Mazita – Sem 1 1112BEL10103

Chapter 4 Circuit Theorems

1. Linearity and Proportionality

2. Source Transformation

3. Superposition Theorem

4. Thevenin’s Theorem and Norton’s Theorem

5. Maximum Power Transfer Theorem

Learning Outcomes...

At the end of this topic, students should be able to:

• Simplify the circuit’s complexity by using Thevenin-Norton equivalent networks and/or source transformation

• Analyse the circuits by using the superposition theorem, etc.

Linearity and Proportionality

• Motivation:

Engineering combines the study of mathematics and natural and social sciences to direct the forces of nature for the benefit of humankind.

An engineer, in the accomplishment of a task,

1. Analyses the problem

2. Synthesises a solution

3. Evaluates the results and possibly re-synthesises a solution

⇒use models to represent the electric circuit element

Source: Richard C. Dorf and James A. Svoboda

Linearity and Proportionality (cont.)

• A device or element is said to be linear if its excitation and response satisfy the properties of superposition and homogeneity.

• Mathematically:

Superposition:

Homogeneity:

2121

22

11

vvii

vi

vi

+→+→→

kvki

vi

→→

then

then

Example

Consider the element represented by the relationship between current and voltage as v = Ri. Determine whether this device is linear.

Solution:

( )21

2121

22

11

iiR

RiRivv

Riv

Riv

+=

+=+

=

=

then

⇒satisfy the superposition property

12

1

22

12

kvv

Rki

Riv

kii

=

=

=

=

⇒satisfy the homogeneity property

then

∴This device is linear as it satisfies both superposition

and homogeneity properties.

Source Transformation

A source transformation is a procedure for transforming one source into another while retaining the terminal characteristics of the original source.

⇒ Based on the concept of equivalence, i.e. terminal characteristics still remain identical to those of the origin.

Source Transformation (cont.)

RL = 0 Ω

Fig. i:

Fig. ii:

RL = ∞∞∞∞ Ω

Fig. i:

Fig. ii:

Consider two extreme values of RL

s

s

R

vi =

sii =

sabvv =

PsabRiv =

⇒ For both circuits to be equivalent, vabmust be equal.

∴RS = Rp

Source Transformation (cont.)

Thevenin → Norton

Norton → Thevenin

Example

Find the source transformation for the circuits shown below.

Solution

For circuit i;

A 47

28

7

===

Ω==

S

S

S

SP

R

vi

RR

Solution (cont.)

For circuit ii;

V 24122

12

=×==

Ω==

PSS

PS

Riv

RR

Current 2A is flowing down ∴ reverse the terminal polarity for voltage source

Exercise

Using source transformation, determine the current i for the circuit shown below.

(Answer: 1.125 A)

Superposition Theorem

• Linear element holds superposition theorem, i.e.

The superposition principle requires that the total effect of several causes acting simultaneously is equal to the sum of the effects of the individual causes acting one at a time.

2121

22

11

vvii

vi

vi

+→+→→

then

How to apply the principle of Superposition Theorem?

The principle of superposition is only applicable for linear circuits consisting of linear elements and independent sources.

Steps in applying the superposition theorem:

1. Activate only one independent source at one time and deactivate the rest of independent sources. If the dependent source is available, it should remain active.

2. Determine the current or voltage where necessary.

3. Repeat steps 1-2 until the effects of all the independent sources in the network have been analysed.

4. Add the total currents or voltages.

Example 1

Find the current i in the 6 Ω resistor using the principle of superposition for the circuit shown below.

Solution

Consider the effect of 6 V voltage source:

A 3

2

63

61 =

+==

TR

vi

Note:

1. Set the current source to zero ⇒appears as an open circuit.

2. Label portion of current due to excitation by 6V source as i1.

3. Do circuit analysis.

Solution (cont.)

Consider the effect of 2 A current source:

A 3

22

63

3A2

63

32 =×

+=×

+=

ΩΩ

Ω

RR

Ri

Note:

1. Set the voltage source to zero ⇒appears as a short circuit.

2. Label portion of current due to excitation by 2 A source as i2.

3. Do circuit analysis.

Apply current divider:

Solution (cont.)

The total current, i = i1 + i2 :

Note:

1. Check whether all the independent sources have been analysed.

2. If yes, total up the current value.

A 3

43

2

3

221

=

+=

+= iii

Example 2

Find the current i using the principle of superposition for the circuit shown below.

