elec 3202 chap 4

7/30/2019 Elec 3202 Chap 4

http://slidepdf.com/reader/full/elec-3202-chap-4 1/73

UAE University Department of Electrical Engineering Dr.Hazem N.Nounou

Chapter (4)

Additional Analysis TechniquesHere we will study four additional Techniques

• Superposition

• Source transformation• Thevenin and Norton Theorems

• Maximum power principle

1. Superposition :Definition :

Whenever a linear circuit is excited by more than one independent

source, the total response is the algebraic sum of individual responses

The idea is to activate one independent source at a time to

get individual response.

Then add the individual response to get total response

7/30/2019 Elec 3202 Chap 4



Note:

1. Dependent source are Never deactivated (always active)

2. When an independent voltage source is deactivated, it is set to zero.replaced by short circuit

3. When an independent current source is deactivated, it is set to zero.

replaced by open circuit

Example:

Use superposition to find i1,i2,i3,i4 ?

+

-

Ω4

Ω2Ω6

Ω3 A12120 V

i1

i2

i3

i4

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+

-

Ω4

Ω2Ω6

Ω3120 V

'i1

'i2

'i3

'i4

V1

open circuit for

current source

•Activate independent voltage source 120 V only

•Using KCL at V1 (nodal analysis)

0i'i'i' 321 =−−

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V30V

06

1

3

1

6

1V20

0

42

V

3

V

6

V120

1

1

111

=⇒

=

++−

=

+

−−−

A56

Vii'

A10

3

30

3

Vi'

A156

90

6

V120i'

143

12

11

===

===

==−

=

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Ω4

Ω2Ω6

Ω3

''i1

''i2

''i3

''i4

V3

short circuit for

voltage source 12 A

V 4

* Activate the independent current source only

(1)0V3V6-

0)V(V3V2V

02

VV

3

V

6

V

0i"i"i"

43

333

4333

321

KK

=+

=−−−−=

−

−−

−

=−−KCL at V3:

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0124

V2

VV

012i"i"

443

43

=−−−

=−−

(2)48V3V2

48VV2V2

43

443

KK=−

=−−

KCL at V4:

V24V

V12V

4

3

−=−=

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A1-65i"i'i

A1165i"i'i

A6410i"i'i

A17215i"i'i

444

333

222

111

=−=+=

=+=+=

=−=+=

=+=+=

A6

4

24

4

Vi"

A62

2412

2

VVi"

A43

12

3

Vi"

A26

12

6

Vi"

44

433

3

2

31

−=−

==

=+−

=−

=

−=−==

==−

=

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Use super position to find V0 ?

Activate voltage source only:

Example :

Ω01Ω02 A5

10 V+

-

Ω5

Φi

ΦV0.4

ΦV

+

-

+

-

oV

Φi2

Ω01Ω02

10 V+

-

Ω5

Φi'

ΦV'0.4

ΦV'

+

-

+

-

oV'

Φi'2

X

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0V'V'4V'

V'4)V'(0.410V'

ΦΦΦ

ΦΦΦ

=⇒=

==

V8V'

25

2010V'

0

0

=

=

Dependent current source is open

Ω5

+

-oV'

+

-

Ω20

10 V

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Activate independent current source only:

Ω01Ω

02

Ω5

Φi''

ΦV''0.4

ΦV''

+

-

+

-

oV''

Φi''2

y

5 A

Z

KCL at node (y):

(1)0V"8V"5-

0V"8V"V"4-

0V"0.420

V"

5

V-"

Φ0

Φ00

Φ

00

KK=+

=+−=+−

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V8-168V"V'V

V16

5

80-V"

5

8V"

V10V"5V"0.5

0V"0.410

V"5

000

Φ0

ΦΦ

Φ

Φ

=−=+=

−===∴

−=⇒−=

=++

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Example:

Consider the independent source only

Use superposition to find V ?

100 V+

-

4 A

Ω12

Ω2Ω5

Ω10+

-

V

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V59.23V'2214

1

10

1

5

1V'

014

V'

10

V'

5

V'22

014

V'

10

V'

5

V'100

0iii 321

=⇒=

++

=−−−

=−−−

=−−

Apply KCL at node (x) :

100 V+

- Ω12

Ω2Ω5

Ω10

+

-

'V

i3i1

i2

X

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Consider the independent source only.

