chapter 5 - circuit theorems

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Problems Section 5-2: Source Transformations P5.2-1 (a)

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Introdução Aos Circuitos Elétricos - 7th ed - Dorf Svoboda - Resolução - Capitulo 5

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Page 1: Chapter 5 - Circuit Theorems

Problems

Section 5-2: Source Transformations P5.2-1 (a)

Page 2: Chapter 5 - Circuit Theorems

= 2

= 0.5 Vt

t

Rv

∴ Ω−

(b) 9 4 2 ( 0.5) 09 ( 0.5) 1.58 A

4 2

i i

i

− − − + − =− + −

= = −+

9 4 9 4( 1.58) 2.67 Vv i= + = + − = (c) 1.58 Aai i= = −

(checked using LNAP 8/15/02) P5.2-2

Finally, apply KVL: 1610 3 4 0 2.19 A3a a ai i i− + + − = ∴ =

(checked using LNAP 8/15/02)

Page 3: Chapter 5 - Circuit Theorems

P5.2-3

Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:

Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top:

Source transformation at left; series resistors at right:

Parallel resistors, then source transformation at left:

Page 4: Chapter 5 - Circuit Theorems

Finally, apply KVL to loop

o6 (9 19) 36 0i v− + + − − =

o5 / 2 42 28 (5 / 2) 28 Vi v= ⇒ = − + =

(checked using LNAP 8/15/02) P5.2-4

4 2000 4000 10 2000 3 0 375 A

a a a

a

i i ii μ

− − − + − − =∴ =

(checked using LNAP 8/15/02)

Page 5: Chapter 5 - Circuit Theorems

P5.2-5

12 6 24 3 3 0 1 Aa a ai i i− − + − − = ⇒ =

(checked using LNAP 8/15/02) P5.2-6 A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor:

Source transformations on both the right side and the left side of the circuit:

Page 6: Chapter 5 - Circuit Theorems

Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source:

Finally, ( ) ( ) ( )50 100 1000.21 0.21 7 V50 100 3av = =

+=

(checked using LNAP 8/15/02)

P5.2-7 Use source transformations to simplify the circuit:

Page 7: Chapter 5 - Circuit Theorems

Label the node voltages. The 8-V source is connected between nodes 1 and 3. Consequently,

1 3 8v v− =

Apply KCL to the supernode corresponding to the 8-V source to get

1 2 31 2 3

24 100 0.125 0.3 0.05 0.02 0.2 0

8 20 50v v v

v v v− −

+ + = ⇒ − + + − =

Apply KCL at node 2 to get

1 2 2 2 31 2 3 =0.04 0.19 0.1 0

25 20 10v v v v v

v v v− −

= + ⇒ + − =

Solving, for example using MATLAB

1 1

2 2

3 3

1 0 1 8 4.78730.125 0.05 0.02 0.5 0.6831

0.04 0.19 0.1 0 3.2127

v vv vv v

⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢= ⇒ = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢− − −⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦ ⎣ ⎦

⎤⎥⎥⎥⎦

The power supplied by the 8-V source is

( )4.7873 0.6831 4.7873 248 4.316 W25 8− −⎛ ⎞−

+ =⎜ ⎟⎝ ⎠

Apply KCL at node 4 of the original circuit to get

( )3 4 4 34

2 30 2 3.2127 300.5 4.71 V

30 20 5 5v v v v

v− + − +

+ = ⇒ = = =

The power supplied by the 0.5 A source is

( )0.5 4.71 2.355 W=

(checked: LNAP 5/31/04)

Page 8: Chapter 5 - Circuit Theorems

P5.2-8 Replace series and parallel resistors by an equivalent resistor.

( )18 12 24 12 + = Ω

Do a source transformation, then replace series voltage sources by an equivalent voltage source.

Do two more source transformations Now current division gives

8 238 8

iR R

⎛ ⎞= =⎜ ⎟+ +⎝ ⎠4

Then Ohm’s Law gives 248

Rv RiR

= =+

( )

( )

24a 2 A8 4

24 8(b) 12 V

8 8

24(c) 1 16 8

24(d) 16 16 8

i

v

RR

R RR

= =+

= =+

= ⇒ =+

= ⇒ =+

Ω

Ω

(checked: LNAP 6/9/04)

Page 9: Chapter 5 - Circuit Theorems

P5.2-9 Use source transformations and equivalent resistances to reduce the circuit as follows

The power supplied by the current source is given by

( )23.1 2 10.3125 2 87.45 Wp = + =⎡ ⎤⎣ ⎦

Page 10: Chapter 5 - Circuit Theorems
Page 11: Chapter 5 - Circuit Theorems

Section 5-3 Superposition P5.3–1 Consider 6 A source only (open 9 A source)

Use current division:

11

15 6 40 V20 15 30v v⎡ ⎤= ⇒⎢ ⎥+⎣ ⎦

=

Consider 9 A source only (open 6 A source)

Use current division:

22

10 9 40 V20 10 35v v⎡ ⎤= ⇒⎢ ⎥+⎣ ⎦

=

1 2 40 40 80 Vv v v∴ = + = + =

(checked using LNAP 8/15/02) P5.3-2 Consider 12 V source only (open both current sources)

KVL:

1 1 1

1

20 12 4 12 0 1/ 3 mA

i i ii+ + + =

⇒ = −

Consider 3 mA source only (short 12 V and open 9 mA sources)

Current Division:

216 43 mA

16 20 3i ⎡ ⎤= =⎢ ⎥+⎣ ⎦

Consider 9 mA source only (short 12 V and open 3 mA sources)

Page 12: Chapter 5 - Circuit Theorems

Current Division:

3129 3

24 12i ⎡ ⎤= − = −⎢ ⎥+⎣ ⎦

mA

1 2 3 1/ 3 4 / 3 3 2 mAi i i i∴ = + + = − + − = −

(checked using LNAP 8/15/02)

P5.3–3 Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to the 30 mA current source.

