electric circuits i theorems.pdffor the circuit in fig., find i o when v s = 12 v and v s = 24 v....

Electric Circuits I

Dr. Eng.

Hassan M. Ahmad [email protected], [email protected]

كــلــيـــة هندسة الحاسوب والمعلوماتية

Computer and Informatics Engineering

Faculty

Dr. H

assan A

hmad

2

Chapter 4 Circuit Theorems

4.1 Motivation 4.2 Linearity Property 4.3 Superposition 4.4 Source Transformation 4.5 Thevenin’s Theorem

4.6 Norton’s Theorem

4.7 Maximum Power Transfer

23-Sep-18 Dr. Eng. Hassan Ahmad Dr. H

assan A

hmad

3

If you are given the following circuit, are there any other alternative(s) to

determine the voltage across 2Ω resistor?

What are they? And how?

Can you work it out by inspection?

4.1 Motivation


assan A

hmad

4

4.2 Linearity Property

A linear circuit is one whose output (i) is linearly related (or directly

proportional) to its input (vs), as shown in Fig.

Ohm’s law relates the input i to the output v,

23-Sep-18 Dr. Eng. Hassan Ahmad

1v iR i v

R

Dr. H

assan A

hmad

5

Example 4.1. By assume Io = 1 A, use linearity to find the actual value of Io in the circuit shown

in Fig.

Solution:

If Io = 1 A, then

Applying KCL at node 1 gives

Applying KCL at node 2 gives

Therefore, Is = I4 = 5 A.

This shows that assuming Io = 1 gives Is = 5 A, the actual source current of 15 A

will give Io = 3 A as the actual value.


1 1 1(3 5) 8 and / 4 2 A.oV I V I V

22 1 2 1 2 33 2 8 6 14 V, 2A

7o

VI I I A V V I I

4 3 2 5 AI I I

Dr. H

assan A

hmad

6

Example 4.2. For the circuit in Fig., find Io when vs = 12 V and vs = 24 V.

Solution:

For loop 1,

For loop 2,

But vx = 2i1 Equation (2) becomes

By solving Eqs. (1) and (3), we have

When,

showing that when the source value is doubled, Io doubles.


1 2

1 1 1 2

12 4 0 (1)

6 2 4( ) 0s

s

i i i

i

i

v

v

i

1 2

2 1 2 2

4 16 3 0 (

4( ) 8 4 3

2)

0

x s

s x

i i v v

v i i i i v

1 210 16 0 (3)si i v

1 2 26 ,76

svi i i

2 2

12 2412V A; 24V A

76 76s o s ov I i v I i

Dr. H

assan A

hmad

7

4.3 Superposition Theorem

It states that the voltage across (or current through) an element in a linear circuit

is the algebraic sum of the voltage across (or currents through) that element due

to each independent source acting alone.

The principle of superposition helps us to analyze a linear circuit with more

than one independent source by calculating the contribution of each

independent source separately.

For example, we consider the effects of 8A and 20V one by one, then add the

two effects together for final vo.


assan A

hmad

8

Steps to apply superposition principle

1. Turn off all independent sources except one source. Find the output

(voltage or current) due to that active source using nodal or mesh analysis.

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution (المساهمة) by adding algebraically all the

contributions due to the independent sources.

Two things have to be keep in mind:

1. When we say turn off all other independent sources:

Independent voltage sources are replaced by 0 V (short circuit) and

Independent current sources are replaced by 0 A (open circuit).

2. Dependent sources are left intact ( سليمة تترك ) because they are controlled by

circuit variables.


assan A

hmad

9

Example 4.3. Use the superposition theorem to find v in the circuit of

Fig. Solution:

Let v1 and v2 are the contributions due to the 6-V voltage

source and the 3-A current source, respectively. Hence,

To obtain v1 we set the current source to zero, Fig.(a).

Applying KVL to the loop in Fig.(a) gives

• We may also use voltage division to get v1 by writing

To get v2 we set the voltage source to zero, as in Fig.(b).

• Using current division,

Finally,


1 2v v v

1 1 1 112 6 0 0.5 A 4 2 Vi i v i

1

4(6) 2V

4 8v

3 2 3

8(3) 2A 4 8V

4 8i v i

1 2 2 8 10 Vv v v Dr. H

assan A

hmad

10 23-Sep-18 Dr. Eng. Hassan Ahmad

Example 4.5. Find io in the circuit of Fig. using superposition.

Solution:

Let and are due to the 4-A current source

and 20-V voltage source respectively. Hence,

To obtain , see Fig.(a),

For loop 1,

For loop 2,

For loop 3,

But at node 0,

Substituting Eqs. (2) and (5) into Eqs. (3) and (4)

gives

By solving Eqs. (6) and (7), we have

'

oi"

oi

' " (1)o o oi i i

1 2)4 A (i

'

1 2 33 6 1 5 0 (3)oi i i i

'

oi

'

1 2 3 (4)5 1 10 5 0oi i i i

' '

3 1 (5)4o oi i i i

'

2

'

2

(6)3

(7)

2 8

5 20

o

o

i i

i i

' 52A

17oi

Dr. H

assan A

hmad

11 23-Sep-18 Dr. Eng. Hassan Ahmad

To obtain , see Fig.(b).

