/Setup/AC Sweep and set Total Pts. = 1, Start Freq = 0.1592, and End Freq = 0.1592. Since we assume that w = 1. The output file includes: FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E-01 1.465 E+00 7.959 E+01 Thus, Io = 1.465∠79.59o A

Chapter 10, Solution 82. The schematic is shown below. We insert PRINT to print Vo in the output file. For AC Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we print out the output file which includes: FREQ VM($N_0001) VP($N_0001) 1.592 E-01 7.684 E+00 5.019 E+01

which means that Vo = 7.684∠50.19o V


he schematic is shown below. The frequency is

C

T 15.1592

10002/f =π

=πω=

When the circuit is saved and simulated, we obtain from the output file

REQ VM(1) VP(1)

hus, vo = 6.611cos(1000t – 159.2o) V

F1.592E+02 6.611E+00 -1.592E+02 T


he schematic is shown below. We set PRINT to print Vo in the output file. In AC

FREQ VM($N_0003)

1.592 E-01 1.664 E+00 -1.646

amely, Vo = 1.664∠-146.4o V

C TSweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes: VP($N_0003) E+02 N


C

The schematic is shown below. We let =ω rad/s so that L=1H and C=1F.

When the circuit is saved and simulated, we obtain from the output file

FREQ VM(1) VP(1)

From this, we conclude that

5.167228.2Vo −∠= V

1

1.591E-01 2.228E+00 -1.675E+02


e insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V1, V2, and V3,

FREQ VM($N_0002)

1.592 E-01 6.000 E+01 3.000

FREQ VM($N_0003)

1.592 E-01 2.367 E+02 -8.483

Winto the output file. Assume that w = 1, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit, we obtain the output file which includes: VP($N_0002) E+01 VP($N_0003) E+01

FREQ VM($N_0001)

1.592 E-01 1.082 E+02 1.254

herefore,

V1 = 60∠30o V

VP($N_0001) E+02 T

V2 = 236.7∠-84.83o V V3 = 108.2∠125.4o V


he schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set otal Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After

VM($N_0004) VP($N_0004)

1.696

FREQ VM($N_0001) VP($N_0001)

-1.386

C TTsimulation, the output file includes: FREQ 1.592 E-01 1.591 E+01 E+02 1.592 E-01 5.172 E+00 E+02

FREQ VM($N_0003) VP($N_0003)

-1.524

ore,

∠ o V

1.592 E-01 2.270 E+00 E+02 Theref

V1 = 15.91 169.6 V2 = 5.172∠-138.6o V V3 = 2.27∠-152.4o V


ow. We insert IPRINT and PRINT to print Io and Vo in the utput file. Since w = 4, f = w/2π = 0.6366, we set Total Pts = 1, Start Freq = 0.6366, nd End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:

6.366 E-01 3.496 E+01 1.261

FREQ IM(V_PRINT2) IP _PRINT2)

6.366 E-01 8.912 E-01

C The schematic is shown beloa FREQ VM($N_0002) VP($N_0002) E+01 (V -8.870 E+01

Therefore, Vo = 34.96∠12.6o V, Io = 0.8912∠-88.7o A

vo = 34.96 cos(4t + 12.6o)V, io = 0.8912cos(4t - 88.7o )A

onsider the circuit below.

Chapter 10, Solution 89. C

At node 1,

in−

2

2

1

in

RVVV

=−0

in

R−

in1

22in R

R- VVV =+ (1)

At node 3,

Cj1R4in

3

in2

ω−

=− VVVV

V

C R1 R2 Vin

+

Iin

in

3

R3 R4

− +

−+

V1

2 4

3

2in4in CRj

-ω−

=+VV

VV (2)

rom (1) and (2),

F

in13

24in RCRj

R-- VVV

ω=+

1R

Thus,

in43

2

4

4inin RCRj

RR

VVV

Iω

=−

=

eq2

431

in

inin Lj

RRRCRj

ω=ω

==IV

Z

where 2

431eq R

CRRRL =


et

LRCj1

RC

1||R

ω+==Z

C

j4 ω

CjRCj1

Cj1

R3 ωω+

=ω

+=Z

onsider the circuit shown below.

R2Z4

+ Vo −Vi + −

Z3 R1

i21

2i

43

4o RR

RVV

ZZZ

V+

−+

=

21

2

i

o

RRR

CjRCj1

Cj1R

Cj1R

+−

ωω+

+ω+

ω+=

VV

21

22 RR

R)RCj1(RCj

RCj+

−ω++ω

ω=

21

2222

i

o

RRR

RC3jCR1RCj

+−

ω+ω−ω

=V

V

For and to be in phase, oV iVi

o

VV

must be purely real. This happens when

0CR1 222 =ω−

f2RC1

π==ω

or RC21

fπ

=

At this freque cy, n

21

2

i

ov RR

R31

+−==

VV

A


(a) Let

=2V voltage at the noninverting terminal of the op amp output=oV voltage of the op amp

Rk10 =Ω op =Z1Cj

LjRs ω+ω+=Z

s in Section 10.9, A

Cj

LjRR

R

o

o

ps

p

o

2

ω−ω++

=+

=ZZ

ZVV

)1LC(j)RR(CCR

2o

o

o

2

−ω++ωω

=VV

For this to be purely real,

LC1

01LC o2o =ω→=−ω

)102)(104.0(21

LC21

f9-3-o

××π=

π=

=of 180 kHz

(b) At oscillation,

o

o

oo

oo

o

2

RRR

)RR(CCR

+=

+ωω

=VV

This must be compensated for by

52080

12

ov =+==

VV

A

==→=+ o

o

o R4R51

RRR

40 kΩ


Let voltage at the noninverting terminal of the op amp =2V=oV output voltage of the op amp

os R=Z

)1LC(jRLRL

Lj1

CjR1

1R||

Cj1

||Lj 2p −ω+ωω

=

ω+ω+

=ω

ω=Z

As in Section 10.9,

)1LC(jRLRL

R

)1LC(jRLRL

2o

2

ps

p

o

2

−ω+ωω

+

−ω+ωω

=+

=ZZ

ZVV

)1LC(RjRLRRLRL

2ooo

2

−ω+ω+ωω

=VV

For this to be purely real,

LC21

f1LC o2o π

=→=ω

(a) At , oω=ω

oooo

o

o

2

RRR

LRRLRL

+=

ω+ωω

=VV

This must be compensated for by

11k100k1000

1RR

1o

f

2

ov =+=+==

VV

A

Hence,

==→=+

R10R111

RRR

oo

100 kΩ

(b) )102)(1010(2

1f

9-6-o××π

=

=of 1.125 MHz Chapter 10, Solution 93. As shown below, the impedance of the feedback is

jωL

1Cj1ω2Cj

1ω

ZT

ω+ω

ω=

21T Cj

1Lj||

Cj1

Z

)CLCCC(j

LC1

Cj-

LjCj-

Cj-

LjCj-

212

21

2

21

21T ω−+

ω−ω=

ω+ω+

ω

ω+ω

ω=Z

In order for to be real, the imaginary term must be zero; i.e. TZ

0CLCCC 212o21 =ω−+

T21

212o LC

1CLCCC

=+

=ω

To LC2

1f

π=

Chapter 10, Solution 94. If we select C

nF20C21 ==

nF102

CCC

CCC 1

21

21T ==

+=

Since T

o LC21

fπ

= ,

mH13.10)1010)(102500)(4(

1C)f2(

1L 9-62

T2 =

××π=

π=

=××π

=ω

=)1020)(1050)(2(

1C1

X 9-32

c 159 Ω

We may select and , say Ω= k20R i if RR ≥ Ω= k20R f . Thus,

== 21 CC 20 nF, =L 10.13 mH == if RR 20 kΩ Chapter 10, Solution 95. First, we find the feedback impedance.

C

L1L2

ZT

ω

+ωω=Cj

1Lj||Lj 21TZ

)1)LL(C(j)L1(CL

Cj

LjLj

Cj

LjLj

212

212

21

21

T −+ωω−ω

=

ω−ω+ω

ω−ωω

=Z

In order for to be real, the imaginary term must be zero; i.e. TZ

01)LL(C 212o =−+ω

)LL(C1

f221

oo +=π=ω

)LL(C21

f21

o +π=


(a) Consider the feedback portion of the circuit, as shown below.

V2V1

+ −

R

R

jωL

Vo jωL

21 LjLjR

LjRLj

VVVV 12 ωω+

=→ω+

ω= (1)

Applying KCL at node 1,

LjRRLj111o

ω++=

ω− VVVV

ω+

+ω=−LjR

1R1

Lj 11o VVV

ω+ω−ω

+=)LjR(RLRL2j

122

1o VV

(2) From (1) and (2),

2

22

o )LjR(RLRL2j

1Lj

LjRVV

ω+ω−ω

+

ωω+

=

RLjLRL2jRLjR 222

2

o

ωω−ω+ω+

=VV

RLjLR

3

1222

o

2

ωω−

+=

VV

=o

2

VV

)LRRL(j31

ω−ω+

(b) Since the ratio o

2

VV

must be real,

0L

RR

L

o

o =ω

−ω

LR

Lo

2

o ω=ω

LR

f2 oo =π=ω

L2R

fo π=

(c) When oω=ω

31

o

2 =VV

This must be compensated for by 3v =A . But

3RR

11

2v =+=A

12 R2R =


)t50cos(160)t(v = )9018030t50cos(2)30t50sin(20-)t(i °−°+°−=°−=

)60t50cos(20)t(i °+=

)60t50cos()t50cos()20)(160()t(i)t(v)t(p °+== [ ] W)60cos()60t100cos(1600)t(p °+°+=

=)t(p W)60t100cos(1600800 °++

)60cos()20)(160(21

)cos(IV21

P ivmm °=θ−θ=

=P W800

Chapter 11, Solution 2. First, transform the circuit to the frequency domain.

°∠→ 030)t500cos(30 , 500=ω 150jLjH3.0 =ω→

100j-)10)(20)(500(

j-Cj

1F20 6- ==

ω→µ

I

I1

I2

+ − 30∠0° V j150 Ω

-j100 Ω

200 Ω

2.0j-902.0150j

0301 =°−∠=

°∠=I

)t500sin(2.0)90t500cos(2.0)t(i1 =°−=

06.0j12.056.261342.0j2

3.0100j200030

2 +=°∠=−

=−

°∠=I

)56.25t500cos(1342.0)t(i2 °+=

°∠=−=+= 49.4-1844.014.0j12.021 III )35t500cos(1844.0)t(i °−=

For the voltage source,

])35t500cos(1844.0[])t500cos(30[)t(i)t(v)t(p °−×== At , s2t = )351000cos()1000cos(532.5p °−=

)935.0)(5624.0)(532.5(p = =p W91.2

For the inductor,

])t500sin(2.0[])t500cos(30[)t(i)t(v)t(p ×== At , s2t = )1000sin()1000cos(6p =

)8269.0)(5624.0)(6(p = =p W79.2

For the capacitor,

°∠== 63.44-42.13)100j-(2c IV )56.25t500cos(1342.0[])44.63500cos(42.13[)t(i)t(v)t(p °+×°−==

At , s2t = )56.261000cos()44.631000cos(18p °+°−=

)1329.0)(991.0)(18(p = =p W37.2

For the resistor,

°∠== 56.2584.26200 2R IV ])56.26t500cos(1342.0[])56.26t500cos(84.26[)t(i)t(v)t(p °+×°+==

At , s2t = )56.251000(cos602.3p 2 °+=

21329.0)(602.3(p = =p W0636.0

Chapter 11, Solution 3. 10 , °∠→°+ 3010)30t2cos( 2=ω

2jLjH1 =ω→

-j2Cj

1F25.0 =

ω→

4 Ω 2 ΩI I1

I2

+ − 10∠30° V j2 Ω -j2 Ω

2j22

)2j2)(2j()2j2(||2j +=

−=−

°∠=++°∠

= 565.11581.12j24

3010I

°∠=== 565.101581.1j22j

1 III

°∠=−

= 565.56236.22

2j22 II

For the source,

)565.11-581.1)(3010(21* °∠°∠== IVS

5.2j5.718.43905.7 +=°∠=S

The average power supplied by the source = W5.7 For the 4-Ω resistor, the average power absorbed is

=== )4()581.1(21

R21

P 22I W5

For the inductor,

5j)2j()236.2(21

21 2

L

2

2 === ZIS

The average power absorbed by the inductor = W0

For the 2-Ω resistor, the average power absorbed is

=== )2()581.1(21

R21

P 22

1I W5.2

For the capacitor,

5.2j-)2j-()581.1(21

21 2

c

2

1 === ZIS

The average power absorbed by the capacitor = W0


20 Ω 10 Ω

I2I1+ − -j10 Ω50 V j5 Ω

For mesh 1,

21 10j)10j20(50 II +−=

21 j)j2(5 II +−= (1) For mesh 2,

12 10j)10j5j10(0 II +−+=

12 2j)j2(0 II +−= (2) In matrix form,

−

−=

2

1

j22jjj2

05

II

4j5−=∆ , )j2(51 −=∆ , -j102 =∆

°∠=−−

=∆∆

= 1.12746.14j5

)j2(511I

°∠==∆∆

= 66.128562.1j4-5

j10-22I

For the source,

°∠== 12.1-65.4321 *

1IVS

The average power supplied =°= )1.12cos(65.43 W68.42 For the 20-Ω resistor,

== R21

P2

1I W48.30

For the inductor and capacitor, =P W0

For the 10-Ω resistor,

== R21

P2

2I W2.12

Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1 Ω 2 Ω

+ − j6

–j2 8∠–40˚

W4159.112

6828.1P

38.256828.1

2j26j)2j2(6j1

408I

21

1

==

°−∠=

−+−

+

°−∠=

Ω

Ω

P3H = P0.25F = 0

W097.522

258.2P

258.238.256828.12j26j

6jI

22

2

==

=°−∠−+

=

Ω

Ω


20 Ω 10 Ω

I2I1+ − -j10 Ω50 V j5 Ω

For mesh 1,

04)604(2j)2j4( o1 =+°∠−+ VI (1) )604(2 2o IV −°∠= (2)

For mesh 2, 04)604(2)j2( o2 =−°∠−− VI (3)

Substituting (2) into (3), 0)604(8608)j2( 22 =−°∠−°∠−− II

j106040

2 −°∠

=I

Hence,

j10608j-

j106040

6042o −°∠

=

−

°∠−°∠=V

Substituting this into (1),

−−

°∠=−

°∠+°∠=+

j10j14

)608j(j10

6032j608j)2j4( 1I

°∠=+

+°∠= 125.06498.2

8j21)14j1)(604(

1I

=== )4()498.2(21

R21

P 22

14 I W48.12


20 Ω 10 Ω

I2I1+ − -j10 Ω50 V j5 Ω

Applying KVL to the left-hand side of the circuit, oo 1.04208 VI +=°∠ (1)

Applying KCL to the right side of the circuit,

05j105j

8 11o =

−++

VVI

But, o11o 105j10

5j1010

VVVV−

=→−

=

Hence, 01050j

5j108 o

oo =+−

+V

VI

oo 025.0j VI = (2)

Substituting (2) into (1), )j1(1.0208 o +=°∠ V

j12080

o +°∠

=V

°∠== 25-2

1010

o1

VI

=

== )10(2

10021

R21

P2

1I W250

Chapter 11, Solution 8. We apply nodal analysis to the following circuit.

At node 1,

Io V2V1

6∠0° A 0.5 Io j10 Ω

-j20 Ω

I2

40 Ω

20j-10j6 211 VVV −

+= 21 120j VV −= (1)

At node 2,

405.0 2

oo

VII =+

But, j20-

21o

VVI

−=

Hence, 40j20-

)(5.1 221 VVV=

−

21 )j3(3 VV −= (2)


0j33360j 222 =+−− VVV

j6)-1(37360

j6360j

2 +=−

=V

j6)-1(379

402

2 +==V

I

=

== )40(379

21

R21

P2

2

2I W78.43


rmsV8)2)(4(V26

1V so ==

+=

=== mW1064

RV

P2o

10 mW4.6

The current through the 2 -kΩ resistor is

mA1k2

Vs =

== RIP 2

2 mW2 Similarly,

== RIP 26 mW6


No current flows through each of the resistors. Hence, for each resistor, =P W0 .


, , 377=ω 410R = -910200C ×=754.0)10200)(10)(377(RC -94 =×=ω

°=ω 02.37)RC(tan -1

Ω°∠=°∠+

= k37.02-375.637.02-)754.0(1

k10Z

2ab

mA)68t377cos(2)22t377sin(2)t(i °−=°+=

°∠= 68-2I

3

2-3

ab2rms 10)37.02-375.6(

2102

ZIS ×°∠

×

==

mVA37.02-751.12S °∠=

== )02.37cos(SP mW181.10


(a) We find using the circuit in Fig. (a). ThZ

Zth

8 Ω -j2 Ω

(a)

882.1j471.0)4j1(178

j28(8)(-j2)

-j2||8Th −=−=−

==Z

== *

ThL ZZ Ω+ 882.1j471.0

We find using the circuit in Fig. (b). ThV

Io +

Vth

−

-j2 Ω 4∠0° A 8 Ω

(b)

)04(2j8

2j-o °∠

−=I

j28j64-

I8 oTh −==V

=

==)471.0)(8(

6864

R8P

2

L

2

Thmax

VW99.15

(b) We obtain from the circuit in Fig. (c). ThZ

5 Ω -j3 Ω

Zth

j2 Ω

4 Ω

(c)

167.1j5.23j9

)3j4)(5(2j)3j4(||52jTh +=

−−

+=−+=Z

== *

ThL ZZ Ω− 167.1j5.2


(a) We find at the load terminals using the circuit in Fig. (a). ThZ

j100 Ω

-j40 ΩZth

80 Ω

(a)

6.1j2.51j6080

j100)(-j40)(80j100)(80||-j40Th −=

++

=+=Z

== *

ThL ZZ Ω+ 6.1j2.51

(b) We find at the load terminals using Fig. (b). ThV

j100 ΩIo

+

Vth

−

3∠20° A -j40 Ω80 Ω

(b)

6j8)203)(8(

)203(40j100j80

80o +

°∠=°∠

−+=I

6j8)2024)(40j(-

40j- oTh +°∠

== IV

=

⋅==

)2.51)(8(

241040

R8P

2

L

2

Thmax

VW5.22

From Fig.(d), we obtain V using the voltage division principle. Th

5 Ω -j3 Ω

+

Vth

−

+ − 10∠30° V

j2 Ω

4 Ω

(d)

°∠

−−

=°∠

−−

= 303

10j33j4

)3010(3j93j4

ThV

=

⋅==

)5.2)(8(3

10105

R8P

2

L

2

Thmax

VW389.1


I

+

VTh

_

16 Ω

–j10 Ω

j8 Ω

j24 Ω

10 Ω

ZTh 40∠90º A

Ω+==

Ω−=++−=+++++

+−=

∗ 3.2j245.8ZZ

3.2j245.87.7j245.810j8j1624j10)8j16)(24j10(10jZ

Th

Th

W6.456245.8)245.8x2(

2

V

245.8IP

V12.158j53.7166.6555.173

)8j16(40j8j1624j10

10)8j16(IV

2

2

2Th

2rmsmax

Th

===

+=°∠=

++++

=+=

Chapter 11, Solution 15. To find Z , insert a 1-A current source at the load terminals as shown in Fig. (a). Th

+

Vo

−

2 1 1 Ω -j Ω

2 Vo j Ω 1 A

(a)At node 1,

2oo2oo j

j-j1VV

VVVV=→

−=+ (1)

At node 2,

o2o2

o )j2(j1j-

21 VVVV

V +−=→−

=+ (2)

Substituting (1) into (2), 222 )j1()j)(j2(j1 VVV −=+−=

j11

2 −=V

5.0j5.02

j11

2Th +=

+==

VV

== *

ThL ZZ Ω− 5.0j5.0

We now obtain from Fig. (b). ThV

1 Ω -j Ω

+

Vth

−

+ − 12∠0° V

+

Vo

−

2 Vo j Ω

(b)

j112

2 ooo

VVV =

−+

j112-

o +=V

0)2j-( Thoo =+×− VVV

j1)2j1)(12(

j2)-(1 oTh ++

=+= VV

=

==)5.0)(8(

2512

R8P

2

L

2

Thmax

VW90


520/14

11F20/1,4H1,4 jxjCj

jLj −==→=→=ω

ωω

We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit shown below. 0.5Vo 2Ω V1 4Ω V2 + + + 10<0o Vo -j5 j4 VTh

- - -

At node 1,

2121

111 25.0)2.01(5

425.0

5210 VjVVVV

jVV

−+=→−

++−

=−

(1)

At node 2,

)25.025.0(5.004

25.04 21

21

21 jVVjVVVV

+−+=→=+− (2)


2 6.1947 7.0796 9.4072 48.81oThV V j= = + = ∠ Chapter 11, Solution 17. We find ThR at terminals a-b following Fig. (a).

ba

j20 Ω

30 Ω

40 Ω

-j10 Ω (a)

10j40-j10))(40(

20j30)20j)(30(

)10j-(||4020j||30Th −+

+=+=Z

41.9j353.285.13j23.9Th −++=Z

Ω+= 44.4j583.11ThZ == *

ThL ZZ Ω− 44.4j583.11

We obtain from Fig. (b). ThV

I1 I2

+ VTh −

j20 Ω

30 Ω

40 Ω

-j10 Ω

j5 A

(b)


3.2j1.1-)5j(10j7020j30

1 +=++

=I

7.2j1.1)5j(10j7010j40

2 +=+−

=I

70j1010j30 12Th +=+= IIV

===)583.11)(8(

5000R8

PL

2

Th

max

VW96.53

Chapter 11, Solution 18. We find Z at terminals a-b as shown in the figure below. Th

a

b

Zth

40 Ω 80 Ω-j10 Ω

40 Ω

j20 Ω

j1080(80)(-j10)

20j20-j10)(||8040||4020jTh −++=++=Z

154.10j23.21Th +=Z

== *

ThL ZZ Ω− 15.10j23.21 Chapter 11, Solution 19. At the load terminals,

j9j)(6)(3

-j2)j3(||62j-Th ++

+=++=Z

561.1j049.2Th −=Z

Ω== 576.2R ThL Z

To get V , let Th 439.0j049.2)j3(||6 +=+=Z . By transforming the current sources, we obtain

756.1j196.8)04(Th +=°∠= ZV

===608.20258.70

R8P

L

2

Thmax

VW409.3

Chapter 11, Solution 20. Combine j20 Ω and -j10 Ω to get

-j20-j10||20j = To find , insert a 1-A current source at the terminals of , as shown in Fig. (a). ThZ LR

Io

-j20 Ω

4 Io

+ −V2V1

-j10 Ω

40 Ω

1 A

(a)At the supernode,

10j-20j-401 211 VVV

++=

21 4j)2j1(40 VV ++= (1)

Also, , where o21 4IVV +=40

- 1o

VI =

1.11.1 2

121

VVVV =→= (2)


22 4j1.1

)2j1(40 VV

+

+=

4.6j144

2 +=V

Ω−== 71.6j05.11

2Th

VZ

== ThLR Z Ω792.6

To find , consider the circuit in Fig. (b). ThV

+ − 120∠0° V

+

Vth

−

Io

-j20 Ω

4 Io

+ −V2V1

-j10 Ω

40 Ω

(b)At the supernode,

j10-j20-40120 211 VVV

+=−

21 4j)2j1(120 VV ++= (3)

Also, , where o21 4IVV +=40

120 1o

VI

−=

1.1122

1

+=

VV (4)


2)818.5j9091.0(82.21j09.109 V+=−

°∠=+−

== 92.43-893.18818.5j9091.082.21j09.109

2Th VV

===)792.6)(8(

)893.18(R8

P2

L

2

Thmax

VW569.6

Chapter 11, Solution 21. We find Z at terminals a-b, as shown in the figure below. Th

100 Ω -j10 Ω

40 Ω

b

a

Zth

j30 Ω50 Ω

])30j40(||10010j-[||50Th ++=Z

where 634.14j707.3130j140

)30j40)(100()30j40(||100 +=

++

=+

634.4j707.81)634.4j707.31)(50(

)634.4j707.31(||50Th ++

=+=Z

73.1j5.19Th +=Z

== ThLR Z Ω58.19


π<<= ttti 0,sin4)(

8)02

(1642sin

216sin161

00

22 =−=

−== ∫

ππππ

ππ tttdtI rms

A 828.28 ==rmsI


<<<<

=6t2,52t0,15

)t(v

[ ]6

550dt5dt15

61

V6

222

022

rms =+= ∫∫

=rmsV V574.9


, 2T =

<<<<

=2t15,-1t0,5

)t(v

[ ] 25]11[225

dt-5)(dt521

V2

121

022

rms =+=+= ∫∫

=rmsV V5


[ ]

266.33

32f

332]16016[

31

dt4dt0dt)4(31dt)t(f

T1f

rms

32

221

10

2T0

22rms

==

=++=

++−== ∫∫∫∫


, 4T =

<<<<

=4t2102t05

)t(v

[ ] 5.62]20050[41

dt)10(dt541

V4

222

022

rms =+=+= ∫∫

=rmsV V906.7

Chapter 11, Solution 27. , 5T = 5t0,t)t(i <<=

333.815

1253t

51

dtt51

I 50

35

022

rms ==⋅== ∫

=rmsI A887.2


[ ]∫∫ +=5

222

022

rms dt0dt)t4(51

V

533.8)8(1516

3t16

51

V 20

32rms ==⋅=

=rmsV V92.2

===2533.8

RV

P2rms W267.4


, 20T =

<<+<<−

=25t152t40-

15t5t220)t(i

[ ]∫∫ ++−=25

15215

522

eff dt2t)-40(dt)t220(201

I

+−++−= ∫∫

25

15

215

5

22eff dt400)t40t(dt)tt20100(

51I

+−+

+−= 2515

23

155

322

eff t400t203t

3t

t10t10051

I

332.33]33.8333.83[51

I2eff =+=

=effI A773.5

== RIP 2

eff W400 Chapter 11, Solution 30.

<<<<

=4t21-2t0t

)t(v

[ ] 1667.1238

41

dt-1)(dtt41

V4

222

022

rms =

+=+= ∫∫

=rmsV V08.1


6667.81634

21)4()2(

21)(

21 2

0

1

0

2

1

222 =

+=

−+== ∫ ∫ ∫ dtdttdttvV rms

V 944.2=rmsV


+= ∫∫

2

1

1

0

222rms dt0dt)t10(

21I

105t

50dtt50I 10

51

042

rms =⋅== ∫

=rmsI A162.3


<<<<−<<

=3t202t1t10201t010

)t(i

[ ]0dt)t1020(dt1031

I2

121

022

rms +−+= ∫∫

33.133)31)(100(100dt)tt44(100100I32

122

rms =+=+−+= ∫

==3

33.133Irms A667.6


[ ]

472.420f

20363t9

31

dt6dt)t3(31dt)t(f

T1f

rms

2

0

3

32

220

2T0

22rms

==

=

+=

+== ∫∫∫


[ ]∫∫∫∫∫ ++++=6

525

424

222

121

022

rms dt10dt20dt30dt20dt1061

V

67.466]1004001800400100[61

V2rms =++++=

=rmsV V6.21


(a) Irms = 10 A

(b) V 528.42916

234

222 =+=→

+= rmsrms VV (checked)

(c) A 055.92

3664 =+=rmsI

(d) V 528.42

16225

=+=rmsV


)302cos(6)10sin(48321oo ttiiii ++++=++=

A 487.9902

362

166432

22

12 ==++=++= rmsrmsrmsrms IIII


08.157j)5.0)(50)(2(jLjH5.0 =π=ω→

08.157j30jXR L +=+=Z

08.157j30)210( 2

*

2

−==

ZV

S

Apparent power ===160

)210( 2

S VA6.275

)19.79cos(36

08.157tancoscospf 1- °=

=θ=

=pf (lagging)1876.0


4j12)8j12)(4j(

)8j12(||4jT −−

=−=Z

°∠=+= 74.7456.4)11j3(4.0TZ

=°= )74.74cos(pf 2631.0

Chapter 11, Solution 40. At node 1,

)1(V4.0)6667.0j4.1(V60j92.103

50VV

30jV

20V30120

21

2111o

−−=+

→−

+=−∠

At node 2,

212221 )25.16(0401050

VjVjVVVV

++−=→−

+=−

(2)


V1 = 45.045 + j66.935, V2 = 9.423 + j9.193

(a) Ω−Ω == 4030 0 jj PP

W665.820/3.173||21 2

22

10 ====Ω RV

RVP rms

W6.034100/1.4603||21 2

2150 ==

−=Ω R

VVP

W86.8740/3514|30120|21 2

120 ==

−∠=Ω R

VPo

(b) 6092.10330120,3467.0944.22030120 1 jVjVI o

s

o

+=∠=−=−∠

=

VA 8.177||,3.1065.14221

==−== • SSjIVS s

(c ) pf = 142.5/177.8 = 0.8015 (leading). Chapter 11, Solution 41.

(a) -j6j

(-j2)(-j3)-j3||-j2)2j5j(||2j- ===−

°∠=−= 56.31-211.76j4TZ

=°= )-56.31cos(pf (leading)5547.0

(b) 52.1j64.03j4

)j4)(2j()j4(||2j +=

++

=+

°∠=++

=−+= 5.214793.044.0j64.144.0j64.0

)j52.1j64.0(||1Z

=°= )5.21cos(pf (lagging)9304.0


°=θ→θ== 683.30cos86.0pf

kVA798.9)683.30sin(

5sin

QSsinSQ =

°=

θ=→θ=

°∠=°∠×

==→= 683.30536.44220

683.3010798.9 3**

VS

IIVS

Peak current =× 536.442= A98.62 Apparent power = S = kVA798.9


(a) V 477.53021

29253

22

21

2 ==++=++= rmsrmsrmsrms VVVV

(b) W310/302

===R

VP rms


opf 46.49cos65.0 =→== θθ

kVA 385.32)7599.065.0(50)sin(cos jjjSS +=+=+= θθ

Thus, Average power = 32.5 kW, Reactive power = 38 kVAR


(a) V 9.4622002

60202

22 =→=+= rmsrms VV

AI rms 061.1125.125.01

22 ==+=

(b) W74.49== rmsrms IVP


(a) °∠=°∠°∠== 30-110)60-5.0)(30220(*IVS=S VA55j26.95 −

Apparent power =110 VA Real power = W26.95 Reactive power = VAR55 pf is leading because current leads voltage

(b) == *IVS °∠=°∠°∠ 151550)252.6)(10-250( =S VA2.401j2.1497 +

Apparent power =1550 VA Real power =1497 W2. Reactive power = VAR2.401 pf is lagging because current lags voltage

(c) °∠=°∠°∠== 15288)154.2)(0120(*IVS=S VA54.74j2.278 +

Apparent power = VA288 Real power = W2.278 Reactive power = VAR54.74 pf is lagging because current lags voltage

(d) °∠=°∠°∠== 135-1360)180-5.8)(45160(*IVS=S VA7.961j7.961- −

Apparent power =1360 VA Real power = W7.961- Reactive power = VAR7.961- pf is leading because current leads voltage


(a) , °∠= 10112V °∠= 50-4I

=°∠== 6022421 *IVS VA194j112+

Average power =112 W Reactive power =194 VAR

(b) , °∠= 0160V °∠= 4525I

=°∠== 45-20021 *IVS VA42.141j42.141 −

Average power =141 W42. Reactive power = VAR42.141-

(c) =°∠=°∠

== 3012830-50

)80( 2

*

2

ZV

S V64j51.90 A+

Average power = W51.90 Reactive power = VAR64

(d) =°∠== )45100)(100(2ZIS kVA071.7j071.7 +

Average power = kW071.7 Reactive power = kVAR071.7


(a) =−= jQPS V150j269 A−

(b) °=θ→=θ= 84.259.0cospf

31.4588)84.25sin(

2000sin

QSsinSQ =

°=

θ=→θ=

48.4129cosSP =θ=

=S VA2000j4129−

(c) 75.0600450

SQ

sinsinSQ ===θ→θ=

59.48=θ , 6614.0pf =

86.396)6614.0)(600(cosSP ==θ= =S VA450j9.396 +

(d) 121040

)220(S

22

===Z

V

8264.012101000

SP

coscosSP ===θ→θ=

°=θ 26.34

25.681sinSQ =θ=

=S VA2.681j1000+


(a) kVA))86.0(sin(cos86.04

j4 1-+=S

=S kVA373.2j4 +

(b) 6.0sincos8.026.1

SP

pf =θ→θ===

=θ−= sin2j6.1S kVA2.1j6.1 −

(c) VA)505.6)(20208(*

rmsrms °∠°∠== IVS=°∠= 70352.1S kVA2705.1j4624.0 +

(d) °∠

=−

==56.31-11.72

1440060j40)120( 2

*

2

ZV

S

=°∠= 56.317.199S VA16.166j77.110 + Chapter 11, Solution 50.

(a) ))8.0(sin(cos8.0

1000j1000jQP 1-−=−=S

750j1000−=S

But, *

2

rms

ZV

S =

23.23j98.30750j1000

)220( 22

rms* +=−

==S

VZ

=Z Ω− 23.23j98.30

(b) ZIS2

rms=

=+

== 22

rms)12(2000j1500

I

SZ Ω+ 89.13j42.10

(c) °∠=°∠

=== 60-6.1)604500)(2(

)120(2

222

rms*

SV

SV

Z

=°∠= 606.1Z Ω+ 386.1j8.0


(a) )6j8(||)5j10(2T +−+=Z

j1820j110

2j18

)6j8)(5j10(2T +

++=

++−

+=Z

°∠=+= 5.382188.8768.0j152.8TZ

=°= )5.382cos(pf (lagging)9956.0

(b) )5.382-188.8)(2(

)16(22

1 2

*

2

*

°∠===

ZV

IVS

°∠= 5.38263.15S

=θ= cosSP W56.15

(c) =θ= sinSQ VAR466.1

(d) == SS VA63.15

(e) =°∠= 382.563.15S VA466.1j56.15 +


749j4200SSSS500j1000S

2749j12009165.0x3000j4.0x3000S

1500j20006.08.0

2000j2000S

CBA

C

B

A

−=++=+=

−=−=

+=+=

(a) .leading9845.07494200

4200pf22=

+=

(b) °−∠=°∠

−=→= ∗∗ 11.5555.35

45120749j4200IIVS rmsrmsrms

Irms = 35.55∠–55.11˚ A.

Chapter 11, Solution 53. S = SA + SB + SC = 4000(0.8–j0.6) + 2400(0.6+j0.8) + 1000 + j500 = 5640 + j20 = 5640∠0.2˚

(a)

A8.2997.9388.2946.66x2I

8.2946.66

230120

2.05640V

SV

SSVSI

rmsrms

CA

rms

Brms

°∠=°∠=

°−∠=°∠°∠

==+

+=∗

(b) pf = cos(0.2˚) ≈ 1.0 lagging.


(a) ))8.0(sin(cos8.0

1000j1000jQP 1-−=−=S

750j1000−=S

But, *

2

rms

ZV

S =

23.23j98.30750j1000

)220( 22

rms* +=−

==S

VZ

=Z Ω− 23.23j98.30

(b) ZIS2

rms=

=+

== 22

rms)12(2000j1500

I

SZ Ω+ 89.13j42.10

(c) °∠=°∠

=== 60-6.1)604500)(2(

)120(2

222

rms*

SV

SV

Z

=°∠= 606.1Z Ω+ 386.1j8.0

Chapter 11, Solution 55. We apply mesh analysis to the following circuit.

40∠0° V rms I2+ −

j10 Ω

20 Ω

I3

I1+ −

-j20 Ω

50∠90° V rms

For mesh 1, 21 I20I)20j20(40 −−=

21 II)j1(2 −−= (1) For mesh 2, 12 I20I)10j20(50j- −+=

21 I)j2(I-2j5- ++= (2) Putting (1) and (2) in matrix form,

+

−=

2

1

II

j22-1-j1

j5-2

j1−=∆ , 3j41 −=∆ , 5j-12 −=∆

°∠=−=−−

=∆∆

= 13.8535.3)j7(21

j13j4

I 11

°∠=−=−−

=∆∆

= 56.31-605.33j2j15j1-

I 22

°∠=+=−−+=−= 8.66808.35.3j5.1)3j2()5.0j5.3(III 213

For the 40-V source,

=

−⋅== )j7(21

)40(-- *1IVS Vj20140- A+

For the capacitor, == c

2

1 ZIS Vj250- A For the resistor,

== R2

3IS V290 A For the inductor,

== L

2

2 ZIS VA130j For the j50-V source,

=+== )3j2)(50j(*2IVS Vj100150- A+


8.1j6.02j6

)2j-)(6(6||2j- −=

−=

j2.23.66||-j2)(4j3 +=++

The circuit is reduced to that shown below.

+

Vo

−

Io

2∠30° A 5 Ω

3.6 + j2.2 Ω

°∠=°∠++

= 47.0895.0)302(2.2j6.82.2j6.3

oI

°∠== 47.0875.45 oo IV

)30-)(247.0875.4(21

21 *

so °∠°∠⋅== IVS

=°∠= 17.0875.4S VA396.1j543.4 +

Chapter 11, Solution 57. Consider the circuit as shown below.

At node o,

+

V2

−

V1Vo

j2 Ω+ − 24∠0° V

4 Ω -j1 Ω 2 Ω

1 Ω 2 Vo

j-1424 1ooo VVVV −

+=−

1o 4j)4j5(24 VV −+= (1)

At node 1, 2j

2j-

1o

1o VV

VV=+

−

o1 )4j2( VV −= (2)


o)168j4j5(24 V−−+=

j41124-

o +=V ,

j411j4)-(-24)(2

1 +=V

The voltage across the dependent source is

o1o12 4)2)(2( VVVVV +=+=

4j11)4j6)(24-(

)44j2(j411

24-2 +

−=+−⋅

+=V

)2(21

21 *

o2*

2 VVIVS ==

)4j6(137576

j4-1124-

4j11)4j6)(24-(

−

=⋅+−

=S

=S VA82.16j23.25 −


4 kΩ

Ix

8 mA

-j3 kΩ j1 kΩ

10 kΩ

From the left portion of the circuit,

mA4.0500

2.0o ==I

mA820 o =I

From the right portion of the circuit,

mAj7

16)mA8(

3jj1044

x −=

−++=I

)1010(50

)1016(R 3

2-32

x ×⋅×

== IS

=S mVA2.51

Chapter 11, Solution 59. Consider the circuit below.

4 kΩ

Ix

8 mA

-j3 kΩ j1 kΩ

10 kΩ

30j40j20-50240

4 ooo

++=

−+

VVV

o)38.0j36.0(88 V+=

°∠=+

= 46.55-13.16838.0j36.0

88oV

°∠== 43.4541.8j20-

o1

VI

°∠=+

= 83.42-363.330j40

o2

VI

Reactive power in the inductor is

=⋅== )30j()363.3(21

21 2

L

2

2 ZIS VAR65.169j

Reactive power in the capacitor is

=⋅== )20j-()41.8(21

21 2

c2

1 ZIS VARj707.3-


15j20))8.0(sin(cos8.0

20j20S 1-1 +=+=

749.7j16))9.0(sin(cos9.0

16j16S 1-

2 +=+=

°∠=+=+= 29.32585.42749.22j36SSS 21

But o

*o V6IVS ==

==6S

Vo °∠ 29.32098.7

=°= )29.32cos(pf (lagging)8454.0

Chapter 11, Solution 61. Consider the network shown below.

I2

So

I1+

Vo

−

Io

S2

S3S1

kVA8.0j2.12 −=S

kVA937.1j4))9.0(sin(cos9.0

4j4 1-

3 +=+=S

Let kVA137.1j2.5324 +=+= SSS

But *2o4 2

1IVS =

104j74.2290100

10)137.1j2.5)(2(2 3

o

4*2 −=

°∠×+

==VS

I

104j74.222 +=I

Similarly, kVA2j2))707.0(sin(cos707.02

j2 1-1 −=−=S

But *1o1 2

1IVS =

40j40-100j

10)4j4(V2 3

o

1*1 −=

×−==

SI

j40-401 +=I

°∠=+=+= 83.96145j144-17.2621o III

*ooo IV

21

=S

VA)96.83-145)(90100(21

o °∠°∠⋅=S

=oS kVA862.0j2.7 −

Chapter 11, Solution 62. Consider the circuit below

0.2 + j0.04 Ω

+

V1

−

+

V2

−

+ −

I2

I1

I 0.3 + j0.15 Ω

Vs

25.11j15))8.0(sin(cos8.0

15j15 1-

2 −=−=S

But *

222 IVS =

12025.11j15

2

2*2

−==

VS

I

09375.0j125.02 +=I

)15.0j3.0(221 ++= IVV )15.0j3.0)(09375.0j125.0(1201 +++=V

0469.0j02.1201 +=V

843.4j10))9.0(sin(cos9.0

10j10 1-

1 +=+=S

But *

111 IVS =

°∠°∠

==02.002.12084.25111.11

1

1*1 V

SI

0405.0j0837.025.82-093.01 −=°∠=I

053.0j2087.021 +=+= III )04.0j2.0(1s ++= IVV

)04.0j2.0)(053.0j2087.0()0469.0j02.120(s ++++=V 0658.0j06.120s +=V

=sV V03.006.120 °∠ Chapter 11, Solution 63. Let S = . 321 SSS ++

929.6j12))866.0(sin(cos866.012

j12 1-1 −=−=S

916.9j16))85.0(sin(cos85.0

16j16 1-

2 +=+=S

20j1520j)6.0(sin(cos

)6.0)(20(1-3 +=+=S

*o2

1987.22j43 IVS =+=

11098.22j442*

o

+==

VS

I

=oI A27.58-4513.0 °∠


I2 I1

+ −

j12

Is

8 Ω 120∠0º V

Is + I2 = I1 or Is = I1 – I2

But,

333.3j83.20Ior

333.3j83.20120

400j2500

VS

IVI

923.6j615.412j8

120I

2

22

1

+=

−=−

==→

−=+

=

∗∗S =

Is = I1 – I2 = –16.22 – j10.256 = 19.19∠–147.69˚ A.


Ω=×

=ω

→= k-j1001010j-

Cj1

nF1C 9-4

At the noninverting terminal,

j14

j100-10004

ooo

+=→=

−°∠V

VV

°∠= 45-2

4oV

)45t10cos(2

4)t(v 4

o °−=

W1050

12

12

4R

VP 3

22rms

×

⋅==

=P W80 µ

Chapter 11, Solution 66. As an inverter,

)454(j34j4)(2--

si

fo °∠⋅

++

== VZZ

V

mAj3)j2)(4-(6

)45j4)(4(2-mA

2j6o

o +°∠+

=−

=V

I

The power absorbed by the 6-kΩ resistor is

36-

22

o 10610540420

21

R21

P ×××

××

⋅== I

=P mW96.0


51.02

11F1.0,6H3,2 jxjCj

jLj −==→=→=ω

ωω

42510

50)5//(10 jj

jj −=−

−=−

The frequency-domain version of the circuit is shown below. Z2=2-j4 Ω

Z1 =8+j6Ω I1 - + Io + + V Vo206.0 ∠ o - -

Ω= 123Z

(a) oo

jj

jI 87.1606.0

682052.05638.0

680206.0

1 −∠=++

=+

−∠=

mV 86.3618mVA 8.104.14)87.1606.0)(203.0(21

1* ooo

s jIVS ∠=+=+∠∠==

(b) ooooso j

jZV

IVZZ 7.990224.0)206.0(

)68(12)42(,

31

2 ∠=∠+−

−==−=V

mW 904.2)12()0224.0(5.0||21 22 === RIP o

A


Let S cLR SSS ++=

where 0jRI21

jQP 2oRRR +=+=S

LI21

j0jQP 2oLLL ω+=+=S

C1

I21

j0jQP 2occc ω⋅−=+=S

Hence, =S

ω−ω+

C1

LjRI21 2

o


(a) Given that 12j10 +=Z

°=θ→=θ 19.501012

tan

=θ= cospf 6402.0

(b) 09.354j12.295)12j10)(2(

)120(2

2

*

2

+=−

==Z

VS

The average power absorbed === )Re(P S W1.295

(c) For unity power factor, °=θ 01

09.354, which implies that the reactive power due

to the capacitor is Qc =

But 2VC21

X2V

Qc

2

c ω==

=π

=ω

= 22c

)120)(60)(2()09.354)(2(

VQ2

C F4.130 µ


6.0sin8.0cospf =θ→=θ= 528)6.0)(880(sinSQ ==θ=

If the power factor is to be unity, the reactive power due to the capacitor is

VAR528QQc ==

But 2c2

c

2rms

VQ2

CVC21

XV

Qω

=→ω==

=π

= 2)220)(50)(2()528)(2(

C F45.69 µ


67.666.0

508.08.0,6.0

,50,065.1067071.0150 22

2211 ======== SPQSQxQP

5067.66,065.106065.106 21 jSjS −=+=

9512.098.17cos,98.176.18106.56735.17221 ==∠=+=+= oo pfjSSS

058.56)098.17(tan735.172)tan(tan 21 =−=−= o

c PQ θθ

F 33.10120602

058.5622 µ

πω===

xxVQC

rms

c


(a) °==θ 54.40)76.0(cos-11

°==θ 84.25)9.0(cos-12

)tan(tanPQ 21c θ−θ=

kVAR])84.25tan()54.40tan()[40(Qc °−°= kVAR84.14Qc =

=π

=ω

= 22rms

c

)120)(60)(2(14840

VQ

C mF734.2

(b) , °=θ 54.401 °=θ 02

kVAR21.34kVAR]0)54.40tan()[40(Qc =−°=

=πω

= 22rms

c

)120)(60)(2(34210

VQ

C mF3.6


(a) kVA7j1022j15j10 +=+−=S

=+== 22 710S S kVA21.12

(b) 240

000,7j000,10** +==→=

VS

IIVS

=−= 167.29j667.41I A35-86.50 °∠

(c) °=

=θ 35107

tan 1-1 , °==θ 26.16)96.0(cos-1

2

])tan(16.26-)35tan([10]tantan[PQ 211c °°=θ−θ=

=cQ kVAR083.4

=π

=ω

= 22rms

c

)240)(60)(2(4083

VQ

C F03.188 µ

(d) , 222 jQP +=S kW10PP 12 ==

kVAR917.2083.47QQQ c12 =−=−=

kVA917.2j102 +=S

But *22 IVS =

2402917j000,102*

2

+==

VS

I

=−= 154.12j667.412I A16.26-4.43 °∠


(a) °==θ 87.36)8.0(cos-11

kVA308.0

24cos

PS

1

11 ==

θ=

kVAR18)6.0)(30(sinSQ 111 ==θ=

kVA18j241 +=S

°==θ 19.18)95.0(cos-12

kVA105.4295.0

40cos

PS

2

22 ==

θ=

kVAR144.13sinSQ 222 =θ=

kVA144.13j402 +=S

kVA144.31j6421 +=+= SSS

°=

=θ 95.2564144.31

tan 1-

=θ= cospf 8992.0

(b) , °=θ 95.252 °=θ 01

kVAR144.31]0)95.25tan([64]tantan[PQ 12c =−°=θ−θ=

=π

=ω

= 22rms

c

)120)(60)(2(144,31

VQ

C mF74.5


(a) VA59.323j75.5175j8

576050j80)240( 2

*1

2

1 −=+

=+

==ZV

S

VA91.208j13.3587j12

576070j120

)240( 2

2 +=−

=−

=S

VA96060

)240( 2

3 ==S

=++= 321 SSSS VA68.114j88.1835 −

(b) °=

=θ 574.388.183568.114

tan 1-

=θ= cospf 998.0

(c) ]0)574.3tan([88.35.18]tantan[PQ 12c −°=θ−θ=

VAR68.114Qc =

=π

=ω

= 22rms

c

)240)(50)(2(68.114

VQ

C F336.6 µ


The wattmeter reads the real power supplied by the current source. Consider the circuit below.

Vo4 Ω

8 Ω

-j3 Ω

j2 Ω12∠0° V + − 3∠30° A

8j23j412

303 ooo VVV+=

−−

+°∠

°∠=+=−+

= 19.86347.11322.11j7547.004.3j28.2

52.23j14.36oV

)30-3)(19.86347.11(21

21 *

oo °∠°∠⋅== IVS

°∠= 19.56021.17S

== )Re(P S W471.9


The wattmeter measures the power absorbed by the parallel combination of 0.1 F and 150 Ω.

°∠→ 0120)t2cos(120 , 2=ω 8jLjH4 =ω→

-j5Cj

1F1.0 =

ω→

Consider the following circuit.

j8 Ω6 Ω I

+ − 120∠0° V Z

5.4j5.15j15

(15)(-j5)-j5)(||15 −=

−==Z

°∠=−++

= 25.02-5.14)5.4j5.1()8j6(

120I

)5.4j5.1()5.14(21

21

21 22* −⋅=== ZIIVS

VA06.473j69.157 −=S

The wattmeter reads

== )Re(P S W69.157


4The wattmeter reads the power absorbed by the element to its right side.

°∠→ 02)t4cos(2 , =ω

4jLjH1 =ω→

-j3Cj

1F

121

=ω

→


10 Ω I

+ − 20∠0° V Z

3j4)3j-)(4(

4j53j-||44j5−

++=++=Z

08.2j44.6 +=Z

°∠=+

= 7.21-207.108.2j44.16

20I

)08.2j44.6()207.1(21

21 22

+⋅== ZIS

== )Re(P S W691.4

Chapter 11, Solution 79. The wattmeter reads the power supplied by the source and partly absorbed by the 40-Ω resistor.

20j10x500x100j

1Cj

1F500,j10x10x100jmH 10

,100

63 −==

ω→µ=→

=ω

−−

The frequency-domain circuit is shown below.

20 Io

I 40 j V1 V2 +1 2 Io 10<0o -j20 - At node 1,

21

21212121o

1

V)40j6(V)40j7(10jVV

20)VV(3

20VV

jVV

I240

V10

+−+−=

→−

+−

=−

+−

+=−

(1)

At node 2,

2122121 )19()20(02020

VjVjjVVV

jVV

+−+=→−

=−

+− (2)

Solving (1) and (2) yields V1 = 1.5568 –j4.1405

0703.22216.421,4141.08443.0

4010 1 jVISjVI −==+=

−= •

P = Re(S) = 4.222 W.


(a) ===4.6

110ZV

I A19.17

(b) 625.18904.6)110( 22

===Z

VS

°=θ→==θ 41.34825.0pfcos

≅=θ= 76.1559cosSP kW6.1


kWh consumed kWh77132464017 =−= The electricity bill is calculated as follows :

(a) Fixed charge = $12 (b) First 100 kWh at $0.16 per kWh = $16 (c) Next 200 kWh at $0.10 per kWh = $20 (d) The remaining energy (771 – 300) = 471 kWh

at $0.06 per kWh = $28.26. Adding (a) to (d) gives $ 26.76


(a) 0,000,5 11 == QP 171,17)82.0sin(cos000,30,600,2482.0000,30 1

22 ==== −QxP 171,17j600,29)QQ(j)PP(SSS 212121 +=+++=+=

AkV 22.34|S|S ==

(b) Q = 17.171 kVAR

(c ) 865.0220,34600,29

===SPpf

[ ] VAR 2833)9.0tan(cos)865.0tan(cos600,29

)tan(tanPQ11

21c

=−=

θ−θ=−−

(d) F 46.130240602

283322 µ

πω===

xxVQC

rms

c


(a) oooVIS 35840)258)(60210(21

21

∠=−∠∠== ∗

W1.68835cos840cos === oSP θ

(b) S = 840 VA (c) VAR 8.48135sin840sin === oS θQ (d) (lagging) 8191.035cos/ === oSPpf


(a) Maximum demand charge 000,72$30400,2 =×= Energy cost= 000,48$10200,104.0$ 3 =××Total charge = $ 000,120

(b) To obtain $120,000 from 1,200 MWh will require a flat rate of

=×

kWhper10200,1

000,120$3 kWhper10.0$


(a) 15 655.51015602 mH 3 jxxxj =→ −π

We apply mesh analysis as shown below. I1 + Ix 120<0o V 10Ω - In 30Ω Iz + 10Ω 120<0o V Iy - j5.655 Ω I2

For mesh x, 120 = 10 Ix - 10 Iz (1) For mesh y, 120 = (10+j5.655) Iy - (10+j5.655) Iz (2) For mesh z, 0 = -10 Ix –(10+j5.655) Iy + (50+j5.655) Iz (3) Solving (1) to (3) gives Ix =20, Iy =17.09-j5.142, Iz =8 Thus, I1 =Ix =20 A I2 =-Iy =-17.09+j5.142 = A 26.16385. o∠17 In =Iy - Ix =-2.091 –j5.142 = A 5.119907.5 o−∠

(b) 5.3085.1025)120(21,12002060)120(

21

21 jISxIS yx −===== ••

VA 5.3085.222521 jSS −=+=S

(c ) pf = P/S = 2225.5/2246.8 = 0.9905 Chapter 11, Solution 86. For maximum power transfer

*LThi

*ThL ZZZZZ ==→=

)104)(1012.4)(2(j75LjR -66L ××π+=ω+=Z

Ω+= 55.103j75LZ

=iZ Ω− 55.103j75 Chapter 11, Solution 87. jXR ±=Z

Ω=×

==→= k6.11050

80RR 3-

RR I

VIV

22222222

)6.1()3(RXXR −=−=→+= ZZ Ω= k5377.2X

°=

=

=θ 77.576.1

5377.2tan

RX

tan 1-1-

=θ= cospf 5333.0


(a) °∠=°∠= 55220)552)(110(S

=°=θ= )55cos(220cosSP W2.126

(b) == SS VA220 Chapter 11, Solution 89.

(a) Apparent power == S kVA12

kW36.9)78.0)(12(cosSP ==θ= kVAR51.7))78.0(sin(cos12sinSQ -1 ==θ=

=+= jQPS kVA51.7j36.9 +

(b) 3

22

**

2

10)51.7j36.9()210(

×+==→=

SV

ZZV

S

=Z Ω+ 6.27j398.34

Chapter 11, Solution 90 Original load :

kW2000P1 = , °=θ→=θ 79.3185.0cos 11

kVA94.2352cos

PS

1

11 =

θ=


Additional load :

kW300P2 = , °=θ→=θ 87.368.0cos 22

kVA375cos

PS

2

22 =

θ=

kVAR225sinSQ 222 =θ=

Total load :

jQP)QQ(j)PP( 212121 +=+++=+= SSS kW23003002000P =+=

kVAR5.14642255.1239Q =+= The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the generator is

9775.094.2352

2300SP

cos1

===θ

or °=θ 177.12

The maximum load kVAR for this condition is

)177.12sin(94.2352sinSQ 1m °=θ= kVAR313.496Qm =

The capacitor must supply the difference between the total load kVAR ( i.e. Q ) and the permissible generator kVAR ( i.e. ). Thus, mQ

=−= mc QQQ kVAR2.968 Chapter 11, Solution 91

θ= cosSP

===θ=)15)(220(

2700SP

cospf 8182.0

3.1897)09.35sin()15(220sinSQ =°=θ=

When the power is raised to unity pf, °=θ 01 and Q 3.1897Qc ==

=π

=ω

= 22rms

c

)220)(60)(2(3.1897

VQ

C F104 µ


(a) Apparent power drawn by the motor is

kVA8075.0

60cos

PSm ==

θ=

kVAR915.52)60()80(PSQ 2222

m =−=−= Total real power

kW8020060PPPP Lcm =++=++= Total reactive power

=+−=++= 020915.52QQQQ Lcm kVAR91.32 Total apparent power

=+= 22 QPS kVA51.86

(b) ===51.86

80SP

pf 9248.0

(c) ===550

86510VS

I A3.157


(a) kW7285.3)7457.0)(5(P1 ==

kVA661.48.0

7285.3pfP

S 11 ===

kVAR796.2))8.0(sin(cosSQ -1

11 == kVA796.2j7285.31 +=S

kW2.1P2 = , VAR0Q2 =

kVA0j2.12 +=S

kW2.1)120)(10(P3 == , VAR0Q3 = kVA0j2.13 +=S

kVAR6.1Q4 = , 8.0sin6.0cos 44 =θ→=θ

kVA2sin

QS

4

44 =

θ=

kW2.1)6.0)(2(cosSP 444 ==θ=

kVA6.1j2.14 −=S

4321 SSSSS +++= kVA196.1j3285.7 +=S

Total real power = 7 kW3285. Total reactive power = 1 kVAR196.

(b) °=

=θ 27.93285.7196.1

tan 1-

=θ= cospf 987.0


°=θ→=θ 57.457.0cos 11 kVA1000MVA1S1 == kW700cosSP 111 =θ=

kVAR14.714sinSQ 111 =θ= For improved pf,

°=θ→=θ 19.1895.0cos 22 kW700PP 12 ==

kVA84.73695.0

700cos

PS

2

22 ==

θ=


P1 = P2 = 700 kW

Q2

Q1

S1

S2

θ1 θ2

Qc

(a) Reactive power across the capacitor

kVAR06.48408.23014.714QQQ 21c =−=−= Cost of installing capacitors =×= 06.48430$ 80.521,14$

(b) Substation capacity released 21 SS −= kVA16.26384.7361000 =−=

Saving in cost of substation and distribution facilities

=×= 16.263120$ 20.579,31$

(c) Yes, because (a) is greater than (b). Additional system capacity obtained by using capacitors costs only 46% as much as new substation and distribution facilities.


(a) Source impedance css XjR −=Z Load impedance 2LL XjR +=Z For maximum load transfer

LcLs*sL XX,RR ==→= ZZ

LC1

XX Lc ω=ω

→=

or f2LC1

π==ω

=××π

=π

=)1040)(1080(2

1LC2

1f

9-3-kHz814.2

(b) ===)10)(4(

)6.4(R4

VP

2

L

2s mW529 (since V is in rms) s


ZTh

+ −

VTh ZL

(a) Hz300,V146VTh =

Ω+= 8j40ZTh

== *ThL ZZ Ω− 8j40

(b) ===)40)(8(

)146(R8

VP

2

Th

2

Th W61.66


Ω+=+++= 22j2.100)20j100()j1.0)(2(ZT

22j2.100240

ZV

IT

s

+==

=+

=== 22

22

L2

)22()2.100()240)(100(

I100RIP W3.547


(a) If , then 400ab =V

=°∠= 30-3

400anV V30-231 °∠

=bnV V150-231 °∠ =cnV V270-231 °∠

(b) For the acb sequence,

°∠−°∠=−= 120V0V ppbnanab VVV

°∠=

−+= 30-3V

23

j21

1V ppabV

i.e. in the acb sequence, lags by 30°. abV anV Hence, if , then 400ab =V

=°∠= 303

400anV V30231 °∠

=bnV V150231 °∠ =cnV V90-231 °∠


Since phase c lags phase a by 120°, this is an acb sequence.

=°+°∠= )120(30160bnV V150160 °∠ Chapter 12, Solution 3. Since V leads by 120°, this is an bn cnV abc sequence.

=°+°∠= )120(130208anV V250208 °∠


=°∠= 120cabc VV V140208 °∠

=°∠= 120bcab VV V260208 °∠

=°∠°∠

=°∠

=303260208

303ab

an

VV V230120 °∠

=°∠= 120-anbn VV V110120 °∠

Chapter 12, Solution 5. This is an abc phase sequence.

°∠= 303anab VV

or =°∠°∠

=°∠

=3030420

303ab

an

VV V30-5.242 °∠

=°∠= 120-anbn VV V150-5.242 °∠

=°∠= 120ancn VV V905.242 °∠


°∠=+= 26.5618.115j10YZ The line currents are

=°∠

°∠==

26.5618.110220

Y

ana Z

VI A26.56-68.19 °∠

=°∠= 120-ab II A146.56-68.19 °∠

=°∠= 120ac II A93.4468.19 °∠

The line voltages are

=°∠= 303200abV V30381 °∠ =bcV V90-381 °∠ =caV V210-381 °∠

The load voltages are

=== anYaAN VZIV V0220 °∠ == bnBN VV V120-220 °∠ == cnCN VV V120220 °∠

Chapter 12, Solution 7. This is a balanced Y-Y system.

+ − 440∠0° V ZY = 6 − j8 Ω

Using the per-phase circuit shown above,

=−

°∠=

8j60440

aI A53.1344 °∠

=°∠= 120-ab II A66.87-44 °∠ =°∠= 120ac II A13.73144 °∠

Chapter 12, Solution 8. , V220VL = Ω+= 9j16YZ

°∠=+

=== 29.36-918.6)9j16(3

2203VV

Y

L

Y

pan ZZ

I

=LI A918.6


=+

°∠=

+=

15j200120

YL

ana ZZ

VI A36.87-8.4 °∠

=°∠= 120-ab II A156.87-8.4 °∠

=°∠= 120ac II A83.138.4 °∠

As a balanced system, =nI A0

Chapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis.

For phase a,

°∠=−

°∠=

+= 36.5355.6

20j270220

2A

ana Z

VI

For phase b,

°∠=°∠

=+

= 120-1022

120-2202B

bnb Z

VI

For phase c,

°∠=+

°∠=

+= 97.3892.16

5j12120220

2C

cnc Z

VI

The current in the neutral line is

)-( cban IIII ++= or cban- IIII ++=

)78.16j173.2-()66.8j5-()9.3j263.5(- n ++−++=I

=−= 02.12j91.1nI A81-17.12 °∠


°∠°∠

=°∠

=°∠

=90-3

1022090-390-3

BCbcan

VVV

=anV V100127 °∠

=°∠= 120BCAB VV V130220 °∠

V110-220120-BCAC °∠=°∠= VV

If , then °∠= 6030bBI

°∠= 18030aAI , °∠= 60-30cCI

°∠=°∠°∠

=°∠

= 21032.1730-3

1803030-3

aAAB

II

°∠= 9032.17BCI , °∠= 30-32.17CAI

== CAAC -II A15032.17 °∠

BCBC VZI =

=°∠

°∠==

9032.170220

BC

BC

IV

Z Ω°∠ 80-7.12

Chapter 12, Solution 12. Convert the delta-load to a wye-load and apply per-phase analysis.

Ia

110∠0° V + −

ZY

Ω°∠== ∆ 45203Y

ZZ

=°∠°∠

=45200110

aI A45-5.5 °∠

=°∠= 120-ab II A165-5.5 °∠ =°∠= 120ac II A755.5 °∠


First we calculate the wye equivalent of the balanced load.

ZY = (1/3)Z∆ = 6+j5

Now we only need to calculate the line currents using the wye-wye circuits.