4 A

Ω12

Ω2Ω5

Ω10

+

-

''V

4 A

Ω2Ω5

Ω10

+- ''V

Ω12

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Current divider

V509.2359.23V"V'V

V9.233

10iV"

A2.769

23

1012

12A4i

x

x

=−=+=

−=

−=

=

++=

4 A

Ω2Ω

310

Ω12

+- ''V

ix

iy

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2. Source Transformation:

A transformation that allow a voltage source in serieswith a resistor to be replaced by a current source in parallel

with the same resistor or vice versa

How?

We need to find Is and Vs such that VL and IL is the same in

both circuits

VL

R

R L

+

-

Vs

IL

+

-

a

bKCT 1

+

-

VL

IL

a

b

IS

KCT 2

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sL

L IR R

R

I +=

ss IR V =

In KCT 2,

For IL to be the same , we need

In KCT 1 ,

R R

VI

L

sL +=

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R

VS

+

-

a

b

R

VI s

s =

Where

ss IR V = or

a

b

IS R

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Example :

Using source transformation, find the power associated with the

6 V source.

1. Consider the 40 V source in series with (5Ω)

+

-

40 V+

-

Ω4 Ω5

Ω02Ω03

Ω6

6 V

Ω01

+ -

Ω4

Ω5Ω02Ω03

Ω6

6 V

Ω01

A85

40=

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2. Take (5// 20 Ω)

3. Consider 8A in parallel with (4Ω)

+

-

Ω4

Ω4

02//5

=Ω03

Ω6

6 V

Ω01

A8

+ -

Ω4

Ω03

Ω6

6 V

Ω01

A4

+ -

(8 A)(4)=

32 V

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4. Take (4+6+10) in series

5. Consider 32 V in series with (20Ω)

+

-

Ω

4

Ω036 V

Ω20

+

- 32 V

+

-

Ω4

Ω02Ω036 V A6.120

32

=

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6. Take (30//20 Ω)

7. Consider 1.6A in parallel with (15 Ω)

+

-

Ω4

Ω12

20//03

=

6 VA6.1

+

-

Ω4

6 V

Ω12

+

-1.6 A (12)

=19.2 V

i

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)(absorbingW4.95P

(0.825)6ivPA0.825124

619.2i

6v

6V

=

==⇒=+

−=

Example :

Use source transformation to find V0

+

-

Ω52

Ω001

Ω5

250 V

A8

+

-

Vo Ω15

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2. Consider (250 V) in series with (25 Ω)

+

-

Ω52

Ω66.16

250 V

A8+

-

Vo

1. Take (5//15)//100 = 6.66 Ω

Ω52 Ω66.16A8+

-

Vo

A10

25

250=

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V20Ω)(10A)(2R iV0 ===

3. Find equivalent

A2810i =−=

Ω10

25//16.66R

=

=+

-

Vo

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Example:

Use source transformation to find V0

+

-

60 V

Ω8

Ω5

Ω6.1

Ω02

Ω6120 V

+

-+

-

Vo

36 A

Ω8

Ω5

Ω6.1

Ω02Ω6

+

-

Vo

36 AA12

5

60=A6

20

120=

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( ) ( ) V48Ω8A6R iV

A6(30)81.62.4

2.4i

R R R

R i

320

321

12

===

=++

=++

=

Ω8

Ω6.1

+

-

VoR 1

i = 36 + 6 - 12= 30 A

Ω2.4

6//5//20

=

R 2

R 3

i1 i2

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Example : Use source transformation to find V0?

Consider (10V) in series with (1Ω)

+

Vo

--

2 A

+

-

Vs

10V 1 ohm

1 ohm1 ohm

1 ohm

1 ohm

1 ohm10 A 2 A1 ohm

1 ohm1 ohm

1 ohm

+

Vo

--

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Take (1//1)=0.5

Consider (10A) in parallel with (0.5 Ω)

1 ohm

1 ohm1 ohm

1/2 ohm 2 A10 A

+

Vo

--

+

-

5 V

1/2 ohm

2 A

1 ohm1 ohm

1 ohm

+

Vo

--

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Take 0.5Ω in series with 1 Ω

Consider 5V in series with 1.5 Ω

1 ohm

1 ohm

2 A

1.5 ohm

+

-

5 V

+

Vo

--

+

Vo

--

1.5 ohm10/3 A 2 A

1 ohm

1 ohm

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Add the current sources

7. Take (4/3A) in parallel with (3/2 Ω)

( ) V4/725.111

1V0 =

++=

4/3 A 1.5 ohm

1 ohm

1 ohm

+

Vo

--

+

-

2 V

1.5 ohm

1 ohm

1 ohm

+

Vo

--

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Thevenin and Norton Theorems