For loop 4, KVL gives

For loop 5,

But . Substituting this in Eqs. (9) and

(10) gives

By solving Eqs. (11) and (12), we have:

Finally,

"

oi

"

4 5 (9)6 5 0oi i i

"

4 510 20 5 0 (10)oi i i "

5 oi i

"

46 (4 10 1)oi i

"

4 2 15 (0 2)oi i

" 60A

17oi

' " 52 60 80.4706A

17 17 17o o oi i i

Dr. H

assan A

hmad

12

4.4 Source Transformation

Source transformation is another tool for simplifying circuits.

Basic to these tools is the concept of equivalence.

An equivalent circuit is one whose v-i characteristics are

identical with the original circuit.

It is the process of replacing a voltage source vS in series with a

resistor R by a current source iS in parallel with a resistor R, or

vice versa.

Source transformation requires that


or s

s s s

vv i R i

R

Dr. H

assan A

hmad

13

Note from Figures:

The arrow of the current source is

directed toward the positive

terminal of the voltage source.

Note from eq.

The source transformation is not

possible when R = 0 for voltage

source and R = ∞ for current

source.


Transformation of dependent sources

Transformation of independent sources

or s

s s s

vv i R i

R

Dr. H

assan A

hmad

14

Solution:

We first transform the current and voltage sources to

obtain the circuit in Fig.(a).

Combining the 4-Ω and 2-Ω resistors in series and

transforming the 12-V voltage source gives us Fig.(b).

We now combine the 3-Ω and 6-Ω resistors in parallel

to get 2-Ω.

We also combine the 2-A and 4-A current sources to

get a 2-A source.

Thus, by repeatedly applying source transformations,

we obtain the circuit in Fig.(c).

We use current division in Fig.(c) to get

Alternatively, since the 8-Ω and 2-Ω and resistors in

Fig.(c) are in parallel, they have the same voltage

across them. Hence,


2(2) 0.4 A 8 8(0.4) 3.2 V

2 8oi v i

8 2(8 || 2)(2A) (2) 3.2 V

8 2ov

Example 4.6. Use source transformation to find vo in the circuit of Fig.

Dr. H

assan A

hmad

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4.5 Thevenin’s Theorem

Thevenin’s theorem states that a linear two-terminal circuit, Fig.(a), can be

replaced by an equivalent circuit, Fig.(b), consisting of a voltage source VTh in

series with a resistor RTh,

where

VTh is the open-circuit voltage at the terminals.

RTh is the input or equivalent resistance at the terminals when the independent

sources are turned off.


assan A

hmad

16

Our major concern right now is how to find the Thevenin equivalent voltage VTh

and resistance RTh .


VTh By removing the load, the terminals are made open-

circuited, then no current flows, so that the open-circuit

voltage across the terminals, Fig.(a) must be equal to the

voltage source in Fig. (b), (see previous slide).

Thus, VTh is the open-circuit voltage across the

terminals, Fig.(a); that is,

RTh Again, with the load disconnected and terminals a-b

open-circuited, we turn off all independent sources.

The input resistance (or equivalent resistance) of the

dead circuit at the terminals a-b must be equal to RTh,

Fig.(b), that is,

NOTE.

If the network has dependent sources, we turn of f all

independent sources and leave the dependent source alone.

We may insert a voltage (vo) or current source (io) at

terminals a-b as shown in Fig.

oTh cV v

Th i eqnR R R

Dr. H

assan A

hmad

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A linear circuit with a variable load can be

replaced by the Thevenin equivalent,

exclusive of the load.

The equivalent network behaves the same

way externally as the original circuit.


Th

Th

Th

Th

;L

L

LL L L

L

VI

R R

RV R I V

R R

Dr. H

assan A

hmad

18

Solution:

RTh

We find by turning off the 32-V voltage source

(replacing it with a short circuit) and the 2-A current

source (replacing it with an open circuit). The circuit

becomes what is shown in Fig.(a).

Thus,

VTh

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

Applying mesh analysis to the two loops, we obtain

Solving for i1, we get: i1 = 0.5 A.

Thus,

Example 4.7. Find the Thevenin equivalent circuit of the circuit shown in Fig., to the left of the

terminals a-b. Then find the current through RL= 6, 16 Ω.


Th

4 12(4 ||12) 1 1 4

4 12R

1 1 2 232 4 12( ) 0, 2Ai i i i

Th 1 212( ) 30 VV i i

(6 ) (16 )

Th

Th

303A, 1.5A

4L L L

L L

VI I I

R R R

Thevenin equivalent circuit

Dr. H

assan A

hmad

19

Solution:

To find RTh, we excite the network with a voltage

source vo = 1 V connected to the terminals as

indicated in Fig. (a).

Our goal is to find the current io through the

terminals, and then obtain RTh = vo ∕ io = 1∕io.

(Alternatively, we may insert a 1-A current source,

find the corresponding voltage vo, and obtain

RTh = vo ∕1).