A07.58471.615j8120110I

A07.178471.615j8120110I

A93.61471.65j610j2

110I

c

b

a

°∠=+

°∠=

°∠=+

°−∠=

°−∠=+++

=

Chapter 12, Solution 14. We apply mesh analysis. Ω+ 2j1 A a + ZL 100 ZV 0o∠ L - I3 n I1 B C - -

100 + - + c I

V 120100 o∠ V 120o∠ Ω+= 1212 jZ L

2 b Ω+ 2j1 Ω+ 2j1 For mesh 1,

0)1212()21()1614(120100100 321 =+−+−++∠+− IjIjjIo or

6.861506.8650100)1212()21()1614( 321 jjIjIjIj −=−+=+−+−+ (1) For mesh 2,

0)1614()1212()21(120100120100 231 =+++−+−−∠−∠ IjIjjIoo or

2.1736.86506.8650)1212()1614()21( 321 jjjIjIjIj −=−+−−=+−+++− (2) For mesh 3,

0)3636()1212()1212( 321 =+++−+− IjIjIj (3) Solving (1) to (3) gives

016.124197.4,749.16098.10,3.19161.3 321 jIjIjI −−=−−=−−= A 3.9958.191

oaA II −∠==

A 8.159392.712

obB III ∠=−=

A 91.5856.192

ocC II ∠=−=


Convert the delta load, , to its equivalent wye load. ∆Z

10j83Ye −== ∆Z

Z

°∠=−

−+== 14.68-076.8

5j20)10j8)(5j12(

|| YeYp ZZZ

047.2j812.7p −=Z

047.1j812.8LpT −=+= ZZZ

°∠= 6.78-874.8TZ

We now use the per-phase equivalent circuit.

Lp

pa

VZZ

I+

= , where 3

210pV =

°∠=°∠

= 78.666.13)6.78-874.8(3

210aI

== aL II A66.13


(a) °∠=°+°∠== 15010)180-30(10- ACCA II This implies that °∠= 3010ABI

°∠= 90-10BCI

=°∠= 30-3ABa II A032.17 °∠ =bI A120-32.17 °∠ =cI A12032.17 °∠

(b) =°∠°∠

==∆ 30100110

AB

AB

IV

Z Ω°∠ 30-11

Chapter 12, Solution 17. Convert the ∆-connected load to a Y-connected load and use per-phase analysis.

Ia

+ −

ZL

Van ZY

4j33Y +== ∆Z

Z

°∠=+++°∠

=+

= 48.37-931.19)5.0j1()4j3(

0120

LY

ana ZZ

VI

But °∠= 30-3ABa II

=°∠

°∠=

30-348.37-931.19

ABI A18.37-51.11 °∠

=BCI A138.4-51.11 °∠ =CAI A101.651.11 °∠

)53.1315)(18.37-51.11(ABAB °∠°∠== ∆ZIV

=ABV V76.436.172 °∠

=BCV V85.24-6.172 °∠ =CAV V8.5416.172 °∠


°∠=°∠°∠=°∠= 901.762)303)(60440(303anAB VV

°∠=+=∆ 36.87159j12Z

=°∠°∠

==∆ 36.8715

901.762ABAB Z

VI A53.1381.50 °∠

=°∠= 120-ABBC II A66.87-81.50 °∠

=°∠= 120ABCA II A173.1381.50 °∠ Chapter 12, Solution 19.

°∠=+=∆ 18.4362.3110j30Z The phase currents are

=°∠

°∠==

∆ 18.4362.310173ab

AB ZV

I A18.43-47.5 °∠

=°∠= 120-ABBC II A138.43-47.5 °∠ =°∠= 120ABCA II A101.5747.5 °∠

The line currents are

°∠=−= 30-3ABCAABa IIII

=°∠= 48.43-347.5aI A48.43-474.9 °∠

=°∠= 120-ab II A168.43-474.9 °∠ =°∠= 120ac II A71.57474.9 °∠


°∠=+=∆ 36.87159j12Z The phase currents are

=°∠

°∠=

36.87150210

ABI A36.87-14 °∠

=°∠= 120-ABBC II A156.87-14 °∠ =°∠= 120ABCA II A83.1314 °∠

The line currents are

=°∠= 30-3ABa II A66.87-25.24 °∠ =°∠= 120-ab II A186.87-25.24 °∠

=°∠= 120ac II A53.1325.24 °∠


(a) )rms(A66.9896.1766.38806.12

1202308j10

120230IAC °−∠=°∠

°∠−=

+°∠−

=

(b)

A34.17110.31684.4j75.30220.11j024.14536.6j729.16

66.3896.1766.15896.178j100230

8j10120230IIIII ABBCBABCbB

°∠=+−=+−−−=

°−∠−°−∠=+

°∠−

+−∠

=−=+=


Convert the ∆-connected source to a Y-connected source.

°∠=°∠=°∠= 30-12030-3

20830-

3

VpanV

Convert the ∆-connected load to a Y-connected load.

j8)5j4)(6j4(

)5j4(||)6j4(3

||Y +−+

=−+== ∆ZZZ

2153.0j723.5 −=Z

Ia

+ −

ZL

Van Z

=−

°∠=

+=

2153.0j723.730120

L

ana ZZ

VI A28.4-53.15 °∠

=°∠= 120-ab II A148.4-53.15 °∠

=°∠= 120ac II A91.653.15 °∠


(a) oAB

AB ZVI

6025208∠

==∆

o

o

oo

ABa II 90411.146025

303208303 −∠=∠

−∠=−∠=

A 41.14|| == aL II

(b) kW 596.260cos25

3208)208(3cos321 =

==+= o

LL IVPPP θ

Chapter 12, Solution 24. Convert both the source and the load to their wye equivalents.

10j32.1730203Y +=°∠== ∆Z

Z

°∠=°∠= 02.24030-3ab

an

VV

We now use per-phase analysis.

Ia

+ −

1 + j Ω

Van 20∠30° Ω

=°∠

=+++

=3137.212.240

)10j32.17()j1(an

a

VI A31-24.11 °∠

=°∠= 120-ab II A151-24.11 °∠

=°∠= 120ac II A8924.11 °∠

But °∠= 30-3ABa II

=°∠°∠

=30-331-24.11

ABI A1-489.6 °∠

=°∠= 120-ABBC II A121-489.6 °∠

=°∠= 120ABCA II A119489.6 °∠


Convert the delta-connected source to an equivalent wye-connected source and consider the single-phase equivalent.

Ya 3

)3010(440Z

I°−°∠

=

where °°∠=−=−++= 78.24-32.146j138j102j3YZ

=°∠

°∠=

)24.78-32.14(320-440

aI A4.7874.17 °∠

=°∠= 120-ab II A115.22-74.17 °∠

=°∠= 120ac II A124.7874.17 °∠

Chapter 12, Solution 26. Transform the source to its wye equivalent.

°∠=°∠= 30-17.7230-3

VpanV

Now, use the per-phase equivalent circuit.

ZV

I anaA = , °∠=−= 32-3.2815j24Z

=°∠°∠

=32-3.2830-17.72

aAI A255.2 °∠

=°∠= 120-aAbB II A118-55.2 °∠

=°∠= 120aAcC II A12255.2 °∠


)15j20(310-220

330-

Y

aba +

°∠=

°∠=

ZV

I

=aI A46.87-081.5 °∠

=°∠= 120-ab II A166.87-081.5 °∠

=°∠= 120ac II A73.13081.5 °∠

Chapter 12, Solution 28. Let °∠= 0400abV

°∠=°∠°∠

=°∠

= 307.7)60-30(3

30-4003

30-

Y

ana Z

VI

== aLI I A7.7

°∠=°∠== 30-94.23030-3an

YaAN

VZIV

== ANpV V V9.230


, θ= cosIV3P pp 3V

V Lp = , pL II =

θ= cosIV3P LL

pL

L I05.20)6.0(3240

5000cosV3

PI ===

θ=

911.6)05.20(3

240I3

VIV

L

L

p

pY ====Z

°=θ→=θ 13.536.0cos

(leading)53.13-911.6Y °∠=Z

=YZ Ω− 53.5j15.4

83336.0

5000pfP

S ===

6667sinSQ =θ=

=S VA6667j5000−


Since this a balanced system, we can replace it by a per-phase equivalent, as shown below. + ZL Vp

-

3,33 *

2L

pp

pp

VVZVSS ===

kVA 454421.14530)208( 2

*

2o

op

L

ZVS ∠=

−∠==

kW 02.1cos == θSP


(a) kVA 5.78.0/6cos

,8.0cos,000,6 =====θ

θ Ppp

PSP

kVAR 5.4sin == θPp SQ

kVA 5.1318)5.46(33 jjSS p +=+== For delta-connected load, Vp = VL= 240 (rms). But

Ω+=+

==→= 608.4144.6,10)5.1318(

)240(3333

22*

*

2

jZxjS

VZZVS P

pp

p

p

(b) A 04.188.02403

6000cos3 ==→=xx

IIVP LLLp θ

(c ) We find C to bring the power factor to unity

F 2.207240602

4500kVA 5.4 22 µπω

===→==xxV

QCQQrms

cpc


θ∠= LLIV3S

3LL 1050IV3S ×=== S

==)440(3

5000IL A61.65

For a Y-connected load,

61.65II Lp == , 03.2543

4403

VV L

p ===

872.361.6503.254

IV

p

p===Z

θ∠= ZZ , °==θ 13.53)6.0(cos-1

)sinj)(cos872.3( θ+θ=Z

)8.0j6.0)(872.3( +=Z

=Z Ω+ 098.3j323.2

Chapter 12, Solution 33. θ∠= LLIV3S

LLIV3S == S

For a Y-connected load,

pL II = , pL V3V =

pp IV3S =

====)208)(3(

4800V3S

IIp

pL A69.7

=×== 2083V3V pL V3.360


3

2203

VV L

p ==

°∠=−

== 5873.6)16j10(3

200V

Y

pa Z

I

== pL II A73.6

°∠××=θ∠= 58-73.62203IV3 LLS

=S VA8.2174j1359−


(a) This is a balanced three-phase system and we can use per phase equivalent circuit. The delta-connected load is converted to its wye-connected equivalent

10203/)3060(31'' jjZZ y +=+== ∆

IL

+ 230 V Z’y Z’’y

-

5.55.13)1020//()1040(//' '' jjjZZZ yyy +=++==

A 953.561.145.55.13

230 jj

I L −=+

=

(b) kVA 368.1361.3* jIVS Ls +== (c ) pf = P/S = 0.9261


(a) S = 1 [0.75 + sin(cos-10.75) ] =0.75 + 0.6614 MVA

(b) 49.5252.5942003

10)6614.075.0(3

36

** jx

xjVSIIVp

ppp +=+

==→=S

kW 19.25)4()36.79(|| 22 === lpL RIP

(c) kV 2.709-4.443 kV 21.04381.4)4( o∠=−=++= jjIV pLsV Chapter 12, Solution 37.

206.0

12pfP

S ===

kVA16j1220S −=θ∠=θ∠=S

But θ∠= LLIV3S

=××

=20831020

I3

L A51.55

p

2

p3 ZIS = For a Y-connected load, pL II = .

2

3

2

L

p )51.55)(3(10)16j12(

I3

×−==

SZ

=pZ Ω− 731.1j298.1


As a balanced three-phase system, we can use the per-phase equivalent shown below.

14j100110

)12j9()2j1(0110

a +°∠

=+++°∠

=I

)12j9()1410(

)110(21

21

22

2

Y

2

ap +⋅+

⋅== ZIS

The complex power is

)12j9(296

)110(23

32

p +⋅⋅== SS

=S VA81.735j86.551 +

Chapter 12, Solution 39. Consider the system shown below.

I2

I1I3

5 Ω

5 Ω

-j6 Ω

10 Ω

j3 Ω

B C

A

8 Ω

4 Ω

b

a

−+

+ −

100∠-120°

100∠120° 100∠0°− +

5 Ω

c

For mesh 1,

321 )6j8(5)6j18(100 III −−−−= (1) For mesh 2,

312 10520120-100 III −−=°∠

321 24-120-20 III −+=°∠ (2)

For mesh 3, 321 )3j22(10)6j8(-0 III −+−−= (3)

To eliminate , start by multiplying (1) by 2, 2I

321 )12j16(10)12j36(200 III −−−−= (4) Subtracting (3) from (4),

31 )15j38()18j44(200 II −−−= (5) Multiplying (2) by 45 ,

321 5.2525.1-120-25 III −+=°∠ (6) Adding (1) and (6),

31 )6j5.10()6j75.16(65.21j5.87 II −−−=− (7) In matrix form, (5) and (7) become

+−+−

=

− 3

1

6j5.10-6j75.1615j38-18j44

65.12j5.87200

II

25.26j5.192 −=∆ , 2.935j25.9001 −=∆ , 6.1327j3.1103 −=∆

144.4j242.538.33-682.67.76-28.194

46.09-1.129811 −=°∠=

°∠°∠

=∆∆

=I

694.6j485.177.49-857.67.76-28.194

85.25-2.133233 −=°∠=

°∠°∠

=∆∆

=I

We obtain from (6), 2I

312 21

41

120-5 III ++°∠=

)347.3j7425.0()0359.1j3104.1()33.4j-2.5(2 −+−+−=I

713.8j4471.0-2 −=I The average power absorbed by the 8-Ω resistor is

W89.164)8(551.2j756.3)8(P22

311 =+=−= II The average power absorbed by the 4-Ω resistor is

W1.188)4()8571.6()4(P 22

32 === I

The average power absorbed by the 10-Ω resistor is

W12.78)10(019.2j1.9321-)10(P 22323 =−=−= II

Thus, the total real power absorbed by the load is

=++= 321 PPPP W1.431 Chapter 12, Solution 40. Transform the delta-connected load to its wye equivalent.

8j73Y +== ∆Z

Z

Using the per-phase equivalent circuit above,

°∠=+++

°∠= 46.75-567.8

)8j7()5.0j1(0100

aI

For a wye-connected load,

567.8II aap === I

)8j7()567.8)(3(3 2p

2

p +== ZIS

=== )7()567.8)(3()Re(P 2S k541.1 W Chapter 12, Solution 41.

kVA25.68.0

kW5pfP

S ===

But LLIV3S =

=××

==40031025.6

V3S

I3

LL A021.9

Chapter 12, Solution 42. The load determines the power factor.

°=θ→==θ 13.53333.13040

tan

(leading)6.0cospf =θ=

kVA6.9j2.7)8.0(6.02.7

j2.7 −=

−=S

But p

2

p3 ZIS =

80)40j30)(3(

10)6.9j2.7(3

3

p

2

p =−

×−==

ZS

I

A944.8Ip =

== pL II A944.8

=×

==)944.8(3

1012I3

SV

3

LL V6.774


p

2

p3 ZIS = , Lp II = for Y-connected loads

)047.2j812.7()66.13)(3( 2 −=S =S kVA145.1j373.4 −

Chapter 12, Solution 44. For a ∆-connected load,

Lp VV = , pL I3I =

LLIV3S =

273.31)240(3

10)512(V3

SI

322

LL =

×+==

At the source,

LLL'L ZIVV +=

)3j1)(273.31(0240'L ++°∠=V

819.93j273.271'L +=V

='LV V04.287

Also, at the source,

*L

'L

' 3 IVS = )273.31)(819.93j273.271(3' +=S

078.19273.271

819.93tan 1- =

=θ

=θ= cospf 9451.0

Chapter 12, Solution 45. θ∠= LLIV3S

LL V3

-I

θ∠=

S, kVA6.635

708.010450

pfP 3

=×

==S

A45-8344403

-)6.635(L °∠=

×θ∠

=I

At the source,

)2j5.0(0440 LL ++°∠= IV

)76062.2)(45-834(440L °∠°∠+=V °∠+= 137.1719440LV

7.885j1.1914L +=V =LV V.8324109.2 °∠

Chapter 12, Solution 46. For the wye-connected load,

pL II = , pL V3V = Zpp VI =

*

2

L*

2

p*pp

3333

ZV

ZV

IVS ===

W121100

)110( 2

*

2

L ===Z

VS

For the delta-connected load,

Lp VV = , pL I3I = , Zpp VI =

*

2

L*

2

p*pp

333

ZV

ZV

IVS ===

W363100

)110)(3( 2

==S

This shows that the delta-connected load will deliver three times more average

power than the wye-connected load. This is also evident from 3Y∆=

ZZ .


°==θ→= 87.36)8.0(cos(lagging)8.0pf -1 kVA150j20087.362501 +=°∠=S

°==θ→= 19.18-)95.0(cos(leading)95.0pf -1

kVA65.93j28519.81-3002 −=°∠=S

°==θ→= 0)1(cos0.1pf -1

kVA4503 =S

kVA45.37.93635.56j935321T °∠=+=++= SSSS

LLT IV3=S

=××

=)108.13(3

107.936I

3

3

L rmsA19.39

=°=θ= )45.3cos(cospf (lagging)9982.0


(a) We first convert the delta load to its equivalent wye load, as shown below. A A ZA 18-j12Ω 40+j15Ω ZC C B C 60Ω

ZB

B

923.1577.73118

)1218)(1540( jjjjZ A −=

+−+

=

105.752.203118

).1540(60 jjjZ B −=

++

=

3303.6992.83118

)1218(60 jjjZC −=

+−

=

The system becomes that shown below.

a 2+j3 A + 240<0o ZA - I1 - - ZB ZC 240<120o 240<-120o + + 2+j3 c I2 b B C 2+j3

We apply KVL to the loops. For mesh 1, 0)()2(120240240 21 =+−+++−∠+− lBBAl

o ZZIZZZI or

85.207360)105.1052.22()13.11097.32( 21 jIjIj +=+−+ (1) For mesh 2,

0)2()(120240120240 21 =++++−−∠−∠ CBllBoo ZZZIZZI

or

69.415)775.651.33()105.1052.22( 21 jIjIj −=+++− (2) Solving (1) and (2) gives

89.11165.15,328.575.23 21 jIjI −=−=

A 6.14281.10,A 64.1234.24 121o

bBo

aA IIIII −∠=−=−∠==

A 9.14127.192o

cC II ∠=−= (b) ooo

aS 64.126.5841)64.1234.24)(0240( ∠=∠∠= ooo

bS 6.224.2594)6.14281.10)(120240( ∠=∠−∠= ooo

bS 9.218.4624)9.14127.19)(120240( −∠=−∠∠= kVA 54.24.12kVA 55.0386.12 o

cba jSSSS ∠=+=++=


(a) For the delta-connected load, (rms) 220,1020 ==Ω+= Lpp VVjZ ,

kVA 56.26943.629045808)1020(

22033 2

*

2o

p

p jj

xZVS ∠=+=

−==

(b) For the wye-connected load, 3/,1020 Lpp VVjZ =Ω+= ,

kVA 56.26164.2)1020(3

22033 2

*

2o

p

p

jx

ZVS ∠=

−==


kVA 3kVA, 4.68.4)8.06.0(8 121 =+=+=+= SjjSSS Hence,

kVA 4.68.112 jSSS +=−=

But p

LLp

p

p

ZVSVV

ZVS *

2

2*

2

2.

3,3

=→==

Ω+=→+

== 34.8346.210)4.68.1(

2403

2

2

** jZ

xjSVZ p

Lp

Chapter 12, Solution 51. Apply mesh analysis to the circuit as shown below.

Za

i2

i1

Zc

Zb

+ −

− +

− +

150∠-120°

150∠0°

n 150∠120°

For mesh 1,

0)(150- 2b1ba =−++ IZIZZ

21 )9j12()j18(150 II +−+= (1) For mesh 2,

0)(120-150- 1b2cb =−++°∠ IZIZZ

12 )9j12()9j27(120-150 II +−+=°∠ (2) From (1) and (2),

+−−+

=

°∠ 2

1

9j279j12-9j12-j18

120-150150

II

27j414 −=∆ , 8.3583j9.37801 +=∆ , 2.1063j9.5792 −=∆

°∠=°∠°∠

=∆∆

= 47.256.123.73-88.414

43.475.520911I

°∠=°∠°∠

=∆∆

= 57.66-919.23.73-88.414

61.39-1.121122I

== 1a II A47.256.12 °∠

∆−

=∆∆−∆

=−=4647j3201-12

12b III

=°∠°∠

=3.73-88.414

235.443.5642bI A239.176.13 °∠

== 2c -II A122.34919.2 °∠

Chapter 12, Solution 52. Since the neutral line is present, we can solve this problem on a per-phase basis.

°∠=°∠°∠

== 6066020120120

AN

ana Z

VI

°∠=°∠°∠

== 040300120

BN

bnb Z

VI

°∠=°∠°∠

== 150-33040120-120

CN

cnc Z

VI

Thus,

cban- IIII ++= °∠+°∠+°∠= 150-304606- nI

)5.1j598.2-()4()196.5j3(- n −+++=I °∠=+= 4075.5696.3j405.4- nI

=nI A22075.5 °∠


3

250Vp =

Since we have the neutral line, we can use per-phase equivalent circuit for each phase.

=°∠

⋅°∠

=6040

13

0250aI A60-608.3 °∠

=°∠

⋅°∠

=45-60

13120-250

bI A75-406.2 °∠

=°∠

⋅°∠

=020

13120250

cI A120217.7 °∠

cban- IIII ++=

)25.6j609.3-()324.2j6227.0()125.3j804.1(- n ++−+−=I

=−= 801.0j1823.1nI A34.12- 428.1 °∠


Iaa

B C

A

j50 Ω

-j50 Ω

50 Ω

IAB

c b

Vp∠0°

+−

Vp∠120° Vp∠-120°

+ −

− +

°∠×== 303100abAB VV

=°∠

==50

303100

AB

ABAB Z

VI A30464.3 °∠

=°∠°∠

==90-50

90-3100

BC

BCBC Z

VI A0464.3 °∠

=°∠

°∠==

90501503100

CA

CACA Z

VI A60464.3 °∠


Iaa A

Ib

I1

I2

c b

220∠0°

+−

220∠120° 220∠-120°

+ −

− +

30 + j40100 – j120

B C

60 + j80

IcFor mesh 1,

0)120j100()40j160(0220120-220 21 =−−−+°∠−°∠ II

21 )6j5()2j8(120-1111 II −−−=°∠− (1) For mesh 2,

0)120j100()80j130(120-220120220 12 =−−−+°∠−°∠ II

21 )4j5.6()6j5(-12011120-11 II −+−=°∠−°∠ (2) From (1) and (2),

+

+−=

+

2

1

j4-6.5j65-j65-2j8

j19.053-526.9j5.16

II

15j55+=∆ , 35.99j04.311 −=∆ , 8.203j55.1012 −=∆

°∠=°∠°∠

=∆∆

= 87.91-8257.115.2601.57

72.65-08.10411I

°∠=°∠°∠

=∆∆

= 78.77-994.315.2601.5763.51-7.2272

2I

°∠== 87.91-8257.11a II

°∠=+−

=∆∆−∆

=−= 71.23-211.215j55

45.104j51.701212b III

°∠== 23.011994.3- 2c II

7.266j99.199)80j60()8257.1( 2AN

2

aA +=+== ZIS

6.586j9.488)120j100()211.2( 2BN

2

bB −=−== ZIS

1.638j6.478)40j30()994.3( 2CN

2

cC +=+== ZIS

=++= CBA SSSS V2.318j5.1167 A+ Chapter 12, Solution 56.

(a) Consider the circuit below.

For mesh 1,

0)(10j0440120-440 31 =−+°∠−°∠ II

°∠=+

=− 60-21.7610j

)866.0j5.1)(440(31 II (1)

For mesh 2,

0)(20120-440120440 32 =−+°∠−°∠ II

1.38j20

)732.1j)(440(23 ==− II (2)

For mesh 3, 05j)(20)(10j 32313 =−−+− IIIII

I2

I1

b −

+

+−

440∠-120°440∠120°

440∠0°

− +

BI3

20 Ω

j10 Ω

a A

-j5 Ω

C c

Substituting (1) and (2) into the equation for mesh 3 gives,

°∠=+

= 6042.152j5

j0.866)-1.5)(440(3I (3)

From (1),

°∠=+=°∠+= 3013266j315.11460-21.7631 II From (2),

°∠=+=−= 50.9493.1209.93j21.761.38j32 II

== 1a II A30132 °∠

=+=−= j27.9-38.10512b III A143.823.47 °∠

== 2c -II A230.99.120 °∠

(b) kVA08.58j)10j(2

31AB =−= IIS

kVA04.29)20(2

32BC =−= IIS

kVA-j116.16-j5)((152.42)-j5)( 22

3CA === IS

kVA08.58j04.29CABCAB −=++= SSSS Real power absorbed = 29 kW04.

(c) Total complex supplied by the source is =S kVA08.58j04.29 −


We apply mesh analysis to the circuit shown below. Ia

+ Va Ω+ 50j80 - I1 - - Ω+ 3020 j

Vc Vb + + Ib

I2 Ic

Ω− 4060 j

263.95165)3020()80100( 21 jVVIjIj ba +=−=+−+ (1) 53.190)1080()3020( 21 jVVIjIj cb −=−=−++− (2)

Solving (1) and (2) gives 722.19088.0,6084.08616.1 21 jIjI −=−= .

55.1304656.11136.1528.0,A 1.189585.1 121 bo

a jIIIII −∠=−−=−=−∠==

A 8.117947.12o

c II ∠=−= Chapter 12, Solution 58. The schematic is shown below. IPRINT is inserted in the neutral line to measure thecurrent through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq0.1592, and End Freq. = 0.1592. After simulation, the output file includes

FREQ IM(V_PRINT4) IP(V_PRINT4) 1.592 E–01 1.078 E+01 –8.997 E+01

A o

. =

i.e. In = 10.78∠–89.97° A

Chapter 12, Solution 59. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, we obtain an output file which includes

FREQ VM(1) VP(1) 6.000 E+01 2.206 E+02 –3.456 E+01 FREQ VM(2) VP(2) 6.000 E+01 2.141 E+02 –8.149 E+01 FREQ VM(3) VP(3) 6.000 E+01 4.991 E+01 –5.059 E+01

i.e. VAN = 220.6∠–34.56°, VBN = 214.1∠–81.49°, VCN = 49.91∠–50.59° V


he schematic is shown below. IPRINT is inserted to give Io. We select Total Pts = 1,

FREQ IM(V_PRINT4) IP(V_PRINT4)

.592 E–01 1.421 E+00 –1.355 E+02

om which, Io = 1.421∠–135.5° A

C TStart Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation, the output file includes

1

fr

Chapter 12, Solution 61. The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to measure IaA and VBN. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit is simulated, we get an output file which includes

FREQ VM(2) VP(2) 1.592 E–01 2.308 E+02 –1.334 E+02 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.115 E+01 3.699 E+01

from which IaA = 11.15∠37° A, VBN = 230.8∠–133.4° V

Chapter 12, Solution 62. Because of the delta-connected source involved, we follow Example 12.12. In the AC Sweep box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, the output file includes

FREQ IM(V_PRINT2) IP(V_PRINT2) 6.000 E+01 5.960 E+00 –9.141 E+01 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000 E+01 7.333 E+07 1.200 E+02

From which Iab = 7.333x107∠120° A, IbB = 5.96∠–91.41° A


Let F 0333.0X1 C and H, 20X/ L that so 1 =====ω

ωω

The schematic is shown below..

.

When the file is saved and run, we obtain an output file which includes the following: FREQ IM(V_PRINT1)IP(V_PRINT1) 1.592E-01 1.867E+01 1.589E+02 FREQ IM(V_PRINT2)IP(V_PRINT2) 1.592E-01 1.238E+01 1.441E+02 From the output file, the required currents are:

A 1.14438.12 A, 9.15867.18 oAC

oaA II ∠=∠=

Chapter 12, Solution 64. We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation the output file includes

FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 4.710 E+00 7.138 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 6.781 E+07 –1.426 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 3.898 E+00 –5.076 E+00 FREQ IM(V_PRINT4) IP(V_PRINT4) 1.592 E–01 3.547 E+00 6.157 E+01 FREQ IM(V_PRINT5) IP(V_PRINT5) 1.592 E–01 1.357 E+00 9.781 E+01 FREQ IM(V_PRINT6) IP(V_PRINT6) 1.592 E–01 3.831 E+00 –1.649 E+02

from this we obtain

IaA = 4.71∠71.38° A, IbB = 6.781∠–142.6° A, IcC = 3.898∠–5.08° A

IAB = 3.547∠61.57° A, IAC = 1.357∠97.81° A, IBC = 3.831∠–164.9° A

Chapter 12, Solution 65. Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it is saved and simulated, we obtain an output file which includes

FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 6.581 E+00 9.866 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.140 E+01 –1.113 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 6.581 E+00 3.866 E+01

Thus, IaA = 6.581∠98.66° A, IbB = 11.4∠–111.3 A, IcC = 6.581∠38.66° A

Chapter 12, Solution 66. Chapter 12, Solution 66.

(a) ===3

2083

VV L

p V120 (a)

===3

2083

VV L

p V120

(b) Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence,

(b) Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence,

A05.248

01201 °∠=

°∠=I

A120-340

120-1202 °∠=

°∠=I

A120260

1201203 °∠=

°∠=I

A05.248

01201 °∠=

°∠=I

A120-340

120-1202 °∠=

°∠=I

A120260

1201203 °∠=

°∠=I

++

−+=++=

23

j5.0-)2(23

j0.5-)3(5.2- 321N IIII

++

−+=++=

23

j5.0-)2(23

j0.5-)3(5.2- 321N IIII

A90866.0866.0j23

jN °∠===I

Hence, =1I A5.2 , =2I A3 , =3I A2 , =NI A866.0

(c) === )48()5.2(RIP 2

1211 W300

=== )40()3(RIP 22

222 W360

=== )60()2(RIP 23

233 W240

(d) =++= 321T PPPP W900


(a) The power to the motor is kW221)85.0)(260(cosSPT ==θ=

The motor power per phase is

kW67.73P31

P Tp ==

Hence, the wattmeter readings are as follows:

=+= 2467.73Wa kW67.97 =+= 1567.73Wb kW67.88 =+= 967.73Wc kW67.83

(b) The motor load is balanced so that 0IN = .

For the lighting loads,

A200120

000,24Ia ==

A125120

000,15Ib ==

A75120000,9

Ic ==

If we let

A02000Iaa °∠=°∠=I A120-125b °∠=I

A12075c °∠=I Then,

cbaN- IIII ++=

++

−+=

23

j0.5-)75(23

j5.0-)125(200- NI

A602.86100- N −=I

=NI A3.132 Chapter 12, Solution 68.

(a) === )4.8)(330(3IV3S LL VA4801

(b) SP

cospfcosSP =θ=→θ=

==24.4801

4500pf 9372.0

(c) For a wye-connected load,

== Lp II A4.8

(d) ===3

3303

VV L

p V53.190


MVA 8.0SMVA, 323.15.1)661.075.0(2SMVA, 72.096.0)6.08.0(2.1 321 =−=−=+=+= jjjS

9833.03153.326.3MVA, 603.026.3321 ===−=++=

SPpfjSSSS

MVA 1379.0)99.0tan(cos)9833.0[tan(cos26.3)tan(tan 11 =−=−= −−

newoldc PQ

mF 28106.6602

101379.031

62

6

==xxx

xxC

π


8004001200PPP 21T =−=+=

1600-1200400-PPQ 12T =−=−=

°=θ→===θ -63.43-28001600-

PQ

tanT

T

=θ= cospf (leading)4472.0

406

240IV

ZL

Lp ===

=pZ Ω°∠ 63.43-40


(a) If , °∠= 0208abV °∠= 120-208bcV , °∠= 120208caV ,

°∠=°∠

== 04.1020

0208

Ab

abAB Z

VI

°∠=°∠°∠

== 75-708.1445-210

120-208

BC

bcBC Z

VI

°∠=°∠°∠

== 97.381622.6213

120208

CA

caCA Z

VI

°∠−°∠=−= 97.381604.10CAABaA III

867.15j055.24.10aA −+=I °∠= 51.87-171.20aAI

°∠−°∠=−= 75-708.1497.8316BCCAcC III

°∠= 101.0364.30cCI

)cos(PaAab IVaAab1 θ−θ= IV

=°+°= )87.510cos()171.20)(208(P1 W2590

)cos(PcCcb IVcCcb2 θ−θ= IV

But °∠== 60208- bccb VV

=°−°= )03.10160cos()64.30)(208(P2 W4808

(b) W17.7398PPP 21T =+=

VAR25.3840)PP(3Q 12T =−= =+= TTT jQPS VA25.3840j17.7398 +

== TTS S V8335 A Chapter 12, Solution 72. From Problem 12.11,

V130220AB °∠=V and A18030aA °∠=I

=°−°= )180130cos()30)(220(P1 W4242

°∠== 190220- BCCB VV °∠= 60-30cCI

=°+°= )60190cos()30)(220(P2 W2257-

Chapter 12, Solution 73. Consider the circuit as shown below.

I1

I2

Z

Z

240∠-120° V − +

Ia

Ib

+ − 240∠-60° V

Z

Ic

°∠=+= 71.5762.3130j10Z

°∠=°∠

°∠= 131.57-59.7

71.5762.3160-240

aI

°∠=°∠°∠

= 191.57-59.771.5762.31120-240

bI

0120-24060-240c =°∠−°∠+ZI

°∠=°∠

= 108.4359.771.5731.62

240-cI

°∠=−= 101.57-146.13ca1 III °∠=+= 138.43146.13cb2 III

[ ] [ ] =°∠°∠== ).57101146.13)(60-240(ReReP *

111 IV W2360

[ ] [ ] =°∠°∠== )138.43-146.13)(120-240(ReReP *222 IV W8.632-


Z = 60 − j30 Ω

For mesh 1,

− + 208∠-60° V I2

I1+ − 208∠0° V

Z

Z

212208 IZIZ −= For mesh 2,

21 2-60-208- IZIZ +=°∠

In matrix form,

=

°∠ 2

1

2--2

60-208-208

II

ZZZZ

23Z=∆ , Z)866.0j5.1)(208(1 +=∆ , Z)732.1j)(208(2 =∆

°∠=−+

=∆∆

= 56.56789.1)30j60)(3(

)866.0j5.1)(208(11I

°∠=−

=∆∆

= 116.5679.1)30j60)(3()732.1j)(208(2

2I

[ ] [ ] =°∠== )56.56-789.1)(208(ReReP *

111 IV W98.208

[ ] [ ] =°∠°∠== )63.4479.1))(60-208(Re)-(ReP *222 IV W65.371


(a) ===60012

RV

I mA20

(b) ===600120

RV

I mA200

Chapter 12, Solution 76. If both appliances have the same power rating, P,

sVP

I =

For the 120-V appliance, 120

PI1 = .

For the 240-V appliance, 240P

I2 = .

Power loss =

=appliance -V240 for the

240RP

appliance -V120 for the120

RP

R2

2

2

2

2I

Since 22 2401

1201

> , the losses in the 120-V appliance are higher.

Chapter 12, Solution 77. , lineloadTg PPPP −−= 85.0pf =

But 3060pf3600cos3600PT =×=θ=

=−−= )80)(3(25003060Pg W320 Chapter 12, Solution 78.

°=θ→==θ 79.3185.06051

cos 11

kVAR61.31)5268.0)(60(sinSQ 111 ==θ=

kW51PP 12 ==

°=θ→=θ 19.1895.0cos 22

kVA68.53cos

PS

2

22 =

θ=


kVAR851.14759.1661.3QQQ 21c =−=−=

For each load,

kVAR95.43

QQ c

1c ==

=π

=ω

= 221c

)440)(60)(2(4950

VQ

C F82.67 µ

Chapter 12, Solution 79. Consider the per-phase equivalent circuit below.

Ia

n

a

Van + −

2 ΩA

ZY = 12 + j5 Ω

N

=+

°∠=

+=

5j140255

2Y

ana Z

VI A19.65-15.17 °∠

Thus,

=°∠= 120-ab II A139.65-15.17 °∠ =°∠= 120ac II A100.3515.17 °∠

=°∠°∠== )22.6213)(19.65-15.17(YaAN ZIV V2.97223 °∠

Thus,

=°∠= 120-ANBN VV V117.63-223 °∠ =°∠= 120ANCN VV V122.97223 °∠


)7071.07071.0(8)]83.0sin(cos83.0[6 21

321 jSjSSSS −+++=++= − kVA 31.26368.10 2SjS +−= (1)

But

kVA j18.28724.383 VA )6.08.0)(6.84)(208(33 +=+=∠= jIVS LL θ (2) From (1) and (2),

kVA 28.5676.246.20746.132 ∠=+= jS Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.


p

p

ZVjjS *

2

21 3SkVA, 240320)6.08.0(400 =+=+=

For the delta-connected load, V pL V=

kVA 93.8427.1053810)2400(3

2

2 jj

xS +=−

=

MVA 0829.13737.121 jSSS +=+=

Let I = I1 + I2 be the total line current. For I1,

3,3 1

*1

Lpp

VVIVS ==

735.5798.76,)2400(310)240320(

3 1

31

1* jIxj

VS

IL

−=+

==

For I2, convert the load to wye.

76.2891.273303810

24003032 jj

II oop −=−∠

+=−∠=

5.34735021 jIII −=+=

kV 372.5||kV 405.1185.5)63(2400 =→+=++=+= slineLs VjjIVVV


kVA 80SkVA, 135.60135.60)707.0707.0(95.0746120 21 =+=+= jjxxS

kVA 135.60135.14021 jSSS +=+=

A 42.1834803

1049.1523

||3||But 3

===→=xx

VSIIVSL

LLL

Chapter 12, Solution 84. We first find the magnitude of the various currents.

For the motor,

A248.53440

4000V3

SI

LL ===

For the capacitor,

A091.4440

1800VQ

IL

cC ===

For the lighting,

V2543

440Vp ==

A15.3254800

VP

Ip

LiLi ===

Consider the figure below.

Ia I1a

-jXC

R In

ILi

I3

I2

Ic

+

Vab

−Ib

IC

b

c

n

If , °∠= 0VpanV °∠= 30V3 pabV °∠= 120VpcnV

°∠== 120091.4X-j C

abC

VI

)30(091.4ab1 °+θ∠==

∆ZV

I

where θ °== 95.43)72.0(cos-1

°∠= 73.95249.51I

°∠= 46.05-249.52I

°∠= 193.95249.53I

°∠== 12015.3R

cnLi

VI

Thus,

°∠+°∠=+= 120091.473.95249.5C1a III =aI A93.96608.8 °∠

°∠−°∠=−= 120091.446.05-249.5C2b III

=bI A52.16-271.9 °∠

°∠+°∠=+= 12015.3193.95249.5Li3c III =cI A167.6827.6 °∠

== Lin -II A60-15.3 °∠

Chapter 12, Solution 85. Let Z RY =

V56.1383

2403

VV L

p ===

RV

kW92

27IVP

2p

pp ====

Ω=== 133.29000

)56.138(P

VR

22p

Thus, =YZ Ω133.2


1 Ω

B

N

A

b

n

I1

I2+ − 120∠0° V rms 15 + j4 Ω

24 – j2 Ω

1 Ω

a

+ − 120∠0° V rms

1 Ω

For the two meshes,

21)2j26(120 II −−= (1)

12)4j17(120 II −+= (2)

In matrix form,

+

−=

2

1

4j171-1-2j26

120120

II

70j449 +=∆ , )4j18)(120(1 +=∆ , )2j27)(120(2 −=∆

°∠=°∠°∠×

=∆∆

= 3.6787.48.8642.454

12.5344.1812011I

°∠=°∠

°∠×=

∆∆

= 13.1-15.78.8642.454

4.24-07.2712022I

== 1aA II A3.6787.4 °∠ == 2bB -II A166.915.7 °∠

∆∆−∆

=−= 1212nN III

=+−

=70j449

)6j9)(120(nNI A42.55-856.2 °∠

Chapter 12, Solution 87. 85.18j)5010)(60)(2(jLjmH50L 3- =π=ω→=Consider the circuit below.

1 Ω

30 Ω

20 ΩI1

I2+ −

2 Ω

+ −

1 Ω

115 V

15 + j18.85 Ω

115 V

Applying KVl to the three meshes, we obtain

11520223 321 =−− III (1) 11530332- 321 =−+ III (2)

0)85.18j65(3020- 321 =++− III (3) In matrix form,

=

+ 0115115

j18.856530-20-30-332-20-2-23

3

2

1

III

232,14j775,12 +=∆ , )8.659j1975)(115(1 +=∆

)3.471j1825)(115(2 +=∆ , )1450)(115(3 =∆

°∠=°∠°∠×

=∆∆

= 29.62-52.1248.0919214

47.18208211511I

°∠=°∠

°∠×=

∆∆

= 33.61-33.1148.0919124

14.489.188411522I

=+

−=

∆∆−∆

=−=75.231,14j775,12

)5.188j-150)(115(1212n III A176.6-448.1 °∠

=°∠== )29.6252.12)(115(IVS *

111 VA6.711j1252 +

=°∠== )33.6133.1)(115(*222 IVS V2.721j1085 A+

Chapter 13, Solution 1. For coil 1, L1 – M12 + M13 = 6 – 4 + 2 = 4

For coil 2, L2 – M21 – M23 = 8 – 4 – 5 = – 1

For coil 3, L3 + M31 – M32 = 10 + 2 – 5 = 7

LT = 4 – 1 + 7 = 10H

or LT = L1 + L2 + L3 – 2M12 – 2M23 + 2M12 LT = 6 + 8 + 10 = 10H Chapter 13, Solution 2. L = L1 + L2 + L3 + 2M12 – 2M23 2M31

= 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H Chapter 13, Solution 3. L1 + L2 + 2M = 250 mH (1) L1 + L2 – 2M = 150 mH (2)

Adding (1) and (2),

2L1 + 2L2 = 400 mH

But, L1 = 3L2,, or 8L2 + 400, and L2 = 50 mH

L1 = 3L2 = 150 mH From (2), 150 + 50 – 2M = 150 leads to M = 25 mH

k = M/ 150x50/5.2LL 21 = = 0.2887

Chapter 13, Solution 4. (a) For the series connection shown in Figure (a), the current I enters each coil from its dotted terminal. Therefore, the mutually induced voltages have the same sign as the self-induced voltages. Thus,

Leq = L1 + L2 + 2M

L2

L1

L2 L1

I1 I2

Is

Vs+–

Leq (a) (b) (b) For the parallel coil, consider Figure (b).

Is = I1 + I2 and Zeq = Vs/Is

Applying KVL to each branch gives,

Vs = jωL1I1 + jωMI2 (1)

Vs = jωMI1 + jω L2I2 (2)

or

ωωωω

=

2

1

2

1

s

s

II

LjMjMjLj

VV

∆ = –ω2L1L2 + ω2M2, ∆1 = jωVs(L2 – M), ∆2 = jωVs(L1 – M)

I1 = ∆1/∆, and I2 = ∆2/∆

Is = I1 + I2 = (∆1 + ∆2)/∆ = jω(L1 + L2 – 2M)Vs/( –ω2(L1L2 – M))

Zeq = Vs/Is = jω(L1L2 – M)/[jω(L1 + L2 – 2M)] = jωLeq

i.e., Leq = (L1L2 – M)/(L1 + L2 – 2M)


(a) If the coils are connected in series,

=++=++= 60x25)5.0(26025M2LLL 21 123.7 mH

(b) If they are connected in parallel,

=−+−

=−+−

= mH 36.19x26025

36.1960x25M2LL

MLLL2

21

221 24.31 mH


V1 = (R1 + jωL1)I1 – jωMI2

V2 = –jωMI1 + (R2 + jωL2)I2

Chapter 13, Solution 7. Applying KVL to the loop,

20∠30° = I(–j6 + j8 + j12 + 10 – j4x2) = I(10 + j6)

where I is the loop current.

I = 20∠30°/(10 + j6)

Vo = I(j12 + 10 – j4) = I(10 + j8)

= 20∠30°(10 + j8)/(10 + j6) = 22∠37.66° V Chapter 13, Solution 8. Consider the current as shown below.

j 6 j 4

1 Ω

-j3

j2

10 + – I2I1

4 Ω

+ Vo –

For mesh 1, 10 = (1 + j6)I1 + j2I2 (1)

For mesh 2, 0 = (4 + j4 – j3)I2 + j2I1 0 = j2I1 +(4 + j)I2 (2)

In matrix form,

+

+=

2

1

II

j42j2j6j1

010

∆ = 2 + j25, and ∆2 = –j20

I2 = ∆2/∆ = –j20/(2 + j25)

Vo = –j3I2 = –60/(2 + j25) = 2.392∠94.57°


2 Ω -j1 2 Ω

For loop 1,

-j2V+ – j 4 j 4

8∠30o + – I2I1

8∠30° = (2 + j4)I1 – jI2 (1)

For loop 2, ((j4 + 2 – j)I2 – jI1 + (–j2) = 0 or I1 = (3 – j2)i2 – 2 (2) Substituting (2) into (1), 8∠30° + (2 + j4)2 = (14 + j7)I2

I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12°

Vx = 2I2 = 2.074∠21.12°


Io

jωL jωL

1/jωC

I2I1

jωM

Iin∠0o

221 LLLkM == = L, I1 = Iin∠0°, I2 = Io

Io(jωL + R + 1/(jωC)) – jωLIin – (1/(jωC))Iin = 0 Io = j Iin(ωL – 1/(ωC)) /(R + jωL + 1/(jωC))


I1

jωL1R1

R2

jωL2 1/jωC

jωM

V2

+ –

V1 + –

I3

I2

For mesh 1, V1 = I1(R1 + 1/(jωC)) – I2(1/jωC)) –R1I3 For mesh 2,

0 = –I1(1/(jωC)) + (jωL1 + jωL2 + (1/(jωC)) – j2ωM)I2 – jωL1I3 + jωMI3

For mesh 3, –V2 = –R1I1 – jω(L1 – M)I2 + (R1 + R2 + jωL1)I3

or V2 = R1I1 + jω(L1 – M)I2 – (R1 + R2 + jωL1)I3


Let .1=ω j4 j2 • + j6 j8 j10 1V - I1 I2

• Applying KVL to the loops,

21 481 IjIj += (1)

21 1840 IjIj += (2) Solving (1) and (2) gives I1 = -j0.1406. Thus

H 111.711

11

==→==jI

LjLI

Z eqeq

We can also use the equivalent T-section for the transform to find the equivalent inductance. Chapter 13, Solution 13. We replace the coupled inductance with an equivalent T-section and use series and parallel combinations to calculate Z. Assuming that ,1=ω

10,101020,81018 21 ===−=−==−=−= MLMLLMLL cba The equivalent circuit is shown below:

12Ω j8Ω j10Ω 2 Ω j10Ω -j6Ω Z j4Ω Z=12 +j8 + j14//(2 + j4) = 13.195 + j11.244Ω Chapter 13, Solution 14. To obtain VTh, convert the current source to a voltage source as shown below.

5 Ω -j3 Ω

+ – 8 V +

VTh –

2 Ω j8 Ω

a

b

j10 V + –

j6 Ω

I

j2 Note that the two coils are connected series aiding.

ωL = ωL1 + ωL2 – 2ωM

jωL = j6 + j8 – j4 = j10 Thus, –j10 + (5 + j10 – j3 + 2)I + 8 = 0

I = (– 8 + j10)/ (7 + j7) But, –j10 + (5 + j6)I – j2I + VTh = 0

VTh = j10 – (5 + j4)I = j10 – (5 + j4)(–8 + j10)/(7 + j7)

VTh = 5.349∠34.11° To obtain ZTh, we set all the sources to zero and insert a 1-A current source at the terminals a–b as shown below.

I2I1 1 A

5 Ω j6 Ω j8 Ω -j3 Ωa

j2

2 Ω

+ Vo –

b Clearly, we now have only a super mesh to analyze.

(5 + j6)I1 – j2I2 + (2 + j8 – j3)I2 – j2I1 = 0 (5 + j4)I1 + (2 + j3)I2 = 0 (1) But, I2 – I1 = 1 or I2 = I1 – 1 (2) Substituting (2) into (1), (5 + j4)I1 +(2 + j3)(1 + I1) = 0

I1 = –(2 + j3)/(7 + j7) Now, ((5 + j6)I1 – j2I1 + Vo = 0

Vo = –(5 + j4)I1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.332∠50°

ZTh = Vo/1 = 2.332∠50° ohms


To obtain IN, short-circuit a–b as shown in Figure (a).

I1

I2 I1

20 Ω

j10 Ω

j20 Ω a

j5

1 + –

20 Ω

j10 Ω

j20 Ω a

j5

IN60∠30o

+ –

I2

b b (a) (b)For mesh 1,

60∠30° = (20 + j10)I1 + j5I2 – j10I2 or 12∠30° = (4 + j2)I1 – jI2 (1) For mesh 2,

0 = (j20 + j10)I2 – j5I1 – j10I1 or I1 = 2I2 (2) Substituting (2) into (1), 12∠30° = (8 + j3)I2

IN = I2 = 12∠30°/(8 + j3) = 1.404∠9.44° A To find ZN, we set all the sources to zero and insert a 1-volt voltage source at terminals a–b as shown in Figure (b). For mesh 1, 1 = I1(j10 + j20 – j5x2) + j5I2 1 = j20I1 + j5I2 (3) For mesh 2, 0 = (20 + j10)I2 + j5I1 – j10I1 = (4 + j2)I2 – jI1 or I2 = jI1/(4 + j2) (4) Substituting (4) into (3), 1 = j20I1 + j(j5)I1/(4 + j2) = (–1 + j20.5)I1

I1 = 1/(–1 + j20.5)

ZN = 1/I1 = (–1 + j20.5) ohms

Chapter 13, Solution 16. To find IN, we short-circuit a-b.

jΩ 8Ω -j2Ω a

••

+ j4Ω j6Ω I2 IN 80 V Io0∠ 1 - b

80)28(0)428(80 2121 =−+→=−+−+− jIIjjIIjj (1)

2112 606 IIjIIj =→=− (2) Solving (1) and (2) leads to

A 91.126246.1362.0584.11148

802

oN j

jII −∠=−=

+==

To find ZN, insert a 1-A current source at terminals a-b. Transforming the current source to voltage source gives the circuit below.

jΩ 8Ω -j2Ω 2Ω a

•• +

j4Ω j6Ω I2 2V I1 - b

28)28(0 2

121 jjIIjIIj+

=→−+= (3)

0)62(2 12 =−++ jIIj (4)

Solving (3) and (4) leads to I2 = -0.1055 +j0.2975, Vab=-j6I2 = 1.7853 +0.6332

Ω∠== oabN 53.19894.1

1V

Z


Z = -j6 // Zo where

7.15j5213.045j2j30j

14420jZo +=++−

+=

Ω−=+−

−= 7.9j1989.0

Z6jxZ6j

Zo

o

Chapter 13, Solution 18. Let 5105.0,20,5.1 2121 ====== xLLkMLLω We replace the transformer by its equivalent T-section.

5,25520,1055)( 11 −=−==+=+==+=−−= MLMLLMLL cba We find ZTh using the circuit below. -j4 j10 j25 j2 -j5 ZTh 4+j6

Ω+=++

+=++= 12.29215.274

)4(627)6//()4(27 jj

jjjjjjZTh

We find VTh by looking at the circuit below.

-j4 j10 j25 j2 + -j5 + VTh 120<0o

4+j6 - -

V 22.4637.61)120(64

4 oTh jj

jV −∠=+++

=


Let H 652540)(.1 1 =+=−−== MLLaω

25LH, 552530 C2 −=−==+=+= MMLLb

Thus, the T-section is as shown below. j65Ω j55Ω -j25Ω

Chapter 13, Solution 20. Transform the current source to a voltage source as shown below.

4 Ω

20∠0o + –

8 Ω

j12 + –

I3

j10 j10

-j5 I2I1

k=0.5

k = M/ 21LL or M = k 21LL

ωM = k 21 LL ωω = 0.5(10) = 5 For mesh 1, j12 = (4 + j10 – j5)I1 + j5I2 + j5I2 = (4 + j5)I1 + j10I2 (1) For mesh 2, 0 = 20 + (8 + j10 – j5)I2 + j5I1 + j5I1 –20 = +j10I1 + (8 + j5)I2 (2)

From (1) and (2),

++++

=

2

1

II

5j810j10j5j4

2012j

∆ = 107 + j60, ∆1 = –60 –j296, ∆2 = 40 – j100

I1 = ∆1/∆ = 2.462∠72.18° A

I2 = ∆2/∆ = 0.878∠–97.48° A

I3 = I1 – I2 = 3.329∠74.89° A

i1 = 2.462 cos(1000t + 72.18°) A

i2 = 0.878 cos(1000t – 97.48°) A

At t = 2 ms, 1000t = 2 rad = 114.6°

i1 = 0.9736cos(114.6° + 143.09°) = –2.445

i2 = 2.53cos(114.6° + 153.61°) = –0.8391 The total energy stored in the coupled coils is

w = 0.5L1i12 + 0.5L2i2

2 – Mi1i2 Since ωL1 = 10 and ω = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH

w = 0.5(10)(–2.445)2 + 0.5(10)(–0.8391)2 – 5(–2.445)(–0.8391)

w = 43.67 mJ Chapter 13, Solution 21. For mesh 1, 36∠30° = (7 + j6)I1 – (2 + j)I2 (1)

For mesh 2, 0 = (6 + j3 – j4)I2 – 2I1 jI1 = –(2 + j)I1 + (6 – j)I2 (2)

Placing (1) and (2) into matrix form,

−−−−−+

=

°∠

2

1

II

j6j2j26j7

03036

∆ = 48 + j35 = 59.41∠36.1°, ∆1 = (6 – j)36∠30° = 219∠20.54°

∆2 = (2 + j)36∠30° = 80.5∠56.56°, I1 = ∆1/∆ = 3.69∠–15.56°, I2 = ∆2/∆ = 1.355∠20.46°

Power absorbed fy the 4-ohm resistor, = 0.5(I2)24 = 2(1.355)2 = 3.672 watts

Chapter 13, Solution 22. With more complex mutually coupled circuits, it may be easier to show the effects of the coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure 13.85 then becomes,

Ix

Ib

Ia

-j50

j30Ib

j60

j20Ia

j10Ia

j80

j30Ic

j40 j10Ib

+ − + − − +

− +

I3

50∠0° V

j20Ic

Io

I2I1

− +

− +

+ −

100 Ω Note the following,

Ia = I1 – I3 Ib = I2 – I1 Ic = I3 – I2

and Io = I3

Now all we need to do is to write the mesh equations and to solve for Io. Loop # 1,

-50 + j20(I3 – I2) j 40(I1 – I3) + j10(I2 – I1) – j30(I3 – I2) + j80(I1 – I2) – j10(I1 – I3) = 0

j100I1 – j60I2 – j40I3 = 50 Multiplying everything by (1/j10) yields 10I1 – 6I2 – 4I3 = - j5 (1)

Loop # 2, j10(I1 – I3) + j80(I2–I1) + j30(I3–I2) – j30(I2 – I1) + j60(I2 – I3) – j20(I1 – I3) + 100I2 = 0

-j60I1 + (100 + j80)I2 – j20I3 = 0 (2)

Loop # 3,

-j50I3 +j20(I1 –I3) +j60(I3 –I2) +j30(I2 –I1) –j10(I2 –I1) +j40(I3 –I1) –j20(I3 –I2) = 0 -j40I1 – j20I2 + j10I3 = 0 Multiplying by (1/j10) yields, -4I1 – 2I2 + I3 = 0 (3) Multiplying (2) by (1/j20) yields -3I1 + (4 – j5)I2 – I3 = 0 (4) Multiplying (3) by (1/4) yields -I1 – 0.5I2 – 0.25I3 = 0 (5) Multiplying (4) by (-1/3) yields I1 – ((4/3) – j(5/3))I2 + (1/3)I3 = -j0.5 (7) Multiplying [(6)+(5)] by 12 yields (-22 + j20)I2 + 7I3 = 0 (8) Multiplying [(5)+(7)] by 20 yields -22I2 – 3I3 = -j10 (9) (8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3 (10) (9) leads to I3 = (j10 – 22I2)/3, substituting (1) into this equation produces, I3 = j3.333 + (-1.2273 – j1.1623)I3 or I3 = Io = 1.3040∠63o amp.

Chapter 13, Solution 23. ω = 10

0.5 H converts to jωL1 = j5 ohms

1 H converts to jωL2 = j10 ohms

0.2 H converts to jωM = j2 ohms

25 mF converts to 1/(jωC) = 1/(10x25x10-3) = –j4 ohms

The frequency-domain equivalent circuit is shown below.

j10

I2I1

+ −

–j4

j5 j2

5 Ω 12∠0°

For mesh 1, 12 = (j5 – j4)I1 + j2I2 – (–j4)I2 –j2 = I1 + 6I2 (1) For mesh 2, 0 = (5 + j10)I2 + j2I1 –(–j4)I1 0 = (5 + j10)I2 + j6I1 (2) From (1), I1 = –j12 – 6I2 Substituting this into (2) produces,

I2 = 72/(–5 + j26) = 2.7194∠–100.89°

I1 = –j12 – 6 I2 = –j12 – 163.17∠–100.89 = 5.068∠52.54°

Hence, i1 = 5.068cos(10t + 52.54°) A, i2 = 2.719cos(10t – 100.89°) A.

At t = 15 ms, 10t = 10x15x10-3 0.15 rad = 8.59°

i1 = 5.068cos(61.13°) = 2.446

i2 = 2.719cos(–92.3°) = –0.1089

w = 0.5(5)(2.446)2 + 0.5(1)(–0.1089)2 – (0.2)(2.446)(–0.1089) = 15.02 J Chapter 13, Solution 24. (a) k = M/ 21LL = 1/ 2x4 = 0.3535 (b) ω = 4 1/4 F leads to 1/(jωC) = –j/(4x0.25) = –j 1||(–j) = –j/(1 – j) = 0.5(1 – j) 1 H produces jωM = j4 4 H produces j16 2 H becomes j8

j4

j8

j16

2 Ω

I2I1

+ −

0.5(1–j) 12∠0°

12 = (2 + j16)I1 + j4I2 or 6 = (1 + j8)I1 + j2I2 (1) 0 = (j8 + 0.5 – j0.5)I2 + j4I1 or I1 = (0.5 + j7.5)I2/(–j4) (2) Substituting (2) into (1),

24 = (–11.5 – j51.5)I2 or I2 = –24/(11.5 + j51.5) = –0.455∠–77.41°

Vo = I2(0.5)(1 – j) = 0.3217∠57.59°

vo = 321.7cos(4t + 57.6°) mV (c) From (2), I1 = (0.5 + j7.5)I2/(–j4) = 0.855∠–81.21°

i1 = 0.885cos(4t – 81.21°) A, i2 = –0.455cos(4t – 77.41°) A At t = 2s,

4t = 8 rad = 98.37°

i1 = 0.885cos(98.37° – 81.21°) = 0.8169

i2 = –0.455cos(98.37° – 77.41°) = –0.4249

w = 0.5L1i12 + 0.5L2i2

2 + Mi1i2

= 0.5(4)(0.8169)2 + 0.5(2)(–.4249)2 + (1)(0.1869)(–0.4249) = 1.168 J

Chapter 13, Solution 25. m = k 21LL = 0.5 H

We transform the circuit to frequency domain as shown below.

12sin2t converts to 12∠0°, ω = 2

0.5 F converts to 1/(jωC) = –j

2 H becomes jωL = j4 j1

–j1

b

a

10 Ω

2 ΩIo 4 Ω

j2j2

1 Ω

+ −

j4 12∠0° Applying the concept of reflected impedance,

Zab = (2 – j)||(1 + j2 + (1)2/(j2 + 3 + j4))

= (2 – j)||(1 + j2 + (3/45) – j6/45)

= (2 – j)||(1 + j2 + (3/45) – j6/45)

= (2 – j)||(1.0667 + j1.8667)

=(2 – j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.5085∠17.9° ohms

Io = 12∠0°/(Zab + 4) = 12/(5.4355 + j0.4636) = 2.2∠–4.88°

io = 2.2sin(2t – 4.88°) A


M = k 21LL ωM = k 21 LL ωω = 0.6 40x20 = 17 The frequency-domain equivalent circuit is shown below.

j17

Io–j30

I2I1

50 Ω

j40j20

+ −

10 Ω 200∠60°

For mesh 1, 200∠60° = (50 – j30 + j20)I1 + j17I2 = (50 – j10)I1 + j17I2 (1)

For mesh 2, 0 = (10 + j40)I2 + j17I1 (2)

In matrix form,

+

−=

°∠

2

1

II

40j1017j17j10j50

060200

∆ = 900 + j100, ∆1 = 2000∠60°(1 + j4) = 8246.2∠136°, ∆2 = 3400∠–30°

I2 = ∆2/∆ = 3.755∠–36.34° Io = I2 = 3.755∠–36.34° A Switching the dot on the winding on the right only reverses the direction of Io. This can be seen by looking at the resulting value of ∆2 which now becomes 3400∠150°. Thus, Io = 3.755∠143.66° A


Zin = –j4 + j5 + 9/(12 + j6) = 0.6 + j.07 = 0.922∠49.4° I1 = 12∠0°/0.922∠49.4° = 13∠–49.4° A


We find ZTh by replacing the 20-ohm load with a unit source as shown below.

j10Ω 8 -jX Ω

•• + j12Ω j15Ω I2 1V - I1 For mesh 1, 0 21 10)128( IjIjjX −+−= (1) For mesh 2, (2) jIIIjIj 1.05.1010151 2112 −=→=−+ Substituting (2) into (1) leads to

XjjXjI

5.18121.08.02.1

2 −+++−

=

XjXjj

IZTh 1.08.02.1

5.18121

2 −−−+

=−

=

6247275.108.0)1.02.1(

)5.18(1220|| 2

22

22

−+=→+−

−+== XX

X

XZTh

Solving the quadratic equation yields X = 6.425


30 mH becomes jωL = j30x10-3x103 = j30 50 mH becomes j50 Let X = ωM Using the concept of reflected impedance,

Zin = 10 + j30 + X2/(20 + j50) I1 = V/Zin = 165/(10 + j30 + X2/(20 + j50)) p = 0.5|I1|2(10) = 320 leads to |I1|2 = 64 or |I1| = 8 8 = |165(20 + j50)/(X2 + (10 + j30)(20 + j50))|

= |165(20 + j50)/(X2 – 1300 + j1100)|

or 64 = 27225(400 + 2500)/((X2 – 1300)2 + 1,210,000) (X2 – 1300)2 + 1,210,000 = 1,233,633 X = 33.86 or 38.13

If X = 38.127 = ωM

M = 38.127 mH k = M/ 21LL = 38.127/ 50x30 = 0.984

j38.127

I2I1

10 Ω

j50j30

+ −

20 Ω 165∠0°

165 = (10 + j30)I1 – j38.127I2 (1) 0 = (20 + j50)I2 – j38.127I1 (2)

In matrix form,

+−

−+=

2

1

II

50j20127.38j127.38j30j10

0165

∆ = 154 + j1100 = 1110.73∠82.03°, ∆1 = 888.5∠68.2°, ∆2 = j6291 I1 = ∆1/∆ = 8∠–13.81°, I2 = ∆2/∆ = 5.664∠7.97° i1 = 8cos(1000t – 13.83°), i2 = 5.664cos(1000t + 7.97°)

At t = 1.5 ms, 1000t = 1.5 rad = 85.94°

i1 = 8cos(85.94° – 13.83°) = 2.457 i2 = 5.664cos(85.94° + 7.97°) = –0.3862 w = 0.5L1i1

2 + 0.5L2i22 + Mi1i2

= 0.5(30)(2.547)2 + 0.5(50)(–0.3862)2 – 38.127(2.547)(–0.3862)

= 130.51 mJ


(a) Zin = j40 + 25 + j30 + (10)2/(8 + j20 – j6) = 25 + j70 + 100/(8 + j14) = (28.08 + j64.62) ohms (b) jωLa = j30 – j10 = j20, jωLb = j20 – j10 = j10, jωLc = j10 Thus the Thevenin Equivalent of the linear transformer is shown below.

j40 j20 j10 25 Ω 8 Ω

Zin

j10

–j6

Zin = j40 + 25 + j20 + j10||(8 + j4) = 25 + j60 + j10(8 + j4)/(8 + j14) = (28.08 + j64.62) ohms


(a) La = L1 – M = 10 H Lb = L2 – M = 15 H Lc = M = 5 H (b) L1L2 – M2 = 300 – 25 = 275

LA = (L1L2 – M2)/(L1 – M) = 275/15 = 18.33 H

LB = (L1L2 – M2)/(L1 – M) = 275/10 = 27.5 H

LC = (L1L2 – M2)/M = 275/5 = 55 H Chapter 13, Solution 32. We first find Zin for the second stage using the concept of reflected impedance.

