CHAPTER 3 NETWORK THEOREM. 2 NETWORK THEOREM Superposition Theorem Source Transformation Thevenin...

CHAPTER 3

NETWORK THEOREM

2

NETWORK THEOREM

Superposition Theorem Source Transformation Thevenin & Norton Equivalent

3

SUPERPOSITION THEOREM

4

Principle

If a circuit has two or more independent sources, the voltage across or current through an element in a linear circuit, is the algebraic sum of voltages across or current through that element due to each independent source acting alone.

5

Steps to apply superposition principle

Turn off all independent sources except one sources.

For voltage source replace by short circuit For current source replace by open circuit.

Find voltage or current due to that active source using any technique.

Repeat the procedure for each of the other source

Find total contribution by adding algebraically all the contribution due to the independent source

6

Practice Problem 10.5Find current in the circuit using the superposition theorem

7

Solution

0j4Ij2)I(8 21

03010j4Ij4)I(6 o12

12 j2)I(0.5I

Let Io = Io’ + Io” where Io’ and Io” are due to voltage source and current source respectively.

For Io’ consider circuit beside where the current source is open circuitFor mesh 1

For mesh 2

…(1)

…(2)

8

Solution

j0.5560.08I'I 1o

o11 3010j4Ij2)Ij4)(0.5(6

j1411

3010I

o

1

Substitute equation (1) into equation (2)

9

Solution

2j81Z

53.0j4164.077.0j846.9

)769.2j846.1)(2(

4j||62 Z

For Io” consider circuit beside where the voltage source is short circuitLet

769.2j846.14j6

24j

)0(2ZZ

Z"I o

21

2o

10

Solution

A45.651939.1 o

"I'II ooo

086.1j4961.0

Therefore

11

Practice Problem 10.6Calculate Vo in the circuit using superposition theorem

12

Solution

o03030sin(5t)

)81.12t4.631sin(5'v oo

oo 81.12-4.631(30)

j1.258

j1.25'V

Let vo = vo’ + vo” where vo’ is due to the voltage source and vo” is due to the current source.

For vo’ we remove current source which is now open circuit

Transfer the circuit to frequency domain ( = 5)

By voltage division

j)(0.2F

5jH1

13

Solution

ot 02)10cos(2

(2)ZZ

ZI

21

2

For vo” we remove voltage source and replace with short circuitTransfer the circuit to frequency domain ( = 10)

By current division

j0.5)(F2.0 10jH1

Let

j0.5)(Z1 j10||8Z2

j3.94.878j108

j80

j0.1382.096

14

Solution

)86.24t1.05cos(10)81.12t4.631sin(5 oo "v'vv ooo

)86.23t1.05cos(10"v oo

Therefore

Thus

1o IZ"V j0.5)j0.138)((2.096

o23.86050.1

15

SOURCE TRANSFORMATION

16

Source Transformation

Source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa.

17

Example 10.7Calculate Vx in the circuit by using source transformation

18

Solution

j4)A(9045

9020I o

o

S

If we transform the voltage source to a current source, we obtain the circuit as shown.

The parallel combination of 5 resistance and (3+j4) impedance gives a new equivalent impedance Z1 j1.25)Ω(2.5

j48

j4)5(3Z1

19

Solution

j1.25)j4)(2.5(ZIV 1SS

V285.519V oX

Convert the current source to a voltage source yields the circuit as below

We then could solve for VX by using voltage division

j10)V(5

j10)(5j134j1.252.510

10VX

20

Practice Problem 10.7Find Io in the circuit by using the concept of source transformation

21

Solution

16j123j)(j4)(4ZIV SSS

If we transform the current source to a voltage source, we obtain the circuit as shown.

22

Solution

j2j34Z

We transform the voltage source to a current source as shown below

j2-6j31

j26

j1612

S

SS Z

VI

j)(13

10

j36

j2)(j5)(6j5||Z

Let Then

Note that

23

Solution

By current division

j3)(1.5j2)(1j)(1

310

j)(13

10

Io

o

o

j

j

1.17602.13

56.11672.44

413

4020

A46.99288.3 o

24

THEVENIN & NORTON EQUIVALENT CIRCUIT

25

Equivalent Circuit

Thevenin Equivalent Circuit

Norton Equivalent Circuit

26

Relationship

NNth IZV Keep in mind that the two equivalent circuit are related as

Nth ZZ and

Vth = VOC = open circuit voltage

IN = ISC = short circuit current

27

Steps to determine equivalent circuit

Zth or ZN

Equivalent impedance looking from the terminals when the independence sources are turn off. For voltage source replace by short circuit and current source replace by open circuit.

Vth

Voltage across terminals when the terminals is open circuit

IN

Current through the terminals when the terminals is short circuit

NoteWhen there is dependent source or sources with difference frequencies, the step to find the equivalent is not straight forward.