12 6 30 6 mA 2 mA

2 8 6 12a ai i i⎛ ⎞ ⎛ ⎞= = ⇒ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i due to the 15 mA current source.

24 6 15 6 mA 2 mA

4 6 6 12b bi i i⎛ ⎞ ⎛ ⎞= = ⇒ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V voltage source.

Page 13: Chapter 5 - Circuit Theorems

( )36 || 6 3 2.5 10 0.5 mA

6 || 6 12 3 12i

⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

Finally, 1 2 3 2 2 0.5 3.5 mAi i i i= + + = + − =

(checked using LNAP 8/15/02) P5.3–4 Consider 10 V source only (open 30 mA source and short the 8 V source)

Let v1 be the part of va due to the 10 V voltage source.

( ) ( )

( )

1100 ||100 10

100 ||100 10050 1010 V

150 3

v =+

= =

Consider 8 V source only (open 30 mA source and short the 10 V source)

Let v2 be the part of va due to the 8 V voltage source.

( ) ( )

( )

1100 ||100 8

100 ||100 10050 88 V

150 3

v =+

= =

Consider 30 mA source only (short both the 10 V source and the 8 V source)

Let v2 be the part of va due to the 30 mA current source.

Page 14: Chapter 5 - Circuit Theorems

3 (100 ||100 ||100)(0.03)100 (0.03) 1 V

3

v =

= =

Finally, 1 2 310 8 1 7 V3 3av v v v= + + = + + =

(checked using LNAP 8/15/02)

P5.3-5 Consider 8 V source only (open the 2 A source)

Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh:

( ) ( ) ( )1 1 16 3 3 8i i i 0+ + − =

18 2 A

12 3i = =

Consider 2 A source only (short the 8 V source)

Let i2 be the part of ix due to the 2 A current source. Apply KVL to the supermesh:

( ) ( )2 2 26 3 2 3i i i 0+ + + =

26 1 A

12 2i −= = −

Finally, 1 22 1 1 A3 2 6xi i i= + = − =

P5.3-6

Using superposition s 2x a

1 2 1 2

v Ri i

R R R R

⎛ ⎞= + ⎜⎜+ +⎝ ⎠

⎟⎟ . Then

2o s

1 2 1 2

A RAv vR R R R

= ++ + ai

Page 15: Chapter 5 - Circuit Theorems

The equation of the straight line is o s7.5 30v v= + so we require1 2

7.5AR R

=+

. For example,

we can choose Then 1 2 10 , and 150 V/A.R R A= = Ω = o s7.5 75v v ai= + so we require

a30 0.4 A75

i = = .

(Checked: LNAP 6/22/04) P5.3-7

s ax

1

v vi

R−

=

s aa o x

1

v vv v A i A

R−

− = =

1 o sa

1

R v A vv

R A+

=+

Apply KCL to the supernode corresponding to the CCVS to get

a s a oa

1 2 30

v v v vi

R R R−

+ + + =

1 2 s o

a a1 2 1 3

0R R v v

v iR R R R+

− + + =

1 2 1 o s s oa

1 2 1 1 30

R R R v A v v vi

R R R A R R

⎛ ⎞+ +− + + =⎜ ⎟⎜ ⎟+⎝ ⎠

( )( )

( )1 21 2

o s3 12 1 1 2 1

1 1 0R R AR R

v vR RR R A R R R A

⎛ ⎞ ⎛ ⎞++⎜ ⎟ ⎜ ⎟+ + − +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

ai =

( ) ( )

( ) ( )3 1 2 2 1 2

o s2 3 1 2 1

0R R R R R A A R

v vR R R A R R A

+ + + −ai+ + =

+ +

( )

( ) ( )( )

( ) ( )3 2 2 3 1

o s3 1 2 2 1 3 1 2 2 1

R R A R R R Av v

R R R R R A R R R R R A

− += −

+ + + + + +ai

Page 16: Chapter 5 - Circuit Theorems

When 1 2 36 , 12 and 6 R R R= Ω = Ω = Ω

( )o s

12 61224 24

AAv vA A

+−= −

+ + ai

9

Comparing this equation to , we requires o s2v v= +

12 V2 1 24 A

A A 2A

−= ⇔ = −

+

Then so we require s o s2 9 2 6v v v+ = = + ai

a a9 6 1.5 Ai i= ⇒ =

(checked: LNAP 6/22/04)

P5.3-8

o1 1 140 ||10 1 1

8 40 ||10 2 2v v v= = ⇒

+a =

o2 1 210 3 3

8 || 40 10 5 5v v v= − = − ⇒ = −

+b

( )o3 3 38 ||10 || 40 4 4v i i= = ⇒ c =

(checked: LNAP 6/22/04)

Page 17: Chapter 5 - Circuit Theorems

P5.3-9 Using superposition:

x x10v i= and

x x xx

12 cos 24

40 10 10v t v v

i−

+ + =

so

xx x

10 12cos 2 122 cos 2t40 70

i ti i

−= ⇒ = −

Finally,

( )o1 x5 4 3.429 cos 2 Vv i= − = t

x x10v i=

and x x x

x2

440 10 10v v v

i−

+ + =

so

x x0.2 1.75 0.11429 Ai i− = ⇒ = − Finally,

( )o1 x5 4 2.286 Vv i= − =

o o1 o2 3.429 cos 2 2.286 Vv v v t= + = +

(checked: LNAP 6/22/04)