Zin’

LBLb

R Zin’ = jωLb + ω2Mb

2/(R + jωLb) = (jωLbR - ω2Lb2 + ω2Mb

2)/(R + jωLb) (1) For the first stage, we have the circuit below.

Zin

LALa Zin’

Zin = jωLa + ω2Ma2/(jωLa + Zin)

= (–ω2La

2 + ω2Ma2 + jωLaZin)/( jωLa + Zin) (2)

Substituting (1) into (2) gives,

=

b

2b

22b

2b

a

b

2b

22b

2b

a2a

22a

2

LjRMLRLj

Lj

LjR)MLRLj(

LjML

ω+ω+ω−ω

+ω

ω+ω+ω−ω

ω+ω+ω−

–Rω2La

2 + ω2Ma2R – jω3LbLa + jω3LbMa

2 + jωLa(jωLbR – ω2Lb2 + ω2Mb

2) =

jωRLa –ω2LaLb + jωLbR – ω2La2 + ω2Mb

2

ω2R(La2 + LaLb – Ma

2) + jω3(La2Lb + LaLb

2 – LaMb2 – LbMa

2) Zin =

ω2(LaLb +Lb2 – Mb

2) – jωR(La +Lb) Chapter 13, Solution 33.

Zin = 10 + j12 + (15)2/(20 + j 40 – j5) = 10 + j12 + 225/(20 + j35)

= 10 + j12 + 225(20 – j35)/(400 + 1225)

= (12.769 + j7.154) ohms Chapter 13, Solution 34. Insert a 1-V voltage source at the input as shown below. j6Ω 1Ω 8Ω

•• + j12Ω j10Ω j4Ω 1<0o V I1 I2 - -j2Ω

For loop 1, 21 4)101(1 IjIj −+= (1)

For loop 2,

21112 )32(062)21048(0 IjjIIjIjIjjj ++−=→−+−++= (2) Solving (1) and (2) leads to I1=0.019 –j0.1068

Ω∠=+== ojI

Z 91.79219.9077.96154.11

1

Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent T circuit and use series/parallel impedance combinations. This leads to exactly the same result. Chapter 13, Solution 35.

For mesh 1, 21 2)410(16 IjIj ++= (1) For mesh 2, 321 12)2630(20 IjIjIj −++= (2) For mesh 3, 32 )115(120 IjIj ++−= (3) We may use MATLAB to solve (1) to (3) and obtain

A 41.214754.15385.03736.11

ojI −∠=−= A 85.1340775.00549.00547.02

ojI −∠=−−= A 41.110077.00721.00268.03

ojI −∠=−−= Chapter 13, Solution 36.

Following the two rules in section 13.5, we obtain the following: (a) V2/V1 = –n, I2/I1 = –1/n (n = V2/V1) (b) V2/V1 = –n, I2/I1 = –1/n (c) V2/V1 = n, I2/I1 = 1/n (d) V2/V1 = n, I2/I1 = –1/n


(a) 5480

2400

1

2 ===VVn

(b) A 17.104480

000,50000,50 1222111 ==→==== IVISVIS

(c ) A 83.202400

000,502 ==I


Zin = Zp + ZL/n2, n = v2/v1 = 230/2300 = 0.1

v2 = 230 V, s2 = v2I2*

I2

* = s2/v2 = 17.391∠–53.13° or I2 = 17.391∠53.13° A

ZL = v2/I2 = 230∠0°/17.391∠53.13° = 13.235∠–53.13°

Zin = 2∠10° + 1323.5∠–53.13°

= 1.97 + j0.3473 + 794.1 – j1058.8

Zin = 1.324∠–53.05° kohms Chapter 13, Solution 39. Referred to the high-voltage side,

ZL = (1200/240)2(0.8∠10°) = 20∠10°

Zin = 60∠–30° + 20∠10° = 76.4122∠–20.31°

I1 = 1200/Zin = 1200/76.4122∠–20.31° = 15.7∠20.31° A

Since S = I1v1 = I2v2, I2 = I1v1/v2

= (1200/240)( 15.7∠20.31°) = 78.5∠20.31° A


V 60)240(41nVV

VV

n,41

2000500

NN

n 121

2

1

2 ===→====

W30012

60R

VP22===

Chapter 13, Solution 41. We reflect the 2-ohm resistor to the primary side.

Zin = 10 + 2/n2, n = –1/3

Since both I1 and I2 enter the dotted terminals, Zin = 10 + 18 = 28 ohms

I1 = 14∠0°/28 = 0.5 A and I2 = I1/n = 0.5/(–1/3) = –1.5 A Chapter 13, Solution 42. 10Ω 1:4 -j50Ω

+ •• + + + V1 V2 20 VΩ o 120<0o V I1 - - - - I2

Applying mesh analysis,

11 VI10120 += (1)

22 VI)50j20(0 +−= (2)

At the terminals of the transformer,

121

2 V4V4nVV

=→== (3)

211

2 I4I41

n1

II

−=→−=−= (4)

Substituting (3) and (4) into (1) gives 120 22 V25.0I40 +−= (5)

Solving (2) and (5) yields 6877.0j4756.2I2 −−=

V 52.1539.51I20V o

2o ∠=−= Chapter 13, Solution 43. Transform the two current sources to voltage sources, as shown below.

+

v1

−

+

v2

−

1 : 4

12V+ –

10 Ω

20 V + – I2I1

12 Ω Using mesh analysis, –20 + 10I1 + v1 = 0 20 = v1 + 10I1 (1) 12 + 12I2 – v2 = 0 or 12 = v2 – 12I2 (2) At the transformer terminal, v2 = nv1 = 4v1 (3) I1 = nI2 = 4I2 (4) Substituting (3) and (4) into (1) and (2), we get, 20 = v1 + 40I2 (5) 12 = 4v1 – 12I2 (6) Solving (5) and (6) gives v1 = 4.186 V and v2 = 4v = 16.744 V


We can apply the superposition theorem. Let i1 = i1’ + i1” and i2 = i2’ + i2” where the single prime is due to the DC source and the double prime is due to the AC source. Since we are looking for the steady-state values of i1 and i2,

i1’ = i2’ = 0.

For the AC source, consider the circuit below.

+–i2”

+

v1

−

1 : n

+

v2

−

R

i1”

Vn∠0°

v2/v1 = –n, I2”/I1” = –1/n

But v2 = vm, v1 = –vm/n or I1” = vm/(Rn)

I2” = –I1”/n = –vm/(Rn2)

Hence, i1(t) = (vm/Rn)cosωt A, and i2(t) = (–vm/(n2R))cosωt A Chapter 13, Solution 45. 48 Ω

+ −

Z4∠–90˚

4j8Cj8ZL −=

ω−= , n = 1/3

°−∠=°−∠

°−∠=

−+°−∠

=

−===

3.7303193.07.1628.125

90436j7248

904I

36j72Z9n

ZZ L2L

We now have some choices, we can go ahead and calculate the current in the second loop and calculate the power delivered to the 8-ohm resistor directly or we can merely say that the power delivered to the equivalent resistor in the primary side must be the same as the power delivered to the 8-ohm resistor. Therefore,

=== −Ω 7210x5098.072

2IP 3

28 36.71 mW

The student is encouraged to calculate the current in the secondary and calculate the power delivered to the 8-ohm resistor to verify that the above is correct. Chapter 13, Solution 46. (a) Reflecting the secondary circuit to the primary, we have the circuit shown below.

Zin

+ −

+ − 16∠60° I1

10∠30°/(–n) = –5∠30°

Zin = 10 + j16 + (1/4)(12 – j8) = 13 + j14 –16∠60° + ZinI1 – 5∠30° = 0 or I1 = (16∠60° + 5∠30°)/(13 + j14) Hence, I1 = 1.072∠5.88° A, and I2 = –0.5I1 = 0.536∠185.88° A

(b) Switching a dot will not effect Zin but will effect I1 and I2.

I1 = (16∠60° – 5∠30°)/(13 + j14) = 0.625 ∠25 A and I2 = 0.5I1 = 0.3125∠25° A


0.02 F becomes 1/(jωC) = 1/(j5x0.02) = –j10 We apply mesh analysis to the circuit shown below.

–j10

+

vo

−

I3

+

v1

−

+

v2

−

3 : 1 10 Ω

10∠0° + – I2I1

2 Ω For mesh 1, 10 = 10I1 – 10I3 + v1 (1) For mesh 2, v2 = 2I2 = vo (2) For mesh 3, 0 = (10 – j10)I3 – 10I1 + v2 – v1 (3) At the terminals, v2 = nv1 = v1/3 (4) I1 = nI2 = I2/3 (5) From (2) and (4), v1 = 6I2 (6) Substituting this into (1), 10 = 10I1 – 10I3 (7) Substituting (4) and (6) into (3) yields 0 = –10I1 – 4I2 + 10(1 – j)I3 (8) From (5), (7), and (8)

=

−−−−

−

0100

III

10j10410106100333.01

3

2

1

I2 = 33.93j20

100j1002

−−−

=∆∆

= 1.482∠32.9°

vo = 2I2 = 2.963∠32.9° V (a) Switching the dot on the secondary side effects only equations (4) and (5). v2 = –v1/3 (9) I1 = –I2/3 (10) From (2) and (9), v1 = –6I2 Substituting this into (1),

10 = 10I1 – 10I3 – 6I2 = (23 – j5)I1 (11) Substituting (9) and (10) into (3),

0 = –10I1 + 4I2 + 10(1 – j)I3 (12)

From (10) to (12), we get

=

−−−−

0100

III

10j10410106100333.01

3

2

1

I2 = 33.93j20

100j1002

+−−

=∆∆

= 1.482∠–147.1°

vo = 2I2 = 2.963∠–147.1° V Chapter 13, Solution 48. We apply mesh analysis. 8Ω 2:1 10Ω + + •• + V1 V2 I1 - j6Ω 100∠0o V - I2 - Ix -j4Ω

121 4)48(100 VIjIj +−−= (1)

212 4)210(0 VIjIj +−+= (2)

But

211

2 221 VVn

VV

=→== (3)

211

2 5.021 IInI

I−=→−=−= (4)

Substituting (3) and (4) into (1) and (2), we obtain

22 2)24(100 VIj +−−= (1)a

22)410(0 VIj ++= (2)a

Solving (1)a and (2)a leads to I2 = -3.5503 +j1.4793

A 4.157923.15.0 221o

x IIII ∠==+= Chapter 13, Solution 49.

101F 201,2 j

Cj−=→=

ωω

Ix -j10 2 Ω I1 1:3 I2 1 2 + • + + V1 V2 6Ω 12<0o V - - - •

At node 1,

2111211 2.0)2.01(212

10212 VjjVII

jVVV

−++=→+−−

=−

(1)

At node 2,

212221

2 )6.01(6.060610

VjVjIVjVVI +−+=→=

−−

+ (2)

At the terminals of the transformer, 1212 31,3 IIV −=−=V

Substituting these in (1) and (2),

)4.23(60),8.01(612 1212 jVIjVI ++=++−= Adding these gives V1=1.829 –j1.463 and

ox j

VjVVI 34.51937.0

104

10121 ∠=

−=

−−

=

A )34.512cos(937.0 o

x ti += Chapter 13, Solution 50.

The value of Zin is not effected by the location of the dots since n2 is involved.

Zin’ = (6 – j10)/(n’)2, n’ = 1/4 Zin’ = 16(6 – j10) = 96 – j160 Zin = 8 + j12 + (Zin’ + 24)/n2, n = 5 Zin = 8 + j12 + (120 – j160)/25 = 8 + j12 + 4.8 – j6.4 Zin = (12.8 + j5.6) ohms

Chapter 13, Solution 51. Let Z3 = 36 +j18, where Z3 is reflected to the middle circuit.

ZR’ = ZL/n2 = (12 + j2)/4 = 3 + j0.5 Zin = 5 – j2 + ZR’ = (8 – j1.5) ohms I1 = 24∠0°/ZTh = 24∠0°/(8 – j1.5) = 24∠0°/8.14∠–10.62° = 8.95∠10.62° A

Chapter 13, Solution 52. For maximum power transfer,

40 = ZL/n2 = 10/n2 or n2 = 10/40 which yields n = 1/2 = 0.5 I = 120/(40 + 40) = 3/2 p = I2R = (9/4)x40 = 90 watts.


(a) The Thevenin equivalent to the left of the transformer is shown below. 8 Ω

20 V

+ −

The reflected load impedance is ZL’ = ZL/n2 = 200/n2. For maximum power transfer, 8 = 200/n2 produces n = 5.

(b) If n = 10, ZL’ = 200/10 = 2 and I = 20/(8 + 2) = 2

p = I2ZL’ = (2)2(2) = 8 watts. Chapter 13, Solution 54. (a)

I1

+ −

ZTh

ZL/n2 VS

For maximum power transfer,

ZTh = ZL/n2, or n2 = ZL/ZTh = 8/128

n = 0.25 (b) I1 = VTh/(ZTh + ZL/n2) = 10/(128 + 128) = 39.06 mA (c) v2 = I2ZL = 156.24x8 mV = 1.25 V But v2 = nv1 therefore v1 = v2/n = 4(1.25) = 5 V Chapter 13, Solution 55. We reflect Zs to the primary side.

ZR = (500 – j200)/n2 = 5 – j2, Zin = Zp + ZR = 3 + j4 + 5 – j2 = 8 + j2

I1 = 120∠0°/(8 + j2) = 14.552∠–14.04°

I2I1

+ −

1 : n Zp Vp Zs

Since both currents enter the dotted terminals as shown above,

I2 = –(1/n)I1 = –1.4552∠–14.04° = 1.4552∠166° S2 = |I2|2Zs = (1.4552)(500 – j200) P2 = Re(S2) = (1.4552)2(500) = 1054 watts

Chapter 13, Solution 56. We apply mesh analysis to the circuit as shown below.

+

v1

−

5 Ω

+

v2

− I2I1

+ −

1 : 2 2 Ω 10 Ω 46V For mesh 1, 46 = 7I1 – 5I2 + v1 (1) For mesh 2, v2 = 15I2 – 5I1 (2) At the terminals of the transformer, v2 = nv1 = 2v1 (3) I1 = nI2 = 2I2 (4) Substituting (3) and (4) into (1) and (2), 46 = 9I2 + v1 (5) v1 = 2.5I2 (6) Combining (5) and (6), 46 = 11.5I2 or I2 = 4 P10 = 0.5I2

2(10) = 80 watts.


(a) ZL = j3||(12 – j6) = j3(12 – j6)/(12 – j3) = (12 + j54)/17

Reflecting this to the primary side gives Zin = 2 + ZL/n2 = 2 + (3 + j13.5)/17 = 2.3168∠20.04° I1 = vs/Zin = 60∠90°/2.3168∠20.04° = 25.9∠69.96° A(rms) I2 = I1/n = 12.95∠69.96° A(rms)

(b) 60∠90° = 2I1 + v1 or v1 = j60 –2I1 = j60 – 51.8∠69.96°

v1 = 21.06∠147.44° V(rms) v2 = nv1 = 42.12∠147.44° V(rms) vo = v2 = 42.12∠147.44° V(rms)

(c) S = vsI1

* = (60∠90°)(25.9∠–69.96°) = 1554∠20.04° VA Chapter 13, Solution 58. Consider the circuit below. 20 Ω

+

vo

−

I3

+

v1

−

+

v2

−

1 : 5 20 Ω

80∠0° + – I2I1

100 Ω

For mesh1, 80 = 20I1 – 20I3 + v1 (1) For mesh 2, v2 = 100I2 (2)

For mesh 3, 0 = 40I3 – 20I1 which leads to I1 = 2I3 (3) At the transformer terminals, v2 = –nv1 = –5v1 (4) I1 = –nI2 = –5I2 (5) From (2) and (4), –5v1 = 100I2 or v1 = –20I2 (6) Substituting (3), (5), and (6) into (1),

4 = I1 – I2 – I 3 = I1 – (I1/(–5)) – I1/2 = (7/10)I1 I1 = 40/7, I2 = –8/7, I3 = 20/7

p20(the one between 1 and 3) = 0.5(20)(I1 – I3)2 = 10(20/7)2 = 81.63 watts p20(at the top of the circuit) = 0.5(20)I3

2 = 81.63 watts p100 = 0.5(100)I2

2 = 65.31 watts Chapter 13, Solution 59. We apply nodal analysis to the circuit below. 2 Ω

I3

v2 v1 I1

4 Ω

+

v1

−

+

v2

−

2 : 1 8 Ω

20∠0° + –

I2

20 = 8I1 + V1 (1) V1 = 2I3 + V2 (2)

V2 = 4I2 (3) At the transformer terminals, v2 = 0.5v1 (4) I1 = 0.5I2 (5) Solving (1) to (5) gives I1 = 0.833 A, I2 = 1.667 A, I3 = 3.333 A V1 = 13.33 V, V2 = 6.667 V.

P8Ω = 0.5(8)|(20 – V1)/8|2 = 2.778 W P2Ω = 0.5(2)I3

2 = 11.11 W, P4Ω = 0.5V22/4 = 5.556 W

Chapter 13, Solution 60. (a) Transferring the 40-ohm load to the middle circuit,

ZL’ = 40/(n’)2 = 10 ohms where n’ = 2 10||(5 + 10) = 6 ohms We transfer this to the primary side. Zin = 4 + 6/n2 = 4 + 96 = 100 ohms, where n = 0.25 I1 = 120/100 = 1.2 A and I2 = I1/n = 4.8 A

I2’

120∠0° + –

5 Ω

10 Ω

I1

+

v1

−

+

v2

−

1 : 4 4 Ω I2 10 Ω

Using current division, I2’ = (10/25)I2 = 1.92 and I3 = I2’/n’ = 0.96 A (b) p = 0.5(I3)2(40) = 18.432 watts

Chapter 13, Solution 61. We reflect the 160-ohm load to the middle circuit.

ZR = ZL/n2 = 160/(4/3)2 = 90 ohms, where n = 4/3

Io’

24∠0° + –

14 Ω

60 Ω

I1

+

v1

−

+

vo

−

1 : 5 2 Ω Io

90 Ω

14 + 60||90 = 14 + 36 = 50 ohms

We reflect this to the primary side.

ZR’ = ZL’/(n’)2 = 50/52 = 2 ohms when n’ = 5 I1 = 24/(2 + 2) = 6A 24 = 2I1 + v1 or v1 = 24 – 2I1 = 12 V vo = –nv1 = –60 V, Io = –I1 /n1 = –6/5 = –1.2 Io‘ = [60/(60 + 90)]Io = –0.48A I2 = –Io’/n = 0.48/(4/3) = 0.36 A

Chapter 13, Solution 62. (a) Reflect the load to the middle circuit.

ZL’ = 8 – j20 + (18 + j45)/32 = 10 – j15 We now reflect this to the primary circuit so that Zin = 6 + j4 + (10 – j15)/n2 = 7.6 + j1.6 = 7.767∠11.89°, where n = 5/2 = 2.5 I1 = 40/Zin = 40/7.767∠11.89° = 5.15∠–11.89° S = 0.5vsI1

* = (20∠0°)(5.15∠11.89°) = 103∠11.89° VA

(b) I2 = –I1/n, n = 2.5

I3 = –I2/n’, n = 3

I3 = I1/(nn’) = 5.15∠–11.89°/(2.5x3) = 0.6867∠–11.89°

p = 0.5|I2|2(18) = 9(0.6867)2 = 4.244 watts Chapter 13, Solution 63. Reflecting the (9 + j18)-ohm load to the middle circuit gives,

Zin’ = 7 – j6 + (9 + j18)/(n’)2 = 7 – j6 + 1 + j12 = 8 + j4 when n’ = 3 Reflecting this to the primary side, Zin = 1 + Zin’/n2 = 1 + 2 – j = 3 – j, where n = 2 I1 = 12∠0°/(3 – j) = 12/3.162∠–18.43° = 3.795∠18.43A I2 = I1/n = 1.8975∠18.43° A I3 = –I2/n2 = 632.5∠161.57° mA

Chapter 13, Solution 64. We find ZTh at the terminals of Z by considering the circuit below. 10Ω 1:n + •• + V1 V2 + – – I1 I2 1 V – 20Ω

For mesh 1, 02030 121 =++ VII (1) For mesh 2, 12020 221 =++ VII (2)

At the terminals, nIInVV 1

212 , −==

Substituting these in (1) and (2) leads to

1)1(20,0)3020( 1212 =+−=+− nVInVIn Solving these gives

5.72040301204030

1 2

222 =+−==→

+−= nn

IZ

nnI Th

Solving the quadratic equation yields n=0.5 or 0.8333 Chapter 13, Solution 65.

40Ω 10Ω I1 1:2 I2 50Ω I2 1:3 I3 1 • + •• 2 + + + + 200 V V3 V4 (rms) V1 V2 - - 20 Ω- - - • At node 1,

1411411 1025.025.1200

4010200 IVVIVVV

+−=→+−

=− (1)

At node 2,

3413441 403

2040IVVIVVV

+=→+=− (2)

At the terminals of the first transformer,

121

2 22 VVVV

−=→−= (3)

211

2 22/1 IIII

−=→−= (4)

For the middle loop,

223322 50050 IVVVIV −=→=++− (5) At the terminals of the second transformer,

343

4 33 VVVV

=→= (6)

322

3 33/1 IIII

−=→−= (7)

We have seven equations and seven unknowns. Combining (1) and (2) leads to

314 50105.3200 IIV ++= But from (4) and (7), 3321 6)3(22 IIII =−−=−= . Hence

34 1105.3200 IV += (8) From (5), (6), (3), and (7),

3122224 45061503)50(3 IVIVIVV +−=−=−= Substituting for V1 in (2) gives

433344 21019450)403(6 VIIIVV =→++−= (9)


87.14452.13200 44 =→= VV

W05.1120

42

==VP


v1 = 420 V (1) v2 = 120I2 (2) v1/v2 = 1/4 or v2 = 4v1 (3) I1/I2 = 4 or I1 = 4 I2 (4)


v2 = 120[(1/4)I1] = 30 I1 4v1 = 30I1 4(420) = 1680 = 30I1 or I1 = 56 A


(a) V 1604004.04.04.0

112

2

21

2

1 ===→=+

= xVVN

NNVV

(b) A 25.311605000000,5 2222 ==→== IVIS

(c ) A 5.12400

5000000,5 21112 ==→=== IVISS

Chapter 13, Solution 68. This is a step-up transformer.

I1

I2

N12 – j6

+

v1

−

+

v2

−

N2 + −

10 + j40 20∠30°

For the primary circuit, 20∠30° = (2 – j6)I1 + v1 (1) For the secondary circuit, v2 = (10 + j40)I2 (2)

At the autotransformer terminals,

v1/v2 = N1/(N1 + N2) = 200/280 = 5/7,

thus v2 = 7v1/5 (3) Also, I1/I2 = 7/5 or I2 = 5I1/7 (4)

Substituting (3) and (4) into (2), v1 = (10 + j40)25I1/49 Substituting that into (1) gives 20∠30° = (7.102 + j14.408)I1 I1 = 20∠30°/16.063∠63.76° = 1.245∠–33.76° A I2 = 5I1/7 = 0.8893∠–33.76° A Io = I1 – I2 = [(5/7) – 1]I1 = –2I1/7 = 0.3557∠146.2° A p = |I2|2R = (0.8893)2(10) = 7.51 watts

Chapter 13, Solution 69. We can find the Thevenin equivalent.

I1

I2

75 Ω

+

VTh

−

N2

j125 Ω

+

v1

−

+

v2

−

N1 + − 120∠0°

I1 = I2 = 0

As a step up transformer, v1/v2 = N1/(N1 + N2) = 600/800 = 3/4

v2 = 4v1/3 = 4(120)/3 = 160∠0° rms = VTh. To find ZTh, connect a 1-V source at the secondary terminals. We now have a step-down transformer.

I2

I1

75 Ω j125 Ω

+

v2

−

+

v1

−

+ − 1∠0° V

v1 = 1V, v2 =I2(75 + j125)

But v1/v2 = (N1 + N2)/N1 = 800/200 which leads to v1 = 4v2 = 1

and v2 = 0.25

I1/I2 = 200/800 = 1/4 which leads to I2 = 4I1

Hence 0.25 = 4I1(75 + j125) or I1 = 1/[16(75 + j125)

ZTh = 1/I1 = 16(75 + j125) Therefore, ZL = ZTh

* = (1.2 – j2) kΩ Since VTh is rms, p = (|VTh|/2)2/RL = (80)2/1200 = 5.333 watts

Chapter 13, Solution 70. This is a step-down transformer.

I1

I2

+

v2

−

+

v1

−

+ −

30 + j12 120∠0° 20 – j40

I1/I2 = N2/(N1 + N2) = 200/1200 = 1/6, or I1 = I2/6 (1) v1/v2 = (N2 + N2)/N2 = 6, or v1 = 6v2 (2)

For the primary loop, 120 = (30 + j12)I1 + v1 (3) For the secondary loop, v2 = (20 – j40)I2 (4)


120 = (30 + j12)( I2/6) + 6v2 and substituting (4) into this yields

120 = (49 – j38)I2 or I2 = 1.935∠37.79° p = |I2|2(20) = 74.9 watts.


Zin = V1/I1 But V1I1 = V2I2, or V2 = I2ZL and I1/I2 = N2/(N1 + N2) Hence V1 = V2I2/I1 = ZL(I2/I1)I2 = ZL(I2/I1)2I1

V1/I1 = ZL[(N1 + N2)/N2] 2 Zin = [1 + (N1/N2)] 2ZL


(a) Consider just one phase at a time.

1:n A

B 20MVA

Load

C

a b c

n = VL/ )312470/(7200V3 Lp = = 1/3

(b) The load carried by each transformer is 60/3 = 20 MVA.

Hence ILp = 20 MVA/12.47 k = 1604 A ILs = 20 MVA/7.2 k = 2778 A

(c) The current in incoming line a, b, c is

85.1603x3I3 Lp = = 2778 A

Current in each outgoing line A, B, C is 2778/(n 3 ) = 4812 A


(a) This is a three-phase ∆-Y transformer. (b) VLs = nvLp/ 3 = 450/(3 3 ) = 86.6 V, where n = 1/3

As a Y-Y system, we can use per phase equivalent circuit. Ia = Van/ZY = 86.6∠0°/(8 – j6) = 8.66∠36.87° Ic = Ia∠120° = 8.66∠156.87° A ILp = n 3 ILs I1 = (1/3) 3 (8.66∠36.87°) = 5∠36.87° I2 = I1∠–120° = 5∠–83.13° A

(c) p = 3|Ia|2(8) = 3(8.66)2(8) = 1.8 kw.


(a) This is a ∆-∆ connection. (b) The easy way is to consider just one phase.

1:n = 4:1 or n = 1/4

n = V2/V1 which leads to V2 = nV1 = 0.25(2400) = 600

i.e. VLp = 2400 V and VLs = 600 V

S = p/cosθ = 120/0.8 kVA = 150 kVA

pL = p/3 = 120/3 = 40 kw 4:1

Ipp

IL

Ips

ILs VLp VLs

But pLs = VpsIps

For the ∆-load, IL = 3 Ip and VL = Vp Hence, Ips = 40,000/600 = 66.67 A

ILs = 3 Ips = 3 x66.67 = 115.48 A

(c) Similarly, for the primary side

ppp = VppIpp = pps or Ipp = 40,000/2400 = 16.667 A

and ILp = 3 Ip = 28.87 A (d) Since S = 150 kVA therefore Sp = S/3 = 50 kVA


(a) n = VLs/( 3 VLp) 4500/(900 3 ) = 2.887 (b) S = 3 VLsILs or ILs = 120,000/(900 3 ) = 76.98 A ILs = ILp/(n 3 ) = 76.98/(2.887 3 ) = 15.395 A

Chapter 13, Solution 76. (a) At the load, VL = 240 V = VAB

VAN = VL/ 3 = 138.56 V

Since S = 3 VLIL then IL = 60,000/(240 3 ) = 144.34 A

1:n 0.05 Ω j0.1 Ω

0.05 Ω j0.1 Ω

0.05 Ω

A

B

Balanced

Load 60kVA 0.85pf leading C

240V j0.1 Ω2640V

(b) Let VAN = |VAN|∠0° = 138.56∠0° cosθ = pf = 0.85 or θ = 31.79° IAA’ = IL∠θ = 144.34∠31.79° VA’N’ = ZIAA’ + VAN = 138.56∠0° + (0.05 + j0.1)(144.34∠31.79°) = 138.03∠6.69°

VLs = VA’N’ 3 = 137.8 3 = 238.7 V (c) For Y-∆ connections,

n = 3 VLs/Vps = 3 x238.7/2640 = 0.1569

fLp = nILs/ 3 = 0.1569x144.34/ 3 = 13.05 A Chapter 13, Solution 77.

(a) This is a single phase transformer. V1 = 13.2 kV, V2 = 120 V

n = V2/V1 = 120/13,200 = 1/110, therefore n = 110 (b) P = VI or I = P/V = 100/120 = 0.8333 A

I1 = nI2 = 0.8333/110 = 7.576 mA


k = 21LL/M = 3x6/1 = 0.2357 In the AC Sweep box, set Total Pts = 1, Start Freq = 0.1592 and End Freq = 0.1592. After simulation, the output file includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 4.253 E+00 –8.526 E+00 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.564 E+00 2.749 E+01 From this, I1 = 4.253∠–8.53° A, I2 = 1.564∠27.49° A The power absorbed by the 4-ohm resistor = 0.5|I|2R = 0.5(1.564)2x4

= 4.892 watts


k1 = 5000/15 = 0.2121, k2 = 8000/10 = 0.1118 In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the circuit is saved and simulated, the output includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 4.068 E–01 –7.786 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.306 E+00 –6.801 E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 1.336 E+00 –5.492 E+01 Thus, I1 = 1.306∠–68.01° A, I2 = 406.8∠–77.86° mA, I3 = 1.336∠–54.92° A


k1 = 80x40/10 = 0.1768, k2 = 60x40/20 = 0.482

k3 = 60x80/30 = 0.433 In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the simulation, we obtain the output file which includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 1.304 E+00 6.292 E+01 i.e. Io = 1.304∠62.92° A


k1 = 8x4/2 = 0.3535, k2 = 8x2/1 = 0.25 In the AC Sweep box, we let Total Pts = 1, Start Freq = 100, and End Freq = 100. After simulation, the output file includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.000 E+02 1.0448 E–01 1.396 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.000 E+02 2.954 E–02 –1.438 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.000 E+02 2.088 E–01 2.440 E+01

i.e. I1 = 104.5∠13.96° mA, I2 = 29.54∠–143.8° mA,

I3 = 208.8∠24.4° mA.

Chapter 13, Solution 82. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 1.955 E+01 8.332 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 6.847 E+01 4.640 E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 4.434 E–01 –9.260 E+01

i.e. V1 = 19.55∠83.32° V, V2 = 68.47∠46.4° V,

Io = 443.4∠–92.6° mA.


he schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq


FREQ VM($N_0001) VP($N_0001)

i.e. iX = 1.08∠33.91° A

C T= 0.1592, and End Freq = 0.1592. After simulation, the output file includes 1.592 E–01 1.080 E+00 3.391 E+01 1.592 E–01 1.514 E+01 –3.421 E+01

, Vx = 15.14∠–34.21° V.


he schematic is shown below. We set Total Pts = 1, Start Freq = 0.1592, and End




i.e. I1 = 4.028∠–52.38° A

TFreq = 0.1592. After simulation, the output file includes 1.592 E–01 4.028 E+00 –5.238 E+01 1.592 E–01 2.019 E+00 –5.211 E+01 1.592 E–01 1.338 E+00 –5.220 E+01

, I2 = 2.019∠–52.11° A,

I3 = 1.338∠–52.2° A.


For maximum power transfer,

Z1 = ZL/n2 or n2 = ZL/Z1 = 8/7200 = 1/900

n = 1/30 = N2/N1. Thus N2 = N1/30 = 3000/30 = 100 turns.

Z1

+ − ZL/n2

VS


n = N2/N1 = 48/2400 = 1/50 ZTh = ZL/n2 = 3/(1/50)2 = 7.5 kΩ

Chapter 13, Solution 87. ZTh = ZL/n2 or n = 300/75Z/Z ThL = = 0.5 Chapter 13, Solution 88. n = V2/V1 = I1/I2 or I2 = I1/n = 2.5/0.1 = 25 A

p = IV = 25x12.6 = 315 watts Chapter 13, Solution 89. n = V2/V1 = 120/240 = 0.5

S = I1V1 or I1 = S/V1 = 10x103/240 = 41.67 A S = I2V2 or I2 = S/V2 = 104/120 = 83.33 A


(a) n = V2/V1 = 240/2400 = 0.1 (b) n = N2/N1 or N2 = nN1 = 0.1(250) = 25 turns (c) S = I1V1 or I1 = S/V1 = 4x103/2400 = 1.6667 A S = I2V2 or I2 = S/V2 = 4x104/240 = 16.667 A


(a) The kVA rating is S = VI = 25,000x75 = 1875 kVA (b) Since S1 = S2 = V2I2 and I2 = 1875x103/240 = 7812 A


(a) V2/V1 = N2/N1 = n, V2 = (N2/N1)V1 = (28/1200)4800 = 112 V (b) I2 = V2/R = 112/10 = 11.2 A and I1 = nI2, n = 28/1200

I1 = (28/1200)11.2 = 261.3 mA (c) p = |I2|2R = (11.2)2(10) = 1254 watts.

Chapter 13, Solution 93. (a) For an input of 110 V, the primary winding must be connected in parallel, with series-aiding on the secondary. The coils must be series-opposing to give 12 V. Thus the connections are shown below.

12 V 110 V

(b) To get 220 V on the primary side, the coils are connected in series, with series-aiding on the secondary side. The coils must be connected series-aiding to give 50 V. Thus, the connections are shown below.

50 V

220 V

Chapter 13, Solution 94. V2/V1 = 110/440 = 1/4 = I1/I2 There are four ways of hooking up the transformer as an auto-transformer. However it is clear that there are only two outcomes.

V2

V1

V2

V1

V2

V1

V2

V1

(1) (2) (3) (4) (1) and (2) produce the same results and (3) and (4) also produce the same results. Therefore, we will only consider Figure (1) and (3). (a) For Figure (3), V1/V2 = 550/V2 = (440 – 110)/440 = 330/440 Thus, V2 = 550x440/330 = 733.4 V (not the desired result) (b) For Figure (1), V1/V2 = 550/V2 = (440 + 110)/440 = 550/440 Thus, V2 = 550x440/550 = 440 V (the desired result) Chapter 13, Solution 95.

(a) n = Vs/Vp = 120/7200 = 1/60 (b) Is = 10x120/144 = 1200/144

S = VpIp = VsIs

Ip = VsIs/Vp = (1/60)x1200/144 = 139 mA


RCj1RCj

Cj1RR

)(i

o

ω+ω

=ω+

==ωVV

H

=ω)(H0

0

j1j

ωω+ωω

, where RC1

0 =ω

20

0

)(1)(H

ωω+

ωω=ω= H

ωω

−π

=ω∠=φ0

1-tan2

)(H

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that RC10 =ω . Thus, the sketches of H and φ are shown below.

H

ωω0 = 1/RC

1 0.7071

0

0

90°

φ

ωω0 = 1/RC

45°


=ω+

=ω+

=ωRLj1

1LjR

R)(H

0j11ωω+

, where LR

0 =ω

20 )(1

1)(H

ωω+=ω= H

ωω

=ω∠=φ0

1-tan-)(H

The frequency response is identical to the response in Example 14.1 except that

LR0 =ω . Hence the response is shown below.

φ

H

ω ω0 = R/L

0.7071 1

0

ω

ω0 = R/L

-45°

-90°

0°


(a) The Thevenin impedance across the second capacitor where V is taken is o

sRC1R

RsC1||RRTh ++=+=Z

sRC1sC1RsC1 i

iTh +=

+=

VVV

ZTh

sC1

+

Vo

−

+ −

VTh

)sC1)(sRC1(sC1sC1

Th

iTh

Tho Z

VV

ZV

++=⋅

+=

))sRC1(sRCsRC1)(sRC1(1

)sRC1)(sC1(1

)sThi

o

++++=

++==

ZVV

H(

=)s(H1sRC3CRs

1222 ++

(b) 08.01080)102)(1040(RC -3-63 =×=××=

There are no zeros and the poles are at

==RC0.383-

s1 4.787-

==RC2.617-

s2 32.712-


(a) RCj1

RCj

1||R

ω+=

ω

)RCj1(LjRR

RCj1R

Lj

RCj1R

)(i

o

ω+ω+=

ω++ω

ω+==ω

VV

H

=ω)(HLjRRLC-

R2 ω++ω

(b) )LjR(Cj1

)LjR(CjCj1LjR

LjR)(

ω+ω+ω+ω

=ω+ω+

ω+=ωH

=ω)(HRCjLC1

RCjLC-2

2

ω+ω−ω+ω


(a) Cj1LjR

Cj1)(

i

o

ω+ω+ω

==ωVV

H

=ω)(HLCRCj1

12ω−ω+

(b) RCj1

RCj

1||R

ω+=

ω

)RCj1(LjR)RCj1(Lj

)RCj1(RLjLj

)(i

o

ω+ω+ω+ω

=ω++ω

ω==ω

VV

H

=ω)(HRLCLjR

RLCLj2

2

ω−ω+ω−ω


(a) Using current division,

Cj1LjRR

)(i

o

ω+ω+==ω

II

H

)25.0)(10()25.0)(20(j1)25.0)(20(j

LCRCj1RCj

)( 22 ω−ω+ω

=ω−ω+

ω=ωH

=ω)(H 25.25j15j

ω−ω+ω

(b) We apply nodal analysis to the circuit below.

1/jωC+ −

IoVx

0.5 Vx

R Is jωL

Cj1Lj5.0

Rxxx

s ω+ω−

+=VVV

I

But )Cj1Lj(2Cj1Lj

5.0ox

xo ω+ω=→

ω+ω= IV

VI

Cj1Lj5.0

R1

x

s

ω+ω+=

VI

)Cj1Lj(21

R1

)Cj1Lj(2 o

s

ω+ω+=

ω+ωII

1R

)Cj1Lj(2

o

s +ω+ω

=II

)LC1(2RCjRCj

R)Cj1Lj(211)( 2

s

o

ω−+ωω

=ω+ω+

==ωII

H

)25.01(2jj

)( 2ω−+ωω

=ωH

=ω)(H 25.0j2j

ω−ω+ω


(a) Hlog2005.0 10=

Hlog105.2 10-3 =×

== × -3105.210H 005773.1

(b) Hlog206.2- 10=Hlog0.31- 10=

== -0.3110H 4898.0

(c) Hlog207.104 10=Hlog235.5 10=

== 235.510H 510718.1 ×


(a) 05.0H === 05.0log20H 10dB 26.02- , =φ °0

(b) 125H =

== 125log20H 10dB 41.94 , =φ °0

(c) °∠=+

= 43.63472.4j2

10j)1(H

== 472.4log20H 10dB 01.13 , =φ °43.63

(d) °∠=−=+

++

= 23.55-254.47.1j9.3j2

6j1

3)1(H

== 254.4log20H 10dB 577.12 , =φ °23.55-


)10j1)(j1(1

)(ω+ω+

=ωH

10/j1log20j1log20-H 1010dB ω+−ω+= )10/(tan)(tan- -1-1 ω−ω=φ

The magnitude and phase plots are shown below.

HdB

0.1

-40

10/j11

log20 10 ω+

ω+ j11

log20 10

-20

10 1 100 ω

1

j11

arg

φ

-135°

10/j1arg

ω+

ω+

-90°

-180°

0.1 10 1 100 ω-45°


ω+ω

=ω+ω

=ω

5j1j1

10)j5(j

50)j(H

ωj1log20

ω+

5j1

1log20

20 log1

-40

0.1

10

1 100 ω

20

-20

HdB 40

φ

-135°

5/j11argω+

ωj1arg

-90°

-180°

0.1 10

1

100 ω -45°


)2j1(j)10j1(5

)(ω+ωω+

=ωH

2j1log20jlog2010j1log205log20H 10101010dB ω+−ω−ω++=

2tan10tan90- -1-1 ω−ω+°=φ


-40

0.1 10 1 100 ω

20 14

-20

HdB 40 34

φ

-45°

45°

ω1001 10 0.1

-90°

90°


201.0log20,)10/1(

)1(1.0)( −=++

=ωωω

jjjwT

The plots are shown below. |T| (db) 20 0 ω 0.1 1 10 100 -20 -40 arg T 90o

0 ω 0.1 1 10 100 -90o


)10j1()j()j1)(101(

)j10()j(j1

)( 22 ω+ωω+

=ω+ω

ω+=ωG

10j1log20jlog40j1log2020-G 101010dB ω+−ω−ω++=

10tantan-180 -1-1 ω−ω+°=φ The magnitude and phase plots are shown below.

GdB 40

20

0.1 1 10 100 ω-20

-40

φ

90°

0.1 1 10 100 ω-90°

-180°


ω

+ω

+ω

ω+=ω 2

5j

2510j

1j

j12550

)(H

ω−ω++= jlog20j1log202log20H 101010dB

2

10 )5j(52j1log20 ω+ω+−

ω−ω

−ω+°=φ51

2510tantan90- 2

1-1-


φ

-90°

-40

0.1 10 1 100 ω

20 6

40 26

-20

HdB

-180°

0.1 10 1 100 ω

90°


)10j1)(2j1()j1(2

)j10)(j2()j1(40

)(ω+ω+

ω+=

ω+ω+ω+

=ωH

10j1log202j1log20j1log202log20H 10101010dB ω+−ω+−ω++=

10tan2tantan -1-1-1 ω−ω−ω=φ


φ

-45°

6

40

-40

0.1 10 1 100 ω

20

-20

HdB

90°

-90°

0.1 10 1 100 ω

45°


2

10j1)j1(100

j)(G

ω+ω+

ω=ω

GdB

-20

ωjlog20

10jlog40 ω

−

φ

-90° 2

10j1

1arg

ω+

ω+ j11arg

arg(jω)

-180°

0.1 10 1 100ω

90°

20 log(1/100) -60

-40

0.1 10 1 100

ω

20


2)2j1)(j1(j)41(

)(ω+ω+ω

=ωG

2j1log40j1log20jlog204-20logG 10101010dB ω+−ω+−ω+=

2tan2tan--90 -1-1 ω−ω°=φ


GdB

-20

20

ω-12

1001 10 0.1

-40

φ

-90°

-180°

0.1 10 1 100 ω

90°


)10j1)(5j1(j50)2j1(4

)(2

ω+ω+ωω+

=ωG

ω−ω++= jlog202j1log40504log20G 101010dB

10j1log205j1log20 1010 ω+−ω+− where 94.21-504log1020 =

10tan5tan2tan290- -1-1-1 ω−ω−ω+°=φ


GdB

-20

180°

0.1 10 1 100 ω

90°

-90°

φ

-60

-40

0.1 10 1 100 ω

20


)10010j1(100j

)( 2ω−ω+ω

=ωH

10010j1log20100log20jlog20H 2

101010dB ω−ω+−−ω=

ω−ω

−°=φ1001

10tan90 2

1-


φ

-90°

20

-60

-40

0.1 10 1 100 ω

40

-20

HdB

-180°

0.1 10 1 100 ω

90°


)10j1)(j1()j1(10

)(2

ω+ω+ω−ω+

=ωN

2

101010dB j1log2010j1log20j1log2020N ω−ω++ω+−ω+−=

10tantan1

tan 1-1-2

1- ω−ω−

ω−ω

=φ


40

180°

0.1 10 1 100 ω

90°

-90°

φ

20

-40

0.1 10 1 100 ω-20

NdB


)10010j1)(10j1(100)j1(j

)( 2ω−ω+ω+ω+ω

=ωT

100log20j1log20jlog20T 101010dB −ω++ω=

10010j1log2010j1log20 21010 ω−ω+−ω+−

ω−ω

−ω−ω+°=φ1001

10tan10tantan90 2

1-1-1-


180°

-180°

0.1 10 1 100 ω

90°

-90°

φ

-40

-60

0.1 10 1 100 ω

20

-20

TdB

Chapter 14, Solution 22. 10kklog2020 10 =→=

A zero of slope at dec/dB20+ 2j12 ω+→=ω

A pole of slope t dec/dB20- a20j1

120

ω+→=ω


1100

ω+→=ω

Hence,

)100j1)(20j1()2j1(10

)(ω+ω+

ω+=ωH

=ω)(H)j100)(j20(

)j2(104

ω+ω+ω+


A zero of slope at the origindec/dB20+ ω→ j


11

ω+→=ω

A pole of slope at dec/dB40- 2)10j1(1

10ω+

→=ω

Hence,

2)10j1)(j1(j

)(ω+ω+

ω=ωH

=ω)(H 2)j10)(j1(j100

ω+ω+ω

Chapter 14, Solution 24. The phase plot is decomposed as shown below.

φ

-45°

ω+ 100/j11

arg

)10/j1(arg ω+

)j(arg ω

100

90°

-90°

0.1 10 1 1000 ω

45°

)j100(j)j10)(10(k

)100j1(j)10j1(k

)(ω+ωω+′

=ω+ωω+′

=ωG

where k′ is a constant since arg 0k =′ .

Hence, =ω)(G)j100(j

)j10(kω+ωω+

, where 10k constant isk ′=


s/krad5)101)(1040(

1LC1

6-3-0 =××

==ω

==ω R)( 0Z Ωk2

ω−

ω+=ω

C4

L4

jR)4(0

00Z

××

−×⋅×

+=ω)101)(105(

41040

4105

j2000)4( 6-33-

3

0Z

)5400050(j2000)4( 0 −+=ωZ

=ω )4( 0Z Ω− k75.0j2

ω−

ω+=ω

C2

L2

jR)2(0

00Z

××

−××

+=ω)101)(105(

2)1040(

2)105(

j2000)2( 6-33-

3

0Z

)52000100(j2000)4( 0 −+=ωZ

=ω )2( 0Z Ω− k3.0j2

ω−ω+=ω

C21

L2jR)2(0

00Z

××

−××+=ω)101)(105)(2(

1)1040)(105)(2(j2000)2( 6-3

3-30Z

=ω )2( 0Z Ω+ k3.0j2

ω−ω+=ω

C41

L4jR)4(0

00Z

××

−××+=ω)101)(105)(4(

1)1040)(105)(4(j2000)4( 6-3

3-30Z

=ω )4( 0Z Ω+ k75.0j2


(a) kHz 51.2210101052

12

139===

−− xxxLCfo

ππ

(b) krad/s 101010

1003 === −xL

RB

(c ) 142.14101.0

101050

1013

36

====−

xx

RL

LCRL

Q oω

Chapter 14, Solution 27. At resonance,

Ω== 10RZ , LC1

0 =ω

LR

B = and R

LB

Q 00 ω=

ω=

Hence,

H1650

)80)(10(RQL

0==

ω=

F25)16()50(

1L

1C 22

0µ==

ω=

s/rad625.01610

LR

B ===

Therefore, =R Ω10 , =L H16 , =C F25 µ , =B s/rad625.0

Chapter 14, Solution 28. Let = 10R . Ω

H5.02010

BR

L ===

F2)5.0()1000(

1L

1C 22

0µ==

ω=

5020

1000B

Q 0 ==ω

=

Therefore, if then Ω= 10R =L H5.0 , =C F2 µ , =Q 50


jω 1/jω

1 Z

jω

ω+ω

+ω

+ω=j1

jj1

jZ

2

2

1j1

jω+ω+ω

+

ω−ω=Z

Since and are in phase, )t(v )t(i

211

0)Im(ω+ω

+ω

−ω==Z

0124 =−ω+ω

618.02

411-2 =+±

=ω

=ω s/rad7861.0

Chapter 14, Solution 30. Select = 10R . Ω

mH5H05.0)20)(10(

10Q

RL

0===

ω=

F2.0)05.0)(100(

1L

1C 2

0==

ω=

s/rad5.0)2.0)(10(

1RC1

B ===

Therefore, if then Ω= 10R =L mH5 , =C F2.0 , =B s/rad5.0


ω=→ω= L

LX

LLX

rad/s 10x796.810x40

10x6.5x10x10x2X

RLRB 6

3

36

L=

π=

ω==

Chapter 14, Solution 32. Since 10Q > ,

2B

01 −ω=ω , 2B

02 +ω=ω

=×

=ω

=120

106Q

B6

0 s/krad50

=−=ω 025.061 s/rad10975.5 6×

=+=ω 025.062 s/rad10025.6 6×


pF 84.5610x40x10x6.5x2

80Rf2

QCRCQ36o

o =π

=π

=→ω=

H 21.1480x10x6.5x2

10x40Qf2

RLL

RQ6

3

ooµ=

π=

π=→

ω=


(a) krad/s 443.110x60x10x8

1LC1

63o ===ω−−

(b) rad/s 33.310x60x10x5

1RC1B

63===

−

(c) 9.43210x60x10x5x10x443.1RCQ 633

o ==ω= −

Chapter 14, Solution 35. At resonance,

=×

==→= 3-10251

Y1

RR1

Y Ω40

=×

=ω

=→ω=)40)(10200(

80R

QCRCQ 3

00 F10 µ

=××

=ω

=→=ω)1010)(104(

1C

1L

LC1

6-1020

0 H5.2 µ

=×

=ω

=80

10200Q

B3

0 s/krad5.2

=−=−ω=ω 5.22002B

01 s/krad5.197

=+=+ω=ω 5.22002B

01 s/krad5.202


s/rad5000LC1

0 ==ω

==ω→=ω R)(R1

)( 00 ZY Ωk2

kS75.18j5.0L

4C

4j

R1

)4(0

00 −=

ω−

ω+=ωY

=−

=ω01875.0j0005.0

1)4( 0Z Ω+ 3.53j4212.1

kS5.7j5.0L

2C

2j

R1

)2(0

00 −=

ω−

ω+=ωY

=−

=ω0075.0j0005.0

1)2( 0Z Ω+ 74.132j85.8

kS5.7j5.0C2

1L2j

R1

)2(0

00 +=

ω−ω+=ωY

=ω )2( 0Z Ω− 74.132j85.8

kS75.18j5.0C4

1L4j

R1

)4(0

00 +=

ω−ω+=ωY

=ω )4( 0Z Ω− 3.53j4212.1


22 )C

1L(R

)C

1L(jRLRjCL

LjCj

1R

)Cj

1R(Lj)

Cj1R//(LjZ

ω−ω+

ω−ω+

ω+

=ω+

ω+

ω+ω

=ω

+ω=

1)LCCR(0)

C1L(R

C1L

CLLR

)ZIm( 22222

2

=+ω→=

ω−ω+

ω−ω+ω

=

Thus,

22CRLC

1

+=ω


222 LRLjR

CjCjLjR

1Y

ω+ω−

+ω=ω+ω+

At resonance, , i.e. 0)Im( =Y

0LR

LC 22

02

00 =

ω+ω

−ω

CL

LR 220

2 =ω+

2

3-6-3-2

2

0 104050

)1010)(1040(1

LR

LC1

×−

××=−=ω

=ω0 s/rad4841


(a) s/krad810x)8690(2)ff(2B 31212 π=−π=−π=ω−ω=

π=π=ω+ω=ω 17610x)88(2)(21 3

21o

nF89.1910x2x10x8

1BR1C

RC1B 33 =

π==→=

(b) H4.16410x89.19x)176(

1C

1LLC1

92o

2o =π

=ω

=→=ω−

(c ) s/krad9.552176o =π=ω (d) s/krad13.258B =π=

(e) 228

176B

Q o =ππ

=ω

=


(a) L = 5 + 10 = 15 mH

===ω−− 630

10x20x10x15

1LC1 1.8257 k rad/sec

==ω= −633

0 10x20x10x25x10x8257.1RCQ 912.8

===−63 10x2010x25

1RC1B 2 rad

(b) To increase B by 100% means that B’ = 4.

==′

=′4x10x25

1BR1C 3 10 µF

Since F10CC

C

21

21 µ=+

′ CC = and C1 = 20 µF, we then obtain C2 = 20 µF.

Therefore, to increase the bandwidth, we merely add another 20 µF in series with the first one.


(a) This is a series RLC circuit.

Ω=+= 862R , H1L = , F4.0C =

===ω4.0

1LC1

0 s/rad5811.1

==ω

=8

5811.1R

LQ 0 1976.0

==LR

B s/rad8

(b) This is a parallel RLC circuit.

F263)6)(3(

F6andF3 µ=+

→µµ

F2C µ= , Ω= k2R , mH20L =

=××

==ω)1020)(102(

1LC1

3-6-0 s/krad5

=××

×=

ω=

)1020)(105(102

LR

Q 3-3

3

020

=××

==)102)(102(

1RC1

B 6-3 s/krad250


(a) )LjR(||)Cj1(in ω+ω=Z

RCjLC1LjR

Cj1

LjR

CjLjR

2in ω+ω−ω+

=

ω+ω+

ωω+

=Z

22222

2

in CR)LC1()RCjLC1)(LjR(

ω+ω−ω−ω−ω+

=Z

At resonance, Im( 0)in =Z , i.e.

CR)LC1(L0 22 ω−ω−ω= CRLLC 22 −=ω

=−

=ωLC

CRL 2

0 LR

C1 2

−

(b) )Cj1R(||Ljin ω+ω=Z

RCj)LC1()RCj1(Lj

Cj1LjR)Cj1R(Lj

2in ω+ω−ω+ω

=ω+ω+ω+ω

=Z

22222

22

in CR)LC1(]RCj)LC1[()LjRLC-(

ω+ω−ω−ω−ω+ω

=Z


LCR)LC1(L0 2232 ω+ω−ω=

1)CRLC( 222 =−ω

=ω0 22CRLC1−

(c) )Cj1Lj(||Rin ω+ω=Z

RCj)LC1()LC1(R

Cj1LjR)Cj1Lj(R

2

2

in ω+ω−ω−

=ω+ω+ω+ω

=Z

22222

22

in CR)LC1(]RCj)LC1)[(LC1(R

ω+ω−ω−ω−ω−

=Z


RC)LC1(R0 2 ωω−= 0LC1 2 =ω−

=ω0 LC1


1/jωC

R1 Zin

jωL R2

(a) )Cj1R(||)Lj||R( 21in ω+ω=Z

ω

+

ω+ω

=Cj

1R||

LjRLjR

21

1inZ

LjRLjR

Cj1

R

Cj1

RLjR

LRj

1

12

21

1

in

ω+ω

+ω

+

ω

+⋅ω+

ω

=Z

12

21

21in LCR)CRj1)(LjR(

)CRj1(LRjω−ω+ω+

ω+ω=Z

)CRRL(jLCRLCRRLRjLCRR-

2122

12

1

1212

in +ω+ω−ω−ω+ω

=Z

221

222

21

21

2122

12

11212

in )CRRL()LCRLCRR()]CRRL(jLCRLCRR)[LRjLCRR-(

+ω+ω−ω−+ω−ω−ω−ω+ω

=Z


)LCRLCRR(LR)CRRL(LCRR0 22

12

1121213 ω−ω−ω++ω=

CLRLRLCRR0 221

321

222

21

3 ω−ω+ω= LC1CR0 222

22 ω−+ω=

1)CRLC( 222

2 =−ω

222

0 CRLC1−

=ω

26-26-0 )109()1.0()109)(02.0(1

×−×=ω

=ω0 s/krad357.2 (b) At , s/krad357.20 =ω=ω

14.47j)1020)(10357.2(jLj -33 =××=ω

0212.0j9996.014.47j1

14.47jLj||R1 +=

+=ω

14.47j1.0)109)(10357.2(j

11.0

Cj1

R 6-32 −=××

+=ω

+

)Cj1R(||)Lj||R()( 210in ω+ω=ωZ

)14.47j1.0()0212.0j9996.0()14.47j1.0)(0212.0j9996.0(

)( 0in −++−+

=ωZ

=ω )( 0inZ Ω1

Chapter 14, Solution 44. We find the input impedance of the circuit shown below.

jω(2/3) 1/jωZ

1

1/jωC

Cj111

j2j

31ω+

+ω+ω

=Z

, 1=ω

2

2

C1jCC

-j0.5jC1

jCj-j1.5

1++

+=+

++=Z

)t(v and i are in phase when Z is purely real, i.e. )t(

1)1C(C1

C-0.50 2

2 =−→+

+= or =C F1

Ω=→=+

= 221

C1C1

2

2

ZZ

20)10)(2( === IZV

V)tsin(20)t(v = , i.e. =oV V20


(a) ω+

ω=ω

j1j

j||1 , ω+

=ω+

ω=

ω j11

j11j1

j1

||1

Transform the current source gives the circuit below.

ω+ j11

ω+ωj1

j

1 I

j1jω+

ω +

−

+

Vo

−

IVω+

ω⋅

ω+ω

+ω+

+

ω+=

j1j

j1j

j11

1

j11

o

==ωI

VH o)( 2)j1(2

jω+

ω

(b) 2)j1(21

)1(+

=H

==2)2(2

1)1(H 25.0


(a) This is an RLC series circuit.

nF26.1110x10x)10x15x2(

1L

1CLC1

323o

2o =π

=ω

=→=ω−

(b) Z = R, I = V/Z = 120/20 = 6 A

(c ) 12.471520

10x10x10x15x2R

LQ

33o =π=

π=

ω=

−


RLj1

1LjR

R)(

i

o

ω+=

ω+==ω

VV

H

1)0(H = and 0)(H =∞ showing that this circuit is a lowpass filter.

At the corner frequency, 2

1)(H c =ω , i.e.

RL

1

RL

1

12

1 c

2c

ω=→

ω

+

= or LR

c =ω

Hence,

cc f2LR

π==ω

=××

⋅π

=⋅π

= 3-

3

c 1021010

21

LR

21

f kHz796


Cj1

||RLj

Cj1

||R)(

ω+ω

ω=ωH

Cj1RCjR

Lj

Cj1RCjR

)(

ω+ω

+ω

ω+ω

=ωH

=ω)(HRLCLjR

R2ω−ω+

1)0(H = and showing that 0)(H =∞ this circuit is a lowpass filter.


At dc, 224

)0( =H = .

Hence, 2

2)0(H

21

)(H ==ω

2c1004

42

2ω+

=

2.081004 c

2c =ω→=ω+

10j12

20j24

)2(H+

=+

=

199.01012

)2(H ==

In dB, =)2(Hlog1020 14.023-

== 10-tan)2(Harg -1 °84.3- Chapter 14, Solution 50.

LjR

Lj)(

i

o

ω+ω

==ωVV

H

0)0(H = and showing that 1)(H =∞ this circuit is a highpass filter.

LR

1

LR

1

12

1)(

c2

c

c ω=→

ω+

==ωH

or cc f2LR

π==ω

=⋅π

=⋅π

=1.0

20021

LR

21

fc Hz3.318


RC1j

jRCj1

RCj)(

+ωω

=ω+

ω=ω′H (from Eq. 14.52)

This has a unity passband gain, i.e. 1)(H =∞ .

50RC1

c =ω=

ω+ω

=ω′=ωj50

10j)(10)( HH ^

=ω)(Hω+ωj50

10j

Chapter 14, Problem 52.

Design an RL lowpass filter that uses a 40-mH coil and has a cut-off frequency of 5 kHz.


cc f2LR

π==ω

=×π=π= )1040)(10)(2(Lf2R -35

c Ωk13.25 Chapter 14, Solution 54.

311 1020f2 ×π=π=ω

322 1022f2 ×π=π=ω

3

12 102B ×π=ω−ω=

3120 1021

2×π=

ω+ω=ω

=ππ

=ω

=221

BQ 0 5.11

C1

LLC1

20

0 ω=→=ω

=××π

=)1080()1021(

1L 12-23 H872.2

BLRLR

B =→=

=×π= )872.2)(102(R 3 Ωk045.18


Ω=π

=π

=→=π=ω−

k3.26510x300x10x2x2

1Cf2

1RRC1f2

123ccc


s/krad10)104.0)(1025(

1LC1

63o =××

==ω−−

s/krad4.01025

10LR

B 3- =×

==

==4.0

10Q 25

s/krad8.92.0102Bo1 =−=−ω=ω or kHz56.12

8.91 =

π=f

s/krad2.102.0102Bo2 =+=+ω=ω or kHz62.12

2.10f2 =

π=

Therefore,

kHz62.1fkHz56.1 <<


(a) From Eq 14.54,

LC1

LRss

LRs

LCssRC1sRC

sC1sLR

R)s(2

2++

=++

=++

=H

Since LR

B = and LC1

0 =ω ,

=)s(H 20

2 sBssB

ω++

(b) From Eq. 14.56,

LC1

LR

ss

LC1

s

sC1

sLR

sC1

sL)s(

2

2

++

+=

++

+=H

=)s(H 20

2

20

2

sBss

ω++ω+


(a) Consider the circuit below.