Thevenin Theorem:A portion of the circuit at a pair of nodes can be replaced

by a voltage source Voc in series with a resistor R TH, where Voc

is the open circuit voltage and R TH is the Thevenin’s equivalentresistance obtained by considering the open circuit with all

independent sources made zero

R Lcircuit

a

b

+

-R L

a

b

VOC

R TH

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Norton Theorem :

A portion of the circuit at pair of nodes can be replaced by a

current source Isc in a parallel with a resistor R TH. Isc is the shortcircuit current at the terminals, and R TH is the Thevenin’s

equivalent resistance

R Lcircuit

a

b

R L

a

b

ISC

R TH

Here we will consider ( 3 ) cases :

1. Circuit containing only independent sources.

2. Circuit containing only dependent sources.

3. Circuit containing both independent and dependent sources.

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Case (1): Circuit containing only independent sources:

• Procedure of Thevenin’s Theorm:a. Find the open circuit voltage at the terminals , Voc.

b. Find the Thevenin’s equivalent resistance, RTH at the

terminals when all independent sources are zero:

¾Replacing independent voltage sources by short circuit

¾ Replacing independent current sources by open circuit

c. Reconnect the load to the Thevenin equivalent circuit

• Procedure of Norton’s Theorm:

a. Find the short circuit current at the terminals, Isc. b. Find Thevenin’s equivalent resistance, R TH (as before).

c. Reconnect the load to Norton’s equivalent circuit.

R

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Example :

Use Thevenin’s and Norton Theorms to find V0

+

-R L

VOC

R TH

R L

ISC

R TH

Using Thevenin Theorm:

+ - + -

Ωk 2

6 V 12 V

Ωk 2Ωk 4

+

-

Vo

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V4k)(4iVAm1k 4k 2

V6i 14k Ω1 −==⇒=+=

V8V4V12Voc =−=

First find VOC:

+ - + -

Ωk 2

6 V 12 V

Ωk 4

+

-

Voc

+

-

i1

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Second, find R TH

R TH = 2k//4k = 4/3 k Ω

Thevenin equivalent circuit is

( )

V4.8V

V8k

310

k 2

VR k 2

Ω

k 2V

0

oc

TH

0

=

=

+=

Ωk 2Ωk 4

R TH

+

-

VOC

R TH

+

-

Vo Ωk 2

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+ - + -

Ωk 2

6 V 12 V

Ωk 4ISC

+ -

Ωk 2

6 V

Ωk 4 +

-

12 V

i1i

i2

+

-V 2 k

-

12 V

+

X

Using Norton Theorm

First find Isc

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R TH is the same as before:

Am2.4m)(6k 2k

34

k 34)(I

k 2R

R I sc

TH

TH0 =

+=

+=

V4.8k)(2m)(2.4k)(2IV 00 ===

2 k

ISC

R TH

Io +

-

Vo

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Example :

Use Thevenin and Norton to find V0

Using Thevenin Theorm:

+

-Ω4

Ω8Ω5

Ω20

+

-

Vo

Ω72

Ω12

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KVL around the upper loop :

1. Find Voc :

+

-

Ω8Ω5

Ω20Ω72

Ω12

i1

a

b

(1)0i5i25

0)i(i5i8i12

21

2111

KK=−

=−++

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A3iA,0.6i

(2)72i25i572i20)i(i5

21

21

212

==

=+− =+−KK

V64.8V

(3)20(0.6)8

i20i8V

oc

21oc

=

+=

+=

KCL around lower loop :

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2. Find R TH

R TH = (8+4) // 12 = 12 // 12 = 6Ω

Ω8Ω5

Ω20

Ω12

a

b

Ω8( )

Ω4

Ω//205

=

Ω12

a b

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V25.92V

(64.8)

64

4

VR 4

4V

o

oc

TH

o

=+

=

+=

3. Reconnect the load :

+

-

VOC

R TH

+

-

Vo Ωk 44Ω

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Using Norton Theorm:

1. Find ISC :

KVL around upper loop :

(1)0i8i5i25

0)i(i5)i(i8i12

321

21311

KK=−−

=−+−+

+

-

Ω8Ω5

Ω20

Ω72

Ω12

i1

i3

i2

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KVL around lower loop :

KVL around right loop :

(2)72i20i25i572)i(i20)i(i5

321

3212

KK=−+− =−+−

(3)0i28i20i8

0)i(i20)i(i8

321

2313

KK=+−−

=−+−

A10.8I

A10.8i,A12.72i,A6i

SC

321

=⇒

===

i d

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2. Find R TH

From before , R TH = 6 Ω3. Reconnect the load

( ) V25.9210.846

64V

I4R

R Ω)(4

iΩ

)(4V

o

SC

TH

TH

2o

=

+=

+=

=

a

b

ISC

R TH

i1 i2

Ω4

+

-

Vo

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Case(2) : Circuits containing only dependent sources