Page 18: Chapter 5 - Circuit Theorems

P5.3-10 Using superposition:

( )1 24 0.3 7.2 Vov = =

230 20 4 V

120 30ov = − = −+

1 2 3.2 Vo o ov v v= + =

(checked: LNAP 5/24/04) P5.3-11 (a)

3 1 2 2 and 1R R R R nR= =

( )

21

3 111 1

nR nR Rn R n

⎛ ⎞= = ⎜ ⎟+ +⎝ ⎠

1 1

1 3

1 1

12 1

1

nR Rnn

1R R Rn nR R

n

⎛ ⎞⎜ ⎟+ ⎛ ⎞⎝ ⎠= = ⎜ ⎟+⎛ ⎞ ⎝ ⎠+ ⎜ ⎟+⎝ ⎠

1 1

2 3 1

1 1

1 11 21

11

n nnR Rnn n

1R R Rn nnR R

nn

⎛ ⎞⎜ ⎟+ ⎛ ⎞⎝ ⎠ += = = ⎜ ⎟+⎛ ⎞ ⎝ ⎠++ ⎜ ⎟ ++⎝ ⎠

R

Page 19: Chapter 5 - Circuit Theorems

12 3

1 2 31 1

22 2

2

n RR R nnanR R R nR R

n

⎛ ⎞⎜ ⎟+⎝ ⎠= = =

+ +⎛ ⎞+ ⎜ ⎟+⎝ ⎠

11 3

2 1 31 1

112 1 2 1

1 2 212 12 1

n RR R n nbnR R R nnR R

nn

⎛ ⎞⎜ ⎟+⎝ ⎠ += = = =

+ +⎛ ⎞ ++ ⎜ ⎟ ++⎝ ⎠

a nb

∴ =

(b) From (a), we require n =4, i.e. R2 = 4R1 and 3 1 245 1R R R R= = . For example

1 2 310 , 40 and 8 .R R R= Ω = Ω = Ω

(checked: LNAP 6/22/04)

P5.3-12 Using superposition

( ) ( )o 1|| 4 42 2

6 || 4 2 || 4 4Rv i

R R⎛ ⎞ ⎛

= − +⎜ ⎟ ⎜⎜ ⎟ ⎜+ +⎝ ⎠ ⎝2i

⎞⎟⎟+ ⎠

Comparing to , we require o 10.5 4v i= − +

( ) ( ) ( )|| 42 0.5 4 || 4 6 || 4 || 4 2 4 6 || 4

R R R R RR

⎛ ⎞− = − ⇒ = + ⇒ = ⇒⎜ ⎟⎜ ⎟+⎝ ⎠

= Ω

and

( ) ( )2 24 42 4 2 4

2 || 4 4 2 4 || 4 4i i

R⎛ ⎞ ⎛ ⎞

= ⇒ = ⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠ ⎝ ⎠2 4 Ai

(checked LNAP 6/12/04)

Page 20: Chapter 5 - Circuit Theorems

P5.3-13 Use units of mA, kΩ and V.

4 + (5||20) = 8 kΩ (a) Using superposition

( )8 82 7 2 8 48 16 k8 8

R RR R

⎛ ⎞= − ⇒ + = ⇒ =⎜ ⎟+ +⎝ ⎠Ω

(b) Using superposition again

a5 16 8 4 2 17 7

5 20 8 16 8 16 5 3 3i ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = × +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

4 mA=

P5.3-14

( ) ( )

( ) ( )

1o 2

3

10 10 2010 40 20 12 40 10 10 40 20 12 40 10

20 1240 20 12 10 40 20 12

vi i

v

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= − + − ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ + + + + +⎡ ⎤⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦⎝ ⎠⎛ ⎞⎛ ⎞+

+ − ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟+ + + +⎡ ⎤⎝ ⎠ ⎣ ⎦⎝ ⎠

o 1 21 1 1

200 10 62.5i v i⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠3v

So 0.05, 0.1 and 0.016a b c= − = − = −

(checked: LNAP 6/19/04) P5.3-15

( )25 3 5 5 3 2 A3 2 2 3mi = − = − =+ +

P5.3-16

( ) ( )3 33 5 18 5 63 (3 3) 3 (3 3)mv ⎡ ⎤

= − = −⎢ ⎥+ + + +⎣ ⎦1 A= −

Page 21: Chapter 5 - Circuit Theorems
Page 22: Chapter 5 - Circuit Theorems

Section 5-4: Thèvenin’s Theorem P5.4-1

(checked using LNAP 8/15/02)

Page 23: Chapter 5 - Circuit Theorems

P5.4-2 The circuit from Figure P5.4-2a can be reduced to its Thevenin equivalent circuit in four steps:

(a)

(b)

(c)

(d)

Comparing (d) to Figure P5.4-2b shows that the Thevenin resistance is Rt = 16 Ω and the open circuit voltage, voc = −12 V.

Page 24: Chapter 5 - Circuit Theorems

P5.4-3 The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps:

(a)

(b)

(c)

(d)

(e) Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V.