I I1

R1/sC+

Vo

−

+ −

R 1/sC

Vs

sC2

R

sC1

RsC1

RsC1

R||sC1

R)s(+

++=

++=Z

)sRC2(sCsRC1

R)s(+

++=Z

)sRC2(sCCRssRC31

)s(222

+++

=Z

ZV

I s=

)sRC2(RsC2sC1 s

1 +=

+=

ZV

II

222s

1o CRssRC31)sRC2(sC

sRC2R

R+++

⋅+

==V

IV

222s

o

CRssRC31sRC

)s(++

==VV

H

++=

222

CR1

sRC3

s

sRC3

31

)s(H

Thus, 2220 CR

1=ω or ==ω

RC1

0 s/rad1

==RC3

B s/rad3

(b) Similarly,

sLR2)sLR(R

sL)sLR(||RsL)s(++

+=++=Z

sLR2LssRL3R

)s(222

+++

=Z

ZV

I s= , )sLR2(

RsLR2

R s1 +

=+

=Z

VII

222s

1o LssRL3RsLR2

sLR2sLR

sL++

+⋅

+=⋅=

VIV

2

22

222s

o

LR

sLR3

s

sLR3

31

LssRL3RsRL

)s(++

=++

==VV

H

Thus, ==ωLR

0 s/rad1

==LR3

B s/rad3


(a) =×

==ω)1040)(1.0(

1LC1

12-0 s/rad105.0 6×

(b) 43

1021.0

102LR

B ×=×

==

250102105.0

BQ 4

60 =

××

=ω

=

As a high Q circuit,

=−=−ω=ω )150(102B 4

01 s/krad490

=+=+ω=ω )150(102B 4

02 s/krad510

(c) As seen in part (b), =Q 250


Ro

+

Vo

−

+ − R

1/sC

Vi

sL

sC1sLR)sC1sL(R

sC1

sL||R)s(+++

=

+=Z

LCssRC1)LCs1(R

)s( 2

2

+++

=Z

LCRsRLCRsCsRRR)LCs1(R

R 2o

2oo

2

oi

o

+++++

=+

==Z

ZVV

H

LCssRC1)LCs1(R

RR 2

2

ooin +++

+=+= ZZ

LCssRC1LCRsRLCRsCsRRR

2

2o

2oo

in ++++++

=Z

ω= js

RCjLC1LCRRLCRCRRjR

2

2o

2oo

in ω+ω−ω−+ω−ω+

=Z

222

2o

2o

2o

in )RC()LC1()RCjLC1)(CRRjLCRLCRRR(

ω+ω−ω−ω−ω+ω−ω−+

=Z

0)Im( in =Z implies that

0)LC1(CRR]LCRLCRRR[RC- 2

o2

o2

o =ω−ω+ω−ω−+ω

0LCRRLCRLCRRR o2

o2

o2

o =ω+−ω−ω−+ RLCR2 =ω

=××

==ω)104)(101(

1LC1

6-3-0 s/krad811.15

LCRLCRRCRRjR)LC1(R

2o

2oo

2

ω−ω−+ω+ω−

=H

RRR

)0(HHo

max +==

or o

oo

2o

2

max RRR

)RR(LCCRRjRR

LC1Rlim)(HH

+=

+−ω

+ω+

−ω=∞=

∞→ω

At and ω , 1ω 2 mzxH2

1=H

CRRj)RR(LCRR)LC1(R

)RR(2R

oo2

o

2

o ω++ω−+ω−

=+

2o

2o

2o

2o

))RR(LCRR()CRR()LC1)(RR(

21

+ω−++ω

ω−+=

28-226-

-92

)10410()10(96)1041(10

21

×⋅ω−+ω×

×⋅ω−=

21

)10410()10(96)1041(10

028-226-

-92

−×⋅ω−+ω×

×⋅ω−=

0)10410()1096()2)(10410( 2-822-6-82 =×⋅ω−+ω×−×⋅ω−

2-822-62-82 )10410()1096()10410)(2( ×⋅ω−+ω×=×⋅ω−

0)10410()1096( 2-822-6 =×⋅ω−−ω×

010010092.8106.1 2-74-15 =+ω×−ω×

01025.610058.5 1684 =×+×−ω

××

=ω 8

82

101471.2109109.2

Hence,

s/krad653.141 =ω

s/krad061.172 =ω

=−=ω−ω= 653.14061.17B 12 s/krad408.2 Chapter 14, Solution 61.

(a) iCj1RCj1

VVω+

ω=+ , oVV =−

Since , −+ = VV

oiRCj11

VV =ω+

==ωi

o)(VV

HRCj1

1ω+

(b) iCj1RR

VVω+

=+ , oVV =−

Since , −+ = VV

oiRCj1RCj

VV =ω+

ω

==ωi

o)(VV

HRCj1

RCjω+

ω

Chapter 14, Solution 62. This is a highpass filter.

RCj11

RCj1RCj

)(ω−

=ω+

ω=ωH

ωω−=ω

cj11

)(H , )1000(2RC1

c π==ω

f1000j11

ffj11

)(c −

=−

=ωH

(a) i

o

5j11

)Hz200f(VV

H =−

==

=−

=5j1

mV120oV mV53.23

(b) i

o

5.0j11

)kHz2f(VV

H =−

==

=−

=5.0j1

mV120oV m3.107 V

(c) i

o

1.0j11

)kHz10f(VV

H =−

==

=−

=1.0j1

mV120oV m4.119 V


For an active highpass filter,

ii

fiRsC1

RsC)s(H

+−= (1)

But

10/s1s10)s(H

+−= (2)

Comparing (1) and (2) leads to:

Ω==→= M10C10R10RC

iffi

Ω==→= k100C

1.0R1.0RCi

iii


ff

f

fff CRj1

RCj1

||RZω+

=ω

=

i

ii

iii Cj

CRj1Cj

1RZ

ωω+

=ω

+=

Hence,

===ωi

f

i

o -)(

ZZ

VV

H)CRj1)(CRj1(

CRj-

iiff

if

ω+ω+ω

This is a bandpass filter. )(ωH is similar to the product of the transfer function of a lowpass filter and a highpass filter.


ii RCj1RCj

Cj1RR

VVVω+

ω=

ω+=+

ofi

i

RRR

VV+

=−

Since , −+ = VV

iofi

i

RCj1RCj

RRR

VVω+

ω=

+

==ωi

o)(VV

H

ω+

ω

+

RCj1RCj

RR

1i

f

It is evident that as ω , the gain is ∞→i

f

RR

+1 and that the corner frequency is RC1

.


(a) Proof

(b) When 3241 RRRR = ,

CR1ss

RRR

)s(243

4

+⋅

+=H

(c) When ∞→3R ,

CR1sCR1-

)s(2

1

+=H


fii

f R4R41

RR

gain DC =→==

s/rad)500(2CR

1frequencyCorner

ffc π==ω=

If we select , then Ω= k20R f Ω= k80R i and

nF915.15)1020)(500)(2(

1C 3 =

×π=

Therefore, if =fR Ωk20 , then =iR Ωk80 and =C nF915.15


ifi

f R5RRR

5gainfrequency High =→==

s/rad)200(2CR

1frequencyCorner

iic π==ω=

If we select , then Ω= k20R i Ω= k100fR and

nF8.39)1020)(200)(2(

1C 3 =

×π=

Therefore, if =iR Ωk20 , then =fR Ωk100 and =C nF8.39

Chapter 14, Solution 69. This is a highpass filter with f 2c = kHz.

RC1

f2 cc =π=ω

3c 104

1f2

1RC

×π=

π=

Hz108 may be regarded as high frequency. Hence the high-frequency gain is

410

RRf −

=−

or R5.2Rf =

If we let =R Ωk10 , then =fR Ωk25 , and =×π

= 41040001

C nF96.7 .


(a) )YYY(YYY

YY)s()s(

)s(321421

21

i

o

+++==

VV

H

where 11

1 GR1

Y == , 22

2 GR1

Y == , 13 sCY = , 24 sCY = .

=)s(H)sCGG(sCGG

GG

121221

21

+++

(b) 1GGGG

)0(H21

21 == , 0)(H =∞

showing that this circuit is a lowpass filter. Chapter 14, Solution 71. L , Ω= 50R , mH F1C40= µ=

)1040(KK

1LKK

L 3-

f

m

f

m ×⋅=→=′

mf KK25 = (1)

fm

-6

fm KK10

1KK

CC =→=′

mf

6

K1

K10 = (2)


ff

6

K251

K10 =

=fK -3102.0 ×

== fm K25K -3105×


CLLC

KKLC

CL 2f2

f ′′=→=′′

8-

-6-32f 104

)2)(1()1020)(104(

K ×=××

=

=fK -4102×

LC

CL

KKCL

CL 2

m2m ⋅

′′

=→=′′

3-

3-

-62m 105.2

)104)(2()1020)(1(

K ×=××

=

=mK -2105×


=×==′ )10800)(12(RKR 3m ΩM6.9

=×==′ )1040(1000800

LKK

L 6-

f

m F32 µ

=×

==′ )1000)(800(10300

KKC

C-9

fmpF375.0


Ω=== 300100x3RK'R 1m1

Ω=== k 1100x10RK'R 2m2

H 200)2(1010L

KK

'L 6

2

f

m µ===

nF 110101

KKC'C 8fm

===


Ω=== 20010x20RK'R m

H 400)4(1010L

KK

'L5f

m µ===

F 110x101

KKC'C

5fmµ===


Ω==→== 5010

10x50R10x50RK'R3

33

m

mH 101010x10x10LH 10L

KK

'L3

66

f

m ==→µ== −

mF 4010x10x10x40CKK

CFp 40'C 6312

fm==→== −

Chapter 14, Solution 77. L and C are needed before scaling.

H25

10BR

LLR

B ===→=

F5.312)2)(1600(

1L

1C

LC1

20

0 µ==ω

=→=ω

(a) ===′ )2)(600(LKL m H1200

=×

==′ 60010125.3

KC

C-4

mF5208.0 µ

(b) ===′ 3f 10

2KL

L mH2

=×

==′ 3

-4

f 1010125.3

KC

C nF5.312

(c) ===′ 5f

m

10)2)(400(

LKK

L mH8

=×

==′ )10)(400(10125.3

KKC

C 5

-4

fmpF81.7


Ω===′ k1)1)(1000(RKR m

H1.0)1(1010

LKK

L 4

3

f

m ===′

F1.0)10)(10(

1KK

CC 43

fmµ===′

The new circuit is shown below.

1 kΩ

1 kΩ 0.1 µF0.1 H1 kΩ +

Vx

−

I


(a) Insert a 1-V source at the input terminals.

Ro

+ −

3Vo

+

Vo

−

Io V2V1

sL + −

R 1/sC

1 V

There is a supernode.

sC1sLR1 21

+=

− VV (1)

But (2) o12o21 33 VVVVVV −=→+=

Also, sC1sLsLsC1sL

sL 2o2o +

=→+

=VV

VV (3)

Combining (2) and (3)

oo12 sLsC1sL

3 VVVV+

=−=

12

2

o LCs41LCs

VV+

= (4)


12o1

LCs41sC

sLR1

VVV

+==

−

12

2

121 LCs41sRCLCs41

LCs41sRC

1 VVV+

++=

++=

sRCLCs41LCs41

2

2

1 +++

=V

)sRCLCs41(RsRC

R1

21

o ++=

−=

VI

sCLCs4sRC11 2

oin

++==

IZ

sC1

RsL4in ++=Z (5)

When , , 5R = 2L = 1.0C = ,

=)s(inZs

105s8 ++

At resonance,

C1

L40)Im( in ω−ω==Z

or ===ω)2)(1.0(2

1LC21

0 s/rad118.1

(b) After scaling,

RKR m→′ Ω→Ω 404 Ω→Ω 505

H2.0)2(10010

LKK

Lf

m ===′

4-

fm10

)100)(10(1.0

KKC

C ===′

From (5),

=)s(inZs

1050s8.0

4

++

===ω)10)(2.0(2

1LC21

4-0 s/rad8.111


(a) Ω===′ 400)2)(200(RKR m

mH2010

)1)(200(K

LKL 4

f

m ===′

F25.0)10)(200(

5.0KK

CC 4

fmµ===′

The new circuit is shown below.

20 mHa

400 Ω

Ix

0.25 µF 0.5 Ix

b

(b) Insert a 1-A source at the terminals a-b.

V2V1

b

a sL

R

Ix

1/(sC) 1 A 0.5 Ix

At node 1,

sL

sC1 211

VVV

−+= (1)

At node 2,

R

5.0sL

2x

21 VI

VV=+

−

But, I . 1x sC V=

RsC5.0

sL2

121 V

VVV

=+−

(2)

Solving (1) and (2),

1sCR5.0LCs

RsL21 ++

+=V

1sCR5.0LCsRsL

1 21

Th +++

==V

Z

At , 410=ω

1)400)(1025.0)(10j(5.0)1025.0)(1020()10j(400)1020)(10j(

6-46-3-24

3-4

Th +×+××+×

=Z

200j6005.0j5.0

200j400Th −=

++

=Z

=ThZ ohms435.18-5.632 °∠

Chapter 14, Solution 81. (a)

1GR)LGRC(jLCRLjZtoleadswhich

LjR1)LjR)(CjG(

LjR1CjG

Z1

2 +++ω+ω−

+ω=

ω++ω+ω+

=ω+

+ω+=

LC1GR

CG

LRj

LCR

Cj

)(Z2 +

+

+ω+ω−

+ω

=ω (1)

We compare this with the given impedance:

25001j2)1j(1000)(Z 2 ++ω+ω−

+ω=ω (2)

Comparing (1) and (2) shows that

LR1 R/LmF, 1C1000C1

=→==→=

1CG2CG

LR

==→=+ mS

L4.0RR10

1R10LC

1GR25013

3==→

+=

+=

−

−

Thus, R = 0.4Ω, L = 0.4 H, C = 1 mF, G = 1 mS

(b) By frequency-scaling, Kf =1000. R’ = 0.4 Ω, G’ = 1 mS

mH4.010

4.0KL'L

3f=== , F1

1010

KC'C 3

3

fµ===

−

−


fmKK

CC =′

2001

200K c

f ==ωω′

=

50002001

101

K1

CC

K 6-f

m =⋅=⋅′

=

==′ RKR m 5 kΩ, thus, ==′ if R2R Ωk10


pF 1.010x100

10CKK

1'CF1 5

6

fm===→µ

−

pF 5.0'CF5 =→µ

Ω=Ω==→Ω M 1k 10x100RK'Rk 10 m

Ω=→Ω M 2'Rk 20 Chapter 14, Solution 84.

The schematic is shown below. A voltage marker is inserted to measure vo. In the AC sweep box, we select Total Points = 50, Start Frequency = 1, and End Frequency = 1000. After saving and simulation, we obtain the magnitude and phase plots in the probe menu as shown below.

Chapter 14, Solution 85. We let A so that Vo

s 01I ∠= .VI/ oso = The schematic is shown below. The circuit is simulated for 100 < f < 10 kHz.


The schematic is shown below. A current marker is inserted to measure I. We set Total Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the AC sweep box. After simulation, the magnitude and phase plots are obtained in the Probe menu as shown below.


The schematic is shown below. In the AC Sweep box, we set Total Points = 50, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude response as shown below. It is evident from the response that the circuit represents a high-pass filter.

Chapter 14, Solution 88. Chapter 14, Solution 88.

The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of Vo as shown below.

The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of Vo as shown below.


The schematic is shown below. In the AC Sweep box, we type Total Points = 101, Start Frequency = 100, and End Frequency = 1 k. After simulation, the magnitude plot of the response Vo is obtained as shown below.


The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.


The schematic is shown below. We type Total Points = 101, Start Frequency = 1, and End Frequency = 100 in the AC Sweep box. After simulating the circuit, the magnitude plot of the frequency response is shown below.


C

R

L

C

2

2

0 LR

LC1

21

f −π

=

610

10240400

L

7

6- =×

=28810

)10120)(10240(1

LC1 16

12-6- =××

= R

,

Since LC1

L<<

R

=π

=π

≅224

10LC2

1f

8

0 kHz938

If R is reduced to 40 Ω, LC1

LR<< .

The result remains the same. Chapter 14, Solution 94.

RC1

c =ω

We make R and C as small as possible. To achieve this, we connect 1.8 kΩ and 3.3 kΩ in parallel so that

Ω=+

= k 164.13.38.13.3x8.1R

We place the 10-pF and 30-pF capacitors in series so that

C = (10x30)/40 = 7.5 pF Hence,

rad/s 10x55.11410x5.7x10x164.1

1RC1 6

123c ===ω−


(a) LC2

1f0 π=

When , pF360C =

MHz541.0)10360)(10240(2

1f

12-6-0 =××π

=

When , pF40C =

MHz624.1)1040)(10240(2

1f

12-6-0 =××π

=

Therefore, the frequency range is MHz624.1fMHz541.0 0 <<

(b) RfL2

Qπ

=

At , MHz541.0f0 =

=××π

=12

)10240)(10541.0)(2(Q

-66

98.67

At , MHz624.1f0 =

=××π

=12

)10240)(10624.1)(2(Q

-66

1.204


C2 C1

+

Vo

−

VoV1

+ −

Ri

RL

L

Vi

Z1Z2

22

L

2L1 CsR1

RsC

1||R

+==Z

+++

=+=2L

2L2

L

11

12 CsR1

LCRsRsL||

sC1

)sL(||sC1

ZZ

2L

2L2

L

1

2L

2L2

L

12

CsR1LCRsRsL

sC1

CsR1LCRsRsL

sC1

+++

+

+++

⋅=Z

21L3

1L12

2L

2L2

L2 CLCRsCsRLCsCsR1

LCRsRsL++++

++=Z

ii2

21 R

VZ

ZV

+=

i1

1

22

21

1

1o sLRsL

VZ

ZZ

ZV

ZZ

V+

⋅+

=+

=

sLR 1

1

22

2

i

o

+⋅

+=

ZZ

ZZ

VV

where

=+ 22

2

RZZ

21Li3

1Li1i2

2Lii2L2

L

2L2

L

CLCRRsCRsRLCRsCRsRRLCRsRsLLCRsRsL

+++++++++

and 2L

2L

L

1

1

LCRssLRR

sL ++=

+ZZ

Therefore,

=i

o

VV

)LCRssLR)(CLCRRsCRsRLCRsCRsRRLCRsRsL(

)LCRsRsL(R

2L2

L21Li3

1Li1i2

2Lii2L2

L

2L2

LL

+++

++++++++

where s . ω= j


C2 C1

+

Vo

−

VoV1

+ −

Ri

RL

L

Vi

Z2 Z1

2L

2L

2L sC1sLR

)sC1R(sLsC

1R||sL

+++

=

+=Z , ω= js

i1i

1 sC1RV

ZZ

V++

=

i1i2L

L1

2L

Lo sC1RsC1R

RsC1R

RV

ZZ

VV++

⋅+

=+

=

)sC1sLR)(sC1R()sC1R(sL)sC1R(sL

sC1RR

)(2L1i2L

2L

2L

L

i

o

++++++

⋅+

==ωVV

H

=ω)(H)1CsR(LCs)1CsRLCs)(1CsR(

CCLRs

2L12

2L22

1i

21L3

+++++

where s . ω= j


π=−π=−π=ω−ω= 44)432454(2)ff(2B 1212

)44)(20(QBf2 00 π==π=ω

==ππ

= )22)(20(2

)44)(20(f0 Hz440


Cf21

C1

Xc π=

ω=

π=

××π=

π=

2010

)105)(102)(2(1

Xf21

C-9

36c

Lf2LXL π=ω=

π×

=×π

=π

=4103

)102)(2(300

f2X

L-4

6L

=

π⋅

π×

π

=π

=

2010

4103

2

1LC2

1f

9-4-0 MHz826.1

=

×π

== 4-1034

)100(LR

B s/rad10188.4 6×


RC1

f2 cc =π=ω

=××π

=π

=)105.0)(1020)(2(

1Cf2

1R 6-3

cΩ91.15


RC1

f2 cc =π=ω

=×π

=π

=)1010)(15)(2(

1Cf2

1R 6-

cΩk061.1


(a) When and 0R s = ∞=LR , we have a low-pass filter.

RC1

f2 cc =π=ω

=××π

=π

=)1040)(104)(2(

1RC21

f 9-3c Hz7.994

(b) We obtain across the capacitor. ThR)RR(||RR sLTh +=

Ω=+= k5.2)14(||5R Th

)1040)(105.2)(2(1

CR21

f 9-3Th

c ××π=

π=

=cf kHz59.1


sC1||RRR

)(12

2

i

o

+==ω

VV

H , ω= js

)sC1(RR)sC1R(R

sC1R)sC1(R

R

R)(

12

12

1

12

2

++

=

++

=ωH

=ω)(H21

12

sCRR)sCR1(R

++


The schematic is shown below. We click Analysis/Setup/AC Sweep and enter Total Points = 1001, Start Frequency = 100, and End Frequency = 100 k. After simulation, we obtain the magnitude plot of the response as shown.


(a) 2

ee)atcosh(

at-at +=

[ ] =

++

−=

as1

as1

21

)atcosh(L 22 ass−

(b) 2

ee)atsinh(

at-at −=

[ ] =

+−

−=

as1

as1

21

)atsinh(L 22 asa−


(a) )sin()tsin()cos()tcos()t(f θω−θω= [ ] [ ])tsin()sin()tcos()cos()s(F ωθ−ωθ= LL

=)s(F 22s)sin()cos(s

ω+θω−θ

(b) )sin()tcos()cos()tsin()t(f θω+θω=

[ ] [ ])tsin()cos()tcos()sin()s(F ωθ+ωθ= LL

=)s(F 22s)cos()sin(s

ω+θω−θ


(a) [ ] =)t(u)t3cos(e-2tL9)2s(

2s2 ++

+

(b) [ ] =)t(u)t4sin(e-2tL16)2s(

42 ++

(c) Since [ ] 22 ass

)atcosh(−

=L

[ ] =)t(u)t2cosh(e-3tL4)3s(

3s2 −+

+

(d) Since [ ] 22 asa

)atsinh(−

=L

[ =)t(u)tsinh(e-4tL ]1)4s(

12 −+

(e) [ ]4)1s(

2)t2sin(e 2

t-

++=L

If )s(F)t(f →←

)s(Fdsd-

)t(ft →←

Thus, [ ] ( )[ ]1-2t- 4)1s(2dsd-

)t2sin(et ++=L

)1s(2)4)1s((

222 +⋅

++=

[ ] =)t2sin(et -tL 22 )4)1s(()1s(4++

+


(a) 16s

se6e4s

s6)s(G 2

ss

22 +=

+=

−−

(b) 3s

e5s2)s(F

s2

2 ++=

−


(a) [ ]4s

)30sin(2)30cos(s)30t2cos( 2 +

°−°=°+L

[ ]

+−°

=°+4s

1)30cos(sdsd

)30t2cos(t 22

22L

( )

+

−=

1-2 4s1s23

dsd

dsd

( ) ( )

+

−−+=

2-21-2 4s1s23s24s

23

dsd

( )

( ) ( ) ( ) ( )32

2

222222 4s

1s23)s8(

4s

23s2

4s

1s232

4s

2s-23

+

−

++

−+

−

−+

=

( ) ( )32

2

22 4s

1s23)s8(

4ss32s3s3-

+

−

++

−+−=

( ) ( )32

23

32

2

4ss8s34

4s)42)(ss3(-3

+

−+

+

++=

[ ] =°+ )30t2cos(t 2L ( ) 32

32

4ss3s6s3128

+

+−−

(b) [ ] =+

⋅= 5t-4

)2s(!4

30et30L 5)2s(720+

(c) =−⋅−=

δ− )01s(4s2

)t(dtd

4)t(ut2 2L s4s22 −

(d) )t(ue2)t(ue2 -t1)--(t =

[ ] =)t(ue2 1)--(tL1s

e2+

(e) Using the scaling property,

[ ] =⋅⋅=⋅⋅=s2

125

)21(s1

211

5)2t(u5Ls5

(f) [ ] =+

=31s

6)t(ue6 3t-L

1s318+

(g) Let f . Then, )t()t( δ= 1)s(F = .

−′−−=

=

δ −− )0(fs)0(fs)s(Fs)t(fdtd

)t(dtd 2n1nn

n

n

n

n

LL

−⋅−⋅−⋅=

=

δ −− 0s0s1s)t(fdtd

)t(dtd 2n1nn

n

n

n

n

LL

=

δ )t(dtd

n

n

L ns


(a) [ ] =−δ )1t(2L -se2

(b) [ ] =− )2t(u10L 2s-es

10

(c) [ ] =+ )t(u)4t(Ls4

s12 +

(d) [ ] [ ] =−=− )4t(uee2)4t(ue2 4)--(t-4-t LL)1s(e

e24

-4s

+


(a) Since [ ] 22 4ss

)t4cos(+

=L , we use the linearity and shift properties to

obtain [ ] =−− )1t(u))1t(4cos(10L16ses10

2

s-

+

(b) Since [ ] 32

s2

t =L , [ ]s1

)t(u =L ,

[ ] 3t2-2

)2s(2

et+

=L , and [ ]s

e)3t(u

s3-

=−L

[ ] =−+ )3t(u)t(uet t2-2Ls

e)2s(

2 -3s

3 ++


(a) [ ]t3-t2- e10e4)t2(u6)t3(2 −++δL

3s10

2s4

2s1

21

631

2+

−+

+⋅⋅+⋅=

=3s

102s

4s6

32

+−

+++

(b) )1t(ue)1t(ue)1t()1t(uet -t-t-t −+−−=−

)1t(uee)1t(uee)1t()1t(uet -11)--(t-11)--(t-t −+−−=−

[ ] =+

++

=−1s

ee)1s(

ee)1t(uet

-s-1

2

-s-1t-L

1se

)1s(e 1)-(s

2

1)-(s

++

+

++

(c) [ ] =−− )1t(u))1t(2cos(L4s

es2

-s

+

(d) Since )t4sin()t4cos()4sin()4cos()t4sin())t(4sin( =π−π=π− )t(u))t(4sin()t(u)t4sin( π−π−=π−

[ ][ ])t(u)t(u)t4sin( π−−L

[ ] [ ])t(u))t(4sin()t(u)t4sin( π−π−−= LL

=+

−+

=π

16se4

16s4

2

s-

2 )e1(16s

4 s-2

π−⋅+


(a) )2t(u2)2t(u)2t()2t(u)4t()t(f −−−−=−−=

=)s(F 2

-2s

2

-2s

se2

se

−

(b) )1t(uee2)1t(ue2)t(g 1)--4(t-4-4t −=−=

=)s(G)4s(e

e24

-s

+

(c) )t(u)1t2cos(5)t(h −=

)Bsin()Asin()Bcos()Acos()BAcos( +=−

)1sin()t2sin()1cos()t2cos()1t2cos( +=−

)t(u)t2sin()1sin(5)t(u)t2cos()1cos(5)t(h +=

4s2

)1sin(54s

s)1cos(5)s(H 22 +

⋅++

⋅=

=)s(H4s

415.84ss702.2

22 ++

+

(d) )4t(u6)2t(u6)t(p −−−=

=)s(P 4s-2s- es6

es6

−


(a) By taking the derivative in the time domain, )tsin(et)tcos()ee-t()t(g -t-t-t −+=

)tsin(et)tcos(et)tcos(e)t(g -t-t-t −−=

++

+

++

++

+++

=1)1s(

1dsd

1)1s(1s

dsd

11)(s1s)s(G 222

=++

+−

+++

−++

+= 2222

2

2 )2s2(s22s

)2s2(s2ss

2s2s1s

)s(G 22

2

)2s2(s2)(ss+++

(b) By applying the time differentiation property,

)0(f)s(sF)s(G −= where , f)tcos(et)t(f -t= 0)0( =

=+++

=

++

+⋅= 22

2

2 )2s2(s2s)(s)(s

1)1s(1s

dsd-

)s()s(G 22

2

)2s2(s2)(ss+++


(a) Since [ ] 22 ass

)atcosh(−

=L

=−+

+=

4)1s()1s(6

)s(F 2 3s2s)1s(6

2 −++

(b) Since [ ] 22 asa

)atsinh(−

=L

[ ]12s4s

1216)2s(

)4)(3()t4sinh(e3 22

2t-

−+=

−+=L

[ ] [ ]1-22t- )12s4s(12dsd-

)t4sinh(e3t)s(F −+=⋅= L

=−++= 2-2 )12s4s)(4s2)(12()s(F 22 )12s4s()2s(24

−++

(c) )ee(21

)tcosh( t-t +⋅=

)2t(u)ee(21

e8)t(f t-t3t- −+⋅⋅=

= )2t(ue4)2t(ue4 -4t-2t −+− = )2t(uee4)2t(uee4 2)--4(t-82)--2(t-4 −+− [ ] [ ])t(ueee4)2t(uee4 -2-2s-42)--2(t-4 LL ⋅=−

[ ]2s

e4)2t(uee4

4)-(2s2)-2(t-4-

+=−

+

L

Similarly, [ ]4s

e4)2t(uee4

8)-(2s2)-4(t-8-

+=−

+

L

Therefore,

=+

++

=++

4se4

2se4

)s(F8)-(2s4)-(2s [ ]

8s6s)e8e16(s)e4e4(e

2

-22-226)-(2s

++++++


)2s(2

)2s(s2

2

2s

)1t(22)1t(222)1t(2

e)2s(3s

2s11

2se

2see

)2s(ee)s(f

)1t(uee)1t(uee)1t()1t(uete)t(f

+−+−−

−−

−

−−−−−−−−−

+

+=

++

+=

++

+=

−+−−=−=


(a) )()( sFdsdt −→←tf

If f(t) = cost, then 22

2

2 )1()12()1)(1()( and

1)(

++−+

=+

=s

ssssFdsd

sssF

22

2

)1(1)cos(

+−+

=sssttL

(b) Let f(t) = e-t sin t.

221

1)1(1)( 22 ++

=++

=sss

sF

22

2

)22()22)(1()0)(22(

+++−++

=ss

sssdsdF

22 )22()1(2)sin(++

+=−=−

sss

dsdFtte tL

(c ) ∫∞

→←s

dssFttf )()(

Let 22)( then ,sin)(β

ββ+

==s

sFttf

sssds

stt

ss

ββ

πββ

ββ

ββ 11122 tantan

2tan1sin −−∞−

∞

=−==+

=

∫L


<<−<<

=2t1t5101t0t5

)t(f

We may write f as )t(

[ ] [ ])2t(u)1t(u)t510()1t(u)t(ut5)t(f −−−−+−−= )2t(u)2t(5)1t(u)1t(10)t(ut5 −−+−−−=

s2-2

s-22 e

s5

es10

s5

)s(F +−=

=)s(F )ee21(s5 s2-s-2 +−

Chapter 15, Solution 15. [ ])2t(u)1t(u)1t(u)t(u10)t(f −+−−−−=

=

+−=s

ee

s2

s1

10)s(Fs2-

s- 2s- )e1(s

10−


)4t(u5)3t(u3)1t(u3)t(u5)t(f −−−+−−=

=)s(F [ ]s4-s3-s- e5e3e35s1

−+−


><<<<

<

=

3t03t111t0t

0t0

)t(f2

[ ] [ ])3t(u)1t(u1)1t(u)t(ut)t(f 2 −−−+−−=

)3t(u)1t(u)1t(u)1t2-()1t(u)1t()t(ut 22 −−−+−++−−−= )3t(u)1t(u)1t(2)1t(u)1t()t(ut 22 −−−−−−−−=

=)s(Fs

ee

s2

)e1(s2 s-3

s-2

s-3 −−−


(a) [ ] [ ])3t(u)2t(u3)2t(u)1t(u2)1t(u)t(u)t(g −−−+−−−+−−= )3t(u3)2t(u)1t(u)t(u −−−+−+=

=)s(G )e3ee1(s1 s3-s2-s- −++

(b) [ ] [ ])3t(u)1t(u2)1t(u)t(ut2)t(h −−−+−−=

[ ])4t(u)3t(u)t28( −−−−+ )3t(u2)1t(u2)1t(u2)1t(u)1t(2)t(ut2 −−−+−−−−−=

)4t(u)4t(2)3t(u2)3t(u)3t(2 −−+−+−−− )4t(u)4t(2)3t(u)3t(2)1t(u)1t(2)t(ut2 −−+−−−−−−=

=+−−= s4-2

s3-2

s-2 e

s2

es2

)e1(s2

)s(H )eee1(s2 s4-s3-s-2 +−−


Since [ ] 1)t( =δL and , 2T = =)s(F s2-e11−


Let 1t0),tsin()t(g1 <<π= [ ])1t(u)t(u)tsin( −−π= )1t(u)tsin()t(u)tsin( −π−π=

Note that sin( )tsin(-)tsin())1t( π=π−π=−π . So, 1)-u(t1))-t(sin(u(t)t)sin()t(g1 π+π=

)e1(s

)s(G s-221 +

π+π

=

=−

= 2s-1

e1)s(G

)s(G)e1)(s(

)e1(2s-22

-s

−π++π


π= 2T

Let [ ])1t(u)t(u2t

1)t(f1 −−

π−=

)1t(u21

1)1t(u)1t(21

)t(u2t

)t(u)t(f1 −

π−−−−

π+

π−=

[ ] [ ]

2

-s-ss-

2

s-

21 s2e1-se)12-(2

s1

e21

1-s2

es2

1s1

)s(Fπ

+++π+π=⋅

π++

π+

π−=

=−

= Ts-1

e1)s(F

)s(F[ ] [ ]

)e1(s2e1-se)12-(2

s2-2

-s-s

π−π+++π+π


(a) Let 1t0,t2)t(g1 <<= [ ])1t(u)t(ut2 −−= )1t(u2)1t(u)1t(2)t(ut2 −+−−−=

s-

2

-s

21 es2

se2

s2

)s(G +−=

1T,e1

)s(G)s(G sT-

1 =−

=

=)s(G)e1(s

)ese1(2s-2

-s-s

−+−

(b) Let h , where is the periodic triangular wave. )t(uh 0 += 0h

Let h be within its first period, i.e. 1 0h

<<−<<

=2t1t241t0t2

)t(h1

)2t(u)2t(2)1t(ut2)1t(u4)1t(ut2)t(ut2)t(h1 −−−−−−+−−=

)2t(u)2t(2)1t(u)1t(4)t(ut2)t(h1 −−−−−−=

2s-

22

-2ss-

221 )e1(s2

se2

es4

s2

)s(H −=−−=

)e1()e1(

s2

)s(H 2s-

2-s

20 −−

=

=)s(H)e1(

)e1(s2

s1

2s-

2-s

2 −−

+


(a) Let

<<<<

=2t11-1t01

)t(f1

[ ] [ ])2t(u)1t(u)1t(u)t(u)t(f1 −−−−−−=

)2t(u)1t(u2)t(u)t(f1 −+−−=

2s-2s-s-1 )e1(

s1

)ee21(s1

)s(F −=+−=

2T,)e1(

)s(F)s(F sT-

1 =−

=

=)s(F)e1(s

)e1(2s-

2-s

−−

(b) Let [ ] )2t(ut)t(ut)2t(u)t(ut)t(h 222

1 −−=−−= )2t(u4)2t(u)2t(4)2t(u)2t()t(ut)t(h 22

1 −−−−−−−−=

2s-2s-2

2s-31 e

s4

es4

)e1(s2

)s(H −−−=

2T,)e1(

)s(H)s(H Ts-

1 =−

=

=)s(H)e1(s

)ss(es4)e1(22s-3

2-2s-2s

−+−−


(a) 5s6s

ss10lim)s(sFlim)0(f 2

4

ss +++

==∞→∞→

==++

+

∞→ 010

s5

s6

s1

s110

lim432

3

s= ∞

=+++

==∞→→ 5s6s

ss10lim)s(sFlim)(f 2

4

0s0s0

(b) =+−

+==

∞→∞→ 6s4sss

lim)s(sFlim)0(f 2

2

ss1

The complex poles are not in the left-half plane.

)(f ∞ existnotdoes

(c) )5s2s)(2s)(1s(

s7s2lim)s(sFlim)0(f 2

3

ss +++++

==∞→∞→

==

++

+

+

+

∞→ 10

s5

s21

s21

s11

s7

s2

lim

2

3

s= 0

==++++

+==∞

→→ 100

)5s2s)(2s)(1s(s7s2

lim)s(sFlim)(f 2

3

0s0s0


(a) )4s)(2s()3s)(1s)(8(

lim)s(sFlim)0(fss ++

++==

∞→∞→

=

+

+

+

+

∞→

s41

s21

s31

s11)8(

lims

= 8

===∞→→ )4)(2(

)3)(1)(8(lim)s(sFlim)(f

0s0s3

(b) 1s

)1s(s6lim)s(sFlim)0(f 4ss −

−==

∞→∞→

==−

−=

∞→ 10

s1

1

s1

s1

6lim)0(f

4

42

s0

All poles are not in the left-half plane.

)(f ∞ existnotdoes Chapter 15, Solution 26.

(a) =++

+==

∞→∞→ 6s4ss3s

lim)s(sFlim)0(f 23

3

ss1

Two poles are not in the left-half plane.

)(f ∞ existnotdoes

(b) )4s2s)(2s(

ss2slim)s(sFlim)0(f 2

23

ss ++−+−

==∞→∞→

=

++

−

+−

∞→

2

2

s

s4

s21

s21

s1

s21

lim= 1

One pole is not in the left-half plane.

)(f ∞ existnotdoes Chapter 15, Solution 27.

(a) =)t(f -te2)t(u +

(b) 4s

113

4s11)4s(3

)s(G+

−=+

−+=

=)t(g -4te11)t(3 −δ

(c) 3s

B1s

A)3s)(1s(

4)s(H

++

+=

++=

2A = , -2B =

3s2

1s2

)s(H+

−+

=

=)t(h -3t-t e2e2 −

(d) 4s

C)2s(

B2s

A)4s()2s(

12)s(J 22 +

++

++

=++

=

62

12B == , 3

)-2(12

C 2 ==

2)2s(C)4s(B)4s()2s(A12 ++++++=

Equating coefficients :

2s : -3-CACA0 ==→+=1s : 6-2ABBA2C4BA60 ==→+=++=0s : 121224-24C4B4A812 =++=++=

4s3

)2s(6

2s3-

)s(J 2 ++

++

+=

=)t(j -2t-2t-4t et6e3e3 +−


)t(ut)e4e2()t(f

5s4

3s2

5s2

)4(2

3s2

)2(2

)s(F

t5t3 −− +−=

++

+−

=+−−

++

−

=

(b)

( ) )t(ut2sine5.1t2cose2e2)t(h5s2s

1s21s

2)s(H

1C;2B;2A3CB,3CB;BA;11CA5

CA5s)CBA2(s)BA()1s)(CBs()5s2s(A11s3

5s2sCBs

1sA

)5s2s)(1s(11s3)s(H

ttt2

22

22

−−− +−=→++

+−+

+=

=−==→−==+−−==+

++++++=+++++=+

++

++

+=

+++

+=


2222

2222

3)2s(3

32

3)2s()2s(2

s2)s(V

6Band2A26s2BsAs26s8s2;3)2s(

BAss2)s(V

++−

++

+−=

−=−=→+=++++++

++=

v(t) = 0t,t3sine32t3cose2)t(u t2t2 ≥−− −−2


(a) 22222213)2s(

332

3)2s()2s(2

3)2s(2)2s(2)s(H

+++

++

+=

++

++=

t3sine32t3cose2)t(h t2t2

1−− +=

(b) )5s2s(

DCs)1s(

B)1s(

A)5s2s()1s(

4s)s( 2222

22

++

++

++

+=

+++

+=H

22222 )1s(D)1s(Cs)5s2s(B)5s2s)(1s(A4s ++++++++++=+

or

)1s2s(D)ss2s(C)5s2s(B)5s7s3s(A4s 2232232 ++++++++++++=+ Equating coefficients:

ACCA0:s3 −=→+=

DBADC2BA31:s2 ++=+++=

2/1C,2/1A2A4D2B2A6D2CB2A70:s =−=→+=++=+++=

4/1D,4/5B1B4A4DB5A54:constant ==→++=++=

++

−++

++

+−

=

++

++

++

+−

=)2)1s(

1)1s(2)1s(

5)1s(

241

)5s2s(1s2

)1s(5

)1s(2

41)s(H 222222

Hence,

( ) )t(ut2sine5.0t2cose2te5e241)t(h tttt

2−−−− −++−=

(c )

++

+=

++

+=

++

+= −−

−

)3s(1

)1s(1e

21

)3s(B

)1s(Ae

)3s)(1s(e)2s()s(H ss

s3

( ) )1t(uee21)t(h )1t(3)1t(

3 −+= −−−−


(a) 3s

C2s

B1s

A)3s)(2s)(1s(

s10)s(F

++

++

+=

+++=

-5210-

)1s()s(FA 1-s ==+= =

201-20-

)2s()s(FB -2s ==+= =

-15230-

)3s()s(FC 3-s ==+= =

3s15

2s20

1s5-

)s(F+

−+

++

=

=)t(f -3t-2t-t e15e20e5- −+

(b) 323

2

)2s(D

)2s(C

2sB

1sA

)2s)(1s(1s4s2

)s(F+

++

++

++

=++++

=

-1)1s()s(FA 1-s =+= =

-1)2s()s(FD -2s3 =+= =

)4s4s)(1s(B)4s4s)(2s(A1s4s2 222 +++++++=++

)1s(D)2s)(1s(C +++++ Equating coefficients :

3s : 1 -ABBA0 ==→+=2s : 3 A2CCACB5A62 =−=→+=++=1s : DC3A4DC3B8A124 ++=+++=

-1A-2DDA64 =−=→++=0s : 116-4D2C4ADC2B4A81 =−+=++=+++=

32 )2s(1

)2s(3

2s1

1s1-

)s(F+

−+

++

++

=

2t-

22t-2t-t- e

2t

et3ee-f(t) −++=

=)t(f 2t-2

t- e2t

t31e-

−++

(c) 5s2s

CBs2s

A)5s2s)(2s(

1s)s(F 22 ++

++

+=

++++

=

51-

)2s()s(FA -2s =+= =

)2s(C)s2s(B)5s2s(A1s 22 ++++++=+


2s : 51

-ABBA0 ==→+=

1s : 1CC0CB2A21 =→+=++=0s : 121-C2A51 =+=+=

222222 2)1s(54

2)1s()1s(51

2s51-

2)1s(1s51

2s51-

)s(F++

++++

++

=+++⋅

++

=

=)t(f )t2sin(e4.0)t2cos(e2.0e0.2- -t-t-2t ++


(a) 4s

C2s

BsA

)4s)(2s(s)3s)(1s(8

)s(F+

++

+=++++

=

3)4)(2(

(8)(3)s)s(FA 0s === =

2)-4(

(8)(-1))2s()s(FB -2s ==+= =

3(-2))-4(

)(8)(-1)(-3)4s()s(FC -4s ==+= =

4s3

2s2

s3

)s(F+

++

+=

=)t(f -4t-2t e3e2)t(u3 ++

(b) 22

2

)2s(C

2sB

1sA

)2s)(1s(4s2s

)s(F+

++

++

=+++−

=

)1s(C)2s3s(B)4s4s(A4s2s 222 +++++++=+−


2s : A1BBA1 −=→+=1s : CA3CB3A42- ++=++= 0s : -6B2B-CB2A44 =→−=++=

7B1A =−= -12A--5C ==

2)2s(12

2s6

1s7

)s(F+

−+

−+

=

=)t(f -2t-t e)t21(6e7 +−

(c) 5s4s

CBs3s

A)5s4s)(3s(

1s)s(F 22

2

+++

++

=+++

+=

)3s(C)s3s(B)5s4s(A1s 222 ++++++=+


2s : A1BBA1 −=→+=1s : -3CACA3CB3A40 =+→++=++=0s : 5A2A-9C3A51 =→+=+=

-4A1B =−= -83A-C =−=

1)2s()2s(4

3s5

1)2s(8s4

3s5

)s(F 22 +++

−+

=++

+−

+=

=)t(f )tcos(e4e5 -2t-3t −


(a) 1s

C1sBAs

)1s)(1s(6

1s)1s(6

)s(F 224 ++

++

=++

=−−

=

)1s(C)1s(B)ss(A6 22 +++++=


2s : -CACA0 =→+=1s : C -ABBA0 ==→+=0s : 3B2BCB6 =→=+=

-3A = , 3B = , 3C =

1s3

1s3s-

1s3

1s33s-

1s3

)s(F 222 ++

++

+=

++

++

=

=)t(f )tcos(3)tsin(3e3 -t −+

(b) 1s

es)s(F 2

s-

+=

π

=)t(f )t(u)tcos( π−π−

(c) 323 )1s(D

)1s(C

1sB

sA

)1s(s8

)s(F+

++

++

+=+

=

8A = , -8D =

sD)ss(C)ss2s(B)1s3s3s(A8 22323 +++++++++=


3s : -ABBA0 =→+=2s : B-ACCACB2A30 ==→+=++=1s : -ADDADCBA30 =→+=+++=0s : 8D,8C,8B,8A −=−=−==

32 )1s(8

)1s(8

1s8

s8

)s(F+

−+

−+

−=

=)t(f [ ] )t(uet5.0ete18 -t2-t-t −−−


(a) 4s

311

4s34s

10)s(F 22

2

+−=

+−+

+=

=)t(f )t2sin(5.1)t(11 −δ

(b) )4s)(2s(

e4e)s(G

-2s-s

+++

=

Let 4s

B2s

A)4s)(2s(

1+

++

=++

21A = 21B =

++

++

++

+=

4s1

2s1

e24s

12s

12

e)s(G 2s-

-s

=)t(g [ ] [ ] )2t(uee2)1t(uee5.0 2)--4(t2)--2(t-1)-4(t-1)-2(t −−+−−

(c) Let 4s

C3s

BsA

)4s)(3s(s1s

++

++=

+++

121A = , 32B = , 43-C =

2s-e

4s43

3s32

s1

121

)s(H

+−

++⋅=

=)t(h )2t(ue43

e32

121 2)-4(t-2)-3(t- −

−+


(a) Let 2s

B1s

A)2s)(1s(

3s)s(G

++

+=

+++

=

2A = , -1B =

2t-t- ee2)t(g

2s1

1s2

)s(G −=→+

−+

=

)6t(u)6t(g)t(f)s(Ge)s(F -6s −−=→=

=)t(f [ ] )6t(uee2 6)--2(t6)--(t −−

(b) Let 4s

B1s

A)4s)(1s(

1)s(G

++

+=

++=

31A = , 31-B =

)4s(31

)1s(31

)s(G+

−+

=

[ ]4t-t- ee31

)t(g −=

)s(Ge)s(G4)s(F -2t−=

)2t(u)2t(g)t(u)t(g4)t(f −−−=

=)t(f [ ] [ ] )2t(uee31

)t(uee34 2)-4(t-2)-(t-4t-t- −−−−

(c) Let 4sCBs

3sA

)4s)(3s(s

)s(G 22 ++

++

=++

=

133-A =

)3s(C)s3s(B)4s(As 22 +++++=


2s : -ABBA0 =→+=1s : CB31 +=0s : C3A40 +=

133-A = , 133B = , 134C =

4s4s3

3s3-

)s(G13 2 ++

++

=

)t2sin(2)t2cos(3e-3)t(g13 -3t ++=

)s(Ge)s(F -s=

)1t(u)1t(g)t(f −−=

=)t(f [ ] )1t(u))1t(2sin(2))1t(2cos(3e3-131 1)-3(t- −−+−+


(a) 3s

D2s

CsB

sA

)3s)(2s(s1

)s(X 22 ++

+++=

++=

61B = , 41C = , 91-D =

)s2s(D)s3s(C)6s5s(B)s6s5s(A1 2323223 +++++++++=


3s : DCA0 ++= 2s : CBA3D2C3BA50 ++=+++= 1s : B5A60 +=0s : 61BB61 =→=

365-B65-A ==

3s91

2s41

s61

s365-

)s(X 2 +−

+++=

=)t(x 3t-2t- e91

e41

t61

)t(u36

5-−++

(b) 22 )1s(C

1sB

sA

)1s(s1

)s(Y+

++

+=+

=

1A = , 1-C =

sC)ss(B)1s2s(A1 22 +++++=


2s : -ABBA0 =→+=1s : -ACCACBA20 =→+=++=0s : -1C-1,B,A1 ===

2)1s(1

1s1

s1

)s(Y+

−+

−=

=)t(y -t-t ete)t(u −−

(c) 10s6s

DCs1s

BsA

)s(Z 2 +++

++

+=

101A = , 51-B =

)ss(D)ss(C)s10s6s(B)10s16s7s(A1 2232323 ++++++++++=


3s : CBA0 ++= 2s : DB5A6DCB6A70 ++=+++= 1s : A 2-BB5A10DB10A160 =→+=++=0s : 101AA101 =→=

101A = , 51--2AB == , 101AC == , 104

A4D ==

10s6s4s

1s2

s1

)s(Z10 2 +++

++

−=

1)3s(1

1)3s(3s

1s2

s1

)s(Z10 22 +++

+++

++

−=

=)t(z [ ] )t(u)tsin(e)tcos(ee211.0 -3t-3t-t ++−


(a) Let 4sCBs

sA

)4s(s12

)s(P 22 ++

+=+

=

3412s)s(PA 0s === =

sCsB)4s(A12 22 +++=


0s : 3AA412 =→=1s : C0 =2s : -3-ABBA0 ==→+=

4ss3

s3

)s(P 2 +−=

)t2cos(3)t(u3)t(p −=

)s(Pe)s(F -2s=

=)t(f [ ] )2t(u))2t(2cos(13 −−−

(b) Let 9sDCs

1sBAs

)9s)(1s(1s2

)s(G 2222 ++

+++

=++

+=

)1s(D)ss(C)9s(B)s9s(A1s2 2323 +++++++=+


3s : -ACCA0 =→+=

2s : -BDDB0 =→+=1s : 82-C,82AA8CA92 ==→=+= 0s : 81-D,81BB8DB91 ==→=+=

++

−

++

=9s1s2

81

1s1s2

81

)s(G 22

9s1

81

9ss

41

1s1

81

1ss

41

)s(G 2222 +⋅−

+⋅−

+⋅+

+⋅=

=)t(g )t3sin(241

)t3cos(41

)tsin(81

)tcos(41

−−+

(c) Let 13s4s

117s369

13s4ss9

)s(H 22

2

+++

−=++

=

2222 3)2s(3

153)2s(

2s369)s(H

++⋅−

+++

⋅−=

=)t(h )t3sin(e15)t3cos(e36)t(9 t2-t2- −−δ


(a) 26s10s

26s626s10s26s10s

s4s)s(F 2

2

2

2

++−−++

=++

+=

26s10s26s6

1)s(F 2 +++

−=

2222 1)5s(4

1)5s()5s(6

1)s(F++

+++

+−=

=)t(f )t5sin(e4)t5cos(e6)t( -t-t +−δ

(b) 29s4s

CBssA

)29s4s(s29s7s5

)s(F 22

2

+++

+=++++

=

sCsB)29s4s(A29s7s5 222 ++++=++


0s : 1AA2929 =→=1s : 3A47CCA47 =−=→+=2s : 4 A5BBA5 =−=→+=

1A = , 4B = , 3C =

22222 5)2s(5

5)2s()2s(4

s1

29s4s3s4

s1

)s(F++

−++

++=

+++

+=

=)t(f )t5sin(e)t5cos(e4)t(u -2t-2t −+


(a) 20s4s

DCs17s2s

BAs)20s4s)(17s2s(

1s4s2)s(F 2222

23

+++

+++

+=

++++++

=

)20s4s(B)s20s4s(A1s4s 22323 +++++=++

)17s2s(D)s17s2s(C 223 ++++++Equating coefficients :

3s : CA2 +=2s : DC2BA44 +++= 1s : D2C17B4A200 +++= 0s : D17B201 +=

Solving these equations (Matlab works well with 4 unknowns),

-1.6A = , -17.8B = , 6.3C = , 21D =

20s4s21s6.3

17s2s17.81.6s-

)s(F 22 +++

+++

−=

22222222 4)2s()4((3.45)

4)2s()2(3.6)(s

4)1s()4((-4.05)

4)1s(1)(-1.6)(s

)s(F++

++++

+++

++++

=

=)t(f )t4sin(e45.3)t4cos(e6.3)t4sin(e05.4)t4cos(e6.1- -2t-2t-t-t ++−

(b) 3s6s

DCs9sBAs

)3s6s)(9s(4s

)s(F 2222

2

+++

+++

=+++

+=

)9s(D)s9s(C)3s6s(B)s3s6s(A4s 232232 +++++++++=+


3s : -ACCA0 =→+=2s : DBA61 ++= 1s : A -CBC6B6C9B6A30 ==→+=++=0s : D9B34 +=

Solving these equations,

121A = , 121B = , 121-C = , 125D =

3s6s5s-

9s1s

)s(F12 22 +++

+++

=

5.449--0.551,2

12-366-03s6s2 =

±→=++

Let 449.5s

F551.0s

E3s6s

5s-)s(G 2 +

++

=++

+=

133.1449.5s5s-

E -0.551s =+

+= =

2.133- 551.0s5s-

F -5.449s =++

= =

449.5s133.2

551.0s133.1

)s(G+

−+

=

449.5s133.2

551.0s133.1

3s3

31

3ss

)s(F12 2222 +−

++

+⋅+

+=

=)t(f -5.449t-0.551t e1778.0e0944.0)t3sin(02778.0)t3cos(08333.0 −++


Let 5s2s

CBs2s

A)5s2s)(2s(

13s7s4)s( 22

2

++

++

+=

+++

++=H

)2s(C)s2s(B)5s2s(A13s7s4 222 ++++++=++

Equating coefficients gives:

BA4:s2 +=

1CCB2A27:s −=→++=

1B 3,Aor 15A5C2A513:constant ===→+=

222 2)1s(2)1s(

2s3

5s2s1s

2s3)s(H

++

−++

+=

++

−+

+=

Hence,

)t2sinsinAt2coscosA(ee3t2sinet2cosee3)t(h tt2ttt2 α−α+=−+= −−−−−

where o45,2A1sinA,1cosA =α=→=α=α Thus,

[ ] )t(ue3)45t2cos(e2)t(h t2ot −− ++= Chapter 15, Solution 41. Let y(t) = f(t)*h(t). For 0 < t < 1, f )t( λ− )(λh 0 t 1 2 λ

2t0

t

0

2 t22d4)1()t(y =λ=λλ= ∫

For 1 <t<3, )t(f λ− )(h λ 0 1 t 2

4t2t8)28(2d)48)(1(d4)1()t(y 2t1

2t0

2t

1

1

0−−=λ−λ+λ=λλ−+λλ= ∫∫

For 3 < t < 4 h )(λ )t(f λ− 0 1 t-2 2 3 t 4 λ

222t

2

2t

2 t2t163228d)48()t(y +−=λ−λ=λλλ−= −−∫

Thus,

<<+<<−

<<

=

otherwise 0,4 t 3 ,2t16t-32

3 t 1 ,42t-8t1 t 0 ,t2

)t(y 2

2

2


(a) For , 1t0 << )t(f1 λ− and f )(2 λ overlap from 0 to t, as shown in Fig. (a).

2t

2d))(1()t(f)t(f)t(y

2t0

2t

021 =λ

=λλ=∗= ∫

1

λt-1 t 1 0

1

λt-1 t 1

f2(λ)f1(t - λ)

0

(a) (b) For 1 , 2t << )t(f1 λ− and f )(2 λ overlap as shown in Fig. (b).

2t

t2

d))(1()t(y2

11t

21

1t−=

λ=λλ= −−

∫

For , there is no overlap. 2t > Therefore,

=)t(y

<<−<<

otherwise,02t1,2tt1t0,2t

2

2

(b) For , the two functions overlap as shown in Fig. (c). 1t0 <<

td)1)(1()t(f)t(f)t(yt

021 =λ=∗= ∫

1

λt-1 t 1 0

1

λt-1 t 1

f2(λ)f1(t - λ)

0

(c) (d)

For 1 , the functions overlap as shown in Fig. (d). 2t <<

t2d)1)(1()t(y 11t

1

1t−=λ=λ= −−

∫ For , there is no overlap. 2t > Therefore,

=)t(y

<<−<<

otherwise,02t1,t21t0,t

(c) For , there is no overlap. For -1-t < 0t1 << , f )t(1 λ− and overlap

as shown in Fig. (e). )(f2 λ

t1-

2t

1-21 2d)1)(1()t(f)t(f)t(y

λ+λ

=λ+λ=∗= ∫

22 )1t(

21

)1t2t(21

)t(y +=++=

-1

1

λ t 1 0-1

1

λt 1

f1(λ)f2(t - λ)

0

(e) (f) For , the functions overlap as shown in Fig. (f). 1t0 <<

∫∫ λλ−+λ+λ=t

0

0

1-d)1)(1(d)1)(1()t(y

t0

201-

2

22)t(y

λ

−λ+

λ+λ

=

)tt21(21

)t(y 2−+=

For , the two functions overlap. 1t >

∫∫ λλ−+λ+λ=1

0

0

1-d)1)(1(d)1)(1()t(y

121

121

221

)t(y 10

2

=−+=

λ

−λ+=

Therefore,

=)t(y

><<++<<++

<

1t1,1t0),1t2t-(5.00t1-),1t2t(5.0

-1t,0

2

2


(a) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (a).

2t

2d))(1()t(h)t(x)t(y

2t0

2t

0=

λ=λλ=∗= ∫

1

λt-1 t 1

h(λ)

x(t - λ)

0

1

λ t-1 t 1 0

(a) (b) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (b).

1t2t21-

2d)1)(1(d))(1()t(y 2t

11

1t

2t

1

1

1t−+=λ+

λ=λ+λλ= −−

∫∫

For , there is a complete overlap so that 2t >

1)1t(td)1)(1()t(y t1t

t

1t=−−=λ=λ= −−

∫

Therefore,

=)t(y

><<−+<<

otherwise,02t,1

2t1,1t2)2t(-1t0,2t

2

2

(b) For , the two functions overlap as shown in Fig. (c). 0t >

t0

-t

0- e-2de2)1()t(h)t(x)t(y λλ =λ=∗= ∫

2

1

h(λ) = 2e-λ

λ0 t

x(t-λ)

(c) Therefore,

=)t(y 0t),e1(2 -t >−

(c) For , 0t1- << )t(x λ− and )(h λ overlap as shown in Fig. (d).

21t0

21t

0)1t(

21

2d))(1()t(h)t(x)t(y +=

λ=λλ=∗= +

+∫

2-1 t+1

1

λ t-1 t 1

h(λ)x(t - λ)

0

(d) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (e).

∫∫ +λλ−+λλ=

1t

1

1

0d)2)(1(d))(1()t(y

21

tt21-

22

2)t(y 21t

1

210

2

++=

λ

−λ+λ

= +

t 2t+1

1

λt-1-1 10

(e) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (f).

∫∫ λλ−+λλ=−

2

1

1

1td)2)(1(d))(1()t(y

21

tt21-

22

2)t(y 22

1

21

1t

2

++=

λ

−λ+λ

= −

1

t 2 λt-1 10 t+1

(f) For , 3t2 << )t(x λ− and )(h λ overlap as shown in Fig. (g).

221t

22

1tt

21

t329

22d)2)(1()t(y +−=

λ

−λ=λλ−= −−∫

1

t2 λ t-110 t+1

(g)

Therefore,

=)t(y

<<+−<<++<<++

otherwise,03t2,29t3)2t(2t0,21t)2t(-0t1-,21t)2t(

2

2

2


(a) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (a).

td)1)(1()t(h)t(x)t(yt

0=λ=∗= ∫

0

h(λ)1x(t - λ)

-1

t 2 λt-1 1

(a) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (b).

t23d)1)(1-(d)1)(1()t(y t1

11t

t

1

1

1t−=λ−λ=λ+λ= −−

∫∫ For , 3t2 << )t(x λ− and )(h λ overlap as shown in Fig. (c).

3t-d)-1)(1()t(y 21t

2

1t−=λ=λ= −−

∫

0 λ1 t-1 2 t

-1

1

0 2 λ1 t-1 t

-1

1

(b) (c) Therefore,

=)t(y

<<−<<−<<

otherwise,03t2,3t2t1,t231t0,t

(b) For , there is no overlap. For 2t < 3t2 << , f )t(1 λ− and overlap,

as shown in Fig. (d). )(f2 λ

∫ λλ−=∗=t

221 d)t)(1()t(f)t(f)t(y

2t22t

2t

2t2

2

+−=

λ−λ=

t-1 t

1

1 5432 λ

f2(λ) f1(t - λ)

0

(d)

0 1

1

2 3 4 5tt-1 λ (e)

For , 5t3 << )t(f1 λ− and )(f2 λ overlap as shown in Fig. (e).

21

2td)t)(1()t(y t

1t

2t

1t=

λ

−λ=λλ−= −−∫

For , the functions overlap as shown in Fig. (f). 6t5 <<

12t5t21-

2td)t)(1()t(y 25

1t

25

1t−+=

λ

−λ=λλ−= −−∫

0 1 λ

1

2 3 4 5 t t-1

(f) Therefore,

=)t(y

<<−+<<<<+−

otherwise,06t5,12t5)2t(-5t3,213t2,2t2)2t(

2

2


(a) t

t

0)(fd)t()(f)t()t(f =λλ=λλ−δλ=δ∗ ∫

)t(f)t()t(f =δ∗

(b) ∫ λλ−λ=∗t

0d)t(u)(f)t(u)t(f

Since

>λ<λ

=λ−t0t1

)t(u

∫ λλ=∗t

od)(f)t(u)t(f

Alternatively,

s

)s(F)t(u)t(f =∗L

∫ λλ=∗=

−t

o1 d)(f)t(u)t(f

s)s(F

L


(a) Let ∫ λλ−=∗=t

0 1221 d)t(x)t(x)t(x)t(x)t(y For , 3t0 << )t(x1 λ− and )(x 2 λ overlap as shown in Fig. (a).

)ee(4dee4dee4)t(y t2-t-t

0-t-t

0)-(t-2- −=λ=λ= ∫∫ λλλ

4

3t0 λ

(a) For , the two functions overlap as shown in Fig. (b). 3t >

( ) e1(e4e-e4dee4)t(y 3-t-30

-t-3

0)-(t-2- −==λ= λλλ∫

λ

4

3 t0

(b)

x1(t-λ)

x2(λ)

)

Therefore,

=)t(y

>−<<−3t),e1(e4

3t0),ee(43-t-

t-2t-

(b) For 1 , 2t << )(x1 λ and )t(x 2 λ− overlap as shown in Fig. (c).

