1

Thevenin’s theorem, Norton’s Thevenin’s theorem, Norton’s theorem and Superposition theorem and Superposition

theorem for AC Circuitstheorem for AC Circuits

3/2/09

2

Problem10-60(b)Problem10-60(b)

20V

D

0

j5

C

4A 0

4

10

0

-j4

1. Find TEC at terminals c-d

TEC: Thevenin Equivalent Circuit

3

Problem10-60(b)Problem10-60(b)

(i) To obtain VTH

20V

0

j5 4A

Vc

0

4

10

0

Vd

-j4

+ Voc -

Voc =VTH

4

Problem10-60(b)Problem10-60(b)

nod c: Vc 20

10

Vc

5j

Vc Vd

4j 0

nod d: Vd Vc

4j

Vd

4 4

Nodal equations:

5

Problem10-60(b)Problem10-60(b)

Vc

Vd

Find Vc Vd Vc

Vd

13.333 13.333i

8 5.333i

Vcd Vc Vd Vcd 5.333 8i

Vcd 9.615 arg Vcd 56.31deg

VTH Vcd VTH 5.333 8i

6

Problem10-60(b)Problem10-60(b)

j5

10

4

-j4

(i) To obtain ZTHZTH

ZTH =[(10//j5)+4]//-j4

7

Problem10-60(b)Problem10-60(b)

In Mathcad:

First we define a function to decribe the result of two parallel impedances:

ZP x y( )x yx y

ZTH ZP ZP 10 5i( ) 4 4i ZTH 2.667 4i

ZTH 4.807

arg ZTH 56.31 deg

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Problem10-60(b)Problem10-60(b)

(iii) Using TEC to find Vo

+- ZL

ZTHVTH

+

Vo

-

ZL = 10+j10Let

THTHL

THo V

ZZ

ZV

ZL 10 10j ZTH 2.667 4i VTH 5.333 8i

Vo

ZL

ZTH ZLVTH Vo 2.353 9.412i

Vo 9.701 arg Vo 75.964deg

Let’s assume,

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Problem10-60(b)Problem10-60(b)

20V

D

0

j5

C

4A 0

4

10

0

-j4

2. Find NEC at terminals c-d

NEC: Norton Equivalent Circuit

10

Problem10-60(b)Problem10-60(b)

(i) To obtain IN

20V

0

j5 4A 0

4

10

0

-j4

IN

V1

11

Problem10-60(b)Problem10-60(b)

(i) To obtain IN

V1 20

10

V1

5j

V1

4 4

V1 Find V1 V1 12.923 7.385i

IN

20 V1

10

V1

5j IN 0.769 1.846i

IN 2arg IN 112.62deg

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Problem10-60(b)Problem10-60(b)

(ii) To obtain ZN

ZN = ZTH

ZN = ZTH= 2.667 – j4

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Problem10-60(b)Problem10-60(b)

(iii) Using NEC to find Vo

ZLZN

IN

+

Vo

-

ZL = 10+j10

Let’s assume,

Vo = (ZN//ZL)IN

IN 0.769 1.846i ZN 2.667 4i ZL 10 10i

Vo ZP ZN ZL IN Vo 2.353 9.412i

Vo 9.701 arg Vo 75.964deg

14

Problem10-60(b)Problem10-60(b)

3. Find Vo using superposition theorem

20V

j10+

0

10

j5 4A 0

4

Vo

10

0

-j4

-

15

Problem10-60(b)Problem10-60(b)

(i) 4A acting alone

j10+

0

10

j5 4A

V2

0

4

Vo1

10

V1

-j4

-

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Problem10-60(b)Problem10-60(b)

Nodal Equations

V11

10

1

5j

V1 V2 1

4j1

10 10j

0

V2 V1 1

4j1

10 10j

V2

4 4

V1

V2

Find V1 V2 V1

V2

3.548 9.267i

7.167 0.869i

Vo1 V1 V2 Vo1 3.62 10.136i

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Problem10-60(b)Problem10-60(b)