(checked using LNAP 8/15/02)

Page 25: Chapter 5 - Circuit Theorems

P5.4-4 Find Rt:

( )( )

12 10 26

12 10 2tR+

= = Ω+ +

Write mesh equations to find voc:

Mesh equations:

( )( )

1 1 2 1

2 1 2

12 10 6 0

6 3 18

i i i i

i i i 0

+ − − =

− + − =

1 2

2 1

28 6

9 6 1

i i

i i 8

=

− =

1 1

2

136 18 A2

14 1 7= A3 2 3

i i

i

= ⇒ =

⎛ ⎞= ⎜ ⎟⎝ ⎠

Finally, 2 17 13 10 3 10 12 V3 2ocv i i ⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(checked using LNAP 8/15/02)

Page 26: Chapter 5 - Circuit Theorems

P5.4-5 Find voc: Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage:

va = −voc Apply KCL at node a:

6 3 08 4 4

oc ococ

v v v−⎛ ⎞ ⎛ ⎞− + + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

V

6 2 6 0 2oc oc oc ocv v v v− + + − = ⇒ = −

Find Rt: We’ll find isc and use it to calculate Rt. Notice that the short circuit forces

va = 0 Apply KCL at node a:

6 0 0 3 0 08 4 4 sci−⎛ ⎞ ⎛ ⎞− + + − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠=

6 3 A8 4sci = =

2 8

3 4 3oc

tsc

vRi

−= = = − Ω

(checked using LNAP 8/15/02)

Page 27: Chapter 5 - Circuit Theorems

P5.4-6 Find voc:

Apply KCL at the top, middle node: 2 3 0 18 V3 6

a a aa

v v v v−= + + ⇒ =

The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Find isc:

Apply KCL at the top, middle node: 2 3 13 6 3

a a a aa

v v v v v 8 V−= + + ⇒ = −

Apply Ohm’s law to the right-hand 3 Ω resistor : 18 6 V3 3a

scvi −

= = = −

Finally: 18 36

oct

sc

vRi

= = = −−

Ω

(checked using LNAP 8/15/02)

Page 28: Chapter 5 - Circuit Theorems

P5.4-7

(a) ( )1 21 0s a av R i d R i− + + + =

( )1 21s

avi

R d R=

+ +

( )( )

2

1 2

11

soc

d R vv

R d R+

=+ +

1

sa

viR

=

( ) ( )1

11 s

sc a

d vi d i

R+

= + =

2

0Ta a T

vi d i iR

− − + − =

1 a TR i v= −

( ) ( )2 1

1 2 1 2

11 T T

T T

R d Rv vi d vR R R R

+ += + + = ×

( )1 2

1 21T

tT

R RvRi R d R

= =+ +

(b) Let R1 = R2 = 1 kΩ. Then

1000 1000625 2 0.4 A/A2 625tR d

dΩ = = ⇒ = − = −

+

and ( )1 0.4 25 5 13.33 V

2 0.4 1s

oc s

d vv v

d+ − +

= = ⇒ = =+ − +

(checked using LNAP 8/15/02)

Page 29: Chapter 5 - Circuit Theorems

P5.4-8

oct

Rv vR R

=+

From the given data:

200062000 1.2 V

1600400024000

oct oc

toc

t

vR v

Rv

R

⎫= ⎪+ =⎧⎪ ⇒⎬ ⎨ = − Ω⎩⎪=⎪+ ⎭

When R = 8000 Ω,

( )8000 1.2 1.5 V1600 8000

v = =− +

P5.4-9

oc

t

viR R

=+

From the given data:

0.0042000 24 V

40000.003

4000

oc

t oc

toc

t

vR v

RvR

⎫= ⎪+ =⎧⎪ ⇒⎬ ⎨ = Ω⎩⎪=⎪+ ⎭

(a) When i = 0.002 A:

240.002 8000 4000

RR

= ⇒ =+

Ω

(b) Maximum i occurs when R = 0:

24 0.006 6 mA 6 mA4000

i= = ⇒ ≤

P5.4-10 The current at the point on the plot where v = 0 is the short circuit current, so isc = 20 mA. The voltage at the point on the plot where i = 0 is the open circuit voltage, so voc = −3 V. The slope of the plot is equal to the negative reciprocal of the Thevenin resistance, so

1 0 0.002 1503 0 t

t

RR

−− = ⇒ = − Ω

− −

Page 30: Chapter 5 - Circuit Theorems

P5.4-11

12 6000 2000 1000 0 4 3000 A

4 1000 V3

a a a

a

oc a

i i ii

v i

− + + +=

= =

=

ia = 0 due to the short circuit

12 6000 0 2 mA43 667

.002

sc sc

oct

sc

i i

vRi

− + = ⇒ =

= = = Ω

43

667bi R=

+

ib = 0.002 A requires

43 667 0

0.002R = − =

(checked using LNAP 8/15/02)

Page 31: Chapter 5 - Circuit Theorems

P5.4-12

10 0 10 A4 2 0

2 20oc

oc

i iv i i

v i V

= + ⇒ =+ − =⇒ = − = −

10 10sc si i i i c+ = ⇒ = −

4 0 2 0 0 10 Asci i i i+ − = ⇒ = ⇒ = 20 2

10oc

tsc

vRi

−= = = − Ω

202 2L L

L

i RR

12−− = = ⇒ = Ω

(checked using LNAP 8/15/02)

Page 32: Chapter 5 - Circuit Theorems

P5.4-13 Replace the part of the circuit that is connected to the variable resistor by its Thevenin equivalent circuit:

( )18 k || 12 k 24 k 18 k || 36 k 12 kΩ Ω+ Ω = Ω Ω = Ω

a3612000

iR

=+

and a 3612000Rv

R=

+

2

a a3612000

p i v RR

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

(a) a36 3 mA

0 12000i = =

+ when R = 0 Ω (a short circuit).

(b) 5

a 510 36 32.14 V

10 12000v = =

+when R is as large as possible, i.e. R = 100 kΩ.