1td)1)(1()t(x)t(x)t(y t1

t

121 −=λ=λ=∗= ∫

1 3

1

λt-1 t 2

x1(λ)x2(t - λ)

0

(c) For , the two functions overlap completely. 3t2 <<

1)1t(td)1)(1()t(y t1t

t

1t=−−=λ=λ= −−

∫ For , the two functions overlap as shown in Fig. (d). 4t3 <<

t4d)1)(1()t(y 31t

3

1t−=λ=λ= −−

∫

1 3

1

λ t-1 t20

(d)Therefore,

=)t(y

<<−<<<<−

otherwise,04t3,t43t2,12t1,1t

(c) For , 0t1- << )t(x1 λ− and )(x 2 λ overlap as shown in Fig. (e).

∫ λ=∗= λt

1-)-(t-

21 de4)1()t(x)t(x)t(y

[ ]1)(t-t

1-t- e14dee4)t(y +λ −=λ= ∫

λ1-1 t-1

1

x1(t - λ)

x2(λ)

(e) For , 1t0 <<

∫∫ λ+λ= λλt

0)-(t-0

1-)-(t- de4)1-(de4)1()t(y

4e4e8ee4ee4)t(y 1)-(t-tt0

-t01-

-t −−=−= +λλ For , the two functions overlap completely. 1t >

∫∫ λ+λ= λλ1

0)-(t-0

1-)-(t- de4)1-(de4)1()t(y

1)-(t1)-(t-t10

-t01-

-t e4e4e8ee4ee4)t(y −+λλ −−=−= Therefore,

=)t(y[ ]

>−−<<−−<<−

−+

+

+

1t,e4e4e81t0,4e4e80t1-,e14

1)(t-1)(t-t-

1)(t-t-

1)-(t


)tcos()t(f)t(f 21 ==

[ ] ∫ λλ−λ=t

0211- d)tcos()cos()s(F)s(FL

[ ])BAcos()BAcos(21

)Bcos()Acos( −++=

[ ] ∫ λλ−+=t

0211- d)]2tcos()t[cos(

21)s(F)s(FL

[ ] t0

t021

1-

2-)2tsin(

21

)tcos(21

)s(F)s(Fλ−

⋅+λ⋅=L

[ ] =)s(F)s(F 21

1-L )tsin(5.0)tcos(t5.0 + Chapter 15, Solution 48.

(a) Let 222 2)1s(2

5s2s2

)s(++

=++

=G

)t2sin(e)t(g -t=

)s(G)s(G)s(F =

[ ] λλ−λ== ∫ d)t(g)(g)s(G)s(G)t(f

t

01-L

∫ λλ−λ= λ−λt

0)t(-- d))t(2sin(e)2sin(e)t(f

[ ])BAcos()BAcos(21

)Bsin()Asin( +−−=

[ ]∫ λλ−−= λt

0-t- d))2t(2cos()t2cos(ee

21

)t(f

∫∫ λλ−−λ= λλt

02-

-tt

02-

-t

d)4t2cos(e2e

de)t2cos(2e

)t(f

[ ]∫ λλ+λ−⋅= λλ t

02-

-tt0

-2-t

d)4sin()t2sin()4cos()t2cos(e2e

2-e

)t2cos(2

e)t(f

∫ λλ−+= λt

02-

-tt2-t- d)4cos(e)t2cos(

2e

)1e-()t2cos(e41

)t(f

∫ λλ− λt

02-

-t

d)4sin(e)t2sin(2e

)e1()t2cos(e41

)t(f t2-t- −=

( ) t0

-2-t

)4sin(4)2cos(4-164

e)t2cos(

2e

λ−λ+

−λ

( ) t0

-2-t

)4cos(4)2sin(4-164

e)t2sin(

2e

λ+λ+

−λ

=)t(f )t4cos()t2cos(20e

)t2cos(20e

)t2cos(4

e)t2cos(

2e -3t-t-3t-t

+−−

)t2sin(10e

)t4sin()t2cos(10e -t-3t

++

)t4cos()t2sin(10e

)t4sin()t2sin(20e -t-t

−+

(b) Let 1s

2)s(X

+= ,

4ss

)s(Y+

=

)t(ue2)t(x -t= , )t(u)t2cos()t(y =

)s(Y)s(X)s(F =

[ ] ∫∞ λλ−λ==

01- d)t(x)(y)s(Y)s(X)t(f L

∫ λ⋅λ= λ−t

0)(t- de2)2cos()t(f

( ) t0

t- )2sin(2)2cos(41

ee2)t(f λ+λ

+⋅=

λ

( )[ ]1)t2sin(2)t2cos(ee52

)t(f tt- −+=

=)t(f t-e52

)t2sin(54

)t2cos(52

−+


Let x(t) = u(t) – u(t-1) and y(t) = h(t)*x(t).

[ ]

+−

=

−

+==

−−

−−−

)2s(s)e1(4)

se

s1()s(X)s(H)t(y

s1

s11 L

2s4LL

But

+−=

++=

+ 2s1

s1

21

2sB

sA

)2s(s1

++−

+−=

−−

2se

se

2s1

s12)s(Y

ss

)1t(u]e1[4)t(u]e1[2)t(y )1t(2t2 −−−−= −−−


Take the Laplace transform of each term.

[ ] [ ]4s

s3)s(V10)0(v)s(Vs2)0(v)0(vs)s(Vs 2

2

+=+−+′−−

4ss3

)s(V102)s(Vs22s)s(Vs 22

+=+−++−

4ss7s

4ss3

s)s(V)10s2s( 2

3

22

++

=+

+=++

10s2sDCs

4sBAs

)10s2s)(4s(s7s

)s(V 2222

3

+++

+++

=+++

+=

)4s(D)s4s(C)10s2s(B)s10s2s(As7s 232233 +++++++++=+


3s : A1CCA1 −=→+=2s : DBA20 ++= 1s : 4B2A6C4B2A107 ++=++=

0s : B-2.5DD4B100 =→+=

Solving these equations yields

269

A = , 2612

B = , 2617

C = , 2630-

D =

++−

+++

=10s2s

30s174s

12s9261

)s(V 22

++

−++

+⋅+

+⋅+

+= 222222 3)1s(

473)1s(

1s17

4s2

64s

s9261

)s(V

=)t(v )t3sin(e7847

)t3cos(e2617

)t2sin(266

)t2cos(269 t-t- −++


Taking the Laplace transform of the differential equation yields

[ ] [ ]1s

10)s(V6)]0(v)s(sV5)0('v)0(sv)s(Vs2+

=+−+−−

or ( ))3s)(2s)(1s(

24s16s2)s(V1s

10104s2)s(V6s5s2

2+++

++=→

+=−−−++

Let 3C,0B,5A,3s

C2s

B1s

A)s( −===+

++

++

=V

Hence,

)t(u)e3e5()t(v t3t −− −=


Take the Laplace transform of each term. [ ] [ ] 01)s(I2)0(i)s(Is3)0(i)0(is)s(Is2 =++−+′−−

0133s)s(I)2s3s( 2 =+−−−++

2sB

1sA

)2s)(1s(5s

)s(I+

++

=++

+=

4A = , 3-B =

2s3

1s4

)s(I+

−+

=

=)t(i )t(u)e3e4( -2t-t −



[ ] [ ]4s

s)s(V6)0(y)s(Ys5)0(y)0(ys)s(Ys 2

2

+=+−+′−−

4ss

54s)s(Y)6s5s( 22

+=−−−++

4s)4s)(9s(s

4ss

9s)s(Y)6s5s( 2

2

22

++++

=+

++=++

4sDCs

3sB

2sA

)4s)(3s)(2s(36s5s9s

)s(Y 22

23

++

++

++

=+++

+++=

427

)s(Y)2s(A -2s =+= = , 1375-

)s(Y)3s(B -3s =+= =

When , 0s =

265

D4D

3B

2A

)4)(3)(2(36

=→++=

When , 1s =

521C

5D

5C

4B

3A

)5)(12(546

=→+++=+

Thus,4s

265s5213s

13752s427

)s(Y 2 ++⋅

++

−+

=

=)t(y )t2sin(525

)t2cos(521

e1375

e4

27 3t-2t- ++−


Taking the Laplace transform of the differential equation gives

[ ] [ ]3s

5)s(V2)0(v)s(Vs3)0(v)0(vs)s(Vs2

+=+−+′−−

3ss2

13s

5)s(V)2s3s( 2

+−

=−+

=++

)3s)(2s)(1s(s2

)2s3s)(3s(s2

)s(V 2 +++−

=+++

−=

3sC

2sB

1sA

)s(V+

++

++

=

23A = , -4B = , 25C =

3s25

2s4

1s23

)s(V+

++

−+

=

=)t(v )t(u)e5.2e4e5.1( -3t-2t-t +−


Take the Laplace transform of each term. [ ] [ ])0(y)0(ys)s(Ys6)0(y)0(ys)0(ys)s(Ys 223 ′−−+′′−′−−

[ ] 22 2)1s(1s)0(y)s(Ys8++

+=−+

Setting the initial conditions to zero gives

5s2s1s

)s(Y)s8s6s( 223

+++

=++

5s2sEDs

4sC

2sB

sA

)5s2s)(4s)(2s(s)1s(

)s(Y 22 +++

++

++

+=++++

+=

401

A = , 201

B = , 104

3-C = ,

653-

D = , 657-

E =

22 2)1s(7s3

651

4s1

1043

2s1

201

s1

401

)s(Y++

+⋅−

+⋅−

+⋅+⋅=

2222 2)1s(4

651

2)1s()1s(3

651

4s1

1043

2s1

201

s1

401

)s(Y++

⋅−++

+⋅−

+⋅−

+⋅+⋅=

=)t(y )t2sin(e652

)t2cos(e653

e104

3e

201

)t(u401 t-t-4t-2t- −−−+


Taking the Laplace transform of each term we get:

[ ] 0)s(Vs

12)0(v)s(Vs4 =+−

8)s(Vs

12s4 =

+

3ss2

12s4s8

)s(V 22 +=

+=

=)t(v ( )t3cos2



[ ]4s

s)s(Y

s9

)0(y)s(Ys 2 +=+−

4s4ss

4ss

1)s(Ys

9s2

2

2

2

+++

=+

+=

+

9sDCs

4sBAs

)9s)(4s(s4ss

)s(Y 2222

23

++

+++

=++

++=

)4s(D)s4s(C)9s(B)s9s(As4ss 232323 +++++++=++


0s : D4B90 +=1s : C4A94 +=2s : DB1 +=3s : CA1 +=

Solving these equations gives

0A = , 54-B = , 1C = , 59D =

9s59

9ss

4s54-

9s59s

4s54-

)s(Y 22222 ++

++

+=

++

++

=

=)t(y )t3sin(6.0)t3cos()t2sin(0.4- ++


We take the Laplace transform of each term and obtain

10s6sse)s(Ve)s(V

s10)]0(v)s(sV[)s(V6 2

s2s2

++=→=+−+

−−

1)3s(e3e)3s()s(V

2

s2s2

++

−+=

−−

Hence, 3( 2) 3( 2)( ) cos( 2) 3 sin( 2) ( 2)t tv t e t e t u t− − − − = − − − −


Take the Laplace transform of each term of the integrodifferential equation.

[ ]2s

6)s(Y

s3

)s(Y4)0(y)s(Ys+

=++−

−+

=++ 12s

6s)s(Y)3s4s( 2

)3s)(2s)(1s(s)s4(

)3s4s)(2s()s4(s

)s(Y 2 +++−

=+++

−=

3sC

2sB

1sA

)s(Y+

++

++

=

5.2A = , 6B = , -10.5C =

3s5.10

2s6

1s5.2

)s(Y+

−+

++

=

=)t(y -3t-2t-t e5.10e6e5.2 −+


Take the Laplace transform of each term of the integrodifferential equation.

[ ]16s

4s4

)s(Xs3

)s(X5)0(x)s(Xs2 2 +=+++−

16s64s36s4s2

16ss4

4s2)s(X)3s5s2( 2

23

22

+−+−

=+

+−=++

)16s)(5.1s)(1s(32s18s2s

)16s)(3s5s2(64s36s4s2

)s(X 2

23

22

23

+++−+−

=+++−+−

=

16sDCs

5.1sB

1sA

)s(X 2 ++

++

++

=

-6.235)s(X)1s(A -1s =+= =

329.7)s(X)5.1s(B -1.5s =+= =

When , 0s =

2579.0D16D

5.1B

A(1.5)(16)

32-=→++=

)16s16ss(B)24s16s5.1s(A32s18s2s 232323 +++++++=−+−

)5.1s5.2s(D)s5.1s5.2s(C 223 ++++++

Equating coefficients of the s terms, 3

0935.0-CCBA1 =→++=

16s0.25790.0935s-

5.1s329.7

1s6.235-

)s(X 2 ++

++

++

=

=)t(x )t4sin(0645.0)t4cos(0935.0e329.7e6.235- -1.5t-t +−+


Consider the s-domain form of the circuit which is shown below.

I(s)

+ −

1

1/s 1/s

s

222 )23()21s(1

1ss1

s1s1s1

)s(I++

=++

=++

=

= t

23sine

32)t(i 2t-

=)t(i A)t866.0(sine155.1 -0.5t


8/s s

s4 2

+ −

+

Vx

−

4

V)t(u)e2e24(v

38j

34s

125.0

38j

34s

125.0s25.016

)8s8s3(s2s16V

s32s16)8s8s3(V

0VsV)s4s2(s

)32s16()8s4(V

0

s84

0V2

0Vs

s4V

t)9428.0j3333.1(t)9428.0j3333.1(x

2x

2x

x2

x2

x

xxx

−−+− ++−=

−+

−+

++

−+−=

++

+−=

+=++

=++++

−+

=+

−+

−+

−

vx = Vt3

22sine2

6t3

22cose)t(u4 3/t43/t4

−

− −−

Chapter 16, Solution 3. s

5/s 1/2 +

Vo

−

1/8

Current division leads to:

)625.0s(165

s16105

s81

21

21

s5

81Vo +

=+

=

++=

vo(t) = ( ) V)t(ue13125. t625.0−−0


The s-domain form of the circuit is shown below.

6 s

10/s1/(s + 1) + −

+

Vo(s)

−


+++

=

+++

=1s

110s6s

101s

1s106s

s10)s(V 2o

10s6sCBs

1sA

)10s6s)(1s(10

)s(V 22o +++

++

=+++

=

)1s(C)ss(B)10s6s(A10 22 ++++++=


2s : -ABBA0 =→+=1s : A5-CCA5CBA60 =→+=++=0s : -10C-2,B,2AA5CA1010 ===→=+=

22222o 1)3s(4

1)3s()3s(2

1s2

10s6s10s2

1s2

)s(V++

−++

+−

+=

+++

−+

=

=)t(vo V)tsin(e4)tcos(e2e2 -3t-3t-t −−


s2

2s1+

2

Io

s

( )

( ) A)t(ut3229.1sin7559.0e

or

A)t(ueee3779.0eee3779.0e)t(i

3229.1j5.0s)646.2j)(3229.1j5.1(

)3229.1j5.0(

3229.1j5.0s)646.2j)(3229.1j5.1(

)3229.1j5.0(

2s1

)3229.1j5.0s)(3229.1j5.0s)(2s(s

2VsI

)3229.1j5.0s)(3229.1j5.0s)(2s(s2

2sss2

2s1

2s

21

s1

12s

1V

t2

t3229.1j2/t90t3229.1j2/t90t2o

22

2o

2

−=

++=

−+++

+−

+++

−−−−

++

=

−++++==

−++++=

+++=

+++=

−

−°−−°−−


2

2s5+

Io

10/s

s

Use current division.

t3sine35t3cose5)t(i

3)1s(5

3)1s()1s(5

10s2ss5

2s5

s102s

2sI

tto

22222o

−− −=

++−

++

+=

++=

+++

+=

Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 Ix + 2s

1

2+s

– Z

2

2

2 s211s2s2

s21s21

s2s1

)s2(s1

1s2//s11Z

+

++=

++=

++=+=

)5.0ss(CBs

)1s(A

)5.0ss)(1s(1s2

1s2s2s21x

1s2

ZVI 22

2

2

2x

++

++

+=

+++

+=

++

++

==

)1s(C)ss(B)5.0ss(A1s2 222 ++++++=+

BA2:s2 +=

2CC2CBA0:s −=→+=++=

-4B ,6A3 0.5A or CA5.01:constant ==→=+=

222x866.0)5.0s()5.0s(4

1s6

75.0)5.0s(2s4

1s6I

++

+−

+=

++

+−

+=

[ ] A)t(ut866.0cose46)t(i t5.0x

−−=


(a) )1s(s

1s5.1ss22)s21(

s1)s21//(1

s1Z

2

+++

=++

+=++=

(b) )1s(s2

2s3s3

s11

1s1

21

Z1 2

+++

=+

++=

2s3s3)1s(s2Z 2 ++

+=


(a) The s-domain form of the circuit is shown in Fig. (a).

=+++

=+=s1s2)s1s(2

)s1s(||2Zin 1s2s)1s(2

2

2

+++

1

1

2 s 2/s 1/s

s

2

(a) (b)(b) The s-domain equivalent circuit is shown in Fig. (b).

2s3)2s(2

s23)s21(2

)s21(||2++

=++

=+

2s36s5

)s21(||21++

=++

=

++

+

++

⋅=

++

=

2s36s5

s

2s36s5

s

2s36s5

||sZin 6s7s3)6s5(s

2 +++

Chapter 16, Solution 10. To find ZTh, consider the circuit below. 1/s Vx + 1V 2 Vo 2Vo - Applying KCL gives

s/12VV21 x

o +=+

But xo Vs/12

2V+

= . Hence

s3)1s2(V

s/12V

s/12V41 x

xx +−=→

+=

++

s3)1s2(

1VZ x

Th+

−==

To find VTh, consider the circuit below. 1/s Vy +

1

2+s

2 Vo 2Vo

- Applying KCL gives

)1s(34V

2VV2

1s2

oo

o +−=→=+

+

But 0Vs1V2V ooy =+•+−

)1s(s3)2s(4

s2s

)1s(34)

s21(VVV oyTh +

+−=

+

+−=+==

Chapter 16, Solution 11. The s-domain form of the circuit is shown below.

4/s s

I2I1+ −

+ −

2 4/(s + 2) 1/s

Write the mesh equations.

21 I2Is4

2s1

−

+= (1)

21 I)2s(I-22s

4-++=

+ (2)

Put equations (1) and (2) into matrix form.

+

+=

+ 2

1

II

2s2-2-s42

2)(s4-s1

)4s2s(s2 2 ++=∆ ,

)2s(s4s4s2

1 ++−

=∆ , s6-

2 =∆

4s2sCBs

2sA

)4s2s)(2s()4s4s(21

I 22

21

1 +++

++

=++++−⋅

=∆∆

=

)2s(C)s2s(B)4s2s(A)4s4s(21 222 ++++++=+−⋅


2s : BA21 += 1s : CB2A22- ++=

0s : C2A42 += Solving these equations leads to A 2= , 23-B = , -3C =

221 )3()1s(3s23-

2s2

I++−

++

=

22221 )3()1s(3

323-

)3()1s()1s(

23-

2s2

I++

⋅++++

⋅++

=

=)t(i1 [ ] A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2 -t-2t −−

2222

2 )3()1s(3-

)4s2s(2s

s6-I

++=

++⋅=

∆∆

=

== )t3sin(e33-

)t(i t-2 A)t(u)t732.1sin(e1.732- -t

Chapter 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below.

Vo

10/(s + 1) + −

1/(2s) 4

s

3/s

oo

osV2

4V

s3

s

V1s

10

+=+−

+

1s15s1510

151s

10V)ss25.01( o

2

+++

=++

=++

1s25.0sCBs

1sA

)1s25.0s)(1s(25s15

V 22o +++

++

=+++

+=

740

V)1s(A 1-so =+= =

)1s(C)ss(B)1s25.0s(A25s15 22 ++++++=+


2s : -ABBA0 =→+=1s : C-0.75ACBA25.015 +=++= 0s : CA25 +=

740A = , 740-B = , 7135C =

43

21

s

23

32

7155

43

21

s

21

s

740

1s1

740

43

21

s

7135

s740-

1s740

V 222o

+

+

⋅+

+

+

+−

+=

+

+

++

+=

+

−= t

23

sine)3)(7()2)(155(

t23

cose740

e740

)t(v 2t-2t-t-o

=)t(vo V)t866.0sin(e57.25)t866.0cos(e714.5e714.5 2-t2-t-t +−



Vo

1/(s + 2)

1/s 2s

Io

2 1

Applying KCL at node o,

ooo V

1s21s

s12V

1s2V

2s1

++

=+

++

=+

)2s)(1s(1s2

Vo +++

=

2sB

1sA

)2s)(1s(1

1s2V

I oo +

++

=++

=+

=

1A = , -1B =

2s1

1s1

Io +−

+=

=)t(io ( ) A)t(uee -2t-t −

Chapter 16, Solution 14. We first find the initial conditions from the circuit in Fig. (a).

1 Ω 4 Ω

io

+

vc(0)

−

+ −

5 V

(a)

A5)0(io =− , V0)0(vc =−

We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).

2s 5/s

Io

Vo

+ −

1 4

15/s 4/s

(b)At node o,

0s440V

s5

s2V

1s15V ooo =

+−

+++−

oV)1s(4

ss2

11

s5

s15

+++=−

o

2

o

22

V)1s(s42s6s5

V)1s(s4

s2s2s4s4s

10+++

=+

++++=

2s6s5)1s(40

V 2o +++

=

s5

)4.0s2.1s(s)1s(4

s5

s2V

I 2o

o +++

+=+=

4.0s2.1sCBs

sA

s5

I 2o +++

++=

sCsB)4.0s2.1s(A)1s(4 s2 ++++=+


0s : 10AA4.04 =→=1s : -84-1.2ACCA2.14 =+=→+=2s : -10-ABBA0 ==→+=

4.0s2.1s8s10

s10

s5

I 2o +++

−+=

2222o 2.0)6.0s()2.0(10

2.0)6.0s()6.0s(10

s15

I++

−++

+−=

=)t(io ( )[ ] A)t(u)t2.0sin()t2.0cos(e1015 0.6t- −−


First we need to transform the circuit into the s-domain.

2s5+

Vo

+ −

+ − 5/s

s/4

+ Vx −

10 3Vx

2s5VV

2s5VV,But

2ss5V120V)40ss2(0

2ss5sVVs2V120V40

010

2s5V

s/50V

4/sV3V

xoox

xo2

oo2

xo

ooxo

++=→

+−=

+−−++==

+−++−

=+−

+−

+−

We can now solve for Vx.

)40s5.0s)(2s()20s(5V

2s)20s(10V)40s5.0s(2

02s

s5V1202s

5V)40ss2(

2

2x

2x

2

xx2

−++

+−=

++

−=−+

=+

−−

++++

Chapter 16, Solution 16. We first need to find the initial conditions. For 0t < , the circuit is shown in Fig. (a). To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit.

1 Ω

+ −Vo

1 F

Vo/2 + − 1 H io

+ −

2 Ω

3 V

(a)

Hence,

A1-33-

i)0(i oL === , V1-vo =

V5.221-

)1-)(2(-)0(vc =

−=

We now incorporate the initial conditions for as shown in Fig. (b). 0t >

I2I1 + −

s

2.5/s + −

1

+ −Vo

1/s

Vo/2 + −

Io

− +

2

-1 V

5/(s + 2)

(b)For mesh 1,

02

Vs5.2

Is1

Is1

22s

5- o21 =++−

+++

But, 2oo IIV ==

s5.2

2s5

Is1

21

Is1

2 21 −+

=

−+

+ (1)

For mesh 2,

0s5.2

2V

1Is1

Is1

s1 o12 =−−+−

++

1s5.2

Is1

s21

Is1

- 21 −=

+++ (2)

Put (1) and (2) in matrix form.

−

−+

=

++

−+

1s5.2

s5.2

2s5

I

I

s1

s21

s1-

s1

21

s1

2

2

1

s3

2s2 ++=∆ , )2s(s

5s4

2-2 +++=∆

3s2s2CBs

2sA

)3s2s2)(2s(132s-

II 22

22

2o +++

++

=+++

+=

∆∆

==

)2s(C)s2s(B)3s2s2(A132s- 222 ++++++=+


2s : B A22- +=1s : CB2A20 ++=0s : C2A313 +=

Solving these equations leads to

7143.0A = , , -3.429B = 429.5C =

5.1ss714.2s7145.1

2s7143.0

3s2s2429.5s429.3

2s7143.0

I 22o ++−

−+

=++

−−

+=

25.1)5.0s()25.1)(194.3(

25.1)5.0s()5.0s(7145.1

2s7143.0

I 22o +++

+++

−+

=

=)t(io [ ] A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0 -0.5t-0.5t-2t +−

Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below.

2/(s+1)

I3

+ −

1/s

I2I1

4

s

1 1

For mesh 3,

0IsIs1

Is1

s1s

2213 =−−

++

+ (1)

For the supermesh,

0Iss1

I)s1(Is1

1 321 =

+−++

+ (2)

But (3) 4II 21 −= Substituting (3) into (1) and (2) leads to

+=

+−

++s1

14Is1

sIs1

s2 32 (4)

1s2

s4-

Is1

sIs1

s- 32 +−=

++

+ (5)

Adding (4) and (5) gives

1s2

4I2 2 +−=

1s1

2I2 +−=

== )t(i)t(i 2o A)t(u)e2( -t− Chapter 16, Solution 18.

vs(t) = 3u(t) – 3u(t–1) or Vs = )e1(s3

se

s3 s

s−

−−=−

1 Ω

+

Vo

−

+ − 1/s

Vs 2 Ω

V)]1t(u)e22()t(u)e22[()t(v

)e1(5.1s

2s2)e1(

)5.1s(s3V

VV)5.1s(02

VsV

1VV

)1t(5.1t5.1o

sso

soo

oso

−−−−=

−

+−=−

+=

=→=++−

−−−

−−

+

Chapter 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below.

I

1/s

− +

2 I V1 Vo

1/s

s

+ −

2

2 4/(s + 2)

At the supernode,

o11 sV

s1

sV

22

V)2s(4++=+

−+

o1 Vss1

Vs1

21

22s

2++

+=++

(1)

But and I2VV 1o +=s

1VI 1 +=

2s2Vs

s)2s(s2V

Vs

)1V(2VV oo

11

1o +−

=+−

=→+

+= (2)


oo Vs2s

2V

2ss

s1s2

s1

22s

2+

+−

+

+

=−++

oVs2s1s2

)2s(s)1s2(2

s1

22s

2

+

++

=++

+−++

o

22

V2s

1s4s2s9s2

)2s(ss9s2

+++

=++

=++

732.3sB

2679.0sA

1s4s9s2

V 2o ++

+=

+++

=

443.2A = , 4434.0-B =

732.3s4434.0

2679.0s443.2

Vo +−

+=

Therefore,

=)t(vo V)t(u)e4434.0e443.2( -3.732t-0.2679t −

Chapter 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown.

1 s

+ −1/s 2/s

Vo

+ −

1

1/(s + 1) 1/s

At the main non-reference node, KCL gives

s1

sV

1V

s11Vs2)1s(1 ooo ++=

+−−+

s1s

V)s1s)(1s(Vs21s

soo

++++=−−

+

oV)s12s2(2s

1s1s

s++=−

+−

+

)1s2s2)(1s(1s4s2-

V 2

2

o +++−−

=

5.0ssCBs

1sA

)5.0ss)(1s(5.0s2s-

V 22o +++

++

=+++

−−=

1V)1s(A 1-so =+= =

)1s(C)ss(B)5.0ss(A5.0s2s- 222 ++++++=−−

Equating coefficients : 2s : -2BBA1- =→+=1s : -1CCBA2- =→++=0s : -0.515.0CA5.00.5- =−=+=

222o )5.0()5.0s()5.0s(2

1s1

5.0ss1s2

1s1

V++

+−

+=

+++

−+

=

=)t(vo [ ] V)t(u)2tcos(e2e 2-t-t −

Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 s

V1 Vo + 2/s 2 1/s

At n

s

s1

10−

At n

1 −s

V

Subs

10 =

=Vo

=10

cons

=Vo

Taki

(vo

10/

-

ode 1,

ooo VsVsVs

sVVV

)12

()1(102

2

11

1−++=→+

−= (1)

ode 2,

)12

(2

21 ++=→+= ssVVsVVV

oooo (2)

tituting (2) into (1) gives

ooo VsssVsVsss )5.12()12

()12/)(1( 22

2 ++=−++++

5.12)5.12(10

22 +++

+=++ ss

CBssA

sss

CsBsssA ++++ 22 )5.12( BAs +=0:2 CAs += 20:

-40/3C -20/3,B ,3/205.110:tant ===→= AA

++

−+++

−=

+++

− 22222 7071.0)1(7071.0414.1

7071.0)1(11

320

5.1221

320

sss

ssss

s

ng the inverse Laplace tranform finally yields

[ ] V)t(ut7071.0sine414.1t7071.0cose1320)t tt −− −−=

Chapter 16, Solution 22. The s-domain version of the circuit is shown below. 4s V1 V2

1s +

12 1 2 3/s

At node 1,

s4V

s411V

1s12

s4VV

1V

1s12 2

1211 −

+=

+→

−+=

+ (1)

At node 2,

++=→+=

−1s2s

34VVV

3s

2V

s4VV 2

212221 (2)


222

2 V23s

37s

34

s41

s4111s2s

34V

1s12

++=

−

+

++=

+

)89s

47s(

CBs)1s(

A

)89s

47s)(1s(

9V22

2++

++

+=

+++=

)1s(C)ss(B)89s

47s(A9 22 ++++++=

Equating coefficients:

BA0:s2 +=

A43CCA

43CBA

470:s −=→+=++=

-18C -24,B ,24AA83CA

899:constant ===→=+=

6423)

87s(

3

6423)

87s(

)8/7s(24)1s(

24

)89s

47s(

18s24)1s(

24V222

2++

+++

+−

+=

++

+−

+=

Taking the inverse of this produces:

[ ] )t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v t875.0t875.0t2

−−− +−= Similarly,

)89s

47s(

FEs)1s(

D

)89s

47s)(1s(

1s2s349

V22

2

1++

++

+=

+++

++

=

)1s(F)ss(E)89s

47s(D1s2s

349 222 ++++++=

++

Equating coefficients:

ED12:s2 +=

D436FFD

436or FED

4718:s −=→+=++=

0F 4,E ,8DD833or FD

899:constant ===→=+=

6423)

87s(

2/7

6423)

87s(

)8/7s(4)1s(

8

)89s

47s(

s4)1s(

8V222

1++

−++

++

+=

+++

+=

Thus, [ ] )t(u)t5995.0sin(e838.5)t5995.0cos(e4e8)t(v t875.0t875.0t

1−−− −+=


The s-domain form of the circuit with the initial conditions is shown below. V

I

1/sC sL R -2/s

4/s 5C


sCVsLV

RV

C5s2

s4

++=++

++=+

LC1

RCs

ss

CVs

sC56 2

LC1RCssC6s5

V 2 +++

=

But 88010

1RC1

== , 208041

LC1

==

22222 2)4s()2)(230(

2)4s()4s(5

20s8s480s5

V++

+++

+=

+++

=

=)t(v V)t2sin(e230)t2cos(e5 -4t-4t +

)20s8s(s4480s5

sLV

I 2 +++

==

20s8sCBs

sA

)20s8s(s120s25.1

I 22 +++

+=++

+=

6A = , -6B = , -46.75C =

22222 2)4s()2)(375.11(

2)4s()4s(6

s6

20s8s75.46s6

s6

I++

−++

+−=

+++

−=

=)t(i 0t),t2sin(e375.11)t2cos(e6)t(u6 -4t-4t >−−

Chapter 16, Solution 24. At t = 0-, the circuit is equivalent to that shown below. + 9A 4Ω 5Ω vo -

20)9(54

4x5)0(vo =+

=

For t > 0, we have the Laplace transform of the circuit as shown below after transforming the current source to a voltage source. 4Ω 16Ω Vo + 36V 10A 2/s 5

Ω

- Applying KCL gives

8.12B,2.7A,5.0s

BsA

)5.0s(ss206.3V

5V

2sV

1020

V36o

ooo −==+

+=++

=→+=+−

Thus,

[ ] )t(ue8.122.7)t(v t5.0o

−−=

Chapter 16, Solution 25. For , the circuit in the s-domain is shown below. 0t >

s 6 I

Applying KVL,

+

V

−

+ −

(2s)/(s2 + 16)

+ −

9/s

2/s

0s2

Is9

s616ss2

2 =+

++++−

)16s)(9s6s(32s4

I 22

2

++++

=

)16s()3s(s288s36

s2

s2I

s9V 22

2

+++

+=+=

16sEDs

)3s(C

3sB

sA

s2

22 ++

++

++

++=

)s48s16s3s(B)144s96s25s6s(A288s36 2342342 ++++++++=+

)s9s6s(E)s9s6s(D)s16s(C 232343 ++++++++

Equating coefficients : 0s : A144288 = (1) 1s : E9C16B48A960 +++= (2) 2s : E6D9B16A2536 +++= (3) 3s : ED6CB3A60 ++++= (4) 4s : (5) DBA0 ++=

Solving equations (1), (2), (3), (4) and (5) gives

2A = , B , C7984.1-= 16.8-= , D 2016.0-= , 765.2E =

16s)4)(6912.0(

16ss2016.0

)3s(16.8

3s7984.1

s4

)s(V 222 ++

+−

+−

+−=

=)t(v V)t4sin(6912.0)t4cos(2016.0et16.8e7984.1)t(u4 -3t-3t +−−−


Consider the op-amp circuit below.

R2

+Vo −

0

+ −

R1

1/sC

−+

Vs

At node 0,

sC)V0(R

V0R

0Vo

2

o

1

s −+−

=−

( )o2

1s V-sCR1

RV

+=

211s

o

RRCsR1-

VV

+=

But 21020

RR

2

1 == , 1)1050)(1020(CR 6-31 =××=

So, 2s

1-VV

s

o

+=

)5s(3Ve3V s

-5ts +=→=

5)2)(s(s3-

Vo ++=

5sB

2sA

5)2)(s(s3

V- o ++

+=

++=

1A = , -1B =

2s1

5s1

Vo +−

+=

=)t(vo ( ) )t(uee -2t-5t −



For mesh 1,

I2

2s

I1+ −

10/(s + 3) 1

2s s

1

221 IsII)s21(3s

10−−+=

+

21 I)s1(I)s21(3s

10+−+=

+ (1)

For mesh 2,

112 IsII)s22(0 −−+=

21 I)1s(2I)s1(-0 +++= (2) (1) and (2) in matrix form,

++++

=

+

2

1

II

)1s(2)1s(-)1s(-1s2

0)3s(10

1s4s3 2 ++=∆

3s)1s(20

1 ++

=∆

3s)1s(10

2 ++

=∆

Thus

=∆∆

= 11I )1s4s3)(3s(

)1s(202 ++++

=∆∆

= 22I

)1s4s3)(3s()1s(10

2 ++++

2I1=



VI1 +

−

1

2s s I2

s 6/s

For mesh 1,

21 IsI)s21(s6

++=

For mesh 2,

21 I)s2(Is0 ++=

21 Is2

1-I

+=


2

2

22 Is

2)5s(s-IsI

s2

12s)-(1s6 ++

=+

++=

or 2s5s

6-I 22 ++=

+

o

−2

(1)

(2)

4.561)0.438)(s(s12-

2s5s12-

I2V 22o ++=

++==

Since the roots of s are -0.438 and -4.561, 02s52 =++

561.4sB

438.0sA

Vo ++

+=

-2.914.123

12-A == , 91.2

4.123-12-

B ==

561.4s91.2

0.438s2.91-

)s(Vo ++

+=

=)t(vo [ ] V)t(uee91.2 t438.0-4.561t −



1 : 2Io

+ −

10/(s + 1)

1

4/s 8

Let 1s2

8s48)s4)(8(

s4

||8ZL +=

+==

When this is reflected to the primary side,

2n,nZ

1Z 2L

in =+=

1s23s2

1s22

1Zin ++

=+

+=

3s21s2

1s10

Z1

1s10

Iin

o ++

⋅+

=⋅+

=

5.1sB

1sA

)5.1s)(1s(5s10

Io ++

+=

+++

=

-10A = , 20B =

5.1s20

1s10-

)s(Io ++

+=

=)t(io [ ] A)t(uee210 t-1.5t −−


)s(X)s(H)s(Y = , 1s3

1231s

4)s(X

+=

+=

22

2

)1s3(34s8

34

)1s3(s12

)s(Y++

−=+

=

22 )31s(1

274

)31s(s

98

34

)s(Y+

⋅−+

⋅−=

Let 2)31s(s

98-

)s(+

⋅=G

Using the time differentiation property,

+=⋅= 3t-3t-3t- eet31-

98-

)et(dtd

98-

)t(g

3t-3t- e

98

et278

)t(g −=

Hence,

3t-3t-3t- et274

e98

et278

)t(u34

)t(y −−+=

=)t(y 3t-3t- et274

e98

)t(u34

+−


s1

)s(X)t(u)t(x =→=

4ss10

)s(Y)t2cos(10)t(y 2 +=→=

==)s(X)s(Y

)s(H4s

s102

2

+


(a) )s(X)s(H)s(Y =

s1

5s4s3s

2 ⋅++

+=

5s4s

CBssA

)5s4s(s3s

22 +++

+=++

+=

CsBs)5s4s(A3s 22 ++++=+


0s : 53AA53 =→= 1s : 57- A41CCA41 =−=→+= 2s : 53--ABBA0 ==→+=

5s4s7s3

51

s53

)s(Y 2 +++

⋅−=

1)2s(1)2s(3

51

s6.0

)s(Y 2 ++++

⋅−=

=)t(y [ ] )t(u)tsin(e2.0)tcos(e6.06.0 -2t-2t −−

(b) 22t-

)2s(6

)s(Xet6)t(x+

=→=

22 )2s(6

5s4s3s

)s(X)s(H)s(Y+

⋅++

+==

5s4sDCs

)2s(B

2sA

)5s4s()2s()3s(6

)s(Y 2222 +++

++

++

=+++

+=


3s : (1) -ACCA0 =→+=2s : DBA2DC4BA60 ++=+++= (2) 1s : D4B4A9D4C4B4A136 ++=+++= (3) 0s : BA2D4B5A1018 +=++= (4)

Solving (1), (2), (3), and (4) gives

6A = , 6B = , 6-C = , -18D =

1)2s(18s6

)2s(6

2s6

)s(Y 22 +++

−+

++

=

1)2s(6

1)2s()2s(6

)2s(6

2s6

)s(Y 222 ++−

+++

−+

++

=

=)t(y [ ] )t(u)tsin(e6)tcos(e6et6e6 -2t-2t-2t-2t −−+


)s(X)s(Y

)s(H = , s1

)s(X =

16)2s()4)(3(

16)2s(s2

)3s(21

s4

)s(Y 22 ++−

++−

++=

== )s(Ys)s(H20s4s

s1220s4s

s2)3s(2

s4 22

2

++−

++−

++



+

Vo(s)

−

Vo

10/s

2

4+ −

Vs

s

Using nodal analysis,

s10V

4V

2sVV ooos +=

+−

o2

os V)s2s(101

)2s(41

1V10s

41

2s1

)2s(V

++++=

+++

+=

( ) o2

s V30s9s2201

V ++=

=s

o

VV

30s9s220

2 ++



+

Vo

−

I V1

+ −

2/s

2I

s

Vs 3

At node 1,

3sV

II2 1

+=+ , where

s2VV

I 1s −=

3sV

s2VV

3 11s

+=

−⋅

1s1 V

2s3

V2s3

3sV

−=+

s1 V2s3

V2s3

3s1

=

++

s21 V2s9s3

)3s(s3V

+++

=

s21o V2s9s3

s9V

3s3

V++

=+

=

==s

o

VV

)s(H2s9s3

s92 ++


From the previous problem,

s21 V

2s9s3s3

3sV

I3++

=+

=

s2 V2s9s3

sI

++=

But o

2

s Vs9

2s9s3V

++=

2

2

3 9 23 9 2 9 9

ooVs s sI V

s s s+ +

= ⋅+ +

=

==I

V)s(H o 9


(a) Consider the circuit shown below.

3 2s

+

Vx

−

2/s + −

I1 I2+ −

Vs 4Vx

For loop 1,

21s Is2

Is2

3V −

+= (1)

For loop 2,

0Is2

Is2

s2V4 12x =−

++

But,

−=s2

)II(V 21x

So, 0Is2

Is2

s2)II(s8

1221 =−

++−

21 Is2s6

Is6-

0

−+= (2)

In matrix form, (1) and (2) become

−

+=

2

1s

II

s2s6s6-s2-s23

0V

−

−

+=∆

s2

s6

s2s6

s2

3

4s6s

18−−=∆

s1 Vs2s6

−=∆ , s2 Vs6

=∆

s1

1 Vs64s18

)s2s6(I

−−−

=∆∆

=

=−−

−=

32s9ss3

VI

s

1

9s2s33s

2

2

−+−

(b) ∆∆

= 22I

∆∆−∆

=−= 2121x s

2)II(

s2

V

∆=

∆−−

= ssx

V4-)s6s2s6(Vs2V

==s

s

x

2

4V-Vs6

VI

2s3-


(a) Consider the following circuit.

1

+

Vo

−

s Is

1/s

VoV1

1/s+ −

Vs

1 Io

At node 1,

sVV

Vs1

VV o11

1s −+=

−

o1s Vs1

Vs1

s1V −

++= (1)

At node o,

oooo1 V)1s(VVs

sVV

+=+=−

o

21 V)1ss(V ++= (2)


oo2

s Vs1V)1ss)(s11s(V −++++=

o23

s V)2s3s2s(V +++=

==s

o1 V

V)s(H

2s3s2s1

23 +++

(b) o

2o

231ss V)1ss(V)2s3s2s(VVI ++−+++=−=

o23

s V)1s2ss(I +++=

==s

o2 I

V)s(H

1s2ss1

23 +++

(c) 1

VI o

o =

==== )s(HIV

II

)s(H 2s

o

s

o3 1s2ss

123 +++

(d). ==== )s(HVV

VI

)s( 1s

o

s

o4H

2s3s2s1

23 +++



+

Vo

−

Io

+ −

1/sC

R Vb

Va−+

Vs

Since no current enters the op amp, flows through both R and C. oI

+=sC1

RI-V oo

sCI-

VVV osba ===

=+

==sC1

sC1RVV

)s(Hs

o 1sRC +


(a) LRs

LRsLR

RVV

)s(Hs

o

+=

+==

=)t(h )t(ueLR LRt-

(b) s1)s(V)t(u)t(v ss =→=

LRsB

sA

)LRs(sLR

VLRs

LRV so +

+=+

=+

=

1A = , -1B =

LRs1

s1

Vo +−=

=−= )t(ue)t(u)t(v L-Rt

o )t(u)e1( L-Rt− Chapter 16, Solution 41.

)s(X)s(H)s(Y =

1s2

)s(H)t(ue2)t(h t-

+=→=

s5)s(X)s(V)t(u5)t(v ii ==→=

1sB

sA

)1s(s10

)s(Y+

+=+

=

10A = , -10B =

1s10

s10

)s(Y+

−=

=)t(y )t(u)e1(10 -t−


)s(X)s(Y)s(Ys2 =+

)s(X)s(Y)1s2( =+

)21s(21

1s21

)s(X)s(Y

)s(H+

=+

==

=)t(h )t(ue5.0 2-t


i(t)

+ −

1Ω

1F u(t)

1H

First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get: '

CC vi;0'ivi)t(u ==+++−

Thus,

)t(uivi

iv

C'

'C

+−−=

=

Finally we get,

[ ] [ ] )t(u0i

v10)t(i;)t(u

10

iv

1110

iv CCC +

=

+

−−

=

′

′

Chapter 16, Solution 44. 1/8 F 1H

)t(u4 2Ω

+ −

+

vx

−

4Ω

First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get:

LCxxLC

'C

C

'C

Cx

x'L

xL'C

'Cx

L

i3333.1v3333.0vor;v2i4v2

vv

8v

4vv

v)t(u4i

v4i8vor;08

v2

vi

+=−+=+=+=

−=

−==++−

LC

'L

LCLCL'C

i3333.1v3333.0)t(u4i

i666.2v3333.1i333.5v3333.1i8v

−−=

+−=−−=

Now we can write the state equations.

=

+

−−

−=

L

Cx

L

C'L

'C

iv

3333.13333.0

v;)t(u40

iv

3333.13333.0666.23333.1

iv


First select the inductor current iL (current flowing left to right) and the capacitor voltage vC (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get:

2o'L

oL'CL

o'C

vvi

v2i4vor0i2

v4

v

−=

+==++−

1Co vvv +−=

21C

'L

1CL'C

vvvi

v2v2i4v

−+−=

+−=

[ ] [ ]

+

−=

−+

−−

=

′

′

)t(v)t(v

01vi

10)t(v;)t(v)t(v

0211

vi

2410

vi

2

1

C

Lo

2

1

C

L

C

L


First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get:

sC'L

sLC'Cs

C'CL

vvi

iiv25.0vor0i4

vvi

+−=

++−==−++−

Thus,

+

=

+

−

−=

s

s

L

Co

s

s'L

'C

'L

'C

iv

0000

iv

01

)t(v;iv

0110

iv

01125.0

iv


First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the left) to be the state variables.

Letting i1 = 4

v'C and i2 = iL and applying KVL we get:

Loop 1:

1CL'CL

'C

C1 v2v2i4vor0i4

v2vv +−==

−++−

Loop 2:

21C21CL

L'L

2'L

'C

L

vvvv2

v2v2i4i2i

or0vi4

vi2

−+−=−+−

+−=

=++

−

1CL1CL

1 v5.0v5.0i4

v2v2i4i +−=

+−=

+

−=

−+

−−

=

′

′

)t(v)t(v

0005.0

vi

015.01

)t(i)t(i

;)t(v)t(v

0211

vi

2410

vi

2

1

C

L

2

1

2

1

C

L

C

L


Let x1 = y(t). Thus, )t(zx4x3yxandxyx 21'22

''1 +−−=′′===

This gives our state equations.

[ ] [ ] )t(z0xx

01)t(y;)t(z10

xx

4310

xx

2

1

2

1'2

'1 +

=

+

−−

=


zxyorzyzxxand)t(yxLet 2'''

121 +=−=−== Thus, z3x5x6zz2z)zx(5x6zyx 21

''21

''2 −−−=−+++−−=−′′=

This now leads to our state equations,

[ ] [ ] )t(z0xx

01)t(y;)t(z3

1xx

5610

xx

2

1

2

1'2

'1 +

=

−

+

−−

=


Let x1 = y(t), x2 = .xxand,x '23

'1 =

Thus, )t(zx6x11x6x 321

"3 +−−−=

We can now write our state equations.

[ ] [ ] )t(z0xxx

001)t(y;)t(z100

xxx

6116100010

xxx

3

2

1

3

2

1

'3

'2

'1

+

=

+

−−−=


We transform the state equations into the s-domain and solve using Laplace transforms.

+=−

s1B)s(AX)0(x)s(sX

Assume the initial conditions are zero.

+++

=

−+=

=−

−

)s/2(0

4s24s

8s4s1

s1

20

s244s

)s(X

s1B)s(X)AsI(

2

1

222222

221

2)2s(2

2)2s()2s(

s1

2)2s(4s

s1

8s4s4s

s1

)8s4s(s8)s(X)s(Y

++

−+

++

+−+=

++

−−+=

++

−−+=

++==

y(t) = ( )( ) )t(ut2sint2cose1 t2 +− −


Assume that the initial conditions are zero. Using Laplace transforms we get,

+−+

++=

+−

+=

−

s/4s/3

2s214s

10s6s1

s/2s/1

0411

4s212s

)s(X 2

1

222211)3s(8.1s8.0

s8.0

)1)3s((s8s3X

++

−−+=

++

+=

2222 1)3s(16.

1)3s(3s8.0

s8.0

+++

++

+−=

)t(u)tsine6.0tcose8.08.0()t(x t3t3

1−− +−=

222221)3s(4.4s4.1

s4.1

1)3s((s14s4X

++

−−+=

++

+=

2222 1)3s(12.0

1)3s(3s4.1

s4.1

++−

++

+−=

)t(u)tsine2.0tcose4.14.1()t(x t3t3

2−− −−=

)t(u)tsine8.0tcose4.44.2(

)t(u2)t(x2)t(x2)t(yt3t3

211−− −+−=

+−−=

)t(u)tsine6.0tcose8.02.1()t(u2)t(x)t(y t3t3

12−− +−−=−=


If is the voltage across R, applying KCL at the non-reference node gives oV

oo

oo

s VsL1

sCR1

sLV

VsCRV

I

++=++=

RLCsRsLIsRL

sL1

sCR1

IV 2

sso ++

=++

=

RsLRLCsIsL

RV

I 2so

o ++==

LC1RCssRCs

RsLRLCssL

II

)s(H 22s

o

++=

++==

The roots

LC1

)RC2(1

RC21-

s 22,1 −±=

both lie in the left half plane since R, L, and C are positive quantities. Thus, the circuit is stable.


(a) 1s

3)s(H1 += ,

4s1

)s(H2 +=

)4s)(1s(3

)s(H)s(H)s(H 21 ++==

[ ]

++

+==

4sB

1sA

)s(H)t(h 1-1- LL

1A = , 1-B = =)t(h )t(u)ee( -4t-t −

(b) Since the poles of H(s) all lie in the left half s-plane, the system is stable.


Let be the voltage at the output of the first op amp. 1oV

sRC1

RsC1

VV

s

1o −=

−= ,

sRC1

VV

1o

o −=

222s

o

CRs1

VV

)s(H ==

22CRt

)t(h =

∞=

∞→)t(hlim

t, i.e. the output is unbounded.

Hence, the circuit is unstable.


LCs1sL

sC1

sL

sC1

sL

sC1

||sL 2+=

+

⋅=

RsLRLCssL

LCs1sL

R

LCs1sL

VV

2

2

2

1

2

++=

++

+=

LC1

RC1

ss

RC1

s

VV

21

2

+⋅+

⋅=

Comparing this with the given transfer function,

RC1

2 = , LC1

6 =

If , Ω= k1R ==R21

C F500 µ

==C61

L H3.333

Chapter 16, Solution 57. The circuit in the s-domain is shown below.

+

Vx

−

V1

+ −

R1

C R2

L

Vi

Z

CsRLCs1sLR

sC1sLR)sLR()sC1(

)sLR(||sC1

Z2

22

2

22 ++

+=

+++⋅

=+=

i1

1 VZR

ZV

+=

i12

21

2

2o V

ZRZ

sLRR

VsLR

RV

+⋅

+=

+=

CsRLCs1sLR

R

CsRLCs1sLR

sLRR

ZRZ

sLRR

VV

22

21

22

2

2

2

12

2

i

o

+++

+

+++

⋅+

=+

⋅+

=

sLRRCRsRLCRsR

VV

212112

2

i

o

++++=

LCRRR

CR1

LR

ss

LCRR

VV

1

21

1

22

1

2

i

o

++

++

=


LCRR

51

2= CR

1L

R6

1

2 += LCR

RR25

1

21 +=

Since and , Ω= 4R1 Ω= 1R 2

201

LCLC41

5 =→= (1)

C41

L1

6 += (2)

201

LCLC45

25 =→=


01C24C80C41

C206 2 =+−→+=

Thus, 201

,41

C =

When 41

C = , 51

C201

L == .

When 201

C = , 1C20

1L == .

Therefore, there are two possible solutions. =C F25.0 =L H2.0 or =C F05.0 =L H1


We apply KCL at the noninverting terminal at the op amp. )YY)(V0(Y)0V( 21o3s −−=−

o21s3 V)YY(-VY +=

21

3

s

o

YYY-

VV

+=

Let , 11 sCY = 12 R1Y = , 23 sCY =

11

12

11

2

s

o

CR1sCsC-

R1sCsC-

VV

+=

+=


1CC

1

2 = , 10CR

1

11=

If , Ω= k1R1

=== 421 101

CC F100 µ

Chapter 16, Solution 59. Consider the circuit shown below. We notice that o3 VV = and o32 VVV == .

Y4

Y3

Y1

V1

V2Y2

+ −

−+ Vo

Vin

At node 1, 4o12o111in Y)VV(Y)VV(Y)VV( −+−=−

)YY(V)YYY(VYV 42o42111in +−++= (1) At node 2,

3o2o1 Y)0V(Y)VV( −=−

o3221 V)YY(YV +=

o2

321 V

YYY

V+

= (2)


)YY(VV)YYY(Y

YYYV 42oo421

2

321in +−++⋅

+=

)YYYYYYYYYYYYYY(VYYV 42

2243323142

2221o21in −−+++++=

43323121

21

in

o

YYYYYYYYYY

VV

+++=

1Y and must be resistive, while and must be capacitive. 2Y 3Y 4Y

Let 1

1 R1

Y = , 2

2 R1

Y = , 13 sCY = , 24 sCY =

212

2

1

1

1

21

21

in

o

CCsRsC

RsC

RR1

RR1

VV

+++=

2121221

212

2121

in

o

CCRR1

CRRRR

ss

CCRR1

VV

+

+⋅+

=

Choose R , then Ω= k11

6

212110

CCRR1

= and 100CRRR

221

21 =R

+

We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is

=2R Ωk1 , C =1 nF50 , =2C F20 µ

Chapter 16, Solution 60. With the following MATLAB codes, the Bode plots are generated as shown below. num=[1 1]; den= [1 5 6]; bode(num,den);

Chapter 16, Solution 61. We use the following codes to obtain the Bode plots below. num=[1 4]; den= [1 6 11 6]; bode(num,den);

Chapter 16, Solution 62. The following codes are used to obtain the Bode plots below. num=[1 1]; den= [1 0.5 1]; bode(num,den);

Chapter 16, Solution 63. We use the following commands to obtain the unit step as shown below. num=[1 2]; den= [1 4 3]; step(num,den);

Chapter 16, Solution 64. With the following commands, we obtain the response as shown below. t=0:0.01:5; x=10*exp(-t); num=4; den= [1 5 6]; y=lsim(num,den,x,t); plot(t,y)

Chapter 16, Solution 65. We obtain the response below using the following commands. t=0:0.01:5; x=1 + 3*exp(-2*t); num=[1 0]; den= [1 6 11 6]; y=lsim(num,den,x,t); plot(t,y)

Chapter 16, Solution 66. We obtain the response below using the following MATLAB commands. t=0:0.01:5; x=5*exp(-3*t); num=1; den= [1 1 4]; y=lsim(num,den,x,t); plot(t,y)


Using the result of Practice Problem 16.14,

)YYY(YYYYY-

VV

321432

21

i

o

+++=

When , 11 sCY = F5.0C1 µ=

12 R

1Y = , Ω= k10R1

23 YY = , 24 sCY = , F1C2 µ=

)RsC2(RsC1RsC-

)R2sC(sCR1RsC-

VV

1112

11

11221

11

i

o

++=

++=

1RC2sRCCsRsC-

VV

122121

211

i

o

+⋅+=

1)10)(1010(1)2(s)10)(1010)(110(0.5s)10)(1010(0.5s-

VV

36-236-6-2

3-6

i

o

+××+×××××

=

42i

o

102s400ss100-

VV

×++=

Therefore,

=a 100- , =b 400 , =c 4102× Chapter 16, Solution 68.

(a) Let 3s

)1s(K)s(Y

++

=

Ks31

)s11(Klim

3s)1s(K

lim)(Yss

=++

=++

=∞∞→∞→

i.e. . K25.0 =

Hence, Y =)s()3s(4

1s++

(b) Consider the circuit shown below.

I

Vs = 8 V + −

t = 0

YS

s8V)t(u8V ss =→=

)3s(s)1s(2

3s1s

s48

)s(V)s(YZV

I ss

++

=++

⋅===

3sB

sA

I+

+=

32A = , 34-B =

=)t(i [ ] A)t(ue4231 3t-−


The gyrator is equivalent to two cascaded inverting amplifiers. Let be the voltage at the output of the first op amp.

1V

ii1 -VVRR-

V ==

i1o VsCR

1V

RsC1-

V ==

CsRV

RV

I 2oo

o ==

CsRIV 2

o

o =

CRLwhen,sLIV 2

o

o ==


(a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic. (c) Since sin A cos B = 0.5[sin(A + B) + sin(A – B)],

g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t which is harmonic or periodic with the fundamental frequency

ω = 1 or T = 2π/ω = 2π. (d) h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and a constant is also periodic, h(t) is periodic. ω = 2 or T = 2π/ω = π. (e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.

ω = 0.2π or T = 2π/ω = 10. (f) p(t) = 10 is not periodic. (g) g(t) is not periodic.


(a) The frequency ratio is 6/5 = 1.2. The highest common factor is 1. ω = 1 = 2π/T or T = 2π.

(b) ω = 2 or T = 2π/ω = π. (c) f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t)

ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600. (d) f4(t) = ej10t = cos 10t + jsin 10t. ω = 10 or T = 2π/ω = 0.2π.


T = 4, ωo = 2π/T = π/2 g(t) = 5, 0 < t < 1 10, 1 < t < 2 0, 2 < t < 4

ao = (1/T) = 0.25[ + ] = ∫T

0dt)t(g ∫

1

0dt5 ∫

2

1dt10 3.75

an = (2/T) = (2/4)[ ∫ ωT

0 o dt)tncos()t(g ∫π1

0dt)t

2ncos(5 + ∫

π2

1dt)t

2ncos(10 ]

= 0.5[1

0

t2

nsinn25 ππ

+ 2

1

t2

nsinn2 ππ

10 ] = (–1/(nπ))5 sin(nπ/2)

an = (5/(nπ))(–1)(n+1)/2, n = odd 0, n = even

bn = (2/T) = (2/4)[ ∫ ωT

0 o dt)tnsin()t(g ∫π1

0dt)t

2nsin(5 + ∫

π2

1dt)t

2nsin(10 ]

= 0.5[1

0

t2

ncosn

5x2 ππ

− – 2

1

t2

ncosn

10x2 ππ

] = (5/(nπ))[3 – 2 cos nπ + cos(nπ/2)]


f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π

ao = (1/T) = (1/2) = ∫T

0dt)t(f ∫ −

2

0dt)t510(

2

0

2 )]2/t5(t10[5.0 − = 5

an = (2/T) = (2/2) ∫ ωT

0 o dt)tncos()t(f ∫ π−2

0dt)tncos()t510(

= – ∫ π2

0dt)tncos()10( ∫ π

2

0dt)tncos()t5(

= 2

022 tncos

n5

ππ

− + 2

0

tnsinn

t5π

π = [–5/(n2π2)](cos 2nπ – 1) = 0

bn = (2/2) ∫ π−2

0dt)tnsin()t510(

= – ∫ π2

0dt)tnsin()10( ∫ π

2

0dt)tnsin()t5(

= 2

022 tnsin

n5

ππ

− + 2

0

tncosn

t5π

π = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ)

Hence f(t) = )tnsin(n110

1nπ

π+ ∑

∞

=

5


1T/2,2T =π=ωπ=

5.0]x2x1[21dt)t(z

T1a

T

0o −=π−π

π== ∫

∫∫∫π

π

ππ

ππ

=π

−π

=π

−π

=ω=2

20

0

T

0on 0ntsin

n2nt..sin

n1ntdtcos21ntdtcos11dtncos)t(z

T2a

∫∫∫π

π

ππ

ππ

=

=π=

π+

π−=

π−

π=ω=

22

00

T

0on

evenn 0,

oddn,n6

ntcosn2ntcos

n1ntdtsin21ntdtsin11dtncos)t(z

T2b

Thus,

ntsinn65.0)t(z

oddn1n∑∞

== π

+−=


.0a,functionoddanisthisSince

326)1x21x4(

21dt)t(y

21a

22,2T

n

20o

o

=

==+==

π=π

=ω=

∫

∑

∫∫∫

∞

==

=

=π

ππ

+=

=π−π

=π−π

−π−π

=

π−ππ

−−ππ

−=π

π−π

π−

=

π+π=ω=

oddn1n

evenn,0

oddn,n4

21

10

21

10

20 on

)tnsin(n143)t(y

))ncos(1(n2))ncos(1(

n2))ncos(1(

n4

))ncos()n2(cos(n2)1)n(cos(

n4)tncos(

n2)tncos(

n4

dt)tnsin(2dt)tnsin(4dt)tnsin()t(y22b


0a,6

T/2,12T 0 =π

=π=ω=

∫∫∫ π−+π=ω=−

10

4

4

2

T

0on ]dt6/tncos)10(dt6/tncos10[

61dtncos)t(f

T2a

[ ]3/n5sin3/nsin3/n2sin2n106/tnsin

n106/tnsin

n10 10

44

2 π−π+ππ

=ππ

−ππ

= −

∫∫∫ π−+π=ω=−

10

4

4

2

T

0on ]dt6/tnsin)10(dt6/tnsin10[

61dtnsin)t(f

T2b

[ ]3/n2sin23/ncos3/n5cosn106/ntncos

n106/tncos

n10 10

44

2 π−π+ππ

=ππ

+ππ

−= −

( )∑∞

=π+π=

1nnn 6/tnsinb6/tncosa)t(f

where an and bn are defined above.


π=π=ω=<<+= T/22,T1, t 1- ),t1(2)t(f o

2ttdt)1t(221dt)t(f

T1a

1

1

1

1

2T

0o =+=+==

−−∫∫

0tnsinn1tnsin

nttncos

n12tdtncos)1t(2

22dtncos)t(f

T2a

1

122

1

1

T

0on =

π

π+π

π+π

π=π+=ω=

−−∫∫

ππ

−=

π

π−π

π−π

π−=π+=ω=

−−∫∫ ncos

n4tncos

n1tncos

nttnsin

n12tdtnsin)1t(2

22dtnsin)t(f

T2b

1

122

1

1

T

0on

∑∞

=π

−π

−=1n

ntncos

n)1(42)t(f

Chapter 17, Solution 9. f(t) is an even function, bn=0.

4//2,8 ππω === TT

183.3104/sin)4(4

1004/cos1082)(1 2

0

2

00

===

+== ∫∫ π

ππ

π tdttdttfT

aT

o

[ ]dtntntdttntdtntfT

aT

on ∫∫∫ −++=+==2

0

2

0

2/

0

4/)1(cos4/)1(cos5]04/cos4/cos10[840cos)(4 ππππω

For n = 1,

102/sin25]12/[cos52

0

2

01 =

+=+= ∫ tdttdtta ππ

π

For n>1,

2)1(sin

)1(20

2)1(sin

)1(20

4)1(sin

)1(20

4)1(sin

)1(20 2

0

−−

++

+=

−−

++

+=

nn

nn

nn

tnn

anπ

ππ

ππ

ππ

π

0sin102sin420,3662.62/sin20sin10

32 =+==+= ππ

ππ

ππ

ππ

aa

Thus,

3213210 0,0,362.6,10,183.3 bbbaaaa ======= Chapter 17, Solution 10.