28

Practice Problem 10.8Find the Thevenin equivalent at terminals a-b of the circuit

29

Solution

j26

j2)j4)(6(10

j3.22.410

j2)(6||j4)(10Z th

To find Zth, set voltage source to zero

j3.212.4

30

Solution

o

oo

43.18324.6

)2030)(904(

o57.5197.18

j26

)20j4)(30()20(30

j4j26

j4V

oo

th

To find Vth, open circuit at terminals a-b

31

Example 10.9Find Thevenin equivalent of the circuit as seen from terminals a-b

32

Solution

oo 0.5II15

To find Vth, we apply KCL at node 1

Applying KVL to the loop on the right-hand side

0Vj3)(40.5Ij4)(2I thoo

j55)(j3)5(4j4)10(2Vth

V9055V oth or

10AIo

Thus the Thevenin voltage is

33

Solution

oo 0.5II3 At the node apply KCL

Applying KVL to the outer loop

j)2(6j4)2j3(4IV oS

j0.6667)Ω(4

3

j62

I

VZ

S

Sth

o

th 46.9055.4Z

The Thevenin impedance is

2AIo

To find Zth, we remove the independent source. Due to the presence of the dependent current source, connect 3A current source to terminals a-b

34

Practice Problem 10.9Determine the Thevenin equivalent of the circuit as seen from the terminals a-b

35

Solution

j48

VV5

j24

V 211

j0.53

50VV 21

0j48

VV)V0.2(V5 21

21

To find Vth, consider circuit besideAt node 1,

Thus node 2 become

)Vj0.5)(V(150j)V(2 211

12 j0.5)V(3j0.5)V(150

0j48

VV0.2V5 21

O

At node 2,

21O VVV but

…(1)

…(2)

36

Solution

V72.97.35VV o2th

Substitute equation (2) into equation (1)

j0.53

j0.53(50)j0.5)V(3j0.5)V(150 22

j12)(3537

50j)V(2500 2

o2 72.97.35

j2

j16.222.702V

37

Solution

j24j48

V0.2VI S

OS

SO Vj24j48

j48V

j212

j0.82.6

j212

1

j212

j48(0.2)IS

To find Zth, we remove the independent source and insert 1V voltage source between terminals a-b

At node a,

So

1VS But

And

38

Solution

oth 7.64473.4Z

Therefore

SS

Sth I

1

I

VZ

o17.102.72

46.9166.12

j0.82.6

j212

o

39

Example 10.10Obtain current Io by using Norton’s Theorem

40

Solution

5ZN

To find ZN,

i) Short circuit voltage source

ii) Open circuit source

As a result, the (8-j2) and (10+j4) impedances are short circuit.

41

Solution

0j4)I(10j2)I(8j2)I(18j40 321

To find IN,

i) Short circuit terminal a-b

ii) Apply mesh analysis

Notice that mesh 2 and 3 form a supermesh.

Mesh 1

0j2)I(18j4)I(10j2)I(13 132 Supermesh

…(1)

…(2)

42

Solution

j833II 23

At node a, due to the current source between mesh 2 and 3

3II 23

j8I05Ij40 22

j8)A3(II 3N

Adding equation (1) and (2) gives

The Norton current is

From equation (3)

…(3)

43

Solution

NO Ij15205

5I

A38.481.465j35

j83 o

By using Norton’s equivalent circuit along with the impedance at terminal a-b, we could solve for Io.

By using current division

44

Practice Problem 10.10Determine the Norton equivalent circuit as seen from terminal a-b. Use the equivalent to find Io

45

Solution

j3)(9||j2)(4

j13

j3)j2)(9(4

To find ZN,

i) Short circuit voltage source

ii) Open circuit source

ZN =

j0.706)Ω(3.176

=

=

46

Solution

To find IN,

i) Short circuit terminal a-b

ii) Solve for IN using mesh

analysis

47

Solution

03j)I(9j3)I(18I20 321

j4II 21

0j3)I(18Ij)I(13 213

Supermesh

Mesh 3

Solving for IN

oo

o

2N 32.688.39618.439.487

51.1179.65

j39

j6250II

…(1)

…(2)

…(3)

48

Solution

Using Norton equivalent, we

could find Io

o

oo

o 18.0513.858

)32.68)(8.39612.53(3.254I

oo 2.101.971I

)32.68(8.396j4.29413.176

j0.7063.176I

j510Z

ZI o

NN

No

49

Problem 11.14

a) Determine Thevenin equivalent circuit looking from the load, Z.

b) Determine load, Z that will produce maximum power transfer.

c) Value of the maximum power.

(is = 5 cos 40t A)

50

Problem 10.36 (Buku Electric Circuit by Nilsson & Riedel)

a) Determine Thevenin equivalent circuit looking from the load, Z.

b) Determine load, Z that will produce maximum power transfer.

c) Value of the maximum power.

51

Problem 11.15

a) Determine Thevenin equivalent circuit looking from the load, ZL.

b) Determine load, ZL that will produce maximum power transfer.

c) Value of the maximum power.