Here there is NO energy source in the circuit.¾ VOC is always zero and ISC is always zero

¾ So we can only find R TH

Procedure for finding R TH

1. Connect an independent voltage ( or current) source at the

terminals ,Vx (or Ix)

2. Find the corresponding current ( or voltage) at the terminal ,Io ( or Vo)

3. Find R TH = Vx /Io or R TH = Vo/Ix a

b

R TH

Example:

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Example:

Find the Thevenin equivalent circuit

1. Apply voltage source at the terminals (Vx=1V)

Ωk 3

2000 Ix

Ωk 2

Ωk 4

Ix

a

b

Ωk 3

2000 Ix

Ωk 2

Ωk 4

Ix

+

-

Vx = 1Vi1i2

V1

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0IIk 3

4

k 3

1I2I

0I

3000

I40001

2000

I4000I2000

Ik)(4Vwhere

XXXX

XXXX

X1

=−−+−

=−−

+−

=

KCL at node V1 :

0Ik 3

V1

k 2

VI2000

0Iii

X11X

X21

=−−

+−

=−+

Am0.1I

30001

342I

X

X

=

=

+

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( ) ( ) ( )k 3

Am0.1k 41

k 3

Ik 4V

k 3

VVi

XX

1X2

−=

−=

−=

Ωk 5

Am0.2

V1

i

VR

Am0.2i

2

XTH

2

===

=

a

b

R TH

Case (3) : Circuits containing both independent and

7/30/2019 Elec 3202 Chap 4



Case (3) : Circuits containing both independent and

dependent sources

Procedure of Thevenin or Norton Theorms:

a. Find the open circuit voltage and the terminals ,VOC

b. Find the short circuit current at the terminals, ISC .

c. Compute R TH = VOC/ISC

Note :

R TH can not be found as in the case of only independent sources

d. Construct the Thevenin or Norton circuits

a

b

ISC

R TH

Norton circuitThevenin circuit

a

b

R TH

+

-Voc

Example :

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Example :

Find the Thevenin equivalent circuit with respect to the terminals a, b

1. Find VOC :

+

-

Voc

Ω20

Ω80Ω60

4 A Ω40

160 ix

ix

a

b

+

-

Voc

Ω20

Ω80

Ω60

Ω40

160 ix

ix+

-

i240 V i1

Z

KVL d h lif l

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KVL around the lift loop :

(1)240i200i80

0i40i160i80240

x

xx

KK=+

=+++−

(2)0ii2

i40i80

x1

x1

KK=−

=

KVL around right loop :

KCL at Z:

(3)0iii x1 KK=−−

A0.75i

A0.375iA1.125i

x

1

=

==

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V30V

Ω)(40(0.75A)

Ω)(40iV

OC

xOC

=

=

=∴

2. Find ISC :

Since we have short circuit , 80 // 40 // 0 = 0

Ω20

Ω80Ω60

4 AΩ40

160 ix

ix

V

i 0

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Current divider

( ) A34

2060

60ISC =

+

=

Ω10A3V30

IVR

SC

OCTH ===

Ω20

Ω60

4 A ISC

3. Find R TH

ix=0

160ix source is zero

a

b

R TH

+

-Voc4.

V

Example :

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Example :

Use Thevenin theorem to find the Thevenin equivalent circuit with

respect to a, b

1. Find VOC

KCL at node z :

+

- Ω1

Ω5V40

ix

2 ix

i1 a

b

Z

8 A

(1)8ii3

0i8ii2

1x

1xx

KK−=−

=−++

KVL d l

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(2)40ii5

0i1i540

1x

1x

KK=+

=++−

KVL around outer loop

V20i1V

A20i,A4i

1OC

1x

==⇒

==⇒

Find ISC :

+

-Ω1

Ω5V40

ix

2 ix

a

b

8 AISC

2 i

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KVL around outer loop :

A8i0i540 xx =⇒=+−

( ) ( ) A32883I

I8i3

I8ii2

SC

SCx

SCxx

=+=⇒

=+=++

Ω0.62532

20

I

V

R SC

OC

TH ===

KCL at z :

3. Find R TH :

+

-

Ω5V40

ix

2 ix

8 AISC

Z

a

b

R TH

+

-Voc

Thevenin equivalent circuit is

4 Maximum Power Transfer

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4. Maximum Power Transfer• A technique in which the load is selected to maximize the

power transfer.• This technique is based on the Thevenin equivalent circuit.

