UNIT I ELECTRICAL CIRCUITS ANALYSIS

Introduction
Electric Potential (E or V):
 It is measured between two points and its unit is volt.
 If the work done in moving a charge of one coulomb the two points is one joule, then the
potential of one point with reference to the second point is one volt.
𝑑𝑊
𝐸 (𝑜𝑟) 𝑉 = 𝑉𝑜𝑙𝑡
𝑑𝑄

Electric Resistance (R):

 The resistance of the circuit is the property by which it opposes the flow of current.
 It is responsible for energy dissipation.
 The resistance of conductor depends on length, cross-sectional area, material of the conductor
and temperature.
 It is given per unit cross-section and unit length. This is called specific resistance or resistivity
of the material (ρ).
𝜌𝑙
𝑅= 𝑂ℎ𝑚
𝑎

 The resistivity of aluminum is 0.0283 µ𝝮-m and copper is 0.0l 73 µ𝝮-m.

Electric Conductance (G):

 The reciprocal of resistance is called conductance. Its unit is Siemen (S).
1
𝐺 = 𝑆𝑖𝑒𝑚𝑒𝑛
𝑅
 The reciprocal of resistivity is called conductivity. Its unit is Siemen per meter (S/m).
1
𝜎 = 𝑆𝑖𝑒𝑚𝑒𝑛 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟
𝜌
 It is the element in which energy is stored in the form of electromagnetic field. It is like a
coil wound on a magnetic core (or) may be air core.
Basic Circuit Components:
The three types of basic circuit components are
1. Resistor 2. Capacitor 3. Inductor
Resistor: It is defined as the property of the material by which it opposes the flow of current through
it.

The relation between voltage and current is given by ohm‘s law.

V = IR

1
Energy dissipated in the resistor in the form of heat.
𝑉2
𝑃 = 𝑉𝐼 = (𝐼𝑅) 𝐼 = 𝐼2 𝑅 = 𝑊𝑎𝑡𝑡𝑠
𝑅
Resistor converts amount of energy into heat during time ‘t’.

𝑡 𝑡
𝑊 = ∫ 𝑃𝑑𝑡 = ∫ 𝐼 2 𝑅𝑑𝑡 = 𝐼 2 𝑅𝑡
0 0

𝑡
𝑉2 𝑉2
Or 𝑊 = ∫ 𝑑𝑡 = 𝑡
0 𝑅 𝑅

𝑡
𝑜𝑟 𝑊 = ∫ 𝑉𝐼𝑑𝑡 = 𝑉𝐼𝑡 𝐽𝑜𝑢𝑙𝑒𝑠
0
Capacitor:
 It is a storage element which can store and deliver energy in electric field.
 Any two metal plates between which an electric field can be maintained constitute a capacitor.

Inductor:
 It is the element in which energy is stored in the form of electromagnetic field.
 It is like a coil wound on a magnetic core (or) may be air core.

Network Definitions:
 Any arrangement of the various electrical energy sources along with the different circuit
elements is called an electric network.
 Any individual circuit element with two terminals which can be connected to other circuit
element is called a network element.
 Network elements are classified into two types i.e..active elements and passive elements.
 Active elements - Voltage and current sources.
 Passive elements - Resistor, capacitor and inductor.

2
 Branch:
A part of the network which connects the various points of the network with one another is called
a branch. The given figure A-C, C-E. E-F. etc., are called the branches of the network.

Junction Point (or) Node:

A point where two (or) more branches meet is called a junction point. The figure shows the
junction points are C and F.

A point at which two or more elements are joined together is called node.
Mesh (or) Loop:

Mesh or loop is a set of branches forming a closed path in a network. The figure shows the
Mesh or loops are ACDBA, ACEFDBA and CEFDC.
Types of Electrical Networks:

Electrical network can be classified according to the network characteristics and behavior.
Linear Network:
A circuit whose parameters are always constant of the change in time. voltage, temperature
etc., is known as linear network. In linear network. the Ohm's law can applied and super position
theorem can used for solving mathematical equations.
Non Linear Network:
A circuit whose parameters are change their values with change in time, voltage,
temperature etc., is known as non linear network. Ohm's law can applied and super position
theorem do not apply for this network.
Bilateral Network:
A circuit whose characteristics, behavior is same irrespective of the direction of current
through various elements of it, is called bilateral network.
Unilateral Network:
A circuit whose operation, behavior is dependent on the direction of the current through
various elements is called unilateral network.
Active Network:
A circuit which contains a source of energy is called active network.