For loop 1,

But

hence,

For loop 2 and 3,

Solving these equations gives

But io = − i3 = 1∕6 A. Hence,

Example 4.8. Find the Thevenin equivalent of the circuit in Fig. at terminals a-b.


1 2 1 22 2( ) 0x xv i i v i i

2 1 24 ;xi v i i

1 2 (1)3i i

2 2 1 2 3

3 2 3

4 2( ) 6( ) 0 (2)

6( ) 2 1 0 (3)

i i i i i

i i i

3

1A

6i

Th 1V 6 .oR i Dr. H

assan A

hmad

20

To get VTh, we find voc in the circuit of Fig.(b).

Applying mesh analysis, we get

For loop 1,

For loop 2,

For loop 3,

But

Solving the equations (5), (6) and (7) we have

Hence,

The Thevenin equivalent is as shown in Fig.


1 4)5 (i

3 2 3 22 2( ) (5)0x xv i i v i i

2 1 2 3 2

2 1 3

4( ) 2( ) 6 0

1 2 (6)2 4 0

i i i i i

i i i

1 24( ) (7)xi i v

2

10A

3i

Th 26 20 VocV v i

Dr. H

assan A

hmad

4.6 Norton’s Theorem

Norton’s theorem states that a linear two-terminal

circuit can be replaced by an equivalent circuit of a

current source IN in parallel with a resistor RN,

where

• IN is the short circuit current (isc) through the

terminals.

• RN is the input or equivalent resistance at the

terminals when the independent sources are

turned off.

Thevenin and Norton resistances are equal; that is,

The Thevenin’s and Norton equivalent circuits are

related by a source transformation; that is,

23-Sep-18 Dr. Eng. Hassan Ahmad 21

ThNR R

Th

Th

N

VI

R

Dr. H

assan A

hmad

22

Solution:

To find RN, we set the independent sources equal to

zero. See Fig.(a). Thus,

To find IN, we short-circuit terminals a and b,

Fig.(b).

We ignore the 5-Ω resistor because it has been short-

circuited. Applying mesh analysis, we obtain

From these equations, we obtain

Example 4.9. Find the Norton equivalent circuit of the circuit in

Fig. at terminals a-b.


20 55 (8 4 8) 4

20 5NR

1 2 12A, 20 4 12 0i i i

2 1A sc Ni i I

Dr. H

assan A

hmad

23

Alternatively, we may determine IN from VTh/RTh.

We obtain VTh as the open-circuit voltage across

terminals a and b in Fig.(c).

Using mesh analysis, we obtain

and

Hence,

Thus, the Norton equivalent circuit is as shown in

Fig.


3

4 3 4

2A

25 4 12 0 0.8A

i

i i i

Th 45 4Vocv V i

Th

Th

41A

4N

VI

R

Dr. H

assan A

hmad

24

Example 4.10. Using Norton’s theorem, find RN and IN of the

circuit in Fig. at terminals a-b.

Solution:

To find RN , we set the independent voltage

source equal to zero and connect a voltage

source of vo = 1 V, Fig.(a).

In Fig.(a), the 4-Ω resistor is short-circuited, so

we ignore it.

The 5- Ω resistor, the voltage source, and the

dependent current source are all in parallel.

Hence, ix = 0.

At nod a,


1V0.2A

5oi

5oN

o

vR

i

Dr. H

assan A

hmad

25

To find IN, we short-circuit terminals a and b

and find the current isc, as in Fig. (b).

Note from this figure that the 4 Ω resistor, the

10-V voltage source, the 5-Ω resistor, and the

dependent current source are all in parallel.

Hence,

At nod a, KCL gives

Thus,


102.5A

4xi

102 2 2(2.5) 7A

5sc xi i

7ANI

Dr. H

assan A

hmad

4.7 Maximum Power Transfer

If the entire circuit is replaced by its Thevenin

equivalent except for the load, the power

delivered to the load is:

Maximum power is transferred to the load when

• This is known as the maximum power

theorem.

• Hence,

23-Sep-18 Dr. Eng. Hassan Ahmad 26

2

Th

2 h

Th

T, withL L

L

L

Vp i R R

RR

RR

ThLR R

2

Thmax

Th4

Vp

R

Dr. H

assan A

hmad

27

Example 4.11. Find the value of RL for maximum power

transfer in the circuit of Fig.

Find the maximum power.

Solution:

To get RTh , we use the circuit in Fig.(a) and

obtain

To get VTh , we applying mesh analysis to the

circuit in Fig.(b).

Applying KVL around the outer loop to get

VTh across terminals a-b, we obtain

For maximum power transfer,

and the maximum power is


Th 2 3 6 12 9R

1 2 2

1

12 18 12 0, 2A

2 3A

i i i

i

1 2 Th12 6 3 2(0) 0i i V Th 22VV

2 2

Thmax

Th

229 13.44W

4 2 9L Th

VR R p

R

Dr. H

assan A

hmad

28


assan A

hmad

electric circuits i theorems.pdffor the circuit in fig., find i o when v s = 12 v and v s = 24 v....

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