(c) Maximum power is delivered to the adjustable resistor when t 12 kR R= = Ω . Then

2

a a36 12000 0.027 27 mW

12000 12000p i v ⎛ ⎞= = = =⎜ ⎟+⎝ ⎠

(checked: LNAP 6/22/04)

Page 33: Chapter 5 - Circuit Theorems

P5.4-14 Replace the source by it’s Thevenin equivalent circuit to get

oco

t LR +Rv

i =

Using the given formation

( ) (oc

tt t

oc

t

0.375R 4

0.375 R 4 0.300 R 80.300

R 8

v

v

⎫= ⎪+ ⎪ ⇒ + =⎬

⎪= ⎪+ ⎭

)+

So ( ) ( ) ( )t oc

0.300 8 0.375 4R 12 and 0.3 12 8 6 V

0.075v

−= = Ω = + =

(a) When L o6R 10 , 0.2727 A.

12 10i= Ω = =

+

(b) . t12 R 48 11R R 16 Ω = = ⇒ = Ω(checked: LNAP 5/24/04)

P5.4-15 (a)

3 2 0.25 Ai i− = Apply KVL to mesh 1 to get

( ) ( )1 2 1 320 20 40 0i i i i− + − − = Apply KVL to the supermesh corresponding to the unspecified resistance to get

Page 34: Chapter 5 - Circuit Theorems

( ) ( )2 3 1 3 1 240 10 20 20 0i i i i i i+ − − − − = Solving, for example using MATLAB, gives

1 1

2 2

3 3

0 1 1 0.25 1.87540 20 20 40 0.75040 60 30 0 1.000

i ii ii i

⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢− − = ⇒ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢−⎣ ⎦ ⎣ ⎦ ⎣⎣ ⎦ ⎣ ⎦

⎤⎥⎥⎥⎦

Apply KVL to mesh 2 to get

( ) ( ) ( )1 2 22 2 3 1 2

2 3

20 4040 20 0 30

i i ii R i i i i R

i i− −

+ − − − = ⇒ = =−

Ω

(b)

oc20 4040 40 12 V

20 20 10 40v ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

t 18 R = Ω

120.25 30 18

RR

= ⇒ = Ω+

(checked: LNAP 5/25/04)

Page 35: Chapter 5 - Circuit Theorems

P5.4-16 Find the Thevenin equivalent circuit for the part of the circuit to the left of the terminals a-b.

Using voltage division twice

oc32 3020 20 5 4 1 V

32 96 120 30v = − = − =

+ +

( ) ( )t 96 || 32 120 || 30 24 24 48 R = + = + = Ω

Replacing the part of the circuit to the left of terminals a-b by its Thevenin equivalent circuit gives

o1 0.0125 A

48 32i = =

+12.5 mA=

(checked: LNAP 5/24/04)

Page 36: Chapter 5 - Circuit Theorems

P5.4-18 Replace the circuit by its Thevenin equivalent circuit:

mm

m

550

Rv

R⎛ ⎞

= ⎜ ⎟⎜ ⎟+⎝ ⎠

(a)

mi mmlim 5 V

Rv v

→∞= =

(b) When so m m1000 , 4.763 VR v= Ω =

% error = 5 4.762 100 4.76%5

−× =

(c)

m

m mm

m

5 550

0.02 0.98 2450 5 50

RR R

RR

⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒

+≥ Ω

(checked: LNAP 6/16/04)

Page 37: Chapter 5 - Circuit Theorems

P5.4-19

( )( )

s oca

1 s 2oc

oc 1 2a a

2

11

v vi

R v R bv

v R R bi bi

R

− ⎫= ⎪

+⎪ ⇒ =⎬ + +⎪+ = ⎪⎭

( ) ( )ssc a

11 1

vi i b b

R= + = +

( )( )

( ) ( )

s 2

oc 1 2 1 2t

ssc 1 2

1

11

11

v R bv R R b R R

R vi RbR

+

+ += = =

+ ++

R b

(checked: LNAP 7/22/04)

Page 38: Chapter 5 - Circuit Theorems
Page 39: Chapter 5 - Circuit Theorems

Section 5-5: Norton’s Theorem P5.5-1 When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off. P5.5-2

(checked using LNAP 8/16/02)

Page 40: Chapter 5 - Circuit Theorems

P5.5-3

P5.5-4 To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.

In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, 2 sci i= . The controlling current of the CCVS is expressed in terms of the mesh currents as

1 2 1a si i i i i c= − = − Apply KVL to mesh 1 to get ( ) ( )1 1 2 1 2 1 23 2 6 10 0 7 4 1i i i i i i i− − + − − = ⇒ − = 0 (1) Apply KVL to mesh 2 to get

( )2 1 2 1 2 1115 6 0 6 11 06

i i i i i i− − = ⇒ − + = ⇒ = 2i

Substituting into equation 1 gives

2 2 2117 4 10 1.13 A 1.13 A6 sci i i i⎛ ⎞ − = ⇒ = ⇒ =⎜ ⎟

⎝ ⎠

Page 41: Chapter 5 - Circuit Theorems

Figure (a) Calculating the short circuit current, isc, using mesh equations.

To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage

source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as

Tt

T

vRi

=

In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.

In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, . The controlling current of the CCVS is expressed in terms of the mesh currents as

2 Ti i=

1 2 1a Ti i i i i= − = + Apply KVL to mesh 1 to get

( ) ( )1 1 2 1 2 1 2 143 2 6 0 7 4 07

i i i i i i i i− − + − = ⇒ − = ⇒ = 2i (2)

Apply KVL to mesh 2 to get

( )2 1 2 1 25 6 0 6 11T Ti v i i i i v+ − − = ⇒ − + = −

Substituting for i1 using equation 2 gives

2 2 246 11 7.577 T Ti i v i⎛ ⎞− + = − ⇒ =⎜ ⎟

⎝ ⎠v−

Finally,

2

7.57T T Tt

T T

v v vRi i i

− −= = = =

−Ω

Page 42: Chapter 5 - Circuit Theorems

Figure (b) Calculating the Thevenin resistance, T

tT

vRi

= , using mesh equations.