L

2

LTH

OC

L

2

LL

R R R

V

R iivP

+=

==

RL

RTH

+

-

Voc

i +

VL

--

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( ) ( )

TH

2

OCmaxL

TH

2

OC

2

TH

TH

2

OC

TH

2

TH

OC

L

2

maxL

R 4

VP

R 4

V

R 4

R V

R

R 2

V

R iP

=

==

=

=

RL

RTH

+

-

Voc i

Example:

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Example:

•Find R L for maximum Power Transfer ?

•Find the maximum Power transfer to R L ?

Ωk 4Am3+

-

10 V+ -

Ωk 01

Ωk 02

Ωk 5.2Ωk 8

R L

10 V

Let’s find Thevenin equivalent circuit

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Ωk 4Am3+

-

10 V

+ -

Ωk 01

Ωk 02

Ωk 5.2Ωk 8

10 V

i1 i4

i3

i2 +

-

Voc

V1 V2

Let’s find Thevenin equivalent circuit .

KCL at node V1 :

0Ωk 8

VV

Ωk 4

VAm3

0iiAm3

211

21

=−

−−

=−−

111

7/30/2019 Elec 3202 Chap 4



(1)m3Vm0.125Vm0.375

m3Vk 8

1

k 8

1

k 4

1V

21

21

KK=−

=

−

+

KCL at node V2:

0k 12.5

V

k 20

10V

k 8

VV0iii

2221

432

=−−

−−

=−−

(2)m0.5Vm0.255Vm0.125

m0.5Vk 12.5

1

k 20

1

k 8

1

k 8

1V

21

21

KK−=−

−=

++−

V7.03V

V10.34V

2

1

=

=

ik1010V +=

7/30/2019 Elec 3202 Chap 4



( )

V4.375V

7.0312.5

1010

k 12.5

V2k 1010

ik 1010V

OC

4OC

−=

+−=

+−=

+−=

To find R TH :

Ωk 8 Ωk 5.2

Ωk 4 Ωk 02Ωk 01

R TH

( )[ ] k10//k2 5k20//k4k8R ++=

7/30/2019 Elec 3202 Chap 4



( )[ ]

( )[ ]

( )

Ωk 5R

k 10//k 10

k 10//k 2.5k 7.5

k 10//k 2.5k 20//k 12

k 10//k 2.5k 20//k 4k 8R

TH

TH

=

=

+=

+=

++=

( )

Wm0.957P

k)(54

4.375

R 4

VP

maxL

2

TH

2

OCmaxL

=

−==

R TH

+

-V

oc

R L

= R TH

Example :

7/30/2019 Elec 3202 Chap 4



Example :

1. Find R L for maximum Power Transfer?

2. Find max. power transfer to R L ?

First , find Thevenin equivalent:

RL

3k ohm

1k ohm

4m A

+ -

2000 Ix

2k ohm

4k ohm

Ix

4k ohm2000 Ix'

1k ohm

7/30/2019 Elec 3202 Chap 4



+

Voc

--

Using source transformation

+

Voc

--

4k ohm

2k ohm

+ -

4m A

1k ohm

3k ohm Ix’

+

-16V

4k ohm

+ -

2000 Ix'

2k ohm

4k ohm

Ix’

KVL around the loop:

7/30/2019 Elec 3202 Chap 4



Now, find Isc:

KVL around the loop:

-16 + 4k Ix’ - 2k Ix’ + 2k Ix’=0

Ix’= 4mA.

Voc= (2k Ω) Ix’= 8V.

4k ohm

2k ohm

+ -

2000 Ix''4k ohm

+

-

16V Ix’’ Isc

I1

V1

KCL at V1:

7/30/2019 Elec 3202 Chap 4



KCL at V1:

I1 - Ix’’ – Isc = 0

04

1

2

1

4

16

=−− k

V

k

V

k

04

1

2

1

4

)''21(16 =−−−−k

V

k

V

k

kIxV

And

Where V1=2k Ix’’

Hence,

Or V1=5.333V

mA

k

V Isc 333.1

4

1==

V8V

7/30/2019 Elec 3202 Chap 4


( )( )

Wm38P

k 24

64

k 64

8

R 4

VP

Ωk 6Am1.333

V8

I

VR

L(max)

2

TH

2

OCL(max)

SC

OCTH

=

===

===

elec 3202 chap 4

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