Solution

Consider the effect of 24 V voltage source:

Note:

1. Set the current source to zero ⇒ appears as an open circuit. A dependent source should remain active.

2. Label current as i1. Dependent source is now referred to i1.

3. Do circuit analysis.A 3

248

032324

1

1

111

=

=

=+++−

i

i

iii

Apply KVL (follow the direction of current, i1)

Solution (cont.)

Consider the effect of 7 A current source:

Note:

1. Set the voltage source to zero ⇒ appears as a short circuit. A dependent source should remain active.

2. Label current as i2. Dependent source is now referred to i2.

3. Do circuit analysis.

(1)equation 145

142372

3

2

2222

L+=

+=−⇒+=−

iv

iiviiv

a

a

a

Apply KCL at node a;

Apply Ohm’s Law across 3 Ω resistor;

(2)equation 32 Liva

−=

Solution (cont.)

Solve equations (1) & (2) to get i2;

Finally, the total current, i: Note:

1. Check whether all the independent sources have been analysed.

2. If yes, total up the current value.

A 4

54

73

21

=

−=

+= iii

A 4

7

8

142 −=−=i

Exercise

Using the principle of superposition, find the voltage v of the circuit shown below.

(Answer: 4 V)

Thevénin’s Theorem

• Motivation:

To reduce the complexity of circuits.

• How?

Reduce some portion of the circuit to an equivalent source and a single element (refer to source transformation for derivation)

Thevenin equivalent circuit

Thevénin’s Theorem (cont.)

• Thevenin’s theorem requires that, for any circuit of resistance elements and energy sources with an identified terminal pair, the circuit can be replaced by a series combination of an ideal source, vTH and a resistance, RTH.

– vTH is the open-circuit voltage at the two terminals

– RTH is the input/equivalent resistance at the terminals when the independent sources are turned off.

Source: Richard C. Dorf & James A. Svoboda and Charles K. Alexander & Matthew N.O. Sadiku

Thevénin’s Theorem (cont.)

Summary of Thevenin Circuit Approach1) Identify circuit A and circuit B

2) Separate circuit A and circuit B

3) Replace circuit A with its Thevenin equivalent

4) Reconnect circuit B and determine the variable of interest

Example 1

Find the Thevenin equivalent circuit between the output terminals A and B of the circuit shown below.

Solution

Determine the voltage, VTH

Note:

1. Since there is no current flowing through R4, therefore no voltage drop across it. (∴Ignore R4)

2. Use voltage divider rule to determine the VTH = VAB.

V08.4

102204701000

220470

=

×++

+=

THV

Solution (cont.)

Determine the resistance, RTH

Note:

1. Turn off the independent source, i.e. replace the voltage source with short circuit.

2. Determine the total equivalent resistance across terminals A-B.

( )( )

Ω=

+++=

++=−

k 41.1

220470

1

1000

11000

//1

3214 RRRRRTH

Solution (cont.)

• The Thevenin equivalent circuit is

Exercise 1

For the circuit shown below, determine the Thevenin equivalent circuit as viewed from the output terminals A and B.

(Answer: RTH = 1.153 kΩ, VTH = 3.06 V)

Exercise 2

Find the Thevenin equivalent of the circuit shown below.

(Answer: RTH = 8 Ω, VTH = 32 V)

Thevenin Theorem (cont.)

When circuit contain one or more dependent sources as well as independent sources:

RTH is determined by:

1. Determine the open-circuit voltage, VOC (i.e. VTH)

2. Determine the short-circuit current, ISC3. RTH = VOC / ISC (which could be justified by writing KVL equation for the loop)

Exercise

Find the Thevenin equivalent circuit for the circuit shown below.

(Answer: RTH = 13.6 Ω, VTH = 12 V)

Thevenin Theorem (cont.)

What about when the circuit contains only dependent sources:

1. Definitely ISC = 0 A and VOC = 0 V.

RTH is determined by:

1. Connect 1 A current source at the terminal A-B

2. Determine the voltage at terminals A-B, VAB3. RTH = VAB / 1

Thevenin Theorem (cont.)

Please take note that:

Thevenin Equivalency Depends on the Viewpoint

Norton’s Theorem

• Motivation:

To reduce the complexity of circuits.

• How?

Reduce some portion of the circuit to an equivalent source and a single element

Norton equivalent circuit

Norton’s Theorem

• Norton and Thevenin equivalent circuits are related by a source transformation.

• To determine Thevenin or Norton equivalent circuit, we need to find

– iN, Norton current equals to the short-circuit current at the terminals of interest

– RN, Norton resistance equals to the Theveninresistance, i.e. resistance at terminals of interest when all the independent sources are off.