To determine the value of the open circuit voltage, voc, we connect an open circuit across

the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.

In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, . The controlling current of the CCVS is expressed in terms of the mesh currents as 2 0 Ai =

1 2 1 0ai i i i i1= − = − =

Apply KVL to mesh 1 to get

( ) ( ) ( ) ( )1 1 2 1 2 1 1 1

1

3 2 6 10 0 3 2 0 6 0 10 0

10 1.43 A7

i i i i i i i i

i

− − + − − = ⇒ − − + − − =

⇒ = =

Apply KVL to mesh 2 to get

( ) ( ) ( )2 1 2 15 6 0 6 6 1.43 8.58 Voc oci v i i v i+ − − = ⇒ = = =

Figure (c) Calculating the open circuit voltage, voc, using mesh equations.

As a check, notice that ( )( )7.57 1.13 8.55t sc ocR i v= = ≈

(checked using LNAP 8/16/02)

Page 43: Chapter 5 - Circuit Theorems

P5.5-5 To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, . The voltage at node 3 is equal to the voltage across a short, 1 24 Vv = − 3 0v = . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. . The voltage at node 3 is equal to the voltage across a short, i.e.

2av v=

3 0v = . Apply KCL at node 2 to get

1 2 2 31 3 22 3 48 3 16 V

3 6 a a

v v v vv v v v v

− −= ⇒ + = ⇒ − = ⇒ = −

Apply KCL at node 3 to get

( )2 32

4 9 9 16 24 A6 3 6 6sc a sc sc

v vv i v i i

−+ = ⇒ = ⇒ = − = −

Figure (a) Calculating the short circuit current, Isc, using mesh equations.

To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage

source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as

Tth

T

vRi

=

Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.

Page 44: Chapter 5 - Circuit Theorems

In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. . The voltage at node 3 is equal to the voltage across the current source, i.e.

1 0v =

2av v=

3 Tv v= . Apply KCL at node 2 to get

1 2 2 31 3 22 3

3 6 T a

v v v vv v v v v

− −= ⇒ + = ⇒ = 3

Apply KCL at node 3 to get

2 32 2 3

4 0 9 6 06 3

9 6 0

3 6 0 2

T T

a T T

T T T T

v vv i v v i

v v i

v v i v i

−+ + = ⇒ − + =

⇒ − + =

⇒ − + = ⇒ = −6 T

Finally,

3Tt

T

vRi

= = − Ω

Figure (b) Calculating the Thevenin resistance, T

thT

vRi

= , using mesh equations.

To determine the value of the open circuit voltage, voc, we connect an open circuit across

the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. . The voltage at node 3 is equal to the open circuit voltage, i.e. .

1 24 Vv = −

2av v= 3 ocv v= Apply KCL at node 2 to get

1 2 2 31 3 22 3 48 3

3 6 oc a

v v v vv v v v v

− −= ⇒ + = ⇒ − + =

Page 45: Chapter 5 - Circuit Theorems

Apply KCL at node 3 to get

2 32 2 3

4 0 9 0 96 3 a o

v vv v v v

−+ = ⇒ − = ⇒ = cv

Combining these equations gives

( )3 48 9 72 Voc a oc ocv v v v− + = = ⇒ =

Figure (c) Calculating the open circuit voltage, voc, using node equations. As a check, notice that

( )( )3 24 72th sc ocR I V= − − = =

(checked using LNAP 8/16/02)

Page 46: Chapter 5 - Circuit Theorems

P5.5-6 (a) Replace the part of the circuit that is connected to the left of terminals a-b by its Norton equivalent circuit:

Apply KCL at the top node of the dependent source to see that b 0 Ai = . Then

( )oc b25 5000 25 Vv i= + =

Apply KVL to the supermesh corresponding to the dependent source to get

( )b b b5000 10000 3 25 0 1 mAi i i− + − = ⇒ =

Apply KCL to get

sc b3 3 mi i A= =

Then oc

tsc

8.33 kv

Ri

= = Ω

Current division gives

83330.5 3 41.67 k8333

RR

= ⇒ =+

Ω

(b)

Notice that bi and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to get

3b b b

10.5 10 4 mA6

i i i−+ × = ⇒ =

Apply KVL to the supermesh corresponding to the dependent source to get

( )( )3b5000 10000 0.5 10 25 0i R −− + + × − =

( )( )3 315000 10 10000 0.5 10 256

R− −⎛ ⎞− × + + × =⎜ ⎟⎝ ⎠

3

1256 41.67 k

0.5 10R −= = Ω

×

Page 47: Chapter 5 - Circuit Theorems

P5.5-7 Use source transformations to reduce the circuit to

Replace the series voltage sources by an equivalent voltage source, the series resistors by an equivalent resistance and do a couple more source transformations to reduce the part of the circuit to the left of the terminals a-b by its Norton equivalent circuit.

Apply KCL at node a to get

220.4 0.8 0

10 2 5v v vv= + ⇒ + − =

so .2 1.8 0.8, -1.0 V

2v − ±= =

Choosing the positive value of v, 20.8 0.32 A

2i = = . Choosing the negative value of v,

21 0.52

i −= = . There are two solutions to this problem. Linear circuits are so much simpler than

nonlinear circuits. (checked: LNAP 5/26/04)

Page 48: Chapter 5 - Circuit Theorems

P5.5-8 Simplify the circuit using a source transformation:

Identify the open circuit voltage and short circuit current.