(i) 20V acting alone

j10+

0

10

j520V

V2

4

Vo2

10

V1

0

-j4

-

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Nodal Equations

V1 20

10

V1

5j V1 V2 1

4j1

10 10j

0

V2 V1 1

4j1

10 10j

V2

4 0

V1

V2

Find V1 V2 V1

V2

7.747 3.91i

1.774 4.633i

Vo2 V1 V2 Vo2 5.973 0.724i

Vo Vo1 Vo2 Vo 2.353 9.412i

Vo 9.701 arg Vo 75.964deg

19

Find vo

2H

10cos 2t

+ vo -

1 4

0.1F 5V2sin 5t

This circuit operates at the different frequencies:

• = 0 ; for the DC Voltage source• = 2 rad/s ; for the AC voltage source• = 5 rad/s ; for the AC current source

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We must use superposition theorem

Different frequencies problem reduces to single-frequency problem.

321 oooo vvvv

Due to 5V Due to 2sin 5t A

Due to 10cos 2t V

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(i) 5V ( = 0) acting alone

+ vo1 -

1 4

5 V

= 0 ; jL = 0 ; Short-circuited

1/jC = infinity ; Open-circuited

Vvo 1)5(41

11

22

(ii) 10V ( = 2 rad/s) acting alone

2H

0.1F

t2cos10 o0104jLj

51

jCj

Convert time-domain quantities to phasor quantities:

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(ii) 10V ( = 2 rad/s) acting alone

+ vo2 - 4

1 j4

-j5 100

951.1439.2

4//5

j

jZ

o

o02

79.30498.2

)010(41

1

Zj

V

In time domain:

V02(t)=2.498cos(2t-30.79)

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(iii) 2A ( = 5 rad/s) acting alone

2H

0.1F

t5sin5 o0510jLj

21

jCj

Convert time-domain quantities to phasor quantities

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(ii) 2A ( = 5 rad/s) acting alone

6.18.0

4//2

j

jZ

o

o1

10328.2

)02(110

10

Zj

jI

In time domain:

V03(t)=2.328sin(2t+10)o

103

10328.2

1

IV

J10

1 4

-j2

+ vo3 -

26

Total response of the circuit:

Note that we can only add the individual responsesin the time domain, not in the phasor.

v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10)

321 oooo vvvv

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Total response of the circuit:

Note that we can only add the individual responsesin the time domain, not in the phasor.

v0(t)= -1 + 2.498cos(-30.79)+2.328sin(5t+10)

321 oooo vvvv

28

Find vo

2H

2cos (4t+30)

+vo

-

6

1/12 F

6 cos 4t

Case #1:

This circuit operates at a single frequency.

The sources are represented by a cosine function.

1

4

29

2H

1/12 F

)304cos(2 ot

o06

8jLj

31

jCj

Convert time-domain quantities to phasor quantities:

t4cos6o302

= 4 rad/s

30

A phasor circuit

230

+vo

-

6

-j3

1

j8 4

60

31

Find vo

2H

2sin (4t+30)

+vo

-

6

1/12 F

6 sin 4t

Case #2:

This circuit operates at a single frequency.

The sources are represented by a sine function.

1

4

32

2H

1/12 F

)304sin(2 ot

o06

8jLj

31

jCj

Convert time-domain quantities to phasor quantities:

t4sin6o302

= 4 rad/s

We use the sine function as the reference for the phasor.

33

A phasor circuit

230

+vo

-

6

-j3

1

j8 4

60

34

Find vo

2H

2sin (4t+30)

+vo

-

6

1/12 F

6 cos 4t

Case #3:

This circuit operates at a single frequency.

One source is represented by a sine function;

Another source is represented by cosine function.

1

4

35

2H

1/12 F

)90304cos(2 oo t

o06

8jLj

31

jCj

Convert time-domain quantities to phasor quantities.Use the cosine function as a reference.

t4cos6o602

= 4 rad/s

36

2H

1/12 F

)304sin(2 ot

o906

8jLj

31

jCj

Convert time-domain quantities to phasor quantities.Use the sine function as a reference.

)904sin(6 oto302

= 4 rad/s

OR