π=π=ω= T/2,2T o

π−−

π−=

−+==

π−π−π−π−ω−∫ ∫ ∫ 2

1

tjn10

tjnT

0

10

21

tjntjntojnn jn

e2jn

e421dte)2(dte4

21dte)t(h

T1c

[ ] ,even n ,0

oddn ,n

j6]6ncos6[

n2je2e24e4

n2jc jnn2jnj

n

=

=π

−=−ππ

=+−−π

= π−π−π−

Thus,

tjn

oddnn

en

6j)t(f π∞

=−∞=∑

π−

=


2/T/2,4T o π=π=ω=

∫ ∫ ∫

++==−

π−π−ω−T

0

01

10

2/tjn2/tjntojnn dte)1(dte)1t(

41dte)t(y

T1c

π−

π−−π−

π−= π−

−π−

π− 10

2/tjn01

2/tjn22

2/tjnn e

jn2e

jn2)12/tjn(

4/ne

41c

π

+π

−π

+−ππ

+π

−π

π−ππjn2e

jn2e

jn2)12/jn(e

n4

jn2

n4

41 2/jn2/jn2/jn

2222=

But

2/nsinj2/nsinj2/ncose,2/nsinj2/nsinj2/ncose 2/jn2/jn π−=π−π=π=π+π= π−π

[ ]2/nsinn2/nsin)12/jn(j1n

1c 22n ππ+π−π+π

=

[ ] 2/tjn22

ne2/nsinn2/nsin)12/jn(j1

n1)t(y π

∞

−∞=ππ+π−π+

π= ∑


A voltage source has a periodic waveform defined over its period as v(t) = t(2π - t) V, for all 0 < t < 2π Find the Fourier series for this voltage. v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1 ao =

(1/T) dt)tt2(21dt)t(f

2

0

2T

0 ∫∫π

−ππ

=3

2)3/21(24)3/tt(

21 23

2

032 π

=−ππ

=−ππ

= π

an = π

π

+π

π=−π∫

2

02

T

0

2 )ntsin(n

t2)ntcos(n21dt)ntcos()tt2(

T2

[ ] π+−

π−

2

0

223 )ntsin(tn)ntsin(2)ntcos(nt2

n1

232 n4)n2cos(n4

n1)11(

n2 −

=πππ

−−=

bn = ∫∫ −π

=− dt)ntsin()tnt2(1dt)ntsin()tnt2(T2 2T

0

2

ππ −+

π−−

π=

2

0

22302 ))ntcos(tn)ntcos(2)ntsin(nt2(

n1))ntcos(nt)nt(sin(

n1n2

0n

4n4

=π

+π−

=

Hence, f(t) = ∑∞

=

−π

1n2

2

)ntcos(n4

32

Chapter 17, Solution 13. T = 2π, ωo = 1

ao = (1/T) dttsin10[21dt)t(h

0

T

0 ∫∫π

π= + ]dt)tsin(20

2

∫π

ππ−

[ ]π

=π−−−π

= π

π

π 30)tcos(20tcos1021 2

0

an = (2/T) ∫ ωT

0 o dt)tncos()t(h

= [2/(2π)]

π−+∫ ∫

π π

π0

2dt)ntcos()tsin(20dt)ntcos(tsin10

Since sin A cos B = 0.5[sin(A + B) + sin(A – B)] sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t] sin(t – π) = sin t cos π – cost sin π = –sin t sin(t – π)cos(nt) = –sin(t)cos(nt)

an =

−++−−++

π ∫ ∫π π

π0

2dt)]t]n1sin([)t]n1[sin([20dt)]t]n1sin([)t]n1[sin([10

21

=

−−

+++

+

−−

−++

−π

π

π

π 2

0 n1)t]n1cos([2

n1)t]n1cos([2

n1)t]n1cos([

n1)t]n1cos([5

an =

−π−

−+

π+−

−+

+π n1)]n1cos([3

n1)]n1cos([3

n13

n135

But, [1/(1+n)] + [1/(1-n)] = 1/(1–n2)

cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ

an = (5/π)[(6/(1–n2)) + (6 cos(nπ)/(1–n2))]

= [30/(π(1–n2))](1 + cos nπ) = [–60/(π(n–1))], n = even = 0, n = odd

bn = (2/T) ∫ ωT

0 o dttnsin)t(h

= [2/(2π)][ + ∫π

0dtntsintsin10 ∫

π

π−

2dtntsin)tsin(20

But, sin A sin B = 0.5[cos(A–B) – cos(A+B)] sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)] bn = (5/π)[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ π+

0)]n1(

+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ π

π+ 2)]n1(

=

+π+

+−

π−−

π n1)]n1sin([

n1)]n1sin([5 = 0

Thus, h(t) = ∑∞

= −π−

π 1k2 )1k4(

)kt2cos(6030

Chapter 17, Solution 14. Since cos(A + B) = cos A cos B – sin A sin B.

f(t) = ∑∞

=

π

+−π

++

1n33 )nt2sin()4/nsin(

1n10)nt2cos()4/ncos(

1n102


(a) Dcos ωt + Esin ωt = A cos(ωt - θ) where A = 22 ED + , θ = tan-1(E/D)

A = 622 n1

)1n(+

+16 , θ = tan-1((n2+1)/(4n3))

f(t) = ∑+10∞

=

−

+−+

+1n3

21

622 n41ntannt10cos

n1

)1n(16

(b) Dcos ωt + Esin ωt = A sin(ωt + θ)

where A = 22 ED + , θ = tan-1(D/E)

f(t) = ∑+10∞

=

−

+

+++1n

2

31

622 1nn4tannt10sin

n1

)1n(16


If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e.

v2*(t) = v2(t) + 1.

1

v2*(t)

2

t

-2 -1 0 1 2 3 4 5

Comparing v2*(t) with v1(t) shows that

v2

*(t) = 2v1((t + to)/2) where (t + to)/2 = 0 at t = -1 or to = 1 Hence v2

*(t) = 2v1((t + 1)/2) But v2

*(t) = v2(t) + 1 v2(t) + 1 = 2v1((t+1)/2) v2(t) = -1 + 2v1((t+1)/2)

= -1 + 1

+

+π+

+

π+

+

ππ

−2

1t5cos251

21t3cos

91

21tcos8

2

v2(t) =

+

π

+π

+

π

+π

+

π

+π

π−

25

2t5cos

251

23

2t3cos

91

22tcos8

2

v2(t) =

+

π

+

π

+

π

π 2t5sin

251

2t3sin

91

2tsin8

2−

Chapter 17, Solution 17. We replace t by –t in each case and see if the function remains unchanged.

(a) 1 – t, neither odd nor even.

(b) t2 – 1, even

(c) cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt, odd

(d) sin2 n(-t) = (-sin πt)2 = sin2 πt, even

(e) et, neither odd nor even.


(a) T = 2 leads to ωo = 2π/T = π

f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric.

(b) T = 3 leads to ωo = 2π/3 f2(t) = f2(-t), showing that f2(t) is even.

(c) T = 4 leads to ωo = π/2

f3(t) is even and half-wave symmetric. Chapter 17, Solution 19. This is a half-wave even symmetric function.

ao = 0 = bn, ωo = 2π/T π/2

an = ∫ ω

−

2/T

0 o dt)tncos(Tt41

T4

= [4/(nπ)2](1 − cos nπ) = 8/(n2π2), n = odd

= 0, n = even

f (t) = ∑∞

=

π

π oddn22 2

tncosn18

Chapter 17, Solution 20. This is an even function.

bn = 0, T = 6, ω = 2π/6 = π/3

ao =

−= ∫∫∫

3

2

2

1

2/T

0dt4dt)4t4(

62dt)t(f

T2

=

−+− )23(4)t4t2(

31 2

1

2 = 2

an = ∫ π4/T

0dt)3/tncos()t(f

T4

= (4/6)[ + ] ∫ π−2

1dt)3/tncos()4t4( ∫ π

3

2dt)3/tncos(4

= 3

2

2

122 3

tnsinn3

616

3tnsin

n3

3tnsin

nt3

3tncos

n9

616

π

π+

π

π−

π

π+

π

π

= [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)]

Thus f(t) = ∑∞

=

π

π−

π

π+

1n22 3

tncos3ncos

3n2cos

n1242

At t = 2,

f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3)

+ (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3) + (1/9)(cos(2π) − cos(π))cos(2π) + -----]

= 2 + 2.432(0.5 + 0 + 0.2222 + -----)

f(2) = 3.756 Chapter 17, Solution 21. This is an even function.

bn = 0, T = 4, ωo = 2π/T = π/2. f(t) = 2 − 2t, 0 < t < 1 = 0, 1 < t < 2

ao = ∫

−=−

1

0

1

0

2

2ttdt)t1(2

42 = 0.5

an = ∫∫

π

−=ω1

0

2/T

0 o dt2

tncos)t1(244dt)tncos()t(f

T4

= [8/(π2n2)][1 − cos(nπ/2)]

f(t) = ∑∞

=

π

π

−π

+1n

22 2tncos

2ncos1

n8

21


Calculate the Fourier coefficients for the function in Fig. 16.54. f(t)

4

t

-5 -4 -3 -2 -1 0 1 2 3 4 5

Figure 17.61 For Prob. 17.22

This is an even function, therefore bn = 0. In addition, T=4 and ωo = π/2.

ao = === ∫∫1

0

21

0

2T

0ttdt4

42dt)t(f

T2 1

an = ∫∫ π=ω1

0

2T

0 o dt)2/tncos(t444dt)ntcos()t(f

T4

1

022 )2/tnsin(

nt2)2/tncos(

n44

π

π+π

π=

an = )2/nsin(n8)1)2/n(cos(

n16

22 ππ

+−ππ

Chapter 17, Solution 23. f(t) is an odd function.

f(t) = t, −1< t < 1 ao = 0 = an, T = 2, ωo = 2π/T = π

bn = ∫∫ π=ω1

0

2/T

0 o dt)tnsin(t24dt)tnsin()t(f

T4

= [ ]1022 )tncos(tn)tnsin(

n2

ππ−ππ

= −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ)

f(t) = ∑∞

=

+

π−

π 1n

1n

)tnsin(n)1(2


(a) This is an odd function.

ao = 0 = an, T = 2π, ωo = 2π/T = 1

bn = ∫ ω2/T

0 o dt)ntsin()t(fT4

f(t) = 1 + t/π, 0 < t < π

bn = ∫π

π+π 0

dt)ntsin()/t1(24

= π

π−

π+−

π 02 )ntcos(

nt)ntsin(

n1)ntcos(

n12

= [2/(nπ)][1 − 2cos(nπ)] = [2/(nπ)][1 + 2(−1)n+1]

a2 = 0, b2 = [2/(2π)][1 + 2(−1)] = −1/π = −0.3183

(b) ωn = nωo = 10 or n = 10

a10 = 0, b10 = [2/(10π)][1 − cos(10π)] = −1/(5π)

Thus the magnitude is A10 = 2

102 ba10+ = 1/(5π) = 0.06366

and the phase is φ10 = tan−1(bn/an) = −90°

(c) f(t) = ∑ π ∞

=

π−π1n

)ntsin()]ncos(21[n2

f(π/2) = ∑∞

=

ππ−π1n

)2/nsin()]ncos(21[n2

π

For n = 1, f1 = (2/π)(1 + 2) = 6/π For n = 2, f2 = 0 For n = 3, f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π) For n = 4, f4 = 0 For n = 5, f5 = 6/(5π), ----

Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) ---------

= (6/π)[1 − 1/3 + 1/5 − 1/7 + --------] f(π/2) ≅ 1.3824 which is within 8% of the exact value of 1.5.

(d) From part (c) f(π/2) = 1.5 = (6/π)[1 − 1/3 + 1/5 − 1/7 + - - -] (3/2)(π/6) = [1 − 1/3 + 1/5 − 1/7 + - - -] or π/4 = 1 − 1/3 + 1/5 − 1/7 + - - - Chapter 17, Solution 25. This is an odd function since f(−t) = −f(t).

ao = 0 = an, T = 3, ωo = 2π/3.

b n = ∫∫ π=ω1

0

2/T

0 o dt)3/nt2sin(t34dt)tnsin()t(f

T4

= 1

022 3

nt2cosn2t3

3nt2sin

n49

34

π

π−

π

π

=

π

π−

π

π 3n2cos

n23

3n2sin

n49

34

22

f(t) = ∑∞

=

π

π

π−

π

π1n22 3

t2sin3n2cos

n2

3n2sin

n3

Chapter 17, Solution 26. T = 4, ωo = 2π/T = π/2

ao =

++= ∫∫∫∫

4

3

3

1

1

0

T

0dt1dt2dt1

41dt)t(f

T1 = 1

an = dt)tncos()t(fT

T

0 o∫ ω2

an =

π+π+π ∫∫∫

4

3

3

2

2

1dt)2/tncos(1dt)2/tncos(2dt)2/tncos(1

42

=

ππ

+π

π+

ππ

4

3

3

2

2

1 2tnsin

n2

2tnsin

n4

2tnsin

n22

=

π

−π

π 2nsin

2n3sin

n4

bn = ∫ ωT

0 o dt)tnsin()t(fT2

=

π

+π

+π

∫∫∫4

3

3

2

2

1dt

2tnsin1dt

2tnsin2dt

2tnsin1

42

=

ππ

−π

π−

ππ

−4

3

3

2

2

1 2tncos

n2

2tncos

n4

2tncos

n22

= [ ]1)ncos(n4

−ππ

Hence f(t) =

[ ]∑∞

=

π−π+ππ−ππ

+1n

)2/tnsin()1)n(cos()2/tncos())2/nsin()2/n3(sin(n41

Chapter 17, Solution 27. (a) odd symmetry. (b) ao = 0 = an, T = 4, ωo = 2π/T = π/2 f(t) = t, 0 < t < 1 = 0, 1 < t < 2

bn = 1

022

1

0 2tncos

nt2

2tnsin

n4dt

2tnsint

4

π

π−

ππ

=π

∫4

= 02

ncosn2

2nsin

n4

22 −π

π−

ππ

= 4(−1)(n−1)/2/(n2π2), n = odd

−2(−1)n/2/(nπ), n = even

a3 = 0, b3 = 4(−1)/(9π2) = 0.045 (c) b1 = 4/π2, b2 = 1/π, b3 = −4/(9π2), b4 = −1/(2π), b5 = (25π2)

Frms = ( )∑ ++ 2n

2n

2o ba

21a

Frms

2 = 0.5Σbn2 = [1/(2π2)][(16/π2) + 1 + (16/(8π2)) + (1/4) + (16/(625π2))]

= (1/19.729)(2.6211 + 0.27 + 0.00259) Frms = 14659.0 = 0.3829

Compare this with the exact value of Frms = 6/1dttT2 1

0

2 =∫ = 0.4082

Chapter 17, Solution 28. This is half-wave symmetric since f(t − T/2) = −f(t).

ao = 0, T = 2, ωo = 2π/2 = π

an = ∫∫ π−=ω1

0

2/T

0 o dt)tncos()t22(24dt)tncos()t(f

T4

= 1

022 )tnsin(

nt)tncos(

n1)tnsin(

n14

π

π−π

π−π

π

= [4/(n2π2)][1 − cos(nπ)] = 8/(n2π2), n = odd

0, n = even

bn = 4 ∫ π−1

0dt)tnsin()t1(

= 1

022 )tncos(

nt)tnsin(

n1)tncos(

n14

π

π+π

π−π

π−

= 4/(nπ), n = odd

f(t) = ∑∞

=

π

π+π

π1k22 )tnsin(

n4)tncos(

n8 , n = 2k − 1

Chapter 17, Solution 29. This function is half-wave symmetric.

T = 2π, ωo = 2π/T = 1, f(t) = −t, 0 < t < π

For odd n, an = ∫π−

0dt)ntcos()t(

T2 = [ ]π+

π−

02 )ntsin(nt)ntcos(n2 = 4/(n2π)

bn = [ ]ππ−

π−=−

π ∫ 020)ntcos(nt)ntsin(

n2dt)ntsin()t(2 = −2/n

Thus,

f(t) = ∑∞

=

−

π1k2 )ntsin(

n1)ntcos(

n22 , n = 2k − 1


∫ ∫ ∫−

− −ω−

ω−ω==

2/T

2/T

2/T2/T

2/T2/T oo

tojnn tdtnsin)t(fjtdtncos)t(f

T1dte)t(f

T1c (1)

(a) The second term on the right hand side vanishes if f(t) is even. Hence

∫ ω=2/T

0on tdtncos)t(f

T2c

(b) The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,

∫ ω−=2/T

0on tdtnsin)t(f

T2jc


If oo /T2

'T2'/T'T),t(f)t(h αω=

απ

=π

=ω→α=α=

∫∫ ωα=ω='T

0o

'T

0on tdt'ncos)t(f

'T2tdt'ncos)t(h

'T2'a

Let T'T,/dtd,,t =ααλ=λ=α

n

T

0on a/dncos)(f

T2'a =αλλωλα

= ∫

Similarly, nn b'b =

Chapter 17, Solution 32. When is = 1 (DC component)

i = 1/(1 + 2) = 1/3

For n 1, ω≥ n = 3n, Is = 1/n2∠0°

I = [1/(1 + 2 + jωn2)]Is = Is/(3 + j6n)

= )n2tan(n41n3

1)3/n6(tann413

0n1

2212

2−∠

+=

∠+

°∠

−

Thus,

i(t) = ∑∞

=

−−+

+1n

1

22))n2(tann3cos(

n41n3

131


For the DC case, the inductor acts like a short, Vo = 0. For the AC case, we obtain the following:

so

ooso

VVn5n5.2j1

04Vjn

n2jV

10VV

=

π−π+

=π

+π

+−

)5n5.2(jn4

n5n5.2j1

1n4A

n5n5.2j1

VV

22nn

so

−π+π=

π−π+π

=Θ∠

π−π+

=

π−π

−=Θ−π+π

= −n

5n5.2tan;)5n5.2(n

4A22

1n22222n

V)tnsin(A)t(v1n

nno ∑∞

=Θ+π=

Chapter 17, Solution 34. For any n, V = [10/n2]∠(nπ/4), ω = n. 1 H becomes jωnL = jn and 0.5 F becomes 1/(jωnC) = −j2/n

2 Ω jn

−j2/n

+ −

+

Vo

−

V Vo = −j(2/n)/[2 + jn − j(2/n)]V = −j2/[2n + j(n2 − 2)][(10/n2)∠(nπ/4)]

)]n2/)2n((tan)2/()4/n[(

4nn

20)n2/)2n((tan)2n(n4n

)2/)4/n((20

21

22

212222

−−π−π∠+

=

−∠−+

π−π∠=

−

−

vo(t) = ∑∞

=

−

−−

π−

π+

+1n

21

22 n22ntan

24nntcos

4nn20


If vs in the circuit of Fig. 17.72 is the same as function f2(t) in Fig. 17.57(b), determine the dc component and the first three nonzero harmonics of vo(t).

1 Ω 1 H

1 F 1 Ω

+ −

+

vo

−

vS

Figure 17.72 For Prob. 17.35

1

f2(t)

2

t

-2 -1 0 1 2 3 4 5

Figure 17.57(b) For Prob. 17.35

The signal is even, hence, bn = 0. In addition, T = 3, ωo = 2π/3. vs(t) = 1 for all 0 < t < 1 = 2 for all 1 < t < 1.5

ao = 34dt2dt1

32 5.1

1

1

0=

+ ∫∫

an =

π+π ∫∫

5.1

1

1

0dt)3/tn2cos(2dt)3/tn2cos(

34

= )3/n2sin(n2)3/tn2sin(

n26)3/tn2sin(

n23

34 5.1

1

1

0π

π−=

π

π+π

π

vs(t) = ∑∞

=

πππ

−1n

)3/tn2cos()3/n2sin(n12

34

Now consider this circuit,

j2nπ/3 1 Ω

-j3/(2nπ) 1 Ω

+ −

+

vo

−

vS

Let Z = [-j3/(2nπ)](1)/(1 – j3/(2nπ)) = -j3/(2nπ - j3) Therefore, vo = Zvs/(Z + 1 + j2nπ/3). Simplifying, we get

vo = )18n4(jn12

v9j22

s

−π+π−

For the dc case, n = 0 and vs = ¾ V and vo = vs/2 = 3/8 V. We can now solve for vo(t)

vo(t) = volts3

tn2cosA83

1nnn

Θ+

π+∑

∞

=

where

π−

π−=Θ

−

π+π

ππ= −

n23

3ntan90and

63

n4n16

)3/n2sin(n6

A 1on222

22

n

where we can further simplify An to this, 81n4n)3/n2sin(9

44n+ππ

A π=


vs(t) = ∑∞

==

θ−

oddn1n

nn )ntcos(A

where θn = tan−1[(3/(nπ))/(−1/(nπ))] = tan−1(−3) = 100.5°

An = 2

nsin9n1

2nsin

n1

n9 22

2222π

+π

=π

π+

π

ωn = n and 2 H becomes jωnL = j2n

Let Z = 1||j2n = j2n/(1 + j2n) If Vo is the voltage at the non-reference node or across the 2-H inductor. Vo = ZVs/(1 + Z) = [j2n/(1 + j2n)]Vs/1 + [j2n/(1 + j2n)] = j2nVs/(1 + j4n) But Vs = An∠−θn Vo = j2n An∠−θn/(1 + j4n)

Io = Vo/j = [2n An∠−θn]/ 2n161 + ∠tan−14n

= 2

2

n161

n22

nsin9n1

+

π+

π∠−100.5° − tan−14n

Since sin(nπ/2) = (−1)(n−1)/2 for n = odd, sin2(nπ/2) = 1

Io = 2

1

n161

n4tan5.100102

+

−°−∠π

−

io(t) = )n4tan5.100ntcos(n161

1102

oddn1n

1

2∑∞

==

−−°−+π

Chapter 17, Solution 37. From Example 15.1,

vs(t) = ∑∞

=

ππ

+1k

),tnsin(n1205 n = 2k − 1

For the DC component, the capacitor acts like an open circuit.

Vo = 5

For the nth harmonic, Vs = [20/(nπ)]∠0°

10 mF becomes 1/(jωnC) = −j/(nπx10x10−3) = −j100/(nπ)

vo = 22

1s

n25nn5tan90100

n20

100jn20100j

20n

100j

Vn

100j

π+ππ

+°−∠=

π−π−

=+

π−

π− −

vo(t) = ∑ π+°−π

π+π− )

n5tan90tnsin(

n25n

1100 1

22


∑∞

=+=π

π+=

1ks 1k2n,tnsin

n12

21)t(v

π=ωω+

ω= n,V

j1jV ns

n

no

For dc, 0V,5.0V,0 osn ===ω

For nth harmonic, os 90

n2V −∠π

=

22

1o

122

oo

n1

ntan290n2

ntann1

90nVπ+

π−∠=∠

π•

π∠π+

∠π=

−

−

1k2n),ntantncos(n1

2)t(v 1

1k 22o −=π−ππ+

= −∞

=∑


Comparing vs(t) with f(t) in Figure 15.1, vs is shifted by 2.5 and the magnitude is 5 times that of f(t). Hence

vs(t) = ∑∞

=

ππ

+1k

),tnsin(n1105 n = 2k − 1

T = 2, ωo = 2π//T = π, ωn = nωo = nπ

For the DC component, io = 5/(20 + 40) = 1/12 For the kth harmonic, Vs = (10/(nπ))∠0°

100 mH becomes jωnL = jnπx0.1 = j0.1nπ

50 mF becomes 1/(jωnC) = −j20/(nπ) 40 Ω

−j20/(nπ)

+ −

I Io

Z

20 Ω

VS j0.1nπ

Let Z = −j20/(nπ)||(40 + j0.1nπ) = π++

π−

π+π

−

n1.0j40n20j

)n1.0j40(n20j

= )20n1.0(jn40

800jn2n1.0jn4020j

n1.0j40(20j2222 −π+π

−π=

π+π+−π+−

Zin = 20 + Z = )20n1.0(jn40)1200n2(jn802

22

22

−π+π−π+π

I = )]1200n2(jn802[n

)200n(jn400ZV

22

22

in

s

−π+ππ−π+π

=

Io = )20n1.0(jn40

I20j

)n1.0j40(n20j

In20j

22 −π+π−

=π++

π−

π−

= )]1200n2(jn802[n

200j22 −π+ππ

−

= 2222

221

)1200n2()802(n)n802/()1200n2(tan90200

−π+π

π−π−°−∠ −

Thus

io(t) = ∑∞

=

θ−ππ

+1k

nn )tnsin(I200201 , n = 2k − 1

where π

−π+°=θ −

n8021200n2tan90

221

n

In = )1200n2()n804(n

1222 −π+π

Chapter 17, Solution 40. T = 2, ωo = 2π/T = π

ao = 2/12ttdt)t22(

21dt)t(v

T

1

0

21

0

T

0=

−=−= ∫∫

1

an = ∫∫ π−=π1

0

T

0dt)tncos()t1(2dt)tncos()t(v

T2

= 1

022 )tnsin(

nt)tncos(

n1)tnsin(

n12

π

π−π

π−π

π

= 2222

22

)1n2(4oddn,

n4

evenn,0)ncos1(

n2

−π==

π

==π−

π

bn = ∫∫ π−=π1

0

T

0dt)tnsin()t1(2dt)tnsin()t(v

T2

= π

=

π

π+π

π−π

π−

n2)tncos(

nt)tnsin(

n1)tncos(

n12

1

022

vs(t) = ∑ ϕ−π+ )tncos(A21

nn

where 4422n

21

n )1n2(16

n4A,

n2)1n2(

−π+

π=

−π−tan=φ

For the DC component, vs = 1/2. As shown in Figure (a), the capacitor acts like an open circuit.

+

Vx

−

i

Vo2Vx

+

Vo

−

Vx

− +

+ − 3 Ω

1 Ω 0.5V

(a)

Vo

(1/4)F

2Vx

+

Vo

−

Vx

− +

+ − 3 Ω

1 Ω VS (b)

Applying KVL to the circuit in Figure (a) gives –0.5 – 2Vx + 4i = 0 (1) But –0.5 + i + Vx = 0 or –1 + 2Vx + 2i = 0 (2) Adding (1) and (2), –1.5 + 6i = 0 or i = 0.25 Vo = 3i = 0.75 For the nth harmonic, we consider the circuit in Figure (b).

ωn = nπ, Vs = An∠–φ, 1/(jωnC) = –j4/(nπ) At the supernode, (Vs – Vx)/1 = –[nπ/(j4)]Vx + Vo/3 Vs = [1 + jnπ/4]Vx + Vo/3 (3) But –Vx – 2Vx + Vo = 0 or Vo = 3Vx Substituting this into (3), Vs = [1 + jnπ/4]Vx + Vx = [2 + jnπ/4]Vx = (1/3)[2 + jnπ/4]Vo = (1/12)[8 + jnπ]Vo

Vo = 12Vs/(8 + jnπ) = )8/n(tann64

A12122

n

π∠π+

φ−∠−

Vo = ))]n2/()1n2((tan)8/n([tan)1n2(

16n

4n64

12 11442222

−π−π∠−π

+ππ+

−−

Thus

vo(t) = ∑∞

=

θ+π+1n

nn )tncos(V43

where Vn = 442222 )1n2(

16n

4

n64

12−π

+ππ+

θn = tan–1(nπ/8) – tan–1(π(2n – 1)/(2n)) Chapter 17, Solution 41. For the full wave rectifier,

T = π, ωo = 2π/T = 2, ωn = nωo = 2n

Hence

vin(t) = ( )∑∞

= −π−

π 1n2 nt2cos

1n4142

For the DC component,

Vin = 2/π

The inductor acts like a short-circuit, while the capacitor acts like an open circuit.

Vo = Vin = 2/π

For the nth harmonic, Vin = [–4/(π(4n2 – 1))]∠0°

2 H becomes jωnL = j4n

0.1 F becomes 1/(jωnC) = –j5/n

Z = 10||(–j5/n) = –j10/(2n – j)

Vo = [Z/(Z + j4n)]Vin = –j10Vin/(4 + j(8n – 10))

=

−π°∠

−−+

−)1n4(

04)10n8(j4

10j2

= 22

1

)10n8(16)1n4()5.2n2(tan9040

−+−π

−−°∠ −

Hence vo(t) = ∑∞

=

θ++π 1n

nn )nt2cos(A2

where

An = 29n40n16)1n4(

2022 +−−π

θn = 90° – tan–1(2n – 2.5)


∑∞

==π

π+=

1ks 1-2kn ,tnsin

n1205v

π=ω=ωω

=→−ω=−

nn ,VRCjV)V0(Cj

R0V

onsn

oons

For n = 0 (dc component), Vo=0. For the nth harmonic,

22

5

9422o

oo

n210

10x40x10xn2090

n20

RCn901V

π=

π=−∠

ππ∠

=−

Hence,

∑∞

==π

π=

1k22

5o 1-2kn ,tncos

n1

210)t(v

Alternatively, we notice that this is an integrator so that

∑∫∞

==π

π=−=

1k22

5so 1-2kn ,tncos

n1

210dtv

RC1)t(v


(a) Vrms = )1020(2130)ba(

21a 222

1n

2n

2n

20 ++=++ ∑

∞

=

= 33.91 V

(b) Irms = )24(216 222 ++ = 6.782 A

(c) P = VdcIdc + ∑ Φ−Θ )cos(IV21

nnnn

= 30x6 + 0.5[20x4cos(45o-10o) – 10x2cos(-45o+60o)]

= 180 + 32.76 – 9.659 = 203.1 W Chapter 17, Solution 44.

(a) [ ] W73.300535.319.27045cos1025cos6021vip oo =++=++==

(b) The power spectrum is shown below.

p 27.19 3.535 0 1 2 3 ω Chapter 17, Solution 45. ωn = 1000n jωnL = j1000nx2x10–3 = j2n 1/(jωnC) = –j/(1000nx40x10–6) = –j25/n

Z = R + jωnL + 1/(jωnC) = 10 + j2n – j25/n

I = V/Z

For n = 1, V1 = 100, Z = 10 + j2 – j25 = 10 – j23 I1 = 100/(10 – j23) = 3.987∠73.89°

For n = 2, V2 = 50, Z = 10 + j4 – j12.5 = 10 – j8.5 I2 = 50/(10 – j8.5) = 3.81∠40.36°

For n = 3, V3 = 25, Z = 10 + j6 – j25/3 = 10 – j2.333 I3 = 25/(10 – j2.333) = 2.435∠13.13° Irms = 0.5 222 435.281.3987.3 ++ = 3.014 A

p = VDCIDC + ∑=

φ−θ3

1nnnnn )cos(IV

21

= 0 + 0.5[100x3.987cos(73.89°) + 50x3.81cos(40.36°) + 25x2.435cos(13.13°)]

= 0.5[110.632 + 145.16 + 59.28] = 157.54 watts Chapter 17, Solution 46. (a) This is an even function

Irms = ∫∫ =2/T

0

2T

0

2 dt)t(fT2dt)t(f

T1

f(t) = 2t1,01t0,t22

<<<<−

T = 4, ωo = 2π/T = π/2

Irms2 =

1

0

321

0

2 )3/ttt(2dt)t1(442

+−=−∫

= 2(1 – 1 + 1/3) = 2/3 or Irms = 0.8165 A (b) From Problem 16.14, an = [8/(n2π2)][1 – cos(nπ/2)], ao = 0.5

a1 = 8/π2, a2 = 4/π2, a3 = 8/(9π2), a4 = 0, a5 = 9/(25π2), a6 = 4/(9π2)

Irms = 66623.08116

62564

81641664

21

41A

21a 4

1n

2no =

++++

π+≅+ ∑

∞

=

Irms = 0.8162 A Chapter 17, Solution 47. Let I = IDC + I1 + I2

For the DC component

IDC = [5/(5 + 10)](3) = 1 A

Is

I

5 Ω

j8 10 Ω

For AC, ω = 100 jωL = j100x80x10–3 = j8

In = 5Is/(5 + 10 + j8)

For Is = 0.5∠–60° I1 = 10∠–60°/(15 + j8) or |I1| = 10/ 22 815 +

For Is = 0.5∠–120°

I2 = 2.5∠–120°/(15 + j8) or |I2| = 2.5/ 22 815 + p10 = (IDC

2 + |I1|2/2 + |I2|2/2)10 = (1 + [100/(2x289)] + [6.25/(2x289)])x10

p10 = 11.838 watts

Chapter 17, Solution 48. (a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that v = Ri(t) = 2x103x20x10–3 = 40

For the AC component, ωn = 10n, n = 1,2

1/(jωnC) = –j/(10nx100x10–6) = (–j/n) kΩ Z = 2||(–j/n) = 2(–j/n)/(2 – j/n) = –j2/(2n – j)

V = ZI = [–j2/(2n – j)]I

For n = 1, V1 = [–j2/(2 – j)]16∠45° = 14.311∠–18.43° mV

For n = 2, V2 = [–j2/(4 – j)]12∠–60° = 5.821∠–135.96° mV

v(t) = 40 + 0.014311cos(10t – 18.43°) + 0.005821cos(20t – 135.96°) V

(b) p = VDCIDC + ∑∞

=

φ−θ1n

nnnn )cos(IV21

= 20x40 + 0.5x10x0.014311cos(45° + 18.43°) +0.5x12x0.005821cos(–60° + 135.96°)

= 800.1 mW Chapter 17, Solution 49.

(a) 5.2)5(21dt4dt1

21dt)t(z

T1Z

0

2T

0

2rms

2 =ππ

=

+

π== ∫ ∫∫

π π

π

581.1Zrms =

(b)

9193.2...251

161

91

411

181

41

n36

21

41)ba(

21aZ 2

1n 1n22n

2n

2o

2rms

2 =

+++++

π+=

π+=++= ∑ ∑

∞

=

∞

= 7086.1rms =Z

(c ) 071.8100x11.581

1.7086 %error =

−=


cn = ∫ π==ωω−T

0 ontj

1n2,dte)t(f

To

1

= ∫−π−1 tjn dtte

21

Using integration by parts,

u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt

cn = ∫−π−

−

π−

π+

π

1

1

tjn1

1

tjn dtejn21e

jn2t

−

= [ ]1

1

tjn222

tjnjn e)j(n2

1eenj

−

π−ππ−

−π++

π

= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)

cn = π−

=ππ

+π−

n)1(j)nsin(

n2j2

n)1( n

22

nj

Thus

f(t) = ∑∞

−∞=

ω

n

tjnn

oec = ∑∞

−∞=

π

π−

n

tjnn enj)1(


π=π=ω= T/2,2T o

( ) 20

T

0

2222

03

tjntjn2tojn

n 2tjn2tn)jn(

e21dtet

21dte)t(f

T1c ∫ ∫ +π+π−

π−===

π−π−ω−

)jn1(n

2)n4jn4(n2j1c 22

2233n π+

π=π+π−

π=

∑∞

−∞=

ππ+π

=n

tjn22

e)jn1(n

2)t(f


cn = ∫ π==ωω−T

0 ontj

1n2,dte)t(f

To

1

= ∫−π−1 tjn dtte

21

Using integration by parts,

u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt

cn = ∫−π−

−

π−

π+

π

1

1

tjn1

1

tjn dtejn21e

jn2t

−

= [ ]1

1

tjn222

tjnjn e)j(n2

1eenj

−

π−ππ−

−π++

π

= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)

cn = π−

=ππ

+π−

n)1(j)nsin(

n2j2

n)1( n

22

nj

Thus

f(t) = ∑∞

−∞=

ω

n

tjnn

oec = ∑∞

−∞=

π

π−

n

tjnn enj)1(

Chapter 17, Solution 53. ωo = 2π/T = 2π

cn = ∫ ∫ ω+−ω−− =1

0

t)jn1(T

0

tjnt dtedtee oo

= [ ]1en2j1

1en2j1

1 )n2j1(1

0

t)n2j1( −π+

−=

π+− π+−π+−

= [1/(j2nπ)][1 – e–1(cos(2πn) – jsin(2nπ))]

= (1 – e–1)/(1 + j2nπ) = 0.6321/(1 + j2nπ

f(t) = ∑∞

−∞=

π

π+n

tn2j

n2j1e6321.0

Chapter 17, Solution 54. T = 4, ωo = 2π/T = π/2

cn = ∫ ω−T

0

ntj dte)t(fT

o1

=

−+ ∫∫∫ π−π−π− 4

2

2/tjn2

1

2/tjn1

0

2/tjn dte1dte1dte241

= [ ]π−π−π−π−π− +−−+−π

jnn2j2/jnjn2/jn eeee2e2n2j

= [ ]π−π− +−π

jn2/jn e23e3n2j

f(t) = ∑∞

−∞=

ω

n

tjnn

oec

Chapter 17, Solution 55. T = 2π, ωo = 2π/T = 1

cn = ∫ ω−T

0

tjn dte)t(iT

o1

But i(t) = π<<π

π<<2t,0

t0),tsin(

cn = ∫∫π −−π π− −

π=

π 0

jntjtjt

0

tjn dte)ee(j2

121dte)tsin(

21

+

−+

−−

−=

+

+−π

=

+π−−π

π+−−

n11e

n11e

41

)n1(je

)n1(je

j41

)1n(j)n1(j

0

)n1(jt)n1(jt

[ ]nne1enne1e)1n(4

1 )n1(j)n1(j)n1(j)n1(j2 +−−+−+−−π

= +π−+π−−π−π

But ejπ = cos(π) + jsin(π) = –1 = e–jπ

cn = [ ])n1(2

e12neneee)1n(4

12

jnjnjnjnjn

2 −π+

=−+−−−−π

π−π−π−π−π−

Thus

i(t) = ∑∞

−∞=

ππ−

−π+

n

tjn2

jn

e)n1(2

e1

Chapter 17, Solution 56. co = ao = 10, ωo = π co = (an – jbn)/2 = (1 – jn)/[2(n2 + 1)]

f(t) = ∑∞

≠−∞=

π

+−

+

0nn

tjn2 e

)1n(2)jn1(10

Chapter 17, Solution 57. ao = (6/–2) = –3 = co cn = 0.5(an –jbn) = an/2 = 3/(n3 – 2)

f(t) = ∑∞

≠−∞= −

+

0nn

nt50j3 e

2n33−

Chapter 17, Solution 58. cn = (an – jbn)/2, (–1) = cos(nπ), ωo = 2π/T = 1

cn = [(cos(nπ) – 1)/(2πn2)] – j cos(nπ)/(2n)

Thus

f(t) = ∑+π

π

−π

−π jnt2 e

n2)ncos(j

n21)ncos(

4

Chapter 17, Solution 59. For f(t), T = 2π, ωo = 2π/T = 1. ao = DC component = (1xπ + 0)/2π = 0.5

For h(t), T = 2, ωo = 2π/T = π. ao = (3x1 – 2x1)/2 = 0.5

Thus by replacing ωo = 1 with ωo = π and multiplying the magnitude by five, we obtain

h(t) = ∑−1 ∞

≠−∞=

π+

π+0n

n

t)1n2(j

)1n2(e5j

2

Chapter 17, Solution 60. From Problem 16.17, ao = 0 = an, bn = [2/(nπ)][1 – 2 cos(nπ)], co = 0 cn = (an – jbn)/2 = [j/(nπ)][2 cos(nπ) – 1], n ≠ 0. Chapter 17, Solution 61.

(a) ωo = 1.

f(t) = ao + ∑ φ−ω )tncos(A non

= 6 + 4cos(t + 50°) + 2cos(2t + 35°)

+ cos(3t + 25°) + 0.5cos(4t + 20°) = 6 + 4cos(t)cos(50°) – 4sin(t)sin(50°) + 2cos(2t)cos(35°)

– 2sin(2t)sin(35°) + cos(3t)cos(25°) – sin(3t)sin(25°) + 0.5cos(4t)cos(20°) – 0.5sin(4t)sin(20°)

= 6 + 2.571cos(t) – 3.73sin(t) + 1.635cos(2t)

– 1.147sin(2t) + 0.906cos(3t) – 0.423sin(3t) + 0.47cos(4t) – 0.171sin(4t)

(b) frms = ∑∞

=

+1n

2n

2o A

21a

frms

2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625 frms = 6.828 Chapter 17, Solution 62.

(a) s3141.0202TT/220o =π

=→π==ω

(b) f ...)90t40cos(1.5)90t20cos(43)tncos(Aa)t( o

1n

onono +++++=φ+ω+= ∑

∞

=

....t80sin8.1t60sin7.2t40sin1.5t20sin43)t(f −−−−−= Chapter 17, Solution 63.

This is an even function.

T = 3, ωo = 2π/3, bn = 0.

f(t) = 5.1t1,2

1t0,1<<<<

ao =

+= ∫∫∫

5.1

1

1

0

2/T

0dt2dt1

32dt)t(f

T2 = (2/3)[1 + 1] = 4/3

an =

π+π=ω ∫∫∫

5.1

1

1

0

2/T

0 o dt)3/tn2cos(2dt)3/tn2cos(134dt)tncos()t(f

T4

=

π

π+

π

π

5.1

1

1

0 3tn2sin

n26

3tn2sin

n23

34

= [–2/(nπ)]sin(2nπ/3)

f2(t) = ∑∞

=

π

π

π−

1n 3tn2cos

3n3sin

n12

34

ao = 4/3 = 1.3333, ωo = 2π/3, an = –[2/(nπ)]sin(2nπt/3)

An =

π

π=+

3n2sin

n2b2

n2na

A1 = 0.5513, A2 = 0.2757, A3 = 0, A4 = 0.1375, A5 = 0.1103

The amplitude spectra are shown below. 1.333

0.1103 0.1378

0

0.275

0.551

4321n

5 0

An


The amplitude and phase spectra are shown below.

An 3.183 2.122 1.591 0.4244 0 2π 4π 6π ω nφ 0 2π 4π 6π ω

-180o

Chapter 17, Solution 65. an = 20/(n2π2), bn = –3/(nπ), ωn = 2n

An = 22442n n

9n400b

π+

π=+2

na

= 22n44.441

n3

π+

π, n = 1, 3, 5, 7, 9, etc.

n An 1 2.24 3 0.39 5 0.208 7 0.143 9 0.109

φn = tan–1(bn/an) = tan–1[–3/(nπ)][n2π2/20] = tan–1(–nx0.4712)

n φn 1 –25.23° 3 –54.73° 5 –67° 7 –73.14° 9 –76.74° ∞ –90°

0 10

φn

ωn

14 62 18

–25.23°

–54.73°–67°

–76.74° –73.14°

–60°

–30°

–90°

2.24

An

ωn

0.0.143 0.109 0.208

0.39

18 14 10 6 2 0

Chapter 17, Solution 66. The schematic is shown below. The waveform is inputted using the attributes of VPULSE. In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, the output plot is shown below. The output file includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)

DC COMPONENT = 5.099510E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 3.184E+00 1.000E+00 1.782E+00 0.000E+00 2 1.000E+00 1.593E+00 5.002E-01 3.564E+00 1.782E+00 3 1.500E+00 1.063E+00 3.338E-01 5.347E+00 3.564E+00 4 2.000E+00 7.978E-01 2.506E-01 7.129E+00 5.347E+00 5 2.500E+00 6.392E-01 2.008E-01 8.911E+00 7.129E+00 6 3.000E+00 5.336E-01 1.676E-01 1.069E+01 8.911E+00 7 3.500E+00 4.583E-01 1.440E-01 1.248E+01 1.069E+01 8 4.000E+00 4.020E-01 1.263E-01 1.426E+01 1.248E+01 9 4.500E+00 3.583E-01 1.126E-01 1.604E+01 1.426E+01 TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT


he Schematic is shown below. In the Transient dialog box, we type “Print step = 0.01s, Final time = 36s, Center frequency = 0.1667, Output vars = v(1),” and click Enable Fourier. After simulation, the output file includes the following Fourier components,

C T

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 2.000396E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED

PHASE (DEG)

.000E+00 -8.996E+01 0.000E+00

NO (HZ) COMPONENT COMPONENT (DEG)

1 1.667E-01 2.432E+00 12 3.334E-01 6.576E-04 2.705E-04 -8.932E+01 6.467E-01 3 5.001E-01 5.403E-01 2.222E-01 9.011E+01 1.801E+02 4 6.668E-01 3.343E-04 1.375E-04 9.134E+01 1.813E+02 5 8.335E-01 9.716E-02 3.996E-02 -8.982E+01 1.433E-01 6 1.000E+00 7.481E-06 3.076E-06 -9.000E+01 -3.581E-02 7 1.167E+00 4.968E-02 2.043E-02 -8.975E+01 2.173E-01 8 1.334E+00 1.613E-04 6.634E-05 -8.722E+01 2.748E+00 9 1.500E+00 6.002E-02 2.468E-02 9.032E+01 1.803E+02

TAL HARMON DISTORT N = 2.280 E+01 PERC T TO IC IO 065 EN

Chapter 17, Solution 68. The schematic is shown below. We set the final time = 6T=12s and the center frequency = 1/T = 0.5. When the schematic is saved and run, we obtain the Fourier series from the output file as shown below.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 1.990000E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.273E+00 1.000E+00 9.000E-01 0.000E+00 2 1.000E+00 6.367E-01 5.001E-01 -1.782E+02 1.791E+02 3 1.500E+00 4.246E-01 3.334E-01 2.700E+00 1.800E+00 4 2.000E+00 3.185E-01 2.502E-01 -1.764E+02 -1.773E+02 5 2.500E+00 2.549E-01 2.002E-01 4.500E+00 3.600E+00 6 3.000E+00 2.125E-01 1.669E-01 -1.746E+0 -1.755E+02 7 3.500E+00 1.823E-01 1.431E-01 6.300E+00 5.400E+00 8 4.000E+00 1.596E-01 1.253E-01 -1.728E+02 -1.737E+02 9 4.500E+00 1.419E-01 1.115E-01 8.100E+00 7.200E+00 Chapter 17, Solution 69.

The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, we obtain V(1) as shown below. We also obtain an output file which includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.048510E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00 2 1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00 3 1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00 4 2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00 5 2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00 6 3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00 7 3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00 8 4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01 9 4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00 TOTAL HARMONIC DISTORTION = 1.214285E+01 PERCENT

Chapter 17, Solution 70. The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier. After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 7.658051E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.070E+00 1.000E+00 1.004E+01 0.000E+00 2 1.000E+00 3.758E-01 3.512E-01 -3.924E+01 -4.928E+01 3 1.500E+00 2.111E-01 1.973E-01 -3.985E+01 -4.990E+01 4 2.000E+00 1.247E-01 1.166E-01 -5.870E+01 -6.874E+01 5 2.500E+00 8.538E-02 7.980E-02 -5.680E+01 -6.685E+01 6 3.000E+00 6.139E-02 5.738E-02 -6.563E+01 -7.567E+01 7 3.500E+00 4.743E-02 4.433E-02 -6.520E+01 -7.524E+01 8 4.000E+00 3.711E-02 3.469E-02 -7.222E+01 -8.226E+01 9 4.500E+00 2.997E-02 2.802E-02 -7.088E+01 -8.092E+01 TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT

Chapter 17, Solution 71. The schematic is shown below. We set Print Step = 0.05, Final Time = 12 s, Center Frequency = 0.5, Output Vars = I(1), and click enable Fourier in the Transient dialog box. After simulation, the output waveform is as shown. The output file includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1) DC COMPONENT = 8.374999E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 2.287E-02 1.000E+00 -6.749E+01 0.000E+00 2 1.000E+00 1.891E-04 8.268E-03 8.174E+00 7.566E+01 3 1.500E+00 2.748E-03 1.201E-01 -8.770E+01 -2.021E+01 4 2.000E+00 9.583E-05 4.190E-03 -1.844E+00 6.565E+01 5 2.500E+00 1.017E-03 4.446E-02 -9.455E+01 -2.706E+01 6 3.000E+00 6.366E-05 2.783E-03 -7.308E+00 6.018E+01 7 3.500E+00 5.937E-04 2.596E-02 -9.572E+01 -2.823E+01 8 4.000E+00 6.059E-05 2.649E-03 -2.808E+01 3.941E+01 9 4.500E+00 2.113E-04 9.240E-03 -1.214E+02 -5.387E+01 TOTAL HARMONIC DISTORTION = 1.314238E+01 PERCENT Chapter 17, Solution 72. T = 5, ωo = 2π/T = 2π/5 f(t) is an odd function. ao = 0 = an

bn = ∫ ∫ π=ω2/T

0

10

0o dt)tn4.0sin(1054dt)tnsin()t(f

T4

= 1

0

)nt4.0cos(n2

5x8π

π− = )]n4.0cos(1[

n20

π−π

f(t) = ∑∞

=

ππ−π 1n

)tn4.0sin()]n4.0cos(1[n120


p = ∑+R

V21

RV 2

n2DC

= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW


(a) An = 2n

2n ba + , φ = tan–1(bn/an)

A1 = 22 86 + = 10, φ1 = tan–1(6/8) = 36.87° A2 = 22 43 + = 5, φ2 = tan–1(3/4) = 36.87°

i(t) = 4 + 10cos(100πt – 36.87°) – 5cos(200πt – 36.87°) A

(b) p = ∑+ RI5.0RI 2n

2DC

= 2[42 +0.5(102 + 52)] = 157 W Chapter 17, Solution 75. The lowpass filter is shown below. R + + vs C vo - -

∑∞

=ω

πτ+

τ=

1nos tncos

Tnsin

n1

TA2

TAv

T/n2n,VRCj1

1V

Cj1R

Cj1

V onsn

s

n

no π=ω=ω

ω+=

ω+

ω=

For n=0, (dc component), TAVsoV τ

== (1)

For the nth harmonic,

o

n122

n2o 90

Tnsin

nTA2

RCtanCR1

1V −∠πτ

•ω∠ω+

=−

When n=1, 22

2o

CRT

41

1T

nsinTA2|V

π+

•πτ

=| (2)

From (1) and (2),

4222

222

10x09.39.30CRT

41

CRT

41

110

sinTA2x50

TA

=τ

=π

+→π

+

π=

τ

mF 59.2410x4

10x09.3x1010R2

TC10CRT

413

4251022

2=

π=

π=→=

π+

−


vs(t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by 10. Hence

vo(t) = ∑∞

=

ππ

+1k

)tnsin(n1205 , n = 2k – 1

T = 2, ωo = 2π/T = 2π, ωn = nωo = 2nπ

jωnL = j2nπ; Z = R||10 = 10R/(10 + R) Vo = ZVs/(Z + j2nπ) = [10R/(10R + j2nπ(10 + R))]Vs

Vo = s2222

1

V)R10(n4R100

)R10)(R5/n(tanR+π+

+π−∠ −10

Vs = [20/(nπ)]∠0°

The source current Is is

Is = )R10(n2jR10

n20)R10(

n2jR10

R10V

n2jZV ss

+π+π

+=

π++

=π+

= 2222

1

)R10(n4R100

)R10)(3/n(tann20)R10(

+π+

+π−∠π

+ −

ps = VDCIDC + ∑ φ−θ )cos(IV21

nnsnsn

For the DC case, L acts like a short-circuit.

Is = R10

)R10(5

R10R

510

+=

+

, Vs = 5 = Vo

++π+

+

π+

π

+

+π+

+π

+

π

++

=

−

−

222

122

222

12

s

)R10(16R100

)R10(5

2tancos)R10(10

)R10(4R100

)R10(5

tancos)R10(20

21

R10)R10(25p

ps = ∑∞

=

+1n

onDC

RV

21

RV

=

+

+π++

+π+

+ 222222 )R10(10R100R100

)R10(4R100R100

21

R25

We want po = (70/100)ps = 0.7ps. Due to the complexity of the terms, we consider only the DC component as an approximation. In fact the DC component has the latgest share of the power for both input and output signals.

R10

)R10(25x107

R25 +

=

100 = 70 + 7R which leads to R = 30/7 = 4.286 Ω


(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest common factor is 2. Hence ωo = 2.

T = 2π/ωo = 2π/2 = π

(b) The average value is the DC component = –2

(c) Vrms = ∑∞

=

++1n

2n

2no )ba(

21a

)136810(21)2( 2222222

rms +++++−=V = 121.5

Vrms = 11.02 V


(a) p = ∑ ∑+=+R

VR

VR

V21

RV 2

rms,n2DC

2n

2DC

= 0 + (402/5) + (202/5) + (102/5) = 420 W

(b) 5% increase = (5/100)420 = 21

pDC = 21 W = 105R21VtoleadswhichR

V 2DC

2DC ==

VDC = 10.25 V

Chapter 17, Solution 79. From Table 17.3, it is evident that an = 0,

bn = 4A/[π(2n – 1)], A = 10.

A Fortran program to calculate bn is shown below. The result is also shown.

C FOR PROBLEM 17.79 DIMENSION B(20) A = 10 PIE = 3.142 C = 4.*A/PIE DO 10 N = 1, 10 B(N) = C/(2.*FLOAT(N) – 1.) PRINT *, N, B(N) 10 CONTINUE STOP END

n bn 1 12.731 2 4.243 3 2.546 4 1.8187 5 1.414 6 1.1573 7 0.9793 8 0.8487 9 0.7498 10 0.67

Chapter 17, Solution 80. From Problem 17.55, cn = [1 + e–jnπ]/[2π(1 – n2)]

This is calculated using the Fortran program shown below. The results are also shown.

C FOR PROBLEM 17.80 COMPLEX X, C(0:20) PIE = 3.1415927

A = 2.0*PIE DO 10 N = 0, 10 IF(N.EQ.1) GO TO 10 X = CMPLX(0, PIE*FLOAT(N)) C(N) = (1.0 + CEXP(–X))/(A*(1 – FLOAT(N*N))) PRINT *, N, C(N)

10 CONTINUE STOP END

n cn 0 0.3188 + j0 1 0 2 –0.1061 + j0 3 0 4 –0.2121x10–1 + j0 5 0 6 –0.9095x10–2 + j0 7 0 8 –0.5052x10–2 + j0 9 0 10 –0.3215x10–2 + j0


A

3T 2TT0

f(t) = ∑∞

=

ω−ππ 1n

o2 )tncos(1n4

1A4A−

2

The total average power is pavg = Frms

2R = Frms2 since R = 1 ohm.

Pavg = Frms2 = ∫

T

0

2 dt)t(fT1 = 0.5A2

(b) From the Fourier series above |co| = 2A/2, |cn| = 4A/[π(4n2 – 1)]

n ωo |cn| 2|cn|2 % power 0 0 2A/π 4A2/(π2) 81.1% 1 2ωo 2A/(3π) 8A2/(9π2) 18.01% 2 4ωo 2A/(15π) 2A2/(225π2) 0.72% 3 6ωo 2A/(35π) 8A2/(1225π2) 0.13% 4 8ωo 2A/(63π) 8A2/(3969π2) 0.04%

(c) 81.1% (d) 0.72% Chapter 17, Solution 82.

P = ∑∞

=1n

2n

2DC

RV

21

R+

V

Assuming V is an amplitude-phase form of Fourier series. But

|An| = 2|Cn|, co = ao |An|2 = 4|Cn|2

Hence,

P = ∑∞

=1n

2n

2o

Rc

R+ 2

c

Alternatively,

P = R

V

2rms

where

∑∑∑∞

−∞=

∞

=

∞

=

=+=+=n

2n

1n

2n

2o

1n

2n

2o

2rms cc2cA

21aV

= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22)

= 100 + 2x94.57 = 289.14

P = 289.14/4 = 72.3 W

Chapter 18, Solution 1. )2t()1t()1t()2t()t('f −δ+−δ−+δ−+δ=

2jjj2j eeee)(Fj ω−ω−ωω +−−=ωω ω−ω= cos22cos2

F(ω) = ω

ω−ωj

]cos2[cos2


<<=

otherwise,01t0,t

)t(f

-δ(t-1)

δ(t)

f ”(t)

t –δ’(t-1)

1

-δ(t-1)

0

1

f ‘(t)

t

f"(t) = δ(t) - δ(t - 1) - δ'(t - 1) Taking the Fourier transform gives

-ω2F(ω) = 1 - e-jω - jωe-jω

F(ω) = 2

j 1e)j1(ω

−ω+ ω

or F ∫ ω−=ω1

0

tj dtet)(

But ∫ +−= c)1ax(aedxex 2

axax

( )=−ω−

ω−=ω

ω−102

j

)1tj(j

e)(F ( )[ ]1ej11 j2 −ω+

ωω−


2t2,21)t('f,2t2,t

21)t(f <<−=<<−=

∫− −

ω−ω −ω−

ω−==ω

2

2

222

tjtj )1tj(

)j(2edtet

21)(F

[ ])12j(e)12j(e2

1 2j2j2 −ω−−ω−

ω−= ωω−

( )[ ]2j2j2j2j2 eeee2j

21 ω−ωωω −++ω−ω

−=

( )ω+ωω−ω

2sin2j2cos4j2

12−=

F(ω) = )2cos22(sinj2 ωω−ω

ω


1

–2δ(t–1)

2δ(t+1)

–1

g’ 2

0 –2

t

–2δ’(t–1)

–2δ(t+1)1

–2δ(t–1)

2δ’(t+1)

–1

g”

4δ(t)

0 –2

t

4sin4cos4ej2e24ej2e2)(G)j(

)1t(2)1t(2)t(4)1t(2)1t(2g

jjjj2

+ωω−ω−=ω−−+ω+−=ωω

−δ′−−δ−δ++δ′++δ−=′′

ω−ω−ωω

)1sin(cos4)(G 2 −ωω+ωω

=ω


1

0

h’(t)

–1

–2δ(t)

1 t

δ(t+1)

–δ(t–1)

1

0

h”(t)

–1

–2δ’(t)

1 t

ω−ω=ω−−=ωω

δ′−−δ−+δ=′′

ω−ω j2sinj2j2ee)(H)j(

)t(2)1t()1t()t(h

jj2

H(ω) = ωω

−ω

sinj2j22


dtetdte)1()(F tj0

1

1

0

tj ω−

−

ω−∫ ∫+−=ω

)1(cos1tsinttcos1tsin1

tdtcosttdtcos)(F Re

2102

01

0

1

1

0

−ωω

=

ω

ω+ω

ω+ω

ω−=

ω+ω−=ω

−

−∫ ∫

Chapter 18, Solution 7. (a) f1 is similar to the function f(t) in Fig. 17.6. )1t(f)t(f1 −=

Since ω−ω

=2F ω

j)1(cos)(

=ω=ω ω )(Fe)(F j1

ω−ωω−

j)1(cose2 j

Alternatively, )

)2t()1t(2)t()t(f '

1 −δ+−δ−δ= e2e(eee21)(Fj jjj2jj

1ωωω−ω−ω− +−=+−=ωω

)2cos2(e j −ω= ω−

F1(ω) = ω

−ωω−

j)1e j (cos2

(b) f2 is similar to f(t) in Fig. 17.14.

f2(t) = 2f(t)

F2(ω) = 2

)cos1(4ω

ω−


(a) 21

tj2

21

tj10

tj

2

1

tj1

0

tj

)1tj(e2ej4e

j2

dte)t24(dte2)(F

−ω−ω−

−ω−

+ω−

=

−+=ω

ω−ω−ω−

ω−ω− ∫∫

ω−ω−ω− ω+

ω−

ω−

ω+

ω+

ω=ω 2j

22jj

2 e)2j1(2ej4

j2e

j22)(F

(b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]

ωω

−ω

ω=ω

sin22sin4)(G


(a) y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]

ωω

+ωω

=ω sin42sin2)(Y

(b) )j1(e22)1tj(e2dte)t2()( 2

j

210

1

02

tjtj ω+

ω−

ω=−ω−

ω−

−=−=ω

ω−ω−ω−∫Z


(a) x(t) = e2tu(t) X(ω) = 1/(2 + jω)

(b)

<

>=

−−

0t,e0t,e

et

t)t(

∫ ∫ ∫− −

ω−−ωω +==ω1

1

0

1

1

0

tjttjttj dteedteedte)t(y)(Y

10

t)j1(0

1

t)j1(

)j1(e

j1e

ω+−+

ω−=

ω+−

−

ω−

ω+

ω−ω+

ω−ω+ω

−ω+

= −

j1sinjcos

j1sinjcose

12 1

2

Y(ω) = [ ])sin(cose11

2 12 ωω−ω−

ω+−

Chapter 18, Solution 11. f(t) = sin π t [u(t) - u(t - 2)]

( )∫ ∫ ω−π−πω− −=π=ω2

0

2

0

tjtjtjtj dteeej2

1dtetsin)(F

+∫ π+ω−π+ω−+2

0

t)(jt)(j dt)ee(j2

1=

π+ω−+

π−ω−

π+ω−π−ω− 2

0

t)(j20

t)(j

)(jee

)(j1

j21

=

ω+π

−+

ω−π− ω−ω− 2j2j e1e1

21

=

( )ω−π+πω−π

2j22 e22)(2

1=

F(ω) = ( )1e 2j22 −

π−ωπ ω−

Chapter 18, Solution 12. (a) F = dtedtee)(

0

2

0

t)j1(tjt∫ ∫∞ ω−ω− =ω

=ω−

= ω− 20

t)j1(ej1

1 ω−−ω−

j11e 2j2

(b) ∫ ∫−

ω−ω− −+=ω0

1

1

0

tjtj dte)1(dte)(H

( ) ( ) )cos22(j11e

j1e1

j1 jj ω+−

ω=−

ω+−

ωω−ω−=

=ωω−

=j

2/sin4 2 2

2/2/sinj

ωω

ω

Chapter 18, Solution 13. (a) We know that )]a()a([]at[cos +ωδ+−ωδπ=F .