3
Passive Network:
A circuit which contains a no energy source is called passive network.
Lumped Network:
A network in which all the network elements are physically separable is known as lumped
network.
Distributed Network:
A network in which the circuit elements like resistance, inductance, etc., cannot be
physically separable for analysis purpose is called distributed network.
SOURCE TRANSFORMATION
Voltage Source to Current Source:
Current Source to Voltage Source:

Problem 1

Convert the given current source into a voltage source. [AU/EEE - MAY 2004]

Solution:

4
Problem 2

Convert the voltage source into a current source for the circuit given below. [AU/EEE - MAY
2008]

Solution:

10
I= = 2A
5

Problem 3
Convert the voltage source into current source for the circuit given below.

Solution:

V 120
I = 𝑅 = 10 = 12A

Problem 4

Convert the current source into a voltage source for the circuit given below.

5
Solution:

V = IR = 5 × 8 = 40

OHM’S LAW

When the temperature remains constant, current flowing through a circuit is directly proportional
to potential difference across the conductor.

𝐸 𝛼 𝐼 (𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑏𝑒𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) 𝐸 = 𝐼𝑅

ELECTRICAL POWER (P)

 The rate at which work is done is power and its unit is joule per second or watt.

 When one coulomb of electric charge is moved through a potential difference of one volt
in one second, the rate of work is one joule per second or one watt.

 Hence power in electric circuit is obtained as a product of the voltage and current.

P = E.I Watts P = I2R (or) P = E2/R Watts

ELECTRICAL ENERGY (W)

 Energy is the total amount of work done and hence is the product of power and time.

W =Pt = E.It = I2Rt = (E2/R)t Joules (Watt-sec)

Power in Watts×Time in sec

 One Kilo Watts Hour is, 1 kWh = 1 unit = 1000×60×60

6
RESISTORS IN SERIES
When resistors are connected in series as shown in figure. Here the same current passes through
all of them, they are said to be in series.

Each resistor has a voltage drop across it given by Ohm’s law.

E1 = IR1; E2 = IR2 ; E3=IR3

The total drop is,

E = E1+ E2 + E3 = I (R1 R2 R3)

The total effective resistance of a number is resistors connected in series is,

Reff= ∑𝑛𝑠𝑒 = R se

The power expended in R1, R2 and R3 is,

E2 E2 E2
P1 = I2 R1 = (R1 ); P2 = I2 R2 = (R2 ); P3 = I2 R3 = (R3 )
1 2 3

The total power,

𝑃 = 𝑃1 + 𝑃2 + 𝑃3

𝑃 = 𝐼2 (𝑅1 + 𝑅2 + 𝑅3) = 𝐼2 𝑅𝑠𝑒

𝐸12 𝐸22 𝐸32

𝑃 = + +
𝑅1 𝑅2 𝑅3
7
 𝐸2
𝑃 =
𝑅𝑠𝑒

𝑃 = 𝐸𝐼

 The same current flows through all the resistances.

 For each resistance, there will be a voltage drop according to ohm's law.
 The sum of the voltage drops will be equal to the applied voltage.

Problem 5

Figure shows 3 reistors RA, RB and RC connected in series to a 250V source: Given RC = 50Ω, and
EB = 80 Volts when the current is 2 amperes, calculate the RA and RB.

Solution:

Current flowing through the circuit I = 2 A.