Apply KVL to the mesh to get:

( ) x x10 2 3 15 0 1 Ai i+ + − = ⇒ = Then

oc x3 3v i V= =

Express the controlling current of the CCVS in terms of the mesh currents:

x 1 si i i c= −

The mesh equations are

( ) ( )1 1 sc 1 sc 1 sc10 2 3 15 0 15 5 15i i i i i i i+ − + − − = ⇒ − = and

( )sc 1 sc 1 sc43 03

i i i i− − = ⇒ = i

so

Page 49: Chapter 5 - Circuit Theorems

sc sc sc415 5 15 1 A3

i i i⎛ ⎞ − = ⇒ =⎜ ⎟⎝ ⎠

The Thevenin resistance is

t3 3 1

R = = Ω

Finally, the Norton equivalent circuit is

(checked: LNAP 6/21/04) P5.5-9 Identify the open circuit voltage and short circuit current.

11 3 1 V3

v ⎛ ⎞= =⎜ ⎟⎝ ⎠

Then

( )oc 1 14 2.5 9 Vv v v= − = −

1 sc13 13

v i⎛ ⎞= − = −⎜ ⎟⎝ ⎠

sc3 i

( )1 sc sc 1

1 sc

4 2.5 5 0

9 9

v i i v

v i

+ + − =

⇒ + = 0

( )sc sc sc19 1 3 9 0 A2

i i i− + = ⇒ =

The Thevenin resistance is

t9 18

0.5R −

= = − Ω

Finally, the Norton equivalent circuit is

Page 50: Chapter 5 - Circuit Theorems

(checked: LNAP 6/21/04) P5.5-10 Replace the circuit by its Norton equivalent circuit:

( )3m

m

1600 1.5 101600

iR

−⎛ ⎞

= ×⎜ ⎟⎜ ⎟+⎝ ⎠

(a)

mmi m0

lim 1.5 mAR

i i→

= =

(b) When Rm = 20 Ω then so m 1.48 mAi =

1.5 1.48% error 100 1.23%1.5−

= × =

(c)

( )m

mm

16000.015 0.0151600 16000.02 0.98 32.65

0.015 1600R

RR

⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠≥ ⇒ ≥ ⇒

+ ≤ Ω

(checked: LNAP 6/18/04)

Page 51: Chapter 5 - Circuit Theorems

P5.5-11

2 12

3 A6

2 6 V

aa a

oc a

ii i

v i

−= ⇒ =

= = −

( )

12 6 2 3 A

23 2 3 23

a a a

sc a sc

i i i

i i i

+ = ⇒ = −

= ⇒ = − = − A

6 32tR −

= = Ω−

P5.5-12

( )

12 24 12 24 8 12 24 36

24 30 20 V12 24

t

oc

R

v

× ×= = =

+

= =+

Ω

20

8i

R=

+

Page 52: Chapter 5 - Circuit Theorems

Section 5-6: Maximum Power Transfer P5.6-1 a) For maximum power transfer, set RL equal to the Thevenin resistance:

100 1 101L tR R= = + = Ω

b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit:

The voltage across RL is ( )101 100 50 V101 101Lv = =

+

Then 2 2

max50 24.75 W101

L

L

vpR

= = =

P5.6-2 Reduce the circuit using source transformations:

Page 53: Chapter 5 - Circuit Theorems

Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value of that maximum power is

2 2( ) (0.03) (60) 54 mWmax RP i R= = =

P5.6-3

LL S

S L

2 2L S L

L 2L S L

( )

Rv v

R R

v v Rp

R R R

⎡ ⎤= ⎢ ⎥

+⎢ ⎥⎣ ⎦

∴ = =+

By inspection, pL is max when you reduce RS to get the smallest denominator.

∴ set RS = 0 P5.6-4 Find Rt by finding isc and voc:

The current in the 3 Ω resistor is zero because of the short circuit. Consequently, isc = 10 ix. Apply KCL at the top-left node to get

0.90.9 10 0.1 A9x x xi i i+ = ⇒ = =

so isc = 10 ix = 1A

Next

Page 54: Chapter 5 - Circuit Theorems

Apply KCL at the top-left node to get

0.90.9 10 0.1 A9x x xi i i+ = ⇒ = =

Apply Ohm’s law to the 3 Ω resistor to get

( ) ( )3 10 30 0.1 3 Voc xv i= = = For maximum power transfer to RL:

3 31

ocL t

sc

vR Ri

= = = = Ω

The maximum power delivered to RL is given by

( )

2 2

max3 3 W

4 4 3 4oc

t

vpR

= = =

P5.6-5

Page 55: Chapter 5 - Circuit Theorems

The required value of R is

( ) ( )( ) ( )

20 120 10 508 50

( ) ( )

20 120 10 50tR R+ +

= = + = Ω+ + +

170 3020 10 20 50170 30 170 30

170(20)(10) 30(20)(50) 4000 20 V200 200

ocv ⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦−

= = =

The maximum power is given by

( )

2 2

max20 2 W

4 4 50oc

t

vpR

= = =

P5.6-6

L so L

AR R

i v=+

( )2 2

s L2L L L 2

o L

A RP R

R R

vi= =

+

(a) so maximizes the power delivered to the load. The corresponding load power is

t oR =R L oR =R 10 = Ω

( )

22

L 2

120 102P 2

10 10

⎛ ⎞⎜ ⎟⎝ ⎠= =+

.5 W .