Using the time shifting property,

)a(e)a(e)]a()a([e)]a3/t(a[cos 3/j3/ja3/j +ωδπ+−ωδπ=+ωδ+−ωδπ=π− ππ−ωπ−F

(b) sin tsinsintcoscostsin)1t( π−=ππ+ππ=+π g(t) = -u(t+1) sin (t+1)

Let x(t) = u(t)sin t, then 22 11

1)j(1)(X

ω−=

+ω=ω

Using the time shifting property,

1ee

11)(G 2

jj

2 −ω=

ω−−=ω

ωω

(c ) Let y(t) = 1 + Asin at, then Y )]a()a([Aj)(2)( −ωδ−+ωδπ+ωπδ=ω

h(t) = y(t) cos bt

Using the modulation property,

)]b(Y)b(Y[21)(H −ω++ω=ω

[ ] [ ])ba()ba()ba()ba(2Aj)b()b()(H −−ωδ−−+ωδ++−ωδ−++ωδπ

+−ωδ++ωδπ=ω

(d) )14j(ej

e1)1tj(ej

edte)t1()(2

4j4j

2402

tj4

0

tjtj +ω

ω−

ω−

ω=−ω−

ω−−

ω−=−=ω

ω−ω−ω−ω−ω−∫I


(a) )t3cos()0(t3sin)1(t3cossint3sincost3cos)t3cos( −=−−=π−π=π+ (f )t(ut3cose)t t−−=

F(ω) = ( )( ) 9j1

j12 +ω+ω+−

(b)

[ ])1t(u)1t(utcos)t('g −−−ππ=

g(t)

t 1 -1

-1

1

-π

-1 1

g’(t)

t

π

)1t()1t()t(g)t("g 2 −πδ++πδ−π−= ω−ω π+π−ωπ−=ωω− jj22 ee)(G)(G

( ) ωπ−=−π−=ωω−π ω−ω sinj2)ee()(G jj22

G(ω) = 22

sinj2π−ωωπ

Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F(ω) = G(ω)e-jω

( ) ( )ωω−ω −π−ω

π=ω=ω jj

22j eee)(FG

22

sin2jπ−ωωπ−

=

G(ω) = 22

sinj2ω−πωπ

(c) tcos)0(tsin)1(tcossintsincostcos)1t(cos π−=π+−π=ππ+ππ=−π Let ex = )t(he)1t(u)1t(cos)t( 2)1t(2 −=−−π−−

and )t(u)tcos(e)t(y t2 π= −

22)j2(j2)(Y

π+ω+ω+

=ω

)1t(x)t(y −= ω−ω=ω je)(X)(Y

( )( ) 22

j

j2ej2)(Xπ+ω+

ω+=ω

ω

)(He)(X 2 ω−=ω )(Xe)(H 2 ω−=ω −

= ( )( ) 22

2j

j2ej2π+ω+

ω+− −ω

(d) Let x )t(y)t(u)t4sin(e)t( t2 −=−−= −

)t(x)t(p −=where )t(ut4sine)t(y t2=

( ) 22 4j2

j2)(Y+ω+ω+

=ω

( ) 16j2

j2)(Y)(X 2 +ω−ω−

=ω−=ω

=ω−=ω )(X)(p( ) 162j

2j2 +−ω

−ω

(e) 2j2j ej1)(23e

j8)(Q ω−ω−

ω

+ωπδ−+ω

=ω

Q(ω) = 2j2j e)(23ej6 ω−ω ωπδ−+ω

Chapter 18, Solution 15. (a) F =−=ω ω−ω 3j3j ee)( ω3sinj2

(b) Let g ω−=ω−δ= je2)(G),1t(2)t(

=ω)(F F

∫ ∞−

tdt)t(g

)()0(Fj

)(Gωδπ+

ωω

=

)()1(2je2 j

ωδ−πδ+ω

=ω−

= ω

ω−

je j2

(c) F [ ] 121)t2( ⋅=δ

=ω−⋅=ω j211

31)(F

2j

31 ω−


(a) Using duality properly

2

2tω−

→

ωπ→− 2t

22

or ωπ−→ 4t42

F(ω) = F =

2t4

ωπ4−

(b) tae− 22aa2ω+

22 taa2+

ω−π ae2

22 ta8+

ω−π 2e4

G(ω) = F =

+ 2t48

ω−π 2e4


(a) Since H(ω) = F ( ) ( ) ([ ]000 FF21)t(ftcos ω−ω+ω+ω=ω )

where F(ω) = F ( )[ ] ( ) 2,j1tu 0 =ωω

+ωπδ=

( ) ( ) ( ) ( ) ( )

−ω

+−ωπδ++ω

++ωπδ=ω2j

122j

1221H

( ) ( )[ ] ( )( )

−ω+ω−ω++ω

−−ωδ++ωδπ

=2222

2j22

2

H(ω) = ( ) ( )[ ]4

j222 2 −ω

ω−−ωδ++ωδ

π

(b) G(ω) = F [ ] ( ) ([ ]000 FF2j)t(ftsin ω−ω−ω+ω=ω )

where F(ω) = F ( )[ ] ( )ω

+ωπδ=j1tu

( ) ( ) ( ) ( ) ( )

−ω

−−ωπδ−+ω

++ωπδ=ω10j

11010j

1102jG

( ) ( )[ ]

+ω−

−ω+−ωδ−+ωδ

π=

10j

10j

2j1010

2j

= ( ) ( )[ ]100

1010102 2

j−ω

−−ωδ−+ωδπ


Let f ( ) ( )tuet t−= ( )ω+

=ωjj

1F

( ) tcostf ( ) ([ ]1F1F21

+ω+−ω )

Hence ( ) ( ) ( )

+ω+

+−ω+

=ω1j1

11j1

121Y

( )[ ] ( )[ ]

+ω+−ω+−ω+++ω+

=1j11j1jj1jj1

21

1jjjj1

j12 +ω−−ω++ω+

ω+=

= 2j2

j12 +ω−ωω+


( ) ( )∫ ∫∞

∞−

ω−π−πω +==ω dteee21dte)t(fF tj1

0

t2jt2jtj

( ) ( ) ( )[ ]∫ π−ω−π+ω− +=ω1

0

t2jt2j dtee21F

( )( )

( )( )

1

0

t2jt2j e2j

1e2j

121

π−ω−

+π+ω−

= π−ω−π+ω−

( )

( )( )

( )

π−ω−

+π+ω−

−=π−ω−π+ω−

2j1e

2j1e

21 2j2j

But π−π ==π+π= 2j2j e12sinj2cose

( )

π−ω+

π+ω

−−=ω

ω−

21

21

j1e

21F

j

= ( )1e4

j j22 −

π−ωω ω−


(a) F (cn) = cnδ(ω) F ( ) ( )on

tjnn ncec o ω−ωδω =

F =

∑∞

−∞=

ω

n

tjnn

oec ( )∑∞

−∞=

ω−ωδn

on nc

(b) π= 2T 1T2

o =π

=ω

( )∫ ∫

+⋅

π==

π −ω−T

0 0

jnttjnn 0dte1

21dtetf

T1c o

( )1en2

jejn1

21 jn

0jnt −

π=

−

π= π−π

But e njn )1(ncosnsinjncos −=π=π+π=π−

( )[ ]

=−−

π= =

≠=π−

evenn,0

0n,oddn,n

jn

n 11n2jc

for n = 0

∫π

=π

=0n 2

1dt121c

Hence

∑∞

=≠−∞= π

−=

oddn0n

n

jntenj

21)t(f

F(ω) = ( )∑∞

=≠−∞=

−ωδπ

−δω

oddn0n

nn

nj

21


Using Parseval’s theorem,

∫∫∞

∞−

∞

∞−ωω

π= d|)(F|

21dt)t(f 22

If f(t) = u(t+a) – u(t+a), then

∫∫ ∫∞

∞−

∞

∞− −ω

ωω

π=== d

aasina4

21a2dt)1(dt)t(f

22a

a22

or

aa4a4d

aasin

2

2 π=

π=ω

ωω

∫∞

∞− as required.


F [ ] ( )∫∞

∞−

ω−ω−ω −

=ω dtej2ee)t(ftsin)t(f tj

tjtj

o

oo

( ) ( )

−= ∫ ∫

∞

∞−

∞

∞−

ω+ω−ω−ω− dtedte)t(fj2

1 tjtj oo

= ( ) ([ ]oo FFj2

1ω+ω−ω−ω )


(a) f(3t) leads to ( )( ) ( )( )ω+ω+=

ω+ω+⋅

j15j630

3/j53/j210

31

F [ ]( ) =− t3f ( )( )ω−ω− j15j630

(b) f(2t) ( )( ) ( )( )ω+ω+=

ω+ω+⋅

j10j420

2/j152/j210

21

f(2t-1) = f [2(t-1/2)] ( )( )ω+ω+

ω−

j10j4e20 2/j

(c) f(t) cos 2t ( ) ( )2F212F

21

+ω++ω

= [ ]( ) ( )[ ] ( ) ( )[ ][ ]2j52j25

2j52j2 −ω+−ω++

+ω++ω+5

(d) F [ ]( ) ( ) ( )( )ω+ω+ω

=ωω=j5j2

10jFjt'f

(e) ( )∫ f ∞−

tdtt ( )

( ) ( ) ( )ωδπ+ωω 0F

jF

( )( ) ( )5x2

10xj5j2j

ωπδ+ω+ω+ω

10=

= ( )( ) ( )ωπδ+ω+ω+ω j5j2j

10

Chapter 18, Solution 24. (a) ( ) ( )ω=ω FX + F [3]

= ( ) ( )1e j

ω+ωπδ ω−j6 −

(b) ( ) ( )2tfty −=

( ) ( ) =ω=ω ω− FeY 2j ( )1eje j2j

−ω

ω−ω−

(c) If h(t) = f '(t)

( ) ( ) ( ) =−ω

ω=ωω=ω ω− 1ejjFjH j ω−− je1

(d) ( )

ω+

ω=ω

=

53F

53x10

23F

23x4)(G,t

35f

3ft +

10t24g

( ) ( )1e

53

j61e

23

j6 5/3j2/3j −ω

+−ω

⋅= ω−ω−

= ( ) ( )1e10j1e4j 5/3j2/3j −ω

+−ω

ω−ω−


(a) ( ) ( ) ω=+

+=+

= js,2s

Bs2ss

s A10F

52

10B,52

10A −=−

===

( )2j

5j5F

+ω−

ω=ω

f(t) = ( ) ( )tue5t2

t2−−sgn5

(b) ( ) ( )( ) 2jB

1jA

2j1j4jF

+ω+

+ω=

+ω+ω−ω

=ω

( ) ( )( ) ω=+

++

=++

−= js,

2sB

1sA

2s1s4ssF

A = 5, B = 6

( )ω+

+ω+

−=ω

j26

j15F

f(t) = ( ) ( )tue6e5 t2t −− +−


(a) )t(ue)t( )2t( −−=f (b) )t(ute)t( t4−=h

(c) If ωω

=ω→−−+=sin2)(X)1t(u)1t(u)t(x

By using duality property,

ttsin2)t(g)1(u2)1(u2)(G

π=→−ω−+ω=ω


(a) Let ( ) ( ) ω=+

+=+

= js,10s

Bs10ss

s A100F

1010

100B,1010100A −=

−===

( )10j

10j10F

+ω−

ω=ω

f(t) = ( ) ( )tue10tsgn t10−−5

(b) ( ) ( )( ) ω=+

+−

=+−

= js,3s

Bs2

As3s2

s10sG

6530B,4

520A −=

−===

( )3j

62j

4G+ω

−+ω−=

=ω

g(t) = ( ) ( )tue6tue t3t2 −−−4

(c) ( )( ) ( ) 90020j

60130040jj

60H 22 ++ω=

+ω+ω=ω

h(t) = ( )tu)t30sin(e t20−2

( ) ( )( )( )∫

∞

∞−

ω

=⋅π=+ωω+ωωδ

π=

21

21

1jj2de

21ty

tj

π41


(a) ∫∫∞

∞−

ω∞

∞−

ω ωω+ω+

ωπδπ

=ωωπ

= d)j2)(j5(

e)(21de)(F

21)t(f

tjtj

===201

)2)(5(1

21 0.05

(b) ∫∞

∞−

−ω

+−−π=ω

+ωω+ωδ

π=

)12j)(2j(e

210de

)1j(j)2(10

21)t(f

t2jtj

=−π

=−

2j1e

25j t2j

π

+− −

2e)j2( t2j

(c) ∫∞

∞−

ω

++π=ω

ω+ω+−ωδ

π=

)j3)(j2(e

220d

)53)(j2(e)1(20

21)t(f

jttj

=+π

=)j55(2

e20 jt

π− jte)j1(

(d) Let )(F)(F)j5(j

5)j5()(5)( 21 ω+ω=

ω+ω+

ω+F ωπδ

=ω

∫∞

∞−

ω =⋅ππ

=ωω+ωπδ

π= 5.0

51

25de

j5)(5

21)t(f tj

1

1B,1A,5s

BsA

)s5(s5)s(2 −==

++=

+=F

5j

1j1)(F2 +ω−

ω=ω

t5t52 e)t(u

21e)tsgn(

21)t( −+−=−= −f

=+= )t(f)t(f)t(f 21 u t5e)t( −−


(a) f(t) = F -1[ +ωδ )]( F -1 [ ])3(4)3(4 −ωδ++ωδ

π

+π

=t3cos4

21 = ( )t3cos81

21

+π

(b) If )2t(u)2t(u)t(h −−+=

ωω

=ω2sin2)(H

)(H4)(G ω=ω t

t2sin821)t( ⋅π

=g

g(t) = t

t2sin4π

(c) Since

cos(at) ↔ )a()a( −ωπδ++ωπδ Using the reversal property,

)2t()2t(2cos2 −πδ++πδ↔ωπ or F -1 [ ] =ω2cos6 )2t(3)2t(3 −δ++δ


(a) ω+

=ωω

=ω→=ja

1)(X,j2)(Y)tsgn()t(y

)]t(u)t(u[a)t(2)t(hja22

j)ja(2

)(X)(Y)(H −−+δ=→

ω+=

ωω+

=ωω

=ω

(b) ω+

=ωω+

=ωj2

1)(Y,j1

1)(X

)t(ue)t()t(hj2

11j2j1)(H t2−−δ=→

ω+−=

ω+ω+

=ω

(c ) In this case, by definition, )t(u btsine)t(y)t(h at−==


(a) ω+

=ωω+

=ωja

1)(H,)ja(

1)(Y 2

)t(ue)t(xja

1)(H)(Y)(X at−=→

ω+=

ωω

=ω

(b) By definition, )1t(u)1t(u )t(y)t(x −−+==

(c ) ω

=ωω+

=ωj2)(H,

)ja(1)(Y

)t(ue2a)t(

21)t(x

)ja(2a

21

)ja(2j

)(H)(Y)(X at−−δ=→

ω+−=

ω+ω

=ωω

=ω


(a) Since 1j

j

+ω

ω−e ( ) )1t(ue 1t −−−

and F f(-t) ( )ω−

( )1j

e j

1 +ω−=ω

ω

F ( ) ( ) ( )1tuetf 1t1 −−= −−−

f1(t) = e ( ) ( )1tu1t −−+

(b) From Section 17.3,

1t

22 +

ω−πe2

If ( ) ,e2F2

ω−=ω then

f2(t) = ( )1t2

2 +π

(b) By partial fractions

( )( ) ( ) ( ) ( ) ( ) 1j

41

1j41

1j41

1j41

1j1j1F 22223 −ω

−−ω

++ω

++ω

=−ω+ω

=ω

Hence ( ) ( ) ( )tueteete41t tttt

3 −++= −−f

= ( ) ( ) ( ) (tue1t41tue1t

4tt −++ − )1

(d) ( ) ( ) ( )∫ ∫∞

∞−

∞

∞−

ωω

ω+ωδ

π=ωω

π=

2j1e

21deF

21tf

tjtj

14 = π21

Chapter 18, Solution 33. (a) Let ( ) ( ) ( )[ ]1tu1tutsin2tx −−+π=

From Problem 17.9(b),

( ) 22

sinj4Xω−πωπ

=ω

Applying duality property,

( ) ( ) ( )22 ttsinj2tX

21tf

−π−

=−π

=

f(t) = 22ttsinj2

π−

(b) ( ) ( ) ( )ω−ωω

−ω−ωω

=ω sinjcosj2sinj2cosjF

( )ω

−ω

=−ω

ωω−ω−ω

je

jeeej 2jj

j2j=

( ) ( ) ( )2tsgn211tsgn

21tf −−−=

But sgn( 1)t(u2)t −=

( ) ( ) ( )212tu

211tutf +−−−−=

= ( )u ( 2tu1t −−− ) Chapter 18, Solution 34. First, we find G(ω) for g(t) shown below.

( ) ( ) ( )[ ] ( ) ( )[ ]1tu1tu102tu2tu10tg −−++−−+= ( ) ( ) ( )[ ] ( ) ( )[ ]1t1t102t2t10t'g −δ−+δ+−δ−+δ= The Fourier transform of each term gives

20

t

g(t)

10

0

–2 –1 1 2

10δ(t+2) 10δ(t+1)

–1 –2 1 2t

g ‘(t)

0

–10δ(t-2) –10δ(t-1)

( ) ( ) ( )ω−ωω−ω −+−=ωω jj2j2j ee10ee10Gj ω+ω= sinj202sinj20

( ) =ω

ω+

ωω

=ωsin202sin20G 40 sinc(2ω) + 20 sinc(ω)

Note that G(ω) = G(-ω).

( ) ( )ω−π=ω G2F

( ) ( )tG21tfπ

=

= (20/π)sinc(2t) + (10/π)sinc(t) Chapter 18, Solution 35. (a) x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,’’

)j6(ee

3/j21

31)(X

3/j3/j

ω+=

ω+=ω

ω−ω−

(b) Using the modulation property,

−ω

++ω

=

−ω+

++ω+

=−ω++ω=ω3j

17j

121

)5(j21

)5(j21

21)]5(F)5(F[

21)(Y

(c ) ω+

ω=ωω=ω

j2j)(Fj)(Z

(d) 2)j2(1)(F)(F)(Hω+

=ωω=ω

(e) 22 )j2(1

)j2()j0(j)(F

ddj)(I

ω+=

ω+

−=ω

ω=ω


( ) ( )( )ωω

=ωi

o

VV

H

( ) ( ) ( ) ( )ω+ω

=ωω=ωj2

V10VHV iio

(a) vi = 4δ(t) Vi(ω) = 4

( )ω+

=ωj2

40Vo

( )tue40)t(v t2

o−=

vo(2) = 40e–4 = 0.7326 V

(b) ( )tue6v ti

−= ( )ω+

=ωj1

6Vi

( ) ( )( )ω+ω+=ω

j1j260Vo

( ) ( )( ) ω=+

++

=++

= js,2s

B1s

A1s2s

60sVo

601

60B,60160A −=

−===

( )ω+

−ω+

=ωj2

60j1

60Vo

[ ] )t(uee60)t(v t2t

o−− −=

[ ] ( )01831.013533.060ee60)2(v 42o −=−= −−

= 7.021 V (c) vi(t) = 3 cos 2t

Vi(ω) = π[δ(ω + 2) + δ(ω- 2)]

( ) ( )[ ]ω+

−ωδ++ωδπ=

j22210Vo

( )∫∞

∞−

ω ωωπ

= deV21)t(v tj

oo

( ) ( )∫∫∞

∞−

ω∞

∞−

ω ωω+

−ωδ+ω

ω++ωδ

= dj2

e25dej225

tjtj

( ) ([ ]°−°−−+−

+=+

+−

= 45t2j45t2jt2jt2j

ee22

52j2

e52j2

e5 )

( )°−= 45t2cos2

5

( ) ( ) ( )°−°=°−= 4518.229cos2

5454cos2

52vo

vo(2) = –3.526 V Chapter 18, Solution 37.

ω+ω

=ωj2

2jj2


( ) ( )( ) ω++ω

ω=

ω+ω

+

ω+ω

=ωω

=ω4j82j

2j

j22j4

j22j

II

Hs

o

H(ω) = ω+

ω3j4

j


( ) ( )ω

+ωπδ=ωj1Vi

( ) ( ) ( )

ω

+ωπδω+

=ωω+

=ωj1

j55V

2j1010V io

Let ( ) ( ) ( ) ( )( )ω+ω

+ω+ωπδ

=ω+ω=ωj5j

5j5

521o VVV

( ) ( ) 5sB

sA

5ss5V2 +

+=+

=ω A = 1, B = -1, s = jω

( )ω+

−ω

=ωj5

1j1V2 ( ) t5

2 e)tsgn(21t −−=v

( )ω+ωπδ

=j5

5V1 ( ) ( )∫∞

∞−

ω ωω+ωπδ

π= de

j55

21tv tj

1

v1(t) 5.051

25

=⋅ππ

=

( ) ( ) ( ) ( ) t5

210 etsgn5.05.0tvtvtv −−+=+= But ( ) ( )tu21tsgn +−=

( ) ( ) ( ) =−+−+= − tuetu5.05.0t t5

ov ( ) ( )tuetu t5−− Chapter 18, Solution 39.

ω−ω−∞

∞− ω−

ω+

ω=−=ω ∫ j

22tj

s e11j1dte)t1()(V

ω−

ω+

ωω+=

ω+

ω=ω ω−

−j

226

3

33s e11

j1

j1010

10xj10

)(V)(I

Chapter 18, Solution 40. )2t()1t(2)t()t(v −δ+−δ−δ= 2jj2 ee21)(V ωω− +−=ωω−

( ) 2

2jj ee21Vω−+−

=ωω−ω−

Now ( )ωω+

=ω

+=ωj

2j1j12Z

( )( ) ω+

ω⋅

ω−−

=ωω

=ωω

2j1j1ee2

ZVI 2

2jj

( ) ( )ω−ω− −+ω+ω

j2j ee5.05.0j5.0j

=1

But 5.0s

BsA

)5.0s(s1

++=

+ A = 2, B = -2

( ) ( ) ( )ω−ω−ω−ω −+ω+

−−+ω

=ω j2jj2j ee5.05.0j5.0

2ee5.05.0j2I

i(t) = )1t(ue2)2t(ue)t(ue)1tsgn()2tsgn(21)tsgn(

21 )1t(5.0)2t(5.0t5.0 −−−−−−−−+ −−−−−


ω+ j21

+ − 1/s 0.5s

+

V

−

2

( )ω+

ω+ω−=

ω+ω

+ω=+ω−ω

=−ω

+ω+ω+

−

j29j4

j2j4jV42j

02jV2Vj

2j2

1V

22

V(ω) = )j24)(j2(

)2j5.4(j22 ω+ω−ω+

ω+ω


By current division, ( )ω⋅ω+

= Ij2

2Io

(a) For i(t) = 5 sgn (t),

( )

)j2(j20

j10

j22I

j10I

o ω+ω=

ω⋅

ω+=

ω=ω

Let 10B,10A,2s

BsA

)2s(s20

o −==+

+=+

I =

( )ω+

−ω

=ωj2

10j10

oI

io(t) = 5 A)t(ue10)tsgn( t2−− (b)

)1t(4)t(4)t('i −δ−δ=

4 i(t)

t 1

–4δ(t–1)

4δ(t) i’(t)

t 1

( ) ω−−=ωω je44Ij

( ) ( )ω−

=ωω−

je14I

j

( ) ( )ω−

ω−

−

ω+

−ω

=ω+ω

−= j

j

o e1j2

1j14

)j2(je18I

ω+

+ω

−ω+

−ω

=ω−ω−

j2e4

je4

j24

j4 jj

io(t) = ( )A1tue4)t(ue4)1tsgn(2)tsgn(2 )1t(2t2 −+−−− −−− Chapter 18, Solution 43.

ω+=→=

ω=

ω=

ω→ −

− j51Ie5i,

j50

10x20j1

Cj1 mF 20 s

ts3

ω=++

=ω

•

ω+

= js,)5s)(25.1s(

50j50I

j5040

40V so

+ω

−+ω

=+

++

=5j

125.1j

1340

5sB

25.1sAVo

)t(u)ee(340)t(v t5t25.1

o−− −=

Chapter 18, Solution 44. 1H jω

We transform the voltage source to a current source as shown in Fig. (a) and then combine the two parallel 2Ω resistors, as shown in Fig. (b).

(a)

+

Vo

−

1 Vs/2

2

Io

+

Vo

−jω2

Vs/2

(b)

Io

jω

,122 Ω= 2

Vj1

1 so ⋅

ω+=I

)j1(2Vj

IjV soo ω+

ω=ω=

( ) )2t(10t10)t(vs −δ−δ= ( ) ω−−=ωω 2j

s e1010Vj

( ) =ωsV ( )ω− ω−

je110 2j

Hence ( ) ω−ω−

ω+−

ω+=

ω+−

= 2j2j

o ej1

5j1

5j1

e15V

)2t(ue5)t(ue5)t(v )2t(t

o −−= −−−

=−−= − 01e5)1(v 1o 1.839 V


)t(ute2)t(v)1j(

2)2(

j1j2

j1

V to2o

−=→+ω

=

ω+ω+

ω=


F41

ω−

=ω

4j

41j

1

2H jω2 3 )t(3δ

)t(ue t−

ω+ j11

The circuit in the frequency domain is shown below: 2 Ω Vo

Io(ω)

+ − 3

+ − 1/(1+jω)

–j4/ω

j2ω

At node Vo, KCL gives

ω

=

ω−−

+−

ω+2jV

4jV3

2

Vj1

1oo

o

ω

−=ω−ω+−ω+

ooo

V2jVj3jV2j1

2

ω−ω+

ω+ω+= 2jj2

3jj1

2

Vo

( )

ω−ω+ω

ω+ω−ω+

=ω

=ω2jj22j

j133j2

2jV

I

2

oo

Io(ω) = )28(j64

3j32

22

ω−ω+ω−ω−ω+2


Transferring the current source to a voltage source gives the circuit below:

1/(jω)2 Ω

+ −Vo

+ − 1 Ω

8 V jω/2

Let ω+ω+

=ω

+

ω

+=ω

+=j23j4

2j1

2j

22j12inZ


( ) ( )ω+ω+ω

+=

ω+=⋅

+ω

ω=ω

j23j4j1

8Zj1

88Z

j1

j1

Vin

in

o

= ( )234jj2

j2ω−ω+ω+

8 ω+

= ( )235j2

j2ω−ω+

8 ω+


0.2F ω

−=ω

5jCj

1

As an integrator,

4.010x20x10x20RC 63 == −

∫−=t

o io dtvRC1v

( )

ωδπ+

ω−= )0(V

jV

RC1V i

io

( ) ( )

ωπδ+

ω+ω−=

j2j2

4.01

( ) ( )

ωδπ+

ω+ω−==

j2j2125.0mA

20VI o

o

( )ωπδ−ω+

+ω

−= 125.0j2

125.0j125.0

( ) ( )∫ ω− ωπδπ

−+−= dte2125.0tue125.0)tsgn(125.0)t(i tjt2

o

2125.0)t(ue125.0)t(u25.0125.0 t2 −++= −

io(t) = mA)t(ue125.0)t(u25.0625.0 t2−+−


Consider the circuit shown below:

2 Ω

jω

i2i1

+ −

j2ω

jω

1 Ω

+

vo

−

VS

( ) ([ ]21Vs −ωδ+ )+ωδπ= For mesh 1, ( ) 0IjI2I2j2V 221s =ω−−ω++− ( ) ( ) 21s Ij2Ij12V ω+−ω+= (1) For mesh 2, ( ) 112 IjI2Ij30 ω−−ω+=

( )( )ω+

ω+=

2I3

I 21 (2)


( )( ) ( ) 22

s Ij2j2

Ij3j122V ω+−ω+

ω+ω+=

( ) ( ) ( )[ ] 2

22s I4j44j322V ω−ω+−ω−ω+=ω+

( )22 4j2I ω−ω+=

( )

ω=++

+= js,

2s4sV2s

I 2s

2

== 2o IV ( ) ( ) ( )[ ]( ) 24jj

112j2 +ω+ω

−ωδ++ωδπ+ω

( )∫∞

∞−

ω ωωπ

= dev21)t(v tj

oo

( ) ( )

( )

( ) ( )

( )∫∞

∞−

ωω

+ω+ω

ω−ωδ+ω+

+ω+ω

ω+ωδ+ω=

24jj

d1e2j21

24jj

d1e2j21

2

tj

2

tj

( ) ( )

24j1

e2j21

24j1

e2j21 jtjt

++−

++

+−−

+−=

( )( ) ( )( )17

e4j1j221

e17

4j1j221

)t(vjt

jto

−−+

+−=

( ) ( ) jtjt e7j6341e7j6

341

−++=

( ) ( )°−°−− += 64.13tj64.13tj e271.0e271.0

vo(t) = 0 V)64.13tcos(542. °− Chapter 18, Solution 50.

Consider the circuit shown below:

jωi2i1

+ −

+

vo

− jω

1 Ω

1 Ωj0.5ω

VS

For loop 1, ( ) 0I5.0jIj12 21 =ω+ω++− (1) For loop 2, ( ) 0I5.0jIj1 12 =ω+ω+ (2) From (2),

( ) ( )

ωω+

−=ω−

ω+=

jIj12

5.0jIj1I 22

1


( )

22 I

2j

jIj12

2 ω+

ωω+−

=

22 I

234j4j2

ω−ω+−=ω

22 5.14j4j2I

ω−ω+ω

=

( )22o j5.14j4

j2IVω+ω+

ω−==

( )2

o

j3

8j38

j34

Vω+

ω+

ω=

2222

38j

34

316

38j

34

j344

+

ω+

+

+

ω+

ω+−

=

=)t(Vo V)t(ut38sine657.5)t(ut

38cose4 3/t43/t4

+

− −−


ω+=

ω+

ω=ω

=j1

1

j11

j1

j1//1Z

ω=++

=

ω+ω+=

ω+∗

ω++

ω+=

+=

js,)5.1s)(1s(

1

j12

j231

j12

j112

j11

V2Z

ZV oo

)t(u)ee(2)t(v5.1s

21s

25.1s

B1s

AV t5.1too

−− −=→+

−+

=+

++

=

∞∞ −−−−−−

∞

∞−

∞−−

∫

∫ ∫

−+

−=+−=

−==

00

t3t5.2t2t3t5.2t2

0

2t5.1t2

3e

5.2e2

2e4dt)ee2e(4

dt)ee(4dt)t(fW

J 1332.0)31

5.22

21(4W =+−=


J = ∫∫∞∞

ωωπ

=0

2

0

2 d)(F1dt)t(f2

= 23

1)3/(tan31d

911

0

1

0 22

ππ

=ωπ

=ωω+π

∞−∞

∫ = (1/6)


J = ∫∫∞

∞−

∞π=ωω dt)t(f2d)(F 2

0

2

f(t) = 0t,e0t,e

2

t2

>

<− t

J =

−+π=

+π

∞−

∞−

∞ −

∞− ∫∫0

t40t4

0

t40 t4

4e

4e2dtedte2 = 2π[(1/4) + (1/4)] = π


W1Ω =

∞−∞ −∞

∞−−== ∫∫ 0

t2

0

t22 e8dte16dt)t(f = 8 J


f(t) = 5e2e–tu(t) F(ω) = 5e2/(1 + jω), |F(ω)|2 = 25e4/(1 + ω2)

W1Ω = ∞

−∞∞ω

π=ω

ω+π=ωω

π ∫∫0

14

0 2

4

0

2 )(tane25d1

1e25d)(F1

= 12.5e4 = 682.5 J

or W1Ω = 12.5e== ∫∫∞ −∞

∞− 0

t242 dtee25dt)t(f 4 = 682.5 J


W1Ω = ∫ ∫∞ −∞

∞−=

0

2t22 dt)t2(sinedt)t(f

But, sin2(A) = 0.5(1 – cos(2A))

W1Ω = ∞−∞−∞ − +−

+−

−=−

0

t2

0

t2

0

t2 )]t4sin(4)t4cos(2[164

e2

e21dt)]t4cos(1[5.0e ∫

= (1/4) + (1/20)(–2) = 0.15 J


W1Ω = 0t20 t22 e2dte4dt)t(i∞−∞−

∞

∞−== ∫∫ = 2 J

or I(ω) = 2/(1 – jω), |I(ω)|2 = 4/(1 + ω2)

W1Ω = 2

4)(tan4d)1(

124d)(I

21

0

12

2 ππ

=ωπ

=ωω+π

=ωωπ

∞−∞

∞−

∞

∞− ∫∫ = 2 J

In the frequency range, –5 < ω < 5,

W = )373.1(4)5(tan4tan4 15

0

1

π=

π=ω

π−− = 1.7487

W/ W1Ω = 1.7487/2 = 0.8743 or 87.43% Chapter 18, Solution 58.

ωm = 200π = 2πfm which leads to fm = 100 Hz (a) ωc = πx104 = 2πfc which leads to fc = 104/2 = 5 kHz (b) lsb = fc – fm = 5,000 – 100 = 4,900 Hz (c) usb = fc + fm = 5,000 + 100 = 5,100 Hz


ω+−

ω+=ω+

−ω+=

ωω

=ωj4

3j2

52

j46

j210

)(V)(V)(H

i

o

ω=++

−++

=

ω+

ω+

−ω+

=ωω=ω

js,)4s)(1s(

12)2s)(1s(

20j1

4j4

3j2

5)(V)(H)(V io

Using partial fraction,

ω++

ω+−

ω+=

++

++

++

+=ω

j44

j220

j116

4sD

1sC

2sB

1sA)(Vo

Thus, ( ) V)t(ue4e20e16)t(v t4t2t

o−−− +−=


1/jω jω +

V

−

Is

ω+ω−

ω=

ω++ω

ωω=

2j1

Ij

j2j1

j1

IjV 2s

s

°−∠=+−

°∠=

π+π−

°∠π= 92.712418.1

566.12j48.389027.50

4j418902V

.componentDCnoisthere,inductortheacrossappearsvoltagetheSince

21

°−∠=+−

°∠=

π+π−

°∠π= 9.803954.0

13.25j91.1569083.62

8j1615904V 22

mV)1.129t4cos(3954.0)92.41t2cos(2418.1)t(v °+π+°−π=


lsb = 8,000,000 – 5,000 = 7,995,000 Hz usb = 8,000,000 + 5,000 = 8,005,000 Hz Chapter 18, Solution 62.

For the lower sideband, the frequencies range from 10,000 – 3,500 Hz = 6,500 Hz to 10,000 – 400 Hz = 9,600 Hz

For the upper sideband, the frequencies range from 10,000 + 400 Hz = 10,400 Hz to 10,000 + 3,500 Hz = 13,500 Hz Chapter 18, Solution 63.

Since fn = 5 kHz, 2fn = 10 kHz

i.e. the stations must be spaced 10 kHz apart to avoid interference. ∆f = 1600 – 540 = 1060 kHz The number of stations = ∆f /10 kHz = 106 stations Chapter 18, Solution 64.

∆f = 108 – 88 MHz = 20 MHz The number of stations = 20 MHz/0.2 MHz = 100 stations Chapter 18, Solution 65.

ω = 3.4 kHz fs = 2ω = 6.8 kHz


ω = 4.5 MHz fc = 2ω = 9 MHz Ts = 1/fc = 1/(9x106) = 1.11x10–7 = 111 ns Chapter 18, Solution 67.

We first find the Fourier transform of g(t). We use the results of Example 17.2 in conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and width T as shown below.

Arect(t) transforms to Atsinc(ω2/2)

–ωm/2

G(ω)

ω

ωm/2

F(ω)

ω

–T/2

A f(t)

t

T/2

According to the duality property, Aτsinc(τt/2) becomes 2πArect(τ) g(t) = sinc(200πt) becomes 2πArect(τ) where Aτ = 1 and τ/2 = 200π or T = 400π i.e. the upper frequency ωu = 400π = 2πfu or fu = 200 Hz

The Nyquist rate = fs = 200 Hz

The Nyquist interval = 1/fs = 1/200 = 5 ms


The total energy is

WT = ∫ ∞

∞−dt)t(v2

Since v(t) is an even function,

WT = ∞−∞ −

−=

0

t4

0

t4

4e5000dte2500∫ = 1250 J

V(ω) = 50x4/(4 + ω2)

W = ∫ ωω+π

=ωωπ

5

1 22

25

1

2 d)4(

)200(21d|)(V

2 ∫ |1

But C)a/x(tana1

axx

a21dx

)xa(1 1

222222 +

+

+=

+−∫

W = 5

1

12

4

)2/(tan21

48110x

ω+

ω+ω

π−2

= (2500/π)[(5/29) + 0.5tan-1(5/2) – (1/5) – 0.5tan–1(1/2) = 267.19

W/WT = 267.19/1250 = 0.2137 or 21.37%


The total energy is

WT = ∫∫∞

∞−

∞

∞−ω

ω+π=ωω

πd

4400

21d)(F

2 2221

= [ ]2

100)4/(tan)4/1(4000

1 ππ

=ωπ

∞− = 50

W = [ ]2

0

12

0

2 )4/(tan)4/1(2400d)(F

21

ωπ

=ωωπ

−∫

= [100/(2π)]tan–1(2) = (50/π)(1.107) = 17.6187

W/WT = 17.6187/50 = 0.3524 or 35.24%

Chapter 19, Solution 1. To get z and , consider the circuit in Fig. (a). 11 21z

1 Ω 4 Ω I2 = 0

+

V2

−

Io+

V1

−

2 Ω6 ΩI1

(a)

Ω=++== 4)24(||611

111 I

Vz

1o 21

II = , 1o2 2 IIV ==

Ω== 11

221 I

Vz

To get z and , consider the circuit in Fig. (b). 22 12z

1 Ω 4 ΩI1 = 0

Io' +

V2

−

+

V1

−

2 Ω6 Ω I2

(b)

Ω=+== 667.1)64(||22

222 I

Vz

22'

o 61

1022

III =+

= , 2o1 '6 IIV ==

Ω== 12

112 I

Vz

Hence,

=][z Ω

667.1114


Consider the circuit in Fig. (a) to get and . 11z 21z

+

V2

−

I2 = 0

1 Ω

1 Ω

+

V1

−

1 Ω

Io

1 Ω

1 Ω

Io'1 Ω 1 Ω

I1

1 Ω 1 Ω 1 Ω 1 Ω (a)

])12(||12[||121

111 +++==

IV

z

733.21511

24111)411)(1(

243

2||1211 =+=+

+=

++=z

'

o'

oo 41

311

III =+

=

11'

o 154

41111

III =+

=

11o 151

154

41

III =⋅=

1o2 151

IIV ==

06667.0151

121

221 ==== z

IV

z

To get z , consider the circuit in Fig. (b). 22

1 Ω 1 Ω 1 Ω1 ΩI1 = 0

+

V2

−

+

V1

−

1 Ω1 Ω1 Ω I2

1 Ω 1 Ω 1 Ω 1 Ω(b)

733.2)3||12(||12 112

222 ==++== z

IV

z

Thus,

=][z Ω

733.206667.0

06667.0733.2


(a) To find and , consider the circuit in Fig. (a). 11z 21z

Io +

V2

−

+

V1

−

1 Ωj Ω

I2 = 0-j Ω

I1

(a)

j1j1j)j1(j

)j1(||j1

111 +=

−+−

=−==IV

z


11o jj1j

jIII =

−+=

1o2 jIIV ==

j1

221 ==

IV

z


I1 = 0

+

V2

−

+

V1

−

1 Ω

-j Ω

j Ω I2

(b)

0)jj(||12

222 =−==

IV

z

21 jIV =

j2

112 ==

IV

z

Thus,

=][z Ω

+0jjj1

(b) To find and , consider the circuit in Fig. (c). 11z 21z

+ V1 −

-j Ω 1 Ω

+ V2 −

1 Ω

I2 = 0j Ω -j Ω

I1

(c)

5.0j5.1j1

j-j1-j)(||11j

1

111 +=

−++=++==

IV

z

12 )5.0j5.1( IV −=

5.0j5.11

221 −==

IV

z

To get z and , consider the circuit in Fig. (d). 22 12z

+ V2 −

-j Ω 1 Ω

+ V1 −

1 Ω

I1 = 0 j Ω -j Ω

I2

(d)

j1.5-1.5(-j)||11-j2

222 =++==

IV

z

21 )5.0j5.1( IV −=

5.0j5.12

112 −==

IV

z

Thus,

=][z Ω

−−−+

5.1j5.15.0j5.15.0j5.15.0j5.1

Chapter 19, Solution 4. Transform the Π network to a T network.

Z1 Z3

Z2

5j12120j

5j10j12)10j)(12(

1 +=

−+=Z

j512j60-

2 +=Z

5j1250

3 +=Z

The z parameters are

j4.26--1.77525144

j5)--j60)(12(22112 =

+=== Zzz

26.4j775.1169

)5j12)(120j(1212111 +=+

−=+= zzZz

739.5j7758.1169

)5j12)(50(2121322 −=+

−=+= zzZz

Thus,

=][z Ω

−−−+

739.5j775.126.4j775.1-26.4j775.1-26.4j775.1

Chapter 19, Solution 5. Consider the circuit in Fig. (a).

s 1 I2 = 0

Io

1/s1/s +

V2

−

+

V1

−

1 I1

(a)

s1

s11s

1s1

s11s

1

s1

s1||

s1

1

s1

s1

s1||s1

||111

+++

+

++

+

=

+++

=

++=z

1s3s2s1ss

23

2

11 +++++

=z

12

11o

1ss1s

s1s

s

s1

s11s

11s

1

s1

s1s1

||1

s1

||1IIII

++++

+=+++

+

+=+++

=

123o 1s3s2ss II

+++=

1s3s2ss1

231

o2 +++==

IIV

1s3s2s123

1

221 +++

==IVz

Consider the circuit in Fig. (b).

s 1 I1 = 0

1/s1/s 1 +

V1

−

+

V2

−

I2

(b)

+++=

++==1s

1s1||

s1

s1

||1s1||s1

2

222 I

Vz

1ss

ss1

1s1

s1

1s1

s1s1

1s1

s1s1

222

++++

+++

=

++++

+++

=z

1s3s2s2s2s

23

2

22 +++++

=z

2112 zz =

Hence,

=][z

+++++

+++

++++++++

1s3s2s2s2s

1s3s2s1

1s3s2s1

1s3s2s1ss

23

2

23

2323

2


To find and , connect a voltage source to the input and leave the output open as in Fig. (a).

11z 21z 1V

Vo

+

V2

−

20 ΩI1

30 Ω0.5 V2+ −

10 Ω

V1

(a)

505.0

10o

2o1 V

VVV

+=−

, where oo2 53

302030

VV =+

V =

oo

oo1 2.455

35 VVVVV =+

+=

ooo1

1 32.010

2.310

VVVV

I ==−

=

Ω=== 125.1332.02.4

o

o

1

111 V

VIV

z

Ω=== 875.132.06.0

o

o

1

221 V

VIV

z

To obtain and , use the circuit in Fig. (b). 22z 12z

I2

0.5 V2 + −

10 Ω

+

V1

−

30 Ω

20 Ω

V2

(b)

22

22 5333.030

5.0 VV

VI =+=

Ω=== 875.15333.01

2

222 I

Vz

2221 -9)5.0)(20( VVVV =−=

Ω=== -16.8755333.0

9-

2

2

2

112 V

VIV

z

Thus,

=][z Ω

1.8751.87516.875-125.13

Chapter 19, Solution 7. To get z11 and z21, we consider the circuit below. I2=0 I1 20 100Ω Ω + + + vx 50Ω 60 Ω V1

- V2 - -

12vx - +

1xxxxx1 V

12140V

160V12V

50V

20VV

=→+

+=−

88.29IVz)

20V(

12181

20VVI

1

111

1x11 ==→=

−=

37.70IVzI37.70

I81

121x20)12140(

857V)

12140(

857V

857V12)

160V13(60V

1

2211

11xxx

2

−==→−=

−=−=−=−=

To get z12 and z22, we consider the circuit below. I2 I1=0 20 100Ω Ω + + + vx 50Ω 60 Ω V1

- V2 - -

12vx - +

2x22

222x V09.060

V12V150VI,V

31V

5010050V =

++==

+=

11.1109.0/1IVz

2

222 ===

704.3IVzI704.3I

311.11V

31VV

2

112222x1 ==→====

Thus,

Ω

−

=11.1137.70

704.388.29]z[

Chapter 19, Solution 8. To get z11 and z21, consider the circuit below.

j4Ω I1 -j2Ω 5 Ω I2 =0 •• + j6Ω j8Ω + V2 V1 10Ω - -

4j10IVzI)6j2j10(V

1

11111 +==→+−=

)4j10(IVzI4jI10V

1

221112 +−==→−−=

To get z22 and z12, consider the circuit below.

j4Ω I1=0 -j2Ω 5 Ω I2 •• + j6Ω j8Ω + V2 V1 10Ω - -

8j15IVzI)8j105(V

2

22222 +==→++=

)4j10(IVzI)4j10(V

2

11221 +−==→+−=

Thus,

Ω

++−+−+

=)8j15()4j10()4j10()4j10(

]z[


It is evident from Fig. 19.5 that a T network is appropriate for realizing the z parameters.

R2 R1 2 Ω 6 Ω

R3 4 Ω

=−=−= 410R 12111 zz Ω6 =−=−= 46R 12222 zz Ω2

=== 21123R zz Ω4 Chapter 19, Solution 10.

(a) This is a non-reciprocal circuit so that the two-port looks like the one shown in Figs. (a) and (b).

+ −

+

V1

−

I1

+ −

+

V2

−

z21 I1 z12 I2

z22 I2 z11

(a)

(b) This is a reciprocal network and the two-port look like the one shown in

Figs. (c) and (d).

+

V1

−

I2I1

+

V2

−

z12

z22 – z12z11 – z12

(c)

+ −

+

V1

−

I2 I1

+ −

+

V2

−

5 I1 20 I2

10 Ω25 Ω

(b)

s5.01

1s2

11211 +=+=− zz

s21222 =− zz

s1

12 =z

1 F +

V1

−

I2I1

+

V2

−

0.5 F 2 H 1 Ω

(d)

Chapter 19, Solution 11. This is a reciprocal network, as shown below. 1+j5 3+j 1Ω j5Ω 3 Ω 5-j2 5Ω -j2Ω Chapter 19, Solution 12. This is a reciprocal two-port so that it can be represented by the circuit in Figs. (a) and (b).

Ω

2 Ω

2 Ω8 ΩI1

+

V1

−

+

V2

−

I2

4 Ω

+

V1

−

I2I1

+

V2

−

(a)

z12

z22 – z12z11 – z12

j1

Io

2 Ω

(b) From Fig. (b),

111 10)4||48( IIV =+=


1o 21

II = , 1o2 2 IIV ==

==1

1

1

2

10II

VV

1.0

Chapter 19, Solution 13. This is a reciprocal two-port so that the circuit can be represented by the circuit below.

40 Ω 50 Ω 20 Ω

10 I2 30 I1 + −

+ − 120∠0° V

rms I1 I2 +

− 100 Ω

We apply mesh analysis. For mesh 1,

2121 91201090120- IIII +=→=++ (1) For mesh 2,

2121 -4012030 IIII =→=+ (2) Substituting (2) into (1),

3512-

-35-3612 2222 =→=+= IIII

=

== )100(3512

21

R21

P2

2

2I W877.5

Chapter 19, Solution 14. To find , consider the circuit in Fig. (a). ThZ

I2I1

+

V1

−

+ −

ZS Vo = 1

(a)

2121111 IzIzV += (1)

2221212 IzIzV += (2) But

12 =V , 1s1 - IZV =

Hence, 2s11

1212121s11

-)(0 I

Zzz

IIzIZz+

=→++=

222s11

1221-1 Iz

Zzzz

+

+=

===22

2Th

1II

VZ

s11

122122 Zz

zzz

+−

To find , consider the circuit in Fig. (b). ThV

+ −

I2 = 0ZS

+

V2 = VTh

−

I1

+

V1

−

VS

(b)

02 =I , V s1s1 ZIV −=

Substituting these into (1) and (2),

s11

s1111s1s Zz

VIIzZIV

+=→=−

s11

s211212 Zz

VzIzV

+==

== 2Th VVs11

s21

ZzVz+


(a) From Prob. 18.12,

24104060x80120

ZzzzzZ

s11

211222Th =

+−=

+−=

Ω== 24ZZ ThL

(b) 192)120(1040

80VZz

zs

s11

21Th =

+=

+=V

W19224x8

192R8

VP2

Th

Th2

max ===

Chapter 19, Solution 16. As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).

+ − j6 Ω

b

a

(a)

15∠0° V

5 Ω 4 – j6 Ω10 – j6 Ω

j4 Ω

At terminals a-b, )6j105(||6j)6j4(Th −++−=Z

6j4.26j415

)6j15(6j6j4Th ++−=

−+−=Z

=ThZ Ω4.6

==°∠−++

= 6j)015(6j1056j

6jThV V906 °∠

The Thevenin equivalent circuit is shown in Fig. (b).

6.4 Ω

+

Vo

−

+ − 6∠90° V j4 Ω

(b)From this,

°∠=+

= 14818.3)6j(4j4.6

4joV

=)t(vo V)148t2cos(18.3 °+

Chapter 19, Solution 17. To obtain z and , consider the circuit in Fig. (a). 11 21z

4 Ω Io I2 = 0

Io' +

V2

−

+

V1

−

2 Ω

8 ΩI1

(a)6 Ω

In this case, the 4-Ω and 8-Ω resistors are in series, since the same current, , passes through them. Similarly, the 2-Ω and 6-Ω resistors are in series, since the same current,

, passes through them.

oI

'oI

Ω===++== 8.420

)8)(12(8||12)62(||)84(

1

111 I

Vz

11o 52

1288

III =+

= 1'

o 53

II =

But 024- '

oo2 =+− IIV

111'

oo2 52-

56

58-

2-4 IIIIIV =+=+=

Ω=== -0.452-

1

221 I

Vz


+

V1

−

I1 = 0

8 Ω

2 Ω +

V2

−

4 Ω

I2

(b)6 Ω

Ω===++== 2.420

)14)(6(14||6)68(||)24(2

222I

Vz

Ω== -0.42112 zz

Thus,

=][z Ω

4.20.4-0.4-8.4

We may take advantage of Table 18.1 to get [y] from [z].

20)4.0()2.4)(8.4( 2z =−=∆

21.020

2.4

z

2211 ==

∆=

zy 02.0

204.0-

z

1212 ==

∆=

zy

02.020

4.0-

z

2121 ==

∆=

zy 24.0

208.4

z

1122 ==

∆=

zy

Thus,

=][y S24.002.002.021.0

Chapter 19, Solution 18. To get y and , consider the circuit in Fig.(a). 11 21y

I2I1

+ −

+

V2 = 0

−

(a)

6 Ω

3 Ω

3 Ω

6 ΩV1

111 8)3||66( IIV =+=

81

1

111 ==

VI

y

12-

832-

366- 11

12

VVII ==

+=

121-

1

221 ==

VI

y

To get y and , consider the circuit in Fig.(b). 22 12y

6 Ω 3 ΩIoI1 I2

+ −

+

V1 = 0

−

3 Ω6 Ω V2

(b)

21

6||31

)6||63(||31

2

222 ==

+==

VI

y

2- o

1

II = , 22o 3

163

3III =

+=

12-

21

61-

6- 2

22

1

VV

II =

==

212

112 12

1-y

VI

y ===

Thus,

=][y S

21

121-

121-

81

Chapter 19, Solution 19. Consider the circuit in Fig.(a) for calculating and y . 11y 21

1/s

I2I1

+ −

+

V2 = 0

−

s 1

1

V1

(a)

1111 1s22

)s1(2s2

2||s1

IIIV+

=+

=

=

5.0s2

1s2

1

111 +=

+==

VI

y

2-

1s2-

2)s1()s1-( 11

12

VIII =

+=

+=

-0.51

221 ==

VI

y

To get y and , refer to the circuit in Fig.(b). 22 12y

1 I1 I2

1/s s 1

+ −

+

V1 = 0

−

V2

(b)

222 2ss2

)2||s( IIV+

==

s1

5.0s22s

2

222 +=

+==

VI

y

2-

s22s

2ss-

2ss- 2

221

VVII =

+⋅

+=

+=

-0.52

112 ==

VI

y

Thus,

=][y Ss10.50.5-

0.5-5.0s

+

+

Chapter 19, Solution 20. To get y11 and y21, consider the circuit below.

3ix I1 2Ω I2 + ix + I1 V1 4 6Ω Ω V2 =0

- -

Since 6-ohm resistor is short-circuited, ix = 0

75.0VIyI

68)2//4(IV

1

111111 ==→==

5.0VIyV

21)V

86(

32I

244I

1

2211112 −==→−=−=

+−=

To get y22 and y12, consider the circuit below.

3ix I1 2Ω + ix + V1=0 4 6Ω Ω V2 I2

-

-

1667.061

VIy

6V

2Vi3iI,

6Vi

2

222

22xx2

2x ===→=+−==

0VIy0

2Vi3I

2

112

2x1 ==→=−=

Thus,

S1667.05.0075.0

]y[

−

=

Chapter 19, Solution 21. To get and , refer to Fig. (a). 11y 21y

At node 1,

10 Ω

I2I1

+ −

+

V2 = 0

−

(a)

5 Ω

0.2 V1V1

V1

4.04.02.05 1

11111

11 ==→=+=

VI

yVVV

I

-0.2-0.21

22112 ==→=

VI

yV

and 12y , refer to the circuit in Fig. (b).

I

o get

ince , the dependent current source can be replaced with an open circuit.

22y

0.2 V1

T

+

V1 0 +

I1 I2

V2 5 Ω

(b)

10 Ω

V1

− =

−

01 =VS

1.0101

102

22222 ===→=

VI

yIV

02

112 ==

VI

y

Thus,

onsequently, the y parameter equivalent circuit is shown in Fig. (c)

=][y S1.02.0-

04.0

C .


and refer to the circuit in Fig. (a).

At node 1,

I I1 2

0.1 S 0.2 V1 + +

(c)

0.4 S V1

−

V2

−

C

(a) To get y11 21y

Ω Ω2

1 Ω

(a)

+

V2 0 V1 =

−

+

I1 I2 V1

+

Vx

−

3

Vx/2

Vo

−

o11o11

1 5.05.121

VVIVVV

−=→−

+ (1)

At node 2,

o1o1o 2.1

322VV

VVV=→=+ (2)

ubstituting (2) into (1) gives,

I =

1V −

S

9.09.06.05.11

1111111 ==→=−=

VI

yVVVI

-0.4-0.43

-

1

2211

o2 ==→==

VI

yVV

I

o get and refer to the circuit in Fig. (b).

so that the dependent current source can be replaced by an

open circuit.

22y 12y

2 Ω 3 Ω

T

1 Ω

(b)

+

V1 0

I1 I2V1

+

x Vx/2 + V2 − V

−

=

−

01x == V

V

2.051

5)023(2

222222 ===→=++=

VI

yIIV

-0.2-0.2-2

112221 ==→==

VI

yVII

Thus,

=][y S0.20.4-0.2-9.0

(b) o get and refer to Fig. (c). 11y 21y

j Ω

T

)5.0j5.1(||jj1

1||j-j)||11(||jin −= −

+=+=Z j-

-j Ω

1 Ω1 Ω

V1

(c)

+

V2 0

I1 I2

+

Io

Io'

Io''

Zin

− =

−

8.0j6.05.0j5.1

)5.0j5.1(j+=

+−

=

8.0j6.08.0j6.0

11

in1

1111in1 −=

+===→=

ZVI

yIZV

1o 5.0j5.1j

II+

= , 1'

o 5.0j5.15.0j5.1

II+−

=

1I

j2)5.0j5.1)(j1(j1j- 1

o''

o −=

+−=

−=

III

111

2''

o'

o2 )4.0j2.1(5

j25.2

)5.0j5.1(- IIIIII −=

++

−=+=

112 )2.1j4.0()8.0j6.0)(4.0j2.1(- VVI −=−−=

121

221 j1.2-0.4 y

VI

y =+==

To get refer to the circuit in Fig.(d).

22y

8.0j.0)j-||11(||jout =+=Z

-j Ω

1 Ω1 Ω

(d)

+

V1 = 0

I1 I2

Zout

+ −

V2

j Ω

−

6 +

8.0j6.01

out22 −==

Zy

Thus,

=][y S8.0j6.0j1.20.4-

j1.20.4-8.0j6.0

−++−


1s1y

1s1

s1//1y 1212 +

−=→+

==−

1s2s

1s11y1y1yy 12111211 +

+=

++=−=→=+

1s

1ss1s

1sysysyy2

12221222 +++

=+

+=−=→=+

+++

+−

+−

++

=

1s1ss

1s1

1s1

1s2s

]y[ 2

(b) Consider the network below. I1 I2 1 + + + Vs V1 V2 2 - - -

[y]

11s VIV += (1)

22 I2V −= (2)

2121111 VyVyI += (3)

2221212 VyVyI += (4) From (1) and (3)

212111s2121111s VyV)y1(VVyVyVV ++=→+=− (5) From (2) and (4),

22221

12221212 V)y5.0(y1VVyVyV5.0 +−=→+=− (6)


++−

=→=

+++

−=

)y5.0)(y1(y1y

s/2Vs2

VyVy

)y5.0)(y1(V

221121

12

2

212221

2211s

)5.3s5.7s6s2(s)1s(2

1s1ss

21

1s3s2)1s(

1s1

s/2V 2322+++

+=

+++

+

++

+++

−

=

Chapter 19, Solution 24. Since this is a reciprocal network, a Π network is appropriate, as shown below.

Y2

4 Ω

8 Ω 4 Ω

1/4 S

1/8 S 1/4 S

(a)

Y3 Y1

(b) (c)

S41

41

21

12111 =−=+= yyY , =1Z Ω4

S41

- 122 == yY , =2Z Ω4

S81

41

83

21223 =−=+= yyY , =3Z Ω8

Chapter 19, Solution 25. This is a reciprocal network and is shown below. 0.5 S 0.5S 1S Chapter 19, Solution 26. To get y and , consider the circuit in Fig. (a). 11 21y

4 Ω

+

Vx

−

2 Ω

2 Vx

1 I2

+ −

+

V2 = 0

−

1 Ω

2

V1

(a)

At node 1,

x1xx

xx1 -2

412

2VV

VVV

VV=→+=+

− (1)

But 5.15.122

2 1

1111

11x11 ==→=

+=

−=

VI

yVVVVV

I

Also, 1x2xx

2 -3.575.124

VVIVV

I ==→=+

-3.51

221 ==

VI

y

To get y and , consider the circuit in Fig.(b). 22 12y

4 Ω

+ −

+

Vx

−

2 Ω

2 Vx

1 I2

1 Ω

2

I1 V2

(b) At node 2,

42 x2

x2

VVVI

−+= (2)

At node 1,

x2xxxx2

x -23

1242 VVV

VVVVV =→=+=

−+ (3)


2xxx2 -1.55.121

2 VVVVI ==−=

-1.52

222 ==

VI

y

5.022

-

2

112

2x1 ==→==

VI

yVV

I

Thus,

=][y S5.1-5.3-5.05.1

Chapter 19, Solution 27. Consider the circuit in Fig. (a).

+

V2 = 0

−

(a)

20 I1 10 Ω 0.1 V2 − +

I1

+ −

4 Ω I2

V1

25.04

41

1

1

11111 ===→=

II

VI

yIV

55201

221112 ==→==

VI

yVII

Consider the circuit in Fig. (b).

4 Ω I1 I2

+

V1 = 0

−

− +

+ −

20 I1 10 Ω0.1 V2 V2

(b)

025.041.0

1.042

11221 ===→=

VI

yVI

6.06.01.05.010

202

222222

212 ==→=+=+=

VI

yVVVV

II

Thus,

=][y S6.05

025.025.0

Alternatively, from the given circuit,

211 1.04 VIV −=

212 1.020 VII += Comparing these with the equations for the h parameters show that

411 =h , -0.112 =h , 2021 =h , 1.022 =h Using Table 18.1,

25.0411

1111 ===

hy , 025.0

41.0-

11

1212 ===

hh

y

54

20

11

2121 ===

hh

y , 6.04

24.0

11

h22 =

+=

∆=

hy

as above. Chapter 19, Solution 28. We obtain and by considering the circuit in Fig.(a). 11y 21y

I2

+

V2 = 0

−

(a)

1 Ω

2 Ω

+

V1

−

4 Ω

6 ΩI1

4.34||61in =+=Z

2941.01

in1

111 ===

ZVI

y

11

12 346-

4.3106-

106-

VV

II =

==

-0.176534

6-

1

221 ===

VI

y

To get y and , consider the circuit in Fig. (b). 22 12y

1 Ω 4 ΩI1 Io

+

V2

−

+

V1 = 0

−

2 Ω6 Ω I2

(b)

2

2

22 IV

2434

)734(2)734)(2(

764||2)1||64(||21

==+

=

+=+=

y

7059.03424

22 ==y

o1 76-

II = 222o 347

4814

)734(22

VIII ==+

=

-0.176534

6-34

6-

2

11221 ===→=

VI

yVI

Thus,

=][y S0.70590.1765-0.1765-2941.0

The equivalent circuit is shown in Fig. (c). After transforming the current source to a voltage source, we have the circuit in Fig. (d).

6/34 S

18/34 S 4/34 S 1 A 2 Ω

(c)

8.5 Ω 5.667 Ω

1.889 Ω+ −

(d)

+

V

−

8.5 V 2 Ω

5454.0167.149714.0

)5.8)(9714.0(667.55.8889.1||2)5.8)(889.1||2(

V =+

=++

=

===2

)5454.0(R

VP

22

W1487.0


(a) Transforming the ∆ subnetwork to Y gives the circuit in Fig. (a).

+

V2

−

+

V1

−

Vo1 Ω 1 Ω

2 Ω10 A -4 A

(a) It is easy to get the z parameters

22112 == zz , 32111 =+=z , 322 =z

54921122211z =−=−=∆ zzzz

22z

2211 5

3y

zy ==

∆= ,

52--

z

122112 =

∆==

zyy

Thus, the equivalent circuit is as shown in Fig. (b).

2/5 S I2I1

1/5 S+

V2

−

+

V1

−

1/5 S 10 A -4 A

(b)

21211 235052

53

10 VVVVI −=→−== (1)

21212 3-220-53

52-

-4 VVVVI +=→+==

2121 5.1105.110 VVVV +=→−= (2) Substituting (2) into (1),

=→−+= 222 25.43050 VVV V8

=+= 21 5.110 VV V22

(b) For direct circuit analysis, consider the circuit in Fig. (a). For the main non-reference node,

122

410 oo =→=− V

V

=+=→−

= o1o1 10

110 VV

VVV22

=−=→−

= 41

4- o2o2 VV

VVV8


(a) Convert to z parameters; then, convert to h parameters using Table 18.1. Ω=== 60211211 zzz , Ω= 10022z

24003600600021122211z =−=−=∆ zzzz

241002400

22

z11 ==

∆=

zh , 6.0

10060

22

1212 ===

zz

h

-0.6-

22

2121 ==

zz

h , 01.01

2222 ==

zh

Thus,

=][h

ΩS0.010.6-

6.024

(b) Similarly,

Ω= 3011z Ω=== 20222112 zzz

200400600z =−=∆

1020200

11 ==h 12020

12 ==h

-121 =h 05.0201

22 ==h

Thus,

=][h

ΩS0.051-

110

Chapter 19, Solution 31. We get h and by considering the circuit in Fig. (a). 11 21h

2 Ω 4 I1

1 Ω1 Ω

+

V1

−

V3 V4

(a)

2 Ω I2

I1

At node 1,

431433

1 2222

VVIVVV

I −=→−

+= (1)

At node 2,

14

24

143 V

IVV

=+−

431431 6-2163-8 VVIVVI +=→+= (2)

Adding (1) and (2),

1441 6.3518 IVVI =→=

1143 8.283 IIVV =−=

1131 8.3 IIVV =+=

Ω== 8.31

111 I

Vh

-3.6-3.61

-

1

2211

42 ==→==

II

hIV

I

To get h and h , refer to the circuit in Fig. (b). The dependent current source can be replaced by an open circuit since

22 12

04 1 =I .

