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## Chapter 5. Additional analysis

techniques

EMLAB

Contents

1. Introduction
2. Superposition
3. Thevenins and Nortons theorems
4. Maximum power transfer
5. Application Examples

EMLAB

1. Introduction

## Examples of equivalent circuits

To simply solution procedures, the
number of nodes or loops should be
minimized by replacing the original
circuits with equivalent ones.

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Linearity

1 (t )

i1 (t )

Linear system
L

i2 (t )

Linear system
L

2 (t )

Linear system
L

A1 (t ) B 2 (t )

Ai1 (t ) Bi2 (t )

## A system satisfying the above statements is called as a linear system. Resistors,

Capacitors, Inductors are all linear systems. An independent source is not a linear
system.
All the circuits in the circuit theory class are linear systems!

EMLAB

## Example of linear system

R1

output

Resistor

1 (t ) -

1k

R1

i1 (t )

1k

1 (t ) i1 (t ) R1

input

2 (t ) -

Capacitor
C1
1n

C1

i2 (t )

1n

1
2 (t )
C

( ) d

EMLAB

Example 5.1
For the circuit shown in the figure, determine the output voltage Vout using linearity.
First, arbitrarily assume that the output voltage is Vout = 1 [V].
Vout V2 1 [V ]

I1

V1
1 [mA]
3k

I2

V2
0.5 [mA]
2k

I 0 I1 I 2 1.5 [mA]

V1 4k I 2 V2 3 [V ]
Vo 2k I o V1 6 [V ]

For the arbitrary assumption that Vout = 1 [V], the source voltage Vo should be 6 V.
Then from linearity, the actual output voltage should satisfy the following relation.
actual
actual
Vo : Vout 12 : Vout
6 : 1 Vout
2 [V ]

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2. Superposition
Source superposition

VS

VS

VL

IS

I L,2

I L ,1

IL

Circuit

## Circuit with voltage source

set to zero (Short circuited)

Circuit

VL ,1

## Circuit with current source

set to zero(Open circuited)

Circuit

VL , 2

IS

I L I L ,1 I L , 2
VL VL ,1 VL , 2

## Superposition is utilized to simplify the original linear circuits.

If a voltage source is eliminated, it is replaced by a short circuit connected
to the original terminals.
If a current source is eliminated, it is replaced by an open circuit.
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Example 5.2

To provide motivation for this subject, let us examine the simple circuit below, in
which two sources contribute to the current in the network. The actual values of the
sources are left unspecified so that we can examine the concept of superposition.

=
1 3k i1 3k (i1 i2 ) 0
3k (i2 i1 ) 6k i2 2 0
6k
3k

i
1
1
i2 15k

+
1 3k i1 3k (i1 i2 ) 0
3k (i2 i1 ) 6k i2 0
6k
3k

3k i1 1

9k i2 2
3 1 1
1

1 2 15k

31 2
2
2
1

3k i1 1

9k i2 0

i
1
1
i2 15k

31

1

3k i1 3k (i1 i2 ) 0
3k (i2 i1 ) 6k i2 2 0
6k
3k

3k i1 0

9k i2 2

i
1
1
i2 15k

2
2
2

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Example 5.3

+
Io

1
6k

2 [mA]

1
1

3k 6k
Vo I o 6k 4 [V ]

2
[mA]
3

Vo

6k
3 [V ] 2 [V ]
3k 6k

Vo Vo Vo 6 [V ]

## To verify the solution, we can apply the loop analysis technique.

I1 2 [mA]
1k ( I1 I 2 ) 2k ( I1 I 2 ) 3 6k I 2 0

I 2 1 [mA], Vo 6 [V ]

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Example 5.4

10

Consider now the network in Fig. 5.4a. Let us use superposition to find Vo.

8k

3 24 [V ]
V1 6
8k 2k 7
3

6k 18
Vo V1
[V ]
6k 2 k 7

30
10

Vo k || 6k 2 [mA]
[V ]
7
3

Vo Vo Vo

48
[V ]
7
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i
a
RV

VS

RI
b

IS

## Single port network

f (i )
Ai B RTh i VTh
1. To find B, measure voltage with
i=0. (open circuit voltage)
2. To find A, measure the variation
of with i changing. (impedance)

i g (i )
i C D YNor I Nor
1. To find D, measure current with =
0. (short circuit current)
2. To find C, measure the variation of
i with changing. (admittance)

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## (1) Measure open circuit voltage (VOC) with a volt meter.

VS
I 0

Input resistance of a
voltmeter is infinite.
RTH

Circuit

VOC

IS

VOC

## (2) Measure resistance (RTH) with sources suppressed. Voltage

sources short circuited and current sources open circuited.
short

Circuit

Ohm meter

open
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## (1) Measure short circuited current (ISH).

VS

Input resistance of an
ammeter is zero.