– vTH, Thevenin voltage equals to the open-circuit voltage across the terminals of interest.

Example

Determine the Norton equivalent circuit as seen by the RL for the circuit shown below.

Solution

Determine the current, INNote:

1. Short circuit terminals A-B.

2. To determine the IN, use current divider rule.

Ω=+

×+= 79

47100

4710047

TR

A05.179

3.83==

TI

A336.010047

47 Therefore,

=

×+

=TN

II

Solution (cont.)

Determine the resistance, RN

Note:

1. Turn off the independent source, i.e. replace the voltage source with short circuit.

2. Determine the total equivalent resistance across terminals A-B.

( )

Ω=

+=

+=

5.123

2

47100

// 213 RRRRN

Solution (cont.)

• The Norton equivalent circuit is

Exercise

Find the Norton equivalent of the circuit shown below.

(Answer: RN = 8 Ω, IN = 4 A)

Maximum Power Transfer

• Many applications of circuits require maximum power available from a source be transferred to a load resistor, RL.

• General problem of power transfer deals with its efficiency and effectiveness issues.

Example:

power utility systems to transport the power to the load with greatest efficiency by reducing the losses on the power lines

Signal transmission (communication): difficulty in attaining the maximum signal strength at the load (e.g. FM radio, mobile telephone).

Maximum Power Transfer (cont.)

• It is known that a complex circuitry could be reduced to its Thevenin equivalent:

Current, I is given by

Hence, the power to the load isL

THL

THR

RR

VP

2

+=

THL

TH

RR

VI

+=

Maximum Power Transfer (cont.)

To obtain maximum power transfer,

Solving this:

( ) ( )( )

022

4

22

=

+

+−+=

THL

THLLTHL

TH

L RR

RRRRRV

dR

dPNote:

Use different-iation :

To confirm the point is maximum:

2v

dx

dvu

dx

duv

v

u

dx

d−

=

THLRR =

02

2

<L

dR

PdPower delivered to the load varies with the changes of RL

values.

Maximum Power Transfer (cont.)

Hence

Maximum power transfer theorem states that the maximum power delivered by a source represented by its Theveninequivalent circuit is attained when the load RL is equal to the Thevenin resistance RTH.

L

TH

L

L

TH

R

VR

R

VP

42

22

max =

=

Maximum Power Transfer (cont.)

• The efficiency of power transfer is defined as the ratio of the power delivered to the load POUT to the power supplied by the source PIN.

We know that

Efficiency is

Therefore, the efficiency obtained at maximum power condition isonly ½ i.e. 50%.

IN

OUT

P

P=η

L

TH

L

TH

THTHIN

R

V

R

VVIVP

22

2

=

==

2

1

2

42

2

===

L

TH

L

TH

IN

OUTMAX

R

V

R

V

P

Pη

Example

Find the load RL that will result in maximum power delivered to the load for the circuit shown below. Also determine the Pmax and its efficiency.

Solution

Determine VTH Determine RTH

First, determine the Thevenin equivalent circuit.

V15018030150

150=×

+=

THV Ω=

+

×= 25

30150

15030TH

R

Therefore, maximum power transfer is obtained when RL = RTH = 25 Ω.

Solution (cont.)

Then, determine maximum power transfer and power transfer efficiency at RL = 25 Ω.

W225254

150

4

22

max =×

==L

TH

R

VP

W45031502525

150150 =×=

+×=

INP

Total power delivered by the Thevenin source is

The power transferred efficiency is

%50100450

225100% max =×=×=

INP

Pη

Note: The actual source of the circuit is 180 V and it delivers a power p = 180i1 where i1 is the current through 25 Ω, i.e. 3.5 A. Actual source delivers 630 W resulting in an efficiency of 35.7 %.

Exercise

Find the load RL that will result in maximum power delivered to the load of the circuit shown below. Determine the maximum power delivered to the load.

(Answer: RL = 12 Ω, Pmax = 3 W)

References

• Alexander Sadiku, Fundamentals of Electric Circuits, 4th

edition, McGraw-Hill, 2009

• Richard C. Dorf and James A. Svoboda, Introduction to Electric Circuits, 3rd edition, John Wiley, 1996

• Thomas L. Floyd and David M. Buchla, Electric Circuits Fundamentals, 8th edition, Pearson, 2010

• James W. Nilsson & Susan A. Riedel, Electric Circuits, 9th

edition, Pearson-Prentice Hall, 2011