1 Ω1 Ω 2 ΩI1 I2

+ −

+

V1

−

4 I1 = 0 2 Ω V2

(b)

4.052

1222

2

112221 ==→=

++=

VV

hVVV

S2.051

5122 2

222

222 ===→=

++=

VI

hVV

I

Thus,

=][h

ΩS0.23.6-

4.038


(a) We obtain and by referring to the circuit in Fig. (a). 11h 21h

1 s s I2

+

V1

−

+

V2 = 0

−

1/s I1

(a)

1211 1ss

s1s1

||ss1 IIV

+++=

++=

1ss

1s 21

111 +

++==IV

h


1s1-

1s-

s1ss1-

21

221

112 +

==→+

=+

=II

hI

II

To get h and h , refer to Fig. (b). 22 12

1 s s I1 = 0 I2

+ −

+

V1

−

1/s V2

(b)

1s1

1ss1ss1

22

1122

221 +

==→+

=+

=VV

hV

VV

1ss

s1s1

s1

s 22

22222 +

=+

==→

+=

VI

hIV

Thus,

=][h

++

++++

1ss

1s1-

1s1

1ss

1s

22

22

(b) To get g and g , refer to Fig. (c). 11 21

1 s s I1 I2 = 0

+ −

+

V2

−

1/s V1

(c)

1sss

s1s11

s1

s1 21

11111 ++

=++

==→

++=VI

gIV

1ss1

1sss1s1s1

21

2212

112 ++

==→++

=++

=VV

gV

VV

To get g and g , refer to Fig. (d). 22 12

I1

+

V2

−

+

V1 = 0

−

I2s 1

1/s

s

I2

(d)

222 s1s1s)1s(

s)1s(||s1

s IIV

+++

+=

++=

1ss1s

s 22

222 ++

++==

IV

g

1ss1-

1ss-

s1s1s1-

22

1122

221 ++

==→++

=++

=II

gI

II

Thus,

=][g

+++

+++

++++

1ss1s

s1ss

11ss

1-1ss

s

22

22

Chapter 19, Solution 33. To get h11 and h21, consider the circuit below.

4 j6Ω Ω I2 + -j3Ω + I1 V1 5 VΩ 2=0 - -

2821.1j0769.3IVh

6j9I)6j4(5I)6j4//(5V

1

111

111 +==

++

=+=

Also, 2564.0j3846.0IIhI

6j95I

1

22112 +−==→

+−=

To get h22 and h12, consider the circuit below. 4 j6Ω Ω I2 I1 + + -j3Ω + V1 5 VΩ 2

- -

2564.0j3846.06j9

5VVhV

6j95V

2

11221 −=

+==→

+=

2821.0j0769.0)6j9(3j

3j9)6j9//(3j

1VIh

I)6j9//(3jV 2

222

22+=

+−+

=+−

==→+−=

Thus,

++−−+

=2821.0j0769.02564.0j3846.02564.0j3846.02821.1j0769.3

]h[

Chapter 19, Solution 34. Refer to Fig. (a) to get h and . 11 21h

300 Ω

100 Ω

10 Ω 1

+

Vx

−

+

V1

−

(a)

10 Vx

50 Ω 2

+

V2 = 0

−

− +

I2

I1

At node 1,

x1xx

1 4300300

0100

VIVV

I =→−

+= (1)

11x 754

300IIV ==

But Ω==→=+= 8585101

1111x11 I

VhIVIV

At node 2,

111xxxx

2 75.1430075

575

300530050100

IIIVVVV

I =−=−=−+

=

75.141

221 ==

II

h


300 Ω

I2

+ −

I1 = 0 2 50 Ω

10 Vx − +

+

V1

−

+

Vx

−

1 10 Ω

100 Ω V2

(b)

At node 2,

x22x22

2 8094005010

400VVI

VVVI +=→

++=

But 4400

100 22x

VVV ==

Hence, 2222 29209400 VVVI =+=

S0725.040029

2

222 ===

VI

h

25.041

4 2

112

2x1 ===→==

VV

hV

VV

=][h

ΩS0725.075.14

25.085

To get g and g , refer to Fig. (c). 11 21

300 Ω

I1 I2 = 0

+ −

2 50 Ω

10 Vx − +

+

V2

−

+

Vx

−

1 10 Ω

100 ΩV1

(c) At node 1,

x1xxx

1 5.14350350

10100

VIVVV

I =→+

+= (2)

But x11x1

1 1010

VVIVV

I −=→−

=

or (3) 11x 10IVV −=


11111 5.144951455.14350 VIIVI =→−=

S02929.0495

5.14

1

111 ===

VI

g

At node 2,

xxx2 -8.42861035011

)50( VVVV =−

=

1111 4955.14)286.84(-8.4286286.84-8.4286 VVIV

+=+=

-5.96-5.961

22112 ==→=

VV

gVV

To get g and g , refer to Fig. (d). 22 12

300 Ω

I1

+

V1 = 0

−

Io Io

+

V2

−

50 Ω

10 Vx − +

+

Vx

−

10 Ω

100 Ω I2

(d)

091.9100||10 =

091.93005010 2x2

2 ++

+=

VVVI

x22 818.611818.7091.309 VVI += (4)

But 22x 02941.0091.309

091.9VVV == (5)


22 9091.309 VI =

Ω== 34.342

222 I

Vg

091.30934.34

091.30922

o

IVI ==

)091.309)(1.1(34.34-

110100- 2

o1

III ==

-0.1012

112 ==

II

g

Thus,

=][g

Ω34.345.96-

0.101-S02929.0

Chapter 19, Solution 35. To get h and h consider the circuit in Fig. (a). 11 21

I2

+

V2 = 0

−

4 Ω1 Ω

(a)

+

V1

−

1 : 2

I1

144

n4

Z 2R ===

Ω==→=+= 22)11(1

111111 I

VhIIV

-0.521-

2-NN-

1

221

1

2

2

1 ===→==II

hII


− +

1 : 2

+

V1

−

I21 Ω 4 ΩI1 = 0

V2

(b) Since , . 01 =I 02 =IHence, . 022 =h At the terminals of the transformer, we have and which are related as 1V 2V

5.021

2nNN

2

112

1

2

1

2 ===→===VV

hVV

Thus,

=][h

Ω05.0-5.02

Chapter 19, Solution 36. We replace the two-port by its equivalent circuit as shown below.

+

V2

−

100 Ω

2 I1

-2 I1 + −

+ −

+

V1

−

I1 4 Ω

3 V2

16 Ω I2

10 V 25 Ω

Ω= 2025||100

112 40)2)(20( IIV == (1)

032010- 21 =++ VI

111 140)40)(3(2010 III =+=

141

1 =I , 1440

2 =V

14136

316 211 =+= VIV

708-)2(

125100

12 =

= II

(a) ==13640

1

2

VV

2941.0

(b) =1

2

II

1.6-

(c) ==136

1

1

1

VI

S10353.7 -3×

(d) ==140

1

2

IV

Ω40


(a) We first obtain the h parameters. To get h and h refer to Fig. (a). 11 21

6 Ω 3 Ω I2

+

V2 = 0

−

+

V1

−

3 Ω6 ΩI1

(a)

26||3 =

Ω==→=+= 88)26(1

111111 I

VhIIV

32-

32-

636-

1

221112 ==→=

+=

II

hIII

To get h and h , refer to the circuit in Fig. (b). 22 12

6 Ω 3 ΩI1 = 0 I2

+ −

+

V1

−

3 Ω6 Ω V2

(b)

49

9||3 =

94

49

2

22222 ==→=

VI

hIV

32

32

366

2

112221 ==→=

+=

VV

hVVV

=][h

Ω

S94

32-

32

8

The equivalent circuit of the given circuit is shown in Fig. (c).

8 Ω I1 I2

-2/3 I1 +

V2

−

2/3 V2 9/4 Ω + −

+ − 10 V 5 Ω

(c)

1032

8 21 =+ VI (1)

1112 2930

2945

32

49

||532

IIIV =

=

=

21 3029

VI = (2)


1032

3029

)8( 22 =+

VV

==252300

2V V19.1

(b) By direct analysis, refer to Fig.(d).

6 Ω 3 Ω

3 Ω

+

V2

−

+ − 6 Ω10 V 5 Ω

(d)

Transform the 10-V voltage source to a 6

10-A current source. Since

, we combine the two 6-Ω resistors in parallel and transform

the current source back to

Ω= 36||6

V536

10=× voltage source shown in Fig. (e).

3 Ω 3 Ω

+

V2

−

+ −

5 V 3 || 5 Ω

(e)

815

8)5)(3(

5||3 ==

==+

=6375

)5(8156

8152V V19.1

Chapter 19, Solution 38. We replace the two-port by its equivalent circuit as shown below.

50 I1 +

V2

−

200 kΩ + −

+ −

800 Ω

10-4 V2

+

V1

−

I1 200 Ω I2

10 V 50 kΩ

1

sin I

VZ = , Ω= k4050||200

1

6312 )10-2()1040(-50 IIV ×=×=

For the left loop,

12

-4s

100010

IVV

=−

11

6-4s 1000)10-2(10 IIV =×−

111s 8002001000 IIIV =−=

==1

sin I

VZ Ω800

Alternatively,

L22

L211211sin 1 Zh

ZhhhZZ

+−+=

=××+

×−+=

)1050)(105.0(1)1050)(50)(10(

800200 35-

3-4

inZ Ω800

Chapter 19, Solution 39. To get g11 and g21, consider the circuit below which is partly obtained by converting the delta to wye subnetwork. I1 R1 R2 I2 + + R3 V2 V1 10Ω - -

2.320

8x8R,R6.1488

8x4R 321 ====++

=

8919.0VVgV8919.0V

6.12.132.13V

1

221112 ==→=

+=

06757.08.14

1VIgI8.14)102.36.1(IV1

111111 ===→=++=

To get g22 and g12, consider the circuit below. 1.6Ω 1.6Ω I1 + V2 V1=0 I2 13.2Ω -

8919.0IIgI8919.0I

6.12.132.13I

2

112221 −==→−=

+−=

027.3IVgI027.3)6.1//2.136.1(IV

2

222222 ==→=+=

−=

027.38919.08919.006757.0

]g[

Chapter 19, Solution 40. To get g and g , consider the circuit in Fig. (a). 11 21

I2 = 0j10 Ω

12 Ω

-j6 ΩI1

+ −

+

V2

−

(a)

V1

S0333.0j0667.06j12

1)6j12(

1

11111 +=

−==→−=

VI

gIV

4.0j8.0j2

2)6j12(

12

1

1

1

221 +=

−=

−==

II

VV

g

To get g and g , consider the circuit in Fig. (b). 12 22

I1 -j6 Ω

12 Ω

j10 Ω

+

V1 = 0

−

I2

I2

(b)

4.0j8.0--j6-12

12-j6-12

12-21

2

11221 −====→= g

II

gII

22 -j6)||1210j( IV +=

Ω+=+== 2.5j4.2j6-12-j6))(12(

10j2

222 I

Vg

=][g

Ω++

−+2.5j4.24.0j8.0

4.0j8.0-S0333.0j0667.0


For the g parameters 2121111 IgVgI += (1)

2221212 IgVgV += (2) But and s1s1 ZIVV −=

222121L22 - IgVgZIV +==

2L22121 )(0 IZgVg ++=

or 221

L221

)(-I

gZg

V+

=


221

122111L11221 -

)(I

ggggZgg

I−+

=

or gL11

21

1

2

Zgg-

II

∆+=

Also, 222s1s212 )( IgZIVgV +−=

2221s21s21 IgIZgVg +−= 2222gL11ss21 )( IgIZgZVg +∆++=

But L

22

-ZV

I =

+∆+−=

L

222sgLs11s212 ][

ZV

gZZZgVgV

s21L

22sgLs11L2 ][Vg

ZgZZZgZV

=+∆++

22sgLs11L

L21

s

2

gZZZgZZg

VV

+∆++=

22s1221s2211Ls11L

L21

s

2

gZggZggZZgZZg

VV

+−++=

=s

2

VV

s2112L22s11

L21

Zgg)Zg)(Zg1(Zg

−++


(a) The network is shown in Fig. (a).

20 ΩI1 I2

100 Ω 0.5 I1 +

V2

−

+

V1

−

-0.5 I2+ −

(a)

(b) The network is shown in Fig. (b).

s 2 ΩI1 I2

10 Ω 12 V1 +

V2

−

+

V1

−

+ −

(b)


(a) To find and , consider the network in Fig. (a). A C

+

V2

−

I2I1

+ −

Z

V1

(a)

12

121 ==→=

VV

AVV

002

11 ==→=

VI

CI

To get B and , consider the circuit in Fig. (b). D

+

V2 = 0

−

I2I1

+ −

Z

V1

(b)

11 IZV = , 12 -II =

ZIIZ

IV

B ===1

1

2

1

---

1-

2

1 ==II

D

Hence,

=][T

10Z1

(b) To find and , consider the circuit in Fig. (c). A C

I1 I2

Z +

V2

−

+ −

V1

(c)

12

121 ==→=

VV

AVV

YZV

ICVIZV ===→==

1

2

1211

To get B and , refer to the circuit in Fig.(d). D

I2

+

V1

−

Y +

V2 = 0

−

I1

(d)

021 == VV 12 -II =

0-

2

1 ==IV

B , 1-

2

1 ==II

D

Thus,

=][T

1Y01

Chapter 19, Solution 44. To determine and , consider the circuit in Fig.(a). A C

j15 Ω

I1

+

V2

−

(a)

+ − 20 Ω

-j20 Ω

Io'

I2 = 0

Io

-j10 Ω

Io

V1

11 ])20j15j(||)10j-(20[ IV −+=

111 310

j20j15-

-j10)(-j5)(20 IIV

−=

+=

1'

o II =

11o 32

5j10j-10j- III

=

−

=

'20-j20)( oo2 IIV += = 11 I20I340j +− = 1I

340j20

−

=

−

−==

1

1

2

1

I340j20

)310j20( IVVA 0.7692 + j0.3461

=−

==

340j20

1

2

1

VIC 0.03461 + j0.023

To find and , consider the circuit in Fig. (b). B D

j15 Ω

+

V2 = 0

−

I2I1 -j10 Ω

+ −

-j20 Ω

20 ΩV1

(b) We may transform the ∆ subnetwork to a T as shown in Fig. (c).

10j20j10j15j

-j10))(15j(1 =

−−=Z

340

-jj15-

)-j10)(-j20(2 ==Z

20jj15--j20))(15j(

3 ==Z

j10 Ω j20 ΩI1 I2

+

V2 = 0

−

+ − 20 – j40/3 ΩV1

(c)

112 j32j3

20j340j20340j20

- III+−

=+−

−=

6923.0j5385.02j3j3-

2

1 +=−+

==II

D

11 20j340j20)340j20)(20j(

10j IV

+−−

+=

)18j24(j])7j9(210j[ 111 −=++= IIV

j55)-15(136

j3j2)-(3-

)18j24(j--

1

1

2

1 +=

+

−==

I

IIV

B

Ω+= j25.385-6.923B

=][T

++

Ω++j0.69230.5385Sj0.0230.03461

j25.3856.923-3461.0j7692.0

Chapter 19, Solution 45. To obtain A and C, consider the circuit below. I1 sL 1/sC I2 =0 + + V1 R1 V2 - - R2

1

21

2

11

21

12 R

sLRRVVAV

sLRRRV ++

==→++

=

12

1112 R

1VICRIV ==→=

To obtain B and D, consider the circuit below.

I1 sL 1/sC I2 + + V1 R1 V2=0 - - R2

CsRCsR1

IIDI

CsR1CsRI

sC1R

RI1

1

2

11

1

11

1

12

+=−=→

+−=

+−=

[ ]2

C1

1

1

1211

1

1

21 IsR

)CsR1(CsR1

R)sLR)(CsR1(I

sC1R

sCR

sLRV ++

+++−=

+++=

[ ])sLR)(CsR1(RCsR

1IVB 211

12

1 +++=−=

Chapter 19, Solution 46. To get and C , refer to the circuit in Fig.(a). A

+

V2

−

I1

+ −

1 Ω

2 Ω 4 Ix

1 Ω

Ix

2 I2 = 01

+

Vo

−

V1

(a)At node 1,

2o12oo

1 23212

VVIVVV

I −=→−

+= (1)

At node 2,

2ooo

x2o -2

24

41

VVVV

IVV

=→===−

(2)

From (1) and (2),

S-2.525-

-522

121 ===→=

VI

CVI

But 21o1

1 1VV

VVI +=

−=

21212 -3.52.5- VVVVV =→+=

-3.52

1 ==VV

A


+

V2 = 0

−

+

Vo

−

Ix

+ −

1 Ω1 2 I2

4 Ix 2 Ω

1 Ω I1

V1 (b) At node 1,

o1oo

1 3212

VIVV

I =→+= (3)

At node 2,

041 x

o2 =++ I

VI

– I (4) o2oo2 -302 VIVV =→=+=

Adding (3) and (4),

2121 -0.502 IIII =→=+ (5)

5.0-

2

1 ==II

D

But o11o1

1 1VIV

VVI +=→

−= (6)


2221 65-

31-

21-

IIIV =+=

Ω=== 8333.065-

2

1

IV

B

Thus,

=][T

Ω0.5-S2.5-

0.83333.5-

Chapter 19, Solution 47. To get A and C, consider the circuit below. 6Ω I1 1Ω 4Ω I2=0 + + + + Vx 2Ω 5Vx V2

V1 - - - -

x1xxxx1 V1.1V

10V5V

2V

1VV

=→−

+=−

3235.04.3/1.1VVAV4.3V5)V4.0(4V

2

1xxx2 ===→=+−=

02941.04.3/1.0VICV1.0VV1.1

1VVI

2

1xxx

x11 ===→=−=

−=


(a) Refer to the circuit below.

I2I1

+ −

[ T ]+

V2

−

V1 ZL

221 304 IVV −= (1)

221 1.0 IVI −= (2) When the output terminals are shorted, 02 =V . So, (1) and (2) become

21 -30IV = and 21 -II = Hence,

==1

1in I

VZ Ω30

(b) When the output terminals are open-circuited, 02 =I .

So, (1) and (2) become 21 4 VV =

21 1.0 VI = or 12 10IV =

11 40IV =

==1

1in I

VZ Ω40

(c) When the output port is terminated by a 10-Ω load, . 22 I-10V =

So, (1) and (2) become 2221 -7030-40 IIIV =−=

2221 -2- IIII =−=

11 35IV =

==1

1in I

VZ Ω35

Alternatively, we may use DZCBZA

+Z

+=

L

Lin

Chapter 19, Solution 49. To get and C , refer to the circuit in Fig.(a). A

1/s

+ − 1 Ω 1/s

(a)

1 Ω 1/s +

V2

−

I1 I2 = 0

V1

1s1

s11s1

s1

||1+

=+

=

12 s1||1s1s1||1

VV+

=

1s2s

1s1

s1

1s1

1

2

+=

++

+==VV

A

++

+

=

++

+

=)1s(s

1s2||

1s1

1s1

s1

||1s

1111 IIV

)1s3)(1s(1s2

)1s(s1s2

1s1

)1s(s1s2

1s1

1

1

+++

=

++

++

++

⋅

+

=IV

But s

1s221

+⋅= VV

Hence, )1s3)(1s(

1s2s

1s2

1

2

+++

=+

⋅IV

s)1s3)(1s(

1

2 ++==

IV

C


+

V2 = 0

−

I2

1 Ω 1/s

I1

+ −

1/s

1/s 1 Ω V1

(b)

1s2s21

||1s1

||s1

||1 1111 +

=

=

=I

IIV

1

1

2 12ss-

s1

1s1

1s1-

II

I+

=+

+

+=

s1

2s

1s2-

2

1 +=+

==II

D

s1-

s-s-1s2

1s21

2

1221 ==→=

+

+=

IV

BI

IV

Thus,

=][T

+++

+

s1

2s

)1s3)(1s(s1

1s22

Chapter 19, Solution 50. To get a and c, consider the circuit below.

I1=0 2 s I2

+ + V1 4/s V2

- -

21

22221 s25.01VVaV

4s4V

s/4ss/4V +==→

+=

+=

4ss25.0s

VIc

s/4sV)s25.01(

s/4sVI

or I)s/4s(V

2

3

1

212

22

22

+

+==→

++

=+

=

+=

To get b and d, consider the circuit below. I1 2 s I2

+ +

V1=0 4/s V2

- -

s5.01IId

2sI2I

s/42s/4I

1

2221 +=−=→

+−=

+−

=

2ss5.0IVbI

2)2s

2s)(4s2s(

I2s

)4s2s(I)s4//2s(V

2

1

21

2

22

22

++=−=→+

+++

−=

+++

=+=

++

++++

=1s5.0

4sss25.0

2ss5.01s25.0]t[

2

222

Chapter 19, Solution 51. To get a and c , consider the circuit in Fig. (a).

j2 Ω j Ω

I1 = 0 I2

+

V1

−

(a)

+ −

j Ω1 Ω -j3 Ω

V2

222 -j2)3jj( IIV =−=

21 -jIV =

2j-

j2-

2

2

1

2 ===II

VV

a

jj-

1

1

2 ===VI

c

To get b and d , consider the circuit in Fig. (b).

+

V1 = 0

−

I1

j2 Ω j Ω

I2

+ −

1 Ω -j3 Ωj Ω

V2

(b) For mesh 1,

21 j)2j1(0 II −+=

or j2j

2j1

1

2 −=+

=II

j-2-

1

2 +==II

d

For mesh 2, 122 j)3jj( IIV −−=

1112 )5j-2(j)2j-)(j2( IIIV −=−−=

5j2-

1

2 +==IV

b

Thus,

=][t

++

j2-j5j22

Chapter 19, Solution 52. It is easy to find the z parameters and then transform these to h parameters and T parameters.

+

+=

322

221

RRRRRR

][z

223221z R)RR)(RR( −++=∆

133221 RRRRRR ++=

(a)

++

++++

=

∆

=

3232

2

32

2

32

133221

2222

21

22

12

22

z

RR1

RRR-

RRR

RRRRRRRR

1-][

zzz

zz

zh

Thus,

32

32111 RR

RRRh

++= , 21

32

212 h-

RRR

=+

=h , 32

22 RR1+

=h

as required.

(b)

+

+++

=

∆

=

2

32

2

2

133221

2

21

21

22

21

21

z

21

11

RRR

R1

RRRRRRR

RRR

1][

zz

z

zzz

T

Hence,

2

1

RR

1A += , )RR(RR

RB 322

13 ++= ,

2R1

C = , 2

3

RR

1D +=

as required. Chapter 19, Solution 53.

For the z parameters, 2121111 IzIzV += (1)

2221122 IzIzV += (2) For ABCD parameters,

221 IBVAV −= (3)

221 IDVCI −= (4) From (4),

21

2 ICD

CI

V += (5)

Comparing (2) and (5),

Cz

121 = ,

CD

z =22


211 IBC

ADI

CA

V

−+=

21 IC

BCADICA −

+= (6)


CA

z =11 CC

BCADz T

12

∆=

−=

Thus,

[Z] =

∆

CD

C1

CCA T


For the y parameters 2121111 VyVyI += (1)

2221212 VyVyI += (2) From (2),

221

22

21

21 V

yy

yI

V −=

or 221

212

221

1-I

yV

yy

V += (3)


221

112122

21

22111

-I

yy

VyVy

yyI ++=

or 221

112

21

y1

-I

yy

Vy

I +∆

= (4)

Comparing (3) and (4) with the following equations

221 IBVAV −=

221 IDVCI −= clearly shows that

21

22

yy-

A = , 21y1-

B = , 21

y

y-

C∆

= , 21

11

yy-

D =


For the z parameters 2121111 IzIzV += (1)

2221212 IzIzV += (2) From (1),

211

121

111

1I

zz

Vz

I −= (3)


211

1221221

11

212 I

zzz

zVzz

V

−+=

or 211

z1

11

212 I

zV

zz

V∆

+= (4)


2121111 IgVgI +=

2221212 IgVgV += indicates that

1111 z

1g = ,

11

1212 z

z-=g ,

11

2121 z

z=g ,

11

z22 z

∆=g


(a) 5j74j)j3)(j2(y +=+−+=∆

Ω

−−−−−

=

∆∆−∆−∆

=0405.0j2568.00676.0j0946.03784.0j2703.02973.0j2162.0

/y/y/y/y

]z[y11y21

y12y22

(b)

++−−−−

=

∆−

=6.0j8.32.0j4.06.1j8.02.0j4.0

y/y/yy/yy/1

]h11y1121

111211[

(c )

++−

+−=

−∆−−−

=75.0j25.075.1j25.1

25.0j5.0j25.0y/yy/

y/1y/y]t

122212y

121211[


1)1)(20()7)(3(T =−=∆

=

∆

=

CD

C

CCA

z 1][T

Ω

7113

=

∆

=

BA

B

BBD

y 1-

-

][T

S

203

201-

201-

207

=

∆

=

DC

D

DDB

h 1-][T

Ω

S71

71-

71

720

=

∆

=

AB

A

AAC

g 1

-

][T

Ω3

2031

31-

S31

=

∆∆

∆∆=

TT

TT][ AC

BD

t

Ω3S1

207


The given set of equations is for the h parameters.

Ω=

S0.42-21

][h 4.4-2))(2()4.0)(1(h =−=∆

(a) =

∆=

11

h

11

21

11

12

11

-1

][

hhh

hh

hy S

4.42-2-1

(b) =

∆

=

2121

22

21

11

21

h

1--

--

]

hhh

hh

hT[

Ω5.0S2.0

5.02.2


2.00.080.12-0.4)(2)()2)(06.0(g =+=−=∆

(a) =

∆=

11

g

11

21

11

12

11

-1

[

ggg

gg

gz] Ω

333.3333.3667.6667.16

(b) =

∆

=

2222

21

22

12

22

g

1-][

ggg

gg

gy S0.50.1-0.2-1.0

(c) =

∆∆

∆∆=

g

11

g

21

g

12

g

22

-

-

][ gg

gg

h

ΩS0.31-

210

(d) =

∆=

21

g

21

11

21

22

21

1

][

ggg

gg

gT

Ω1S3.0

105


28.002.03.021122211y =−=−=∆ yyyy

(a) =

∆∆

∆∆=

y

11

y

21

y

12

y

22

-

-

][ yy

yy

z Ω

143.23571.0

7143.0786.1

(b) =

∆=

11

y

11

21

11

12

11

-1

][

yyy

yy

yh

ΩS0.46670.1667-

3333.0667.1

(c) =

∆=

12

22

12

y

1212

11

--

1--

][

yy

y

yyy

t

Ω5.2S4.1

53


(a) To obtain and , consider the circuit in Fig. (a). 11z 21z

1 ΩIo

+

V1

−

1 Ω 1 Ω I2 = 0

+

V2

−

1 ΩI1

(a)

1111 35

32

1])11(||11[ IIIV =

+=++=

35

1

111 ==

IV

z

11o 31

211

III =+

=

0- 1o2 =++ IIV

1112 34

31

IIIV =+=

34

1

221 ==

IV

z

To obtain and , consider the circuit in Fig. (b). 22z 12z

1 Ω

+

V2

−

I1

+

V1

−

1 Ω 1 Ω

1 Ω I2

(b) Due to symmetry, this is similar to the circuit in Fig. (a).

35

1122 == zz , 34

1221 == zz

=][z Ω

35

34

34

35

(b) =

∆

=

2222

21

22

12

22

z

1-][

zzz

zz

zh

Ω

S53

54-

54

53

(c) =

∆

=

21

22

21

21

z

21

11

1][

zz

z

zzz

T

Ω

45

S43

43

45


20 kΩ

a 40 kΩ10 kΩ

Ib

b

I1

50 kΩ

30 kΩ

+

V1

−

+−

+

V2

−

I2

Since no current enters the input terminals of the op amp,

13

1 10)3010( IV ×+= (1)

But 11ba 43

4030

VVVV ===

133b

b 10803

1020V

VI

×=

×=

which is the same current that flows through the 50-kΩ resistor. Thus, b

32

32 10)2050(1040 IIV ×++×=

133

23

2 10803

10701040 VIV×

⋅×+×=

23

12 1040821

IVV ×+=

2

31

32 104010105 IIV ×+×= (2)

From (1) and (2),

=][z Ω

k

40105040

8

21122211z 1016×=−=∆ zzzz

=

∆

=

=

21

22

21

21

z

21

11

1][

zz

z

zzz

DCBA

T

µ

Ω381.0S52.9

k24.15381.0

Chapter 19, Solution 63. To get z11 and z21, consider the circuit below. I1 1:3 I2=0 •• + + + + V’1 V’2 9Ω V2 V1 4 - - - -

Ω

3n,1n9Z 2R ===

8.0IVzI

54I)Z//4(V

1

11111R1 ==→==

4.2IVzI)5/4(3nV'nV'VV

1

22111122 ==→====

To get z21 and z22, consider the circuit below. I1=0 1:3 I2 •• + + + + V’1 V’2 9Ω V2 V1 4 - - - -

Ω

3n,36)4(n'Z 2R ===

2.7IVzI

4536x9I)'Z//9(V

2

22222R2 ==→==

4.2IVzI4.2

3V

nVV

2

1212

221 ==→===

Thus,

Ω

=

2.74.24.28.0

]z[


Ω==ω

→µ kj-)10)(10(

j-Cj

1F 6-31

Consider the op amp circuit below.

40 kΩ

+

V1

−

1 +

V2

−

I2 I1

2

Vx20 kΩ 10 kΩ

-j kΩ

−+

At node 1,

100

j-20xxx1 −

+=− VVVV

x1 )20j3( VV += (1)

At node 2,

2x2x

41-

400

100

VVVV

=→−

=−

(2)

But 3x1

1 1020×−

=VV

I (3)


26-

16-

321

1 105.121050102025.0

VVVV

I ×+×=×

+= (4)


21 )20j3(41-

VV +=

or (5) 21 )5j75.0(0 VV ++= Comparing (4) and (5) with the following equations

2121111 VyVyI +=

2221212 VyVyI += indicates that and that 02 =I

=][y S5j75.01

105.121050 -6-6

+××

-6-6

y 10)250j65(10)5.12.25j5.77( ×+=×−+=∆

=

∆=

11

y

11

21

11

12

11

-1

][

yyy

yy

yh

+×Ω×

S5j3.11020.25-102

4

4

Chapter 19, Solution 65. The network consists of two two-ports in series. It is better to work with z parameters and then convert to y parameters.

For aN ,

=

2224

][ az

For bN ,

=

1112

][ bz

=+=

3336

][][][ ba zzz

9918z =−=∆

=

∆∆

∆∆=

z

11

z

21

z

12

z

22

-

-

][ zz

zz

y S

32

31-

31-

31

Chapter 19, Solution 66. Since we have two two-ports in series, it is better to convert the given y parameters to z parameters.

-6-3-321122211y 10200)1010)(102( ×=−××=−=∆ yyyy

Ω

Ω=

∆∆

∆∆=

10000500

-

-

][

y

11

y

21

y

12

y

22

a yy

yy

z

=

+

=

200100100600

100100100100

10000500

][z

i.e. 2121111 IzIzV +=

2221212 IzIzV += or (1) 211 100600 IIV +=

212 200100 IIV += (2) But, at the input port,

11s 60IVV += (3) and at the output port,

2o2 -300IVV == (4) From (2) and (4),

221 -300200100 III =+

21 -5II = (5) Substituting (1) and (5) into (3),

121s 60100600 IIIV ++=

22 100-5))(660( II += = (6) 2-3200I

From (4) and (6),

==2

2

2

o

3200-300-

II

VV

09375.0

Chapter 19, Solution 67. The y parameters for the upper network is

=

21-1-2

][y , 314y =−=∆

=

∆∆

∆∆=

32

31

31

32

-

-

][

y

11

y

21

y

12

y

22

a yy

yy

z

=

1111

][ bz

=+=

35343435

][][][ ba zzz

19

16925

z =−=∆

=

∆

=

21

22

21

21

z

21

11

1][

zz

z

zzz

T

Ω25.1S75.0

75.025.1


For the upper network , [ aN

=

42-2-4

]ay

and for the lower network , [ bN

=

211-2

]by

For the overall network,

=+=

63-3-6

][][][ ba yyy

27936y =−=∆

=

∆=

11

y

11

21

11

12

11

-1

][

yyy

yy

yh

Ω

S29

21-

21-

61

Chapter 19, Solution 69. We first determine the y parameters for the upper network . aNTo get y and , consider the circuit in Fig. (a). 11 21y

21

n = , s4

ns12R ==Z

111R1 s4s2

s4

2)2( IIIZV

+

=

+=+=

)2s(2s

1

111 +

==VI

y

2ss-

-2n

- 11

12 +

===V

II

I

2ss-

1

221 +

==VI

y

To get y and , consider the circuit in Fig. (b). 22 12y

2 : 1

+

V1 =0

−

+

V2

−

I1 2 Ω 1/s I2

I2

(b)

21

)2(41

)2)(n( 2'R =

==Z

222'

R2 s22s

21

s1

s1

IIIZV

+

=

+=

+=

2ss2

2

222 +

==VI

y

2221 2ss-

2ss2

21-

n- VVII

+

=

+

==

2ss-

2

112 +

==VI

y

++

++=

2ss2

2ss-

2ss-

)2s(2s

][ ay

For the lower network , we obtain y and by referring to the network in Fig. (c). bN 11 21y

2 I1 I2

+

V2 = 0

−

+ −

s V1

(c)

21

21

11111 ==→=

VI

yIV

21-

2-

-1

221

112 ==→==

VI

yV

II

To get y and , refer to the circuit in Fig. (d). 22 12y

2 I1 I2

+

V2

−

+

V1 = 0

−

s I2

(d)

s22s

2ss2

)2||s(2

222222

+==→

+==

VI

yIIV

2-

s22s

2ss-

2ss-

- 2221

VVII =

+

+

=+

⋅=

21-

2

112 ==

VI

y

+

=s2)2s(21-

21-21][ by

=+= ][][][ ba yyy

+++

++

++

++

)2s(s24s4s5

2)(s22)(3s-

2)(s22)(3s-

2s1s

2


We may obtain the g parameters from the given z parameters.

=

1052025

][ az , 150100250az =−=∆

=

30252550

][ bz , 8756251500bz =−=∆

∆=

11

z

11

21

11

12

11

zzz

zz-

z1

][g

=

60.20.8-04.0

][ ag ,

=

17.50.50.5-02.0

][ bg

=+= ][][][ ba ggg

Ω23.50.7

1.3-S06.0

Chapter 19, Solution 71. This is a parallel-series connection of two two-ports. We need to add their g parameters together and obtain z parameters from there. For the transformer,

2121 I2I,V21V −==

Comparing this with

221221 DICVI,BIAVV −=−=

shows that

=

2005.0

]T[ 1b

To get A and C for Tb2 , consider the circuit below. I1 4 IΩ 2 =0 + + 5Ω V1 V2 - 2Ω -

1211 I5V,I9V ==

2.05/1VIC,8.15/9

VVA

2

1

2

1 ======

Chapter 19, Solution 72. Consider the network shown below.

+

V1

−

Ia1 Ia2

+

V2

−

I2

+ Va2

+ Vb2

Ib2

+ Va1

Nb

Na

Ib1

+ Vb1

I1

2a1a1a 425 VIV += (1)

2aa12a 4- VII += (2)

2b1b1b 16 VIV += (3)

2bb12b 5.0- VII += (4)

1b1a1 VVV +=

2b2a2 VVV ==

2b2a2 III +=

1a1 II = Now, rewrite (1) to (4) in terms of and 1I 2V

211a 425 VIV += (5)

212a 4 - VII += (6)

21b1b 16 VIV += (7)

2b12b 5.0- VII += (8) Adding (5) and (7),

21b11 51625 VIIV ++= (9) Adding (6) and (8),

21b12 5.14- VIII +−= (10)

11a1b III == (11) Because the two networks and are independent, aN bN

212 5.15- VII += or (12) 212 6667.0333.3 IIV += Substituting (11) and (12) into (9),

2111 5.15

5.125

41 IIIV ++=

211 333.367.57 IIV += (13)


2121111 IzIzV +=

2221212 IzIzV += indicates that

=][z Ω

6667.0333.3333.367.57

Alternatively,

=

14-425

][ ah ,

=

0.51-116

][ bh

=+=

1.55-541

][][][ ba hhh 5.86255.61h =+=∆

=

∆

=

2222

21

22

12

22

h

1-][

hhh

hh

hz Ω

6667.0333.3333.367.57

as obtained previously.

Chapter 19, Solution 73. From Example 18.14 and the cascade two-ports,

==

2132

][][ ba TT

=

==

2132

2132

]][[][ ba TTT

Ω7S4

127

When the output is short-circuited, 02 =V and by definition

21 - IBV = , 21 - IDI = Hence,

===DB

IV

Z1

1in Ω

712

Chapter 19, Solution 74. From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are

=

101

][ a

ZT ,

=

1101

][ b ZT

Z

We partition the given circuit into six subcircuits similar to those in Figs. (a) and (b) as shown in Fig. (c) and obtain [ for each. ]T

(b)

Z

(a)

s

T5 T6 T3 T4T1 T2

1 1/s 1 1/s

s

=

1101

][ 1T , ,

=

10s1

][ 2T

=

1s01

][ 3T

][][ 24 TT = , ][][ 15 TT = , ][][ 36 TT =

==

1s01

1101

][][][][][][][][][][][ 4321654321 TTTTTTTTTTT

=

+

=

+ 11s

0110s1

][][][11s01

][][][][ 3214321 TTTTTTT

=

+++

11ss1ss

1s01

][][2

21 TT

=

++++++

1s1s2ss

s1ss10s1

][ 223

2

1T

=

+++++++++

1s1s2sss2s1s2s3ss

1101

223

3234

=][T

++++++++++++

1s2ss2s4s4s2ss2s1s2s3ss

23234

3234

Note that as expected. 1=−CDAB Chapter 19, Solution 75. (a) We convert [za] and [zb] to T-parameters. For Na, 162440 =−=∆ z .

=

∆=

25.125.042

z/zz/1z/z/z

]T[212221

21z2111a

For Nb, . 88880y =+=∆

−−−−

=

−∆−−−

=4445.05

y/yy/y/1y/y

]T[211121y

212122b

−−−−

==125.525.5617186

]T][T[]T[ ba

We convert this to y-parameters. .3BCADT −=−=∆

−=

−

∆−=

94.100588.01765.03015.0

B/AB/1B/B/D

]y[ T

(b) The equivalent z-parameters are

−

=

∆=

0911.00178.00533.03067.3

C/DC/1C/C/A

]z[ T

Consider the equivalent circuit below. I1 z11 z22 I2 + + + + ZL Vi z12 I2 z21 I1 Vo - - - -

212111i IzIzV += (1)

222121o IzIzV += (2)

But (3) Lo2L2o Z/VIZIV −=→−= From (2) and (3) ,

+=→−=

21L

22

21o1

L

o22121o zZ

zz1VI

ZVzIzV (4)


0051.0VV3.194

Zz

Zzzz

zz

VV

i

.o

L

12

L21

2211

21

11

o

i −=→−=−

+=

Chapter 19, Solution 76. To get z11 and z21, we open circuit the output port and let I1 = 1A so that

21

2211

1

111 V

IVz,V

IVz ====

The schematic is shown below. After it is saved and run, we obtain

122.1Vz,849.3Vz 221111 ==== Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that

22

2221

2

112 V

IVz,V

IVz ====


849.3Vz,122.1Vz 222112 ==== Thus,

Ω

=

849.3122.1122.1949.3

]z[

Chapter 19, Solution 77. We follow Example 19.15 except that this is an AC circuit. (a) We set V2 = 0 and I1 = 1 A. The schematic is shown below. In the AC Sweep Box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes

FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 3.163 E–.01 –1.616 E+02 FREQ VM($N_0001) VP($N_0001) 1.592 E–01 9.488 E–01 –1.616 E+02

From this we obtain

h11 = V1/1 = 0.9488∠–161.6°

h21 = I2/1 = 0.3163∠–161.6°.

(b) In this case, we set I1 = 0 and V2 = 1V. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes

FREQ VM($N_0001) VP($N_0001) 1.592 E–01 3.163 E–.01 1.842 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 9.488 E–01 –1.616 E+02

From this,

h12 = V1/1 = 0.3163∠18.42°

h21 = I2/1 = 0.9488∠–161.6°.

Thus, [h] =

°−∠°−∠°∠°−∠6.1619488.06.1613163.0

42.183163.06.1619488.0

Chapter 19, Solution 78 For h11 and h21, short-circuit the output port and let I1 = 1A. 6366.02/ == πωf . The schematic is shown below. When it is saved and run, the output file contains the following: FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 1.202E+00 1.463E+02 FREQ VM($N_0003) VP($N_0003) 6.366E-01 3.771E+00 -1.350E+02 From the output file, we obtain

o1

o2 135771.3V,3.146202.1I −∠=∠=

so that o2

21o1

11 3.146202.11Ih,135771.3

1Vh ∠==−∠==

For h12 and h22, open-circuit the input port and let V2 = 1V. The schematic is shown below. When it is saved and run, the output file includes: FREQ VM($N_0003) VP($N_0003) 6.366E-01 1.202E+00 -3.369E+01 FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 3.727E-01 -1.534E+02 From the output file, we obtain

o1

o2 69.33202.1V,4.1533727.0I −∠=−∠=

so that o2

22o1

12 4.1533727.01Ih,69.33202.1

1Vh −∠==−∠==

Thus,

−∠∠−∠−∠= o

oo

4.1533727.03.146202.169.33202.1135771.3]h[

Chapter 19, Solution 79 We follow Example 19.16. (a) We set I1 = 1 A and open-circuit the output-port so that I2 = 0. The schematic is shown below with two VPRINT1s to measure V1 and V2. In the AC Sweep box, we enter Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes

FREQ VM(1) VP(1) 3.183 E–01 4.669 E+00 –1.367 E+02 FREQ VM(4) VP(4) 3.183 E–01 2.530 E+00 –1.084 E+02

From this, z11 = V1/I1 = 4.669∠–136.7°/1 = 4.669∠–136.7° z21 = V2/I1 = 2.53∠–108.4°/1 = 2.53∠–108.4°.

(b) In this case, we let I2 = 1 A and open-circuit the input port. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes

FREQ VM(1) VP(1) 3.183 E–01 2.530 E+00 –1.084 E+02 FREQ VM(2) VP(2) 3.183 E–01 1.789 E+00 –1.534 E+02

From this, z12 = V1/I2 = 2.53∠–108.4°/1 = 2.53∠–108..4° z22 = V2/I2 = 1.789∠–153.4°/1 = 1.789∠–153.4°.

Thus,

[z] = 2

°−∠°−∠°−∠°−∠

4.153789.14.10853.4.10853.27.136669.4

Chapter 19, Solution 80 To get z11 and z21, we open circuit the output port and let I1 = 1A so that

21

2211

1

111 V

IVz,V

IVz ====


37.70Vz,88.29Vz 221111 −==== Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that

22

2221

2

112 V

IVz,V

IVz ====


11.11Vz,704.3Vz 222112 ==== Thus,

Ω

−

=11.1137.70

704.388.29]z[

Chapter 19, Solution 81 (a) We set V1 = 1 and short circuit the output port. The schematic is shown below. After simulation we obtain

y11 = I1 = 1.5, y21 = I2 = 3.5

(b) We set V2 = 1 and short-circuit the input port. The schematic is shown below. Upon simulating the circuit, we obtain

y12 = I1 = –0.5, y22 = I2 = 1.5

[Y] =

−5.15.35.05.1

Chapter 19, Solution 82 We follow Example 19.15. (a) Set V2 = 0 and I1 = 1A. The schematic is shown below. After simulation, we obtain

h11 = V1/1 = 3.8, h21 = I2/1 = 3.6

(b) Set V1 = 1 V and I1 = 0. The schematic is shown below. After simulation, we obtain

h12 = V1/1 = 0.4, h22 = I2/1 = 0.25

Hence, [h] =

25.06.34.08.3


To get A and C, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 11 and V2 = 34.

02941.0341

VIC,3235.0

VVA

2

1

2

1 =====

Similarly, to get B and D, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 2.5 and I2 = -2.125.

4706.0125.21

IID,1765.1

125.25.2

IVB

2

1

2

1 ==−===−=

Thus,

=

4706.002941.01765.13235.0

]T[


(a) Since A = 0I2

1

2VV

=

and C = 0I2

1

2VI

=

, we open-circuit the output port and let V1

= 1 V. The schematic is as shown below. After simulation, we obtain A = 1/V2 = 1/0.7143 = 1.4 C = I2/V2 = 1.0/0.7143 = 1.4

(b) To get B and D, we short-circuit the output port and let V1 = 1. The schematic is shown below. After simulating the circuit, we obtain B = –V1/I2 = –1/1.25 = –0.8

8

Thus =

D = –I1/I2 = –2.25/1.25 = –1.

DCBA

−−

8.14.18.04.1

lution 85

(a) Since A =

Chapter 19, So

0I2

1

2VV

=

and C = 0I2

1

2VI

=

, we let V1 = 1 V and open-

circuit the output port. The schematic is shown below. In the AC Sweep box, we set n

FREQ IM(V_PRINT1) IP(V_PRINT1) 01

REQ VM($N_0002) VP($N_0002)

From this, we obtain

A

Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtaian output file which includes

1.592 E– 6.325 E–01 1.843 E+01 F1.592 E–01 6.325 E–01 –7.159 E+01

= °∠=°−∠

= 59.71581.159.716325.0

11 V2

C = °∠=°−∠°∠

= 90159.716325.0

43.186325.0I1

V2

= j

(b) Similarly, since B =

0V2

1

2IV

= 0V2

1

2II

=

− , we let V1 = 1 V and short-

circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start

FREQ IM(V_PRINT1) IP(V_PRINT1) 01

REQ IM(V_PRINT3) IP(V_PRINT3) 01

From this,

B

Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get an output file which includes the following results:

and D =

1.592 E– 5.661 E–04 8.997 E+01 F1.592 E– 9.997 E–01 –9.003 E+01

= j901909997.0

11

2

−=°∠−=°−∠

−=− I

D =°−∠°∠

−=−−

909997.097.8910x661.5I 4

2

1

I = 5.561x10–4

=

DCBA

−°∠−410x661.5j

j59.71581.1


(a) By definition, g11 = 0I1

1

2VI

=

, g21 = 0I2

1

2VV

=

.

We let V1 = 1 V and open-circuit the output port. The schematic is shown below. After simulation, we obtain g11 = I1 = 2.7 g21 = V2 = 0.0

(b) Similarly,

g12 = I

, g22 = 0V2

1

1

I

= 0V2

2

1IV

=

g12 = I1 = 0

We let I2 = 1 A and short-circuit the input port. The schematic is shown below. After simulation,

g22 = V2 = 0

Thus [g] =

000S727.2

hapter 19, Solution 87

a =

C

(a) Since 0I1

2

1VV

=

and c = 0I1

2

1VI

=

,

we open-circuit the input port and let V2 = 1 V. The schematic is shown below. In the C Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After

IP(V_PRINT2) 1.592 E–01 5.000 E–01 1.800 E+02

1) ) .592 E–01 5.664 E–04 8.997 E+01

From this,

Asimulation, we obtain an output file which includes

FREQ IM(V_PRINT2)

FREQ VM($N_000 VP($N_00011

a = °= 97.891 −∠°∠− 1765

97.8910x664.5 4

c = °−∠−=°∠

°∠− 97.8928.882

97.8910x664.51805.04

(b) Similarly,

b = 0V1

2

1I

=

V− and d =

0V1

2

1II

=

−

We short-circuit the input port and let V2 = 1 V. The schematic is shown below. After simulation, we obtain an output file which includes

FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 5.000 E–01 1.800 E+02

FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 5.664 E–04 –9.010 E+01

From this, we get

b = °−∠

− = –j1765 − 1.9010x664.514

d = °−∠

°∠− = j888.28 − 1.9010x664.5

1805.04

Thus [t] =

−−2.888j2.888j

1765j1765j

Chapter 19, Solution 88 To get Z , consider the network in Fig. (a). in

I2I1

+ −

(a)Zin

+

V1

−

Vs Two-Port+

V2

−

RL

Rs

2121111 VyVyI += (1)

2221212 VyVyI += (2)

But 222121L

22 R

-VyVy

VI +==

L22

1212 R1

-+

=y

VyV (3)


+⋅+=

L22

121121111 R1

-y

VyyVyI ,

LL R

1=Y

1L22

L11y1 V

YyYy

I

+

+∆= , 21122211y yyyy −=∆

or ==1

1inZ

IV

L11y

L22

YyYy

+∆+

+

+=

+==

L22

121

1

22in21

1

222121

1

2i

-ZA

YyVy

Iyy

IVyVy

II

+

−

+∆+

=+

−L22

212221

L11y

L22

L22

in2122in21

ZZ

Yyyyy

YyYy

Yyyyy=

=iAL11y

L21

YyYy

+∆

From (3),

==1

2vA

VV

L22

21

Yyy-+

To get Z , consider the circuit in Fig. (b). out

Zout

I2I1

(b)

+ −

+

V1

−

Rs Two-Port V2

222121

2

2

2outZ

VyVyV

IV

+== (4)

But 1s1 R- IV = Substituting this into (1) yields

2121s111 R- VyIyI +=

2121s11 )R1( VyIy =+

s

1

s11

2121 R

-R1

Vy

VyI =

+=

or s11

s12

2

1

R1R-

yy

VV

+=

Substituting this into (4) gives

s11

s211222

out

R1R

1Z

yyy

y+

−=

s2221s221122

s11

RRR1

yyyyyy

−++

=

=outZs22y

s11

YyYy

+∆+


Lfereoeieie

Lfev R)hhhh(h

Rh-A

−+=

54-6-

5

v 10)72106.210162640(26401072-

A⋅××−××+

⋅=

=+⋅

=182426401072-

A5

v 1613-

=== )1613(log20Alog20gain dc v 15.64


(a) Loe

Lfereiein Rh1

RhhhZ

+−=

L6-

L-4

R10201R12010

20001500×+×

−=

L5-

-3

R10211012

500×+×

=

L

-3L

-2 R1012R10500 ×=+

L2 R2.010500 =×

=LR Ωk250

(b) Lfereoeieie

Lfev R)hhhh(h

Rh-A

−+=

34-6-

3

v 10250)1012010202000(200010250120-

A×××−××+

××=

=×+×

×= 33

6

v 1071021030-

A 3333-

=×××+

=+

= 36-Loe

fei 1025010201

120Rh1

hA 20

120101020)2000600(2000600

hhh)hR(hR

Z 4-6-fereoeies

iesout ×−××+

+=

−++

=

=Ω= k40

2600Zout Ωk65

(c) =××==→== 3-svc

s

c

b

cv 104-3333AA VV

VV

VV

V13.33-

Chapter 19, Solution 91 Ω= k2.1R s , Ω= k4R L

(a) Lfereoeieie

Lfev R)hhhh(h

Rh-A

−+=

34-6-

3

v 104)80105.110201200(120010480-

A××××−××+

××=

==124832000-

A v 25.64-

(b) =×××+

=+

= 36-Loe

fei 10410201

80Rh1

hA 074.74

(c) ireiein AhhZ −=

≅××−= 074.74105.11200Z -4in Ωk2.1

(d) fereoeies

iesout hhh)hR(

hRZ

−++

=

==××−××

+=

0468.02400

80105.11020240012001200

Z 4-6-out Ωk282.51

Chapter 19, Solution 92 Due to the resistor , we cannot use the formulas in section 18.9.1. We will need to derive our own. Consider the circuit in Fig. (a).

Ω= 240R E

hoe hfe Ib

hie

+ −

hre Vc

Zin

IE

+ −

Ib Ic

+

Vc

−RE

(a)

+

Vb

−

Rs

RL Vs

cbE III += (1)

Ecbcrebieb R)(hh IIVIV +++= (2)

oeE

cbfec

h1R

h+

+=VII (3)

But (4) Lcc R-IV = Substituting (4) into (3),

c

oeE

Lbfec

h1R

Rh III

+−=

or Loe

oeEfe

b

ci R(h1

)hR1(hA

++

==II

(5)

)240000,4(10301)10x30x2401(100A 6-

6

i +×++

=−

=iA 79.18

From (3) and (5),

oeE

cbfeb

ELoe

oeEfec

h1R

h)RR(h1

h)R1(h+

+=++

+=

VIII (6)


EccrebEieb Rh)Rh( IVIV +++=

EL

ccre

feELoe

oeEfe

oeE

Eiecb R

Rh

h)RR(h1

)hR1(hh1R

)Rh( VVVV −+

−

+++

+

+=

L

Ere

feELoe

oeEfe

oeE

Eie

c

b

v RRh

h)RR(h1

)hR1(hh1R

)Rh(A1

−+

−

+++

+

+==

VV

(7)

400024010

100424010301

)10x30x2401(10010x301240

)2404000(A1 4-

6-

6

6v

−+

−

××++

+

+=

−

−

-0.06606.01010x06.6A1 4-3

v

=−+−= −

=vA –15.15

From (5),

bLoe

fec Rh1

hII

+=

We substitute this with (4) into (2) to get

cLreEbEieb )RhR()Rh( IIV −++=

++

+−++= b

ELoe

oeEfeLreEbEieb )RR(h1

)hR1(h)RhR()Rh( IIV

)RR(h1)hR1)(RhR(h

RhZELoe

oeELreEfeEie

b

bin ++

+−++==

IV

(8)

424010301)10x30x2401)(10410240)(100(2404000Z 6-

63-4

in ××++×××

++=−

=inZ 12.818 kΩ

To obtain , which is the same as the Thevenin impedance at the output, we introduce a 1-V source as shown in Fig. (b).

outZ

hie Rs IcIb

(b)

+

Vb

−RE

+

Vc

−

+ −

IE

Zout

hre Vc

+ −

hfe Ib hoe

1 V

From the input loop, 0)(Rh)hR( cbEcreiesb =++++ IIVI

But 1c =VSo,

0Rh)RhR( cEreEiesb =++++ II (9) From the output loop,

bfeoeE

oebfe

oeE

cc h

1hRh

h

h1R

IIVI ++

=++

=

or oeE

fe

oe

fe

cb hR1

hh

h +−=

II (10)


0hR1

hh)hRR(

Rhh

)hRR(oeE

fe

oeieEs

cErefe

cieEs =

+

++

−++

++ II

refe

oe

oeE

ieEscEc

fe

ieEs hhh

hR1hRR

Rh

hRR−

+

++=+

++ II

feieEsE

reoeE

ieEsfeoe

c h)hRR(R

hhR1

hRR)hh(

+++

−

+

++

=I

fereoeoeE

ieEs

ieEsfeE

cout

hhhhR1

hRRhRRhR1Z

−

+

+++++

==I

10010103010x30x240140002401200

)40002401200(100240Z4-6-

6

out

×−××

+++

+++×=

−

=+

=152.0

544024000Zout 193.7 kΩ


We apply the same formulas derived in the previous problem.

L

Ere

feELoe

oeEfe

oeE

Eie

v RRh

h)RR(h1

)hR1(hh1R

)Rh(A1

−+

−

+++

+

+=

3800200105.2

15004.01

)002.01(150)10200(

)2002000(A1 4-

5v

−×+

−

++

+

+=

-0.0563805263.0105.2004.0A1 4-

v

=−×+−=

=vA –17.74

=+×+

+=

+++

=−

)3800200(101)10x2001(150

)RR(h1)hR1(h

A 5-

5

ELoe

oeEfei 5.144

)RR(h1)hR1)(RhR(h

RhZELoe

oeELreEfeEiein ++

+−++=

04.1)002.1)(108.3105.2200)(150(2002000Z

3-4

in×××−

++=

289662200Zin +=

=inZ 31.17 kΩ

fereoeoeE

ieEs

ieEsfeEout

hhhhR1

hRRhRRhR

Z−

+

+++++

=

0.0055-33200

150105.2002.1

10320020002001000150200Z

4-5-out =

××−

×+++×

=

=outZ –6.148 MΩ

Chapter 19, Solution 94 We first obtain the ABCD parameters.

Given ,

= 6-10100

0200][h -4

21122211h 102×=−=∆ hhhh

×=

∆

= 2-8-

6-

2121

22

21

11

21

h

10-10-2-102-

1--

-

][

hhh

hh

hT

The overall ABCD parameters for the amplifier are

××≅

×

×= 4-10-

-2-8

2-8-

-6

2-8-

-6

1010102102

10-10-2-102-

10-10-2-102-

][T

0102102 -12-12

T =×−×=∆

=

∆

= 6-4-

T

1010-0200

1-][

DC

D

DDB

h

Thus, , h200h ie = 0re = , h , h -4fe -10= -6

oe 10=

=××−×+

×= 34-

34

v 104)0102(200)104)(10(

A 5102×

=−=+

−= 0200Rh1Rhh

hZLoe

Lfereiein Ω200


Let s5s

8s10s13

24

22A +

++==

yZ

Using long division,

B13

2

A Lss5s8s5

s ZZ +=++

+=

i.e. and H1L1 = s5s8s5

3

2

B ++

=Z

as shown in Fig (a).

y22 = 1/ZA

L1

ZB

(a)

8s5s5s1

2

3

BB +

+==

ZY

Using long division,

C22B sC8s5

s4.3s2.0 YY +=

++=

where and F2.0C2 = 8s5s4.3

2C +=Y

as shown in Fig. (b).

43

2

CC Cs

1Ls

s4.38

4.3s5

s4.38s51

+=+=+

==Y

Z

i.e. an inductor in series with a capacitor

H471.14.3

5L3 == and F425.0

84.3

C4 ==

Thus, the LC network is shown in Fig. (c).

Chapter 19, Solution 96 This is a fourth order network which can be realized with the network shown in Fig. (a).

1.471 H0.425 F

L1

1 Ω

L3

C4C2

(c)

0.2 F

1 H

C2

(b)

Yc = 1/ZC

L1

(a))s613.2s613.2()1s414.3s()s( 324 ++++=∆

s613.2s613.21s414.3s

1

s613.2s613.21

)s(H

3

24

3

+++

+

+=

which indicates that

s613.22.613s1-

321 +=y

s613.2s613.21s414.3s

3

4

22 +++

=y

We seek to realize . By long division, 22y

A43

2

22 Css613.2s613.2

1s414.2s383.0 Yy +=

++

+=

i.e. and F383.0C4 = s613.2s613.21s414.2

3

2

A ++

=Y

as shown in Fig. (b).

L1

C4C2

L3

YA

y22(b)

1s414.2s613.2s613.21

2

3

AA +

+==

YZ

By long division,

B32A Ls1s414.2

s531.1s082.1 ZZ +=

++=

i.e. and H082.1L3 = 1s414.2s531.1

2B +=Z

as shown in Fig.(c).

L1

C4C2

L3

ZB

(c)

12

BB Ls

1Cs

s531.11

s577.11

+=+==Z

Y

i.e. and F577.1C2 = H531.1L1 = Thus, the network is shown in Fig. (d).

1.531 H 1.082 H

1.577 F 0.383 F 1 Ω

(d) Chapter 19, Solution 97

s12s24s6

1

s12ss

)24s6()s12s(s

)s(H

3

2

3

3

23

3

++

+

+=+++

=

Hence,

A3

3

2

22 Cs1

s12s24s6

Zy +=++

= (1)

where is shown in the figure below. AZ

C1 C3

L2

ZA y22 We now obtain C and using partial fraction expansion. 3 AZ

Let 12sCBs

sA

)12s(s24s6

22

2

++

+=++

CsBs)12s(A24s6 222 +++=+

Equating coefficients : 2 0s : AA1224 =→= : 1s C0 = 2s : 4BBA6 =→+=Thus,

12ss4

s2

)12s(s24s6

22

2

++=

++

(2)


F21

A1

C3 ==

s3

s41

s412s1 2

A+=

+=

Z (3)

But 2

1A Ls

1sC

1+=

Z (4)


F41

C1 = and H31

2 =L

Therefore, =1C F25.0 , =2L H3333.0 , =3C F5.0


2.08.01h =−=∆

−−−−

=

−−−∆−

== − 005.010x5.210001.0

h/1h/hh/hh/

]T[]T[ 6212122

211121hba

== −−

−

58

5ba

10x510x5.106.010x6.2]T][T[]T[

We now convert this to z-parameters

=

∆= 37

3T

10x33.310x667.60267.010x733.1

C/DC/1C/C/A

]z[

1000 I1 z11 z22 I2 + + + + ZL Vs z12 I2 z21 I1 Vo - - - -

212111s IzI)z1000(V ++= (1)

121222o IzIzV += (2)

But (3) Lo2L2o Z/VIZIV −=→−= Substituting (3) into (2) gives

+=

L21

22

21o1 Zz

zz1VI (4)

We substitute (3) and (4) into (1)

V74410x136.210x653.7

VZzV

Zzz

z1)z1000(V

54

oL

12o

L21

22

1111s

µ=−=

−

++=

−−


)(|| abc31ab ZZZZZZ +=+=

cba

bac31

)(ZZZ

ZZZZZ

+++

=+ (1)

)(|| cba32cd ZZZZZZ +=+=

cba

cba32

)(ZZZ

ZZZZZ

+++

=+ (2)

)(|| cab21ac ZZZZZZ +=+=

cba

cab21

)(ZZZ

ZZZZZ

+++

=+ (3)


cba

acb21

)(ZZZ

ZZZZZ

++−

=− (4)

Adding (3) and (4),

=1Zcba

cb

ZZZZZ++

(5)


=2Zcba

ba

ZZZZZ++

(6)


=3Zcba

ac

ZZZZZ++

(7)

Using (5) to (7)

2cba

cbacba133221 )(

)(ZZZ

ZZZZZZZZZZZZ

++++

=++

cba

cba133221 ZZZ

ZZZZZZZZZ

++=++ (8)

Dividing (8) by each of (5), (6), and (7),

=aZ1

133221

ZZZZZZZ ++

=bZ3

133221

ZZZZZZZ ++

=cZ2

133221

ZZZZZZZ ++

as required. Note that the formulas above are not exactly the same as those in Chapter 9 because the locations of and are interchanged in Fig. 18.122. bZ cZ

solucionario sadiku edición 3

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