I SH

Circuit

IS

I SH

RTH

## (2) Measure resistance (RTH) with sources suppressed.

Voltage sources short circuited and current sources open circuited.
short

Circuit

Ohm meter

open
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Example 5.6
Let us use Thvenins and Nortons theorems to find Vo in the network below.

Vo

Vo

1
1
1

3k 6k

6k
9 6 [V ]
3k 6k

3 [mA] 6 [V ]
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Example 5.7
Let us use Thvenins theorem to find Vo in the network in Fig. 5.9a.

VOC1

6k
12 [V ] 8 [V ]
3k 6k

VOC2 8 4k 2 [mA] 16 [V ]

VO

RTh1 2k

3k 6k
4 [ k ]
3k 6k

RTh1 4 [k]

8k
16 8 [V ]
16k

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## Apply an external voltage or current source between the terminals.

Determine the Thvenin equivalent of the network in Fig. 5.11a at the terminals A-B.

V1 V1 2Vx V1 1

0
1k
2k
1k

V1

Vx 1 V1
Io

5V1 2(1 V1 ) 2 0
4
3
, Vx
7
7

Vx 1 2Vx 1
3
3 15

[mA]
1k
1k
2k 2k 7k 14

RTh

1 [V ] 14
[ k ]
Io
15

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Example 5.10
Let us determine at the terminals A-B for the network in Fig. 5.12a.

V1 2000 I x V1 V1 V2

0
2k
1k
3k
V2 V1 V2

1m 0
3k
2k

Ix

V1
1k

## 3V1 6V1 2V1 2V2 5V1 2V2 0

2V2 2V1 3V2 6 2V1 5V2 6 0
V2

10
V
10
[V ] RTh 2 [k]
7
1m 7
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## In these types of circuits we must calculate both the open-circuit voltage

and short-circuit current to calculate the Thvenin equivalent resistance.
Let us use Thvenins theorem to find Vo in the network in Fig. 5.13a.

## KCL for the super-node around the

12-V source is
VOC 12 2000 I x VOC 12 VOC

0
1k
2k
2k
V
I x OC
2k

I x 0

## 4VOC 24 VOC 12 VOC 6VOC 36 0

I SC

Vo

12
2
[ k ]
3
1k

18 [mA]

1
1k 1k k
3

( 6 )

RTh

VOC 6 [V ]
VOC
1
[ k ]
I SC
3

3
18
( 6) [V ]
7
7

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Example 5.12
Let us find Vo in the network in Fig. 5.14a using Thvenins theorem.
I 2 2 [mA]
I1

Vx
2000

Vx 4k ( I1 I 2 )
Vx 2Vx 8 Vx 8 [V ]
VOC 2k I1 3 11 [V ]
V

3 2k I SC x 0, Vx 4k x 2m
2000

2000

Vx 8 [V ]
I SC

RTh

Vo

Vx
3
11

4m 1.5m [mA]
2000 2k
2

VOC
11

2 [ k ]
I Sc 11 [mA]
2

6k
33
11
[V ]
6k 2 k
4
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Example 5.14
Determine Vo in the circuit in Fig. 5.16a using the repeated application of
source transformation.
I SC

12
3k 6k
4 [mA], RTh
2k
3k
3k 6k

VOC 4m 2k 8 [V ], RTh 2k 2k 4k

I SC

8
2 [mA], RTh 4k
4k

VOC 4m 4k 16 [V ], RTh 4k

VO

8k
16k 8 [V ]
4k 4k 8k

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## 4. Maximum Power Transfer

RS
PL i 2 RL , i

PL
Equivalent circuit of
a signal source

RS RL

2 RL
( RS RL ) 2

## We want to determine the value of RL that maximizes

this quantity. Hence, we differentiate this expression
with respect to RL and equate the derivative to zero.
2
PL
2 ( RS RL ) 2 RL ( RS RL )

RL
( RS RL ) 4

## Maximum power transfer condition :

( RS RL )
0
( RS RL )3

RL RS
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Example 5.16
Let us find the value of RL for maximum power transfer in the network in Fig. 5.20a
and the maximum power that can be transferred to this load.

RL RTh 4k

6k 3k
6k
6k 3k

3k ( I 2 I1 ) 6k I 2 3 0, I1 2m
I2

1
[mA]
3

VOC 4k I1 6k I 2 8 2 10 [V ]

10
PL
6k 4.17 [mW ]
12k

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Example 5.17
Let us find RL for maximum power transfer and the maximum power transferred
to this load in the circuit in Fig. 5.21a.

VOC 2000 I x
V
V
4m OC 0, I x OC
4k
2k
2k
VOC 8 [V ]
I x 0 x I SC 4 [mA]

RTh

VOC
2 k
I SC
2

8
PL
6k 2.67 [mW ]
12k

EMLAB
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