(b) Ro = 0 maximizes PL (The numerator of PL does not depend on Ro so PL can be maximized by making the denominator as small as possible.) The corresponding load power is

2

22 2 2 2

s L sL 2

L L

120A R A 2P 12.5 W.R R 8v v

⎛ ⎞⎜ ⎟⎝ ⎠= = = =

(c) PL is proportional to A2 so the load power continues to increase as A increases. The load can safely receive 15 W. This limit corresponds to

Page 56: Chapter 5 - Circuit Theorems

( )

22

2

1A 815215 A 36 49.3 V.818

⎛ ⎞⎜ ⎟⎝ ⎠= ⇒ = =

(checked: LNAP 6/9/04)

P5.6-7 Replace the part of the circuit connected to the variable resistor by its Thevenin equivalent circuit. First, replace the left part of the circuit by its Thevenin equivalent:

oc1150 10 4.545 V

150 180v ⎛ ⎞= =⎜ ⎟+⎝ ⎠

t1 180 150 81.8 R = = Ω

Next, replace the right part of the circuit by its Thevenin equivalent:

oc2470 20 15.932 V

470 120v ⎛ ⎞= =⎜ ⎟+⎝ ⎠

t2 120 470 95.6 R = = Ω

Now, combine the two partial Thevenin equivalents:

oc oc1 oc2 t t1 t210.387 V and 177.4 v v v R R R= − = − = + = Ω

So

The power received by the adjustable resistor will be maximum when R = Rt = 177.4 Ω. The maximum power received by the adjustable

resistor will be ( )( )

211.3870.183 W

4 177.4 p

−= =

Ω.

(checked LNAPDC 7/24/04)

Page 57: Chapter 5 - Circuit Theorems

P5.6-8

( )( )2

10010 10L L

t L t L t L

R Rp i v

R R R R R R

⎛ ⎞ ⎡ ⎤= = =⎜ ⎟ ⎢ ⎥⎜ ⎟+ +⎢ ⎥ +⎝ ⎠ ⎣ ⎦

The power increases as Rt decreases so choose Rt = 1 Ω. Then

( )( )max 2

100 513.9 W

1 5p i v= = =

+

P5.6-9 From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:

Rt = 20 Ω and

( )2

max max 4 5 4 20 204

ococ t

t

vp v p RR

= ⇒ = = = V

Page 58: Chapter 5 - Circuit Theorems

Section 5-8 How Can We Check…?

P5.8-1

Use the data in the first two lines of the table to determine voc and Rt:

0.09720 39.9 V

4100.0438

500

oc

t oc

toc

t

vR v

RvR

⎫= ⎪+ =⎧⎪ ⇒⎬ ⎨ = Ω⎩⎪=⎪+ ⎭

Now check the third line of the table. When R= 5000 Ω:

39.9 7.37 mA410 5000

oc

t

viR R

= = =+ +

which disagree with the data in the table. The data is not consistent.

P5.8-2

Use the data in the table to determine voc and isc: 12 V (line 1 of the table) 3 mA (line 3 of the table)

so 4 k

oc

sc

oct

sc

vi

vRi

==

= = Ω

Next, check line 2 of the table. When R = 10 kΩ:

( ) ( )3 3

12 0.857 mA10 10 5 10

oc

t

viR R

= = =+ +

which agrees with the data in the table.

To cause i = 1 mA requires ( )3

120.001 8000 10 10

oc

t

vi RR R R

= = = ⇒ =+ +

Ω

I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.

Page 59: Chapter 5 - Circuit Theorems

P5.8-3

1 1 1 1 11 62 3 6 1t

t 1RR

R R R R R= + + = ⇒ =

and 2 3 3 4 6 5 18030 20 10

3 2 3 2 3 4 1 6 5 11ocv⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠=

so the prelab calculation isn’t correct.

But then

( )

180 18011 11 163 mA 54.5 mA6 60 40110 40

11

oc

t

viR R

= = = = ≠+ ++

so the measurement does not agree with the corrected prelab calculation.

P5.8-4 ( )6000 3000 500 1500 2000 2000 1000 + = = Ω

12 12 12 mA1000 1000

iR

= ≤ =+

How about that?! Your lab partner is right.

(checked using LNAP 6/21/05)

Page 60: Chapter 5 - Circuit Theorems

P5.8-5 (a)

KVL gives ( )oc tv R R= + i

from row 2 ( )( )oc t 10 1.333v R= +

from row 3 ( )( )oc t 20 0.857v R= +

So ( )( ) ( )( )t t10 1.333 20 0.857R R+ = +

( ) ( )t t28 10 18 20R R+ = +

Solving gives t t10 360 280 80 8 R R= − = ⇒ = Ω

and ( )( )oc 8 10 1.333 24 Vv = + =

(b)

ococ

t t

24 24 and 8 8

v R Ri vR R R R R R

= = = =+ + + +

v

When R = 0, i = 3 A, and v = 0 V.

When R = 40 Ω, 1 A2

i = .

When R = 80 Ω, ( )24 80 240 21.8288 11

v = = = .

These are the values given in the tabulated data so the data is consistent.

(c) When R = 40 Ω, ( )24 4020 V

48v = = .

When R = 80 Ω, 24 0.2727 A88

i = = .

(d) First ( )t 1 18 24 18 12 24 R R R= = + ⇒ = Ω

the, using superposition,

( )( ) ( )( )oc 1 s s s1

2424 12 24 18 12 8 8 2 A24 18 12

v R i iR

= = + + = + ⇒ =+ +

i

(checked using LNAP 6/21/05)

Page 61: Chapter 5 - Circuit Theorems