Voltage across resistor RB, EB = 80V
𝐸𝐵 = 𝐼𝑅𝐵

𝐸𝐵 80
𝑅𝐵 = = = 40𝑊
𝐼 2
𝑉
Also , 𝐼 =
𝑅𝐴 + 𝑅𝐵 + 𝑅𝐶

𝐸 250
𝑅𝐴 + 𝑅𝐵 + 𝑅𝐶 = = = 125𝛺
𝐼 2
𝑅𝐴 + 40 + 50 = 125𝛺

𝑅𝐴 = 35𝛺

Problem 6
Two resistors connected in parallel across 200 V supply take 10A from the mains. If the power
dissipated in one resistor is 800 W, find the value of the other resistor.
Solution:
Total power taken
Power dissipated in one resistor P, = 800 W

8
Therefore power dissipated in the other resistor
P = V x i = 200 x 10 = 2000 W
P2 = 1200 W

Problem 7

A lamp rated 500 W, 100V is to be operated from 220V supply. Find the value of the resistor to
be connected in series with the lamp. What is the power lost in the resistor?

Solution:
𝑃 500
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑎𝑚𝑝 = = = 5𝐴
𝑉 100
Since voltage drop across the lamp is 100 V,
Voltage to be dropped in the series resistor = 220 - 100= 120 V.
120
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟 𝑅 = = 24Ω
5
Power lost in this resistor = I2 x R = 52 x 24 = 600 W.

9
 RESISTORS IN PARALLEL

• When resistors are connected across one another such that the same voltage is applied to each,
then they are said to be in parallel.
• The current in each resistor is different and the current I taken from the supply divide among
all the three resistors.
• Each resistor has a current path is,

If there were only two resistors in parallel, then the effective resistance is,

1 1 1 𝑅1 𝑅2
= + =
𝑅𝑒𝑓𝑓 𝑅1 𝑅2 𝑅1 + 𝑅2

 The voltage across each resistances of the parallel circuit is the same.
 The current in each branch is given by ohm's law.
 The total current is equal to the sum of branch currents.
Division of current in parallel circuit:
I = I1+ I2 and E = I1R1 = I2R2

10
 So current in a branch is equal to total current multiplied by other branch resistance and
divided by sum of branch resistances.
Division of voltage in series circuit:

11
Problem 8
The effective resistance of two resistors connected in series, is 100 Q. When connected in parallel
the effective value is 24Ω. Determine the values of the two resistors.
Solution:

Problem 9
Three loads A, B and C are connected in parallel across a 250 V source. Load A takes 50 A. Load
B is a resistor of 10 Ω and load C takes 6.25 kW. Calculate (i) RA and Rc (ii) the currents IB and Ic
(iii). Power in loads A and B (iv) total current (v) total power and (vi) total effective resistance.

12
Solution:

Problem 10

Calculate the total power supplied by battery in the network shown in figure. (AU/March - Dec
2004)

13
Solution:

Problem 11
Determine the equivalent resistance between terminals A and B of figure shown below.

14
Solution :

15
Problem 12
Determine the total current taken from the source.(MKU/I YEAR - Apr 2002)

Solution:
50 Q and 50 Q are connected in series.

16
So, 50 + 50 = 100 Ω

100 Ω and 100 Ω are connected in series.

So, 100+ 100 = 200 Ω

17
18
Problem 13

Find the resistance across AB

19
Kirchhoff’s Current Law (I Law)

Statement

The sum of the currents flowing towards a junction is equal to the sum of the currents flowing
away from it.

In figure 1.1, A is a junction (or node) formed by six conductors. The currents in these conductors
are I1, I2 ….. I6.

Some of these currents are flowing towards A and other away from it.

According to Kirchhoff’s Law,

I1+I4+I5+I6 = I2+I3

(flowing towards A) = (flowing away from A)

The current at node A is equal to zero.

Kirchhoff’s Voltage Law (II Law)

Statement

In a closed circuit, the sum of the potential drops is equal to the sum of the potential rises.

20
 In figure 1.2 ABCDA form a closed circuit. Assuming the current direction as shown, from A to
B, we have a potential drop of IR1 volts. Writing for the entire loopABCDA, we have,

Sum of the potential drops = IR1 + IR2 + IR3

Potential rise from D to A = V

IR1+ IR2 + IR3 = V

Sign of EMFs

Problem 14

The power supplied to the load R and the voltage across it in figure are 500 W and 100 V.
Determine (i) the value of the VS, (ii) the power dissipated in each resistor. Also confirm that the
power delivered by the source equals the total power dissipated elsewhere.
(AU/EEE NOV 2003)

Solution: -

The currents and voltages are marked as shown in figure.

𝑉2 1002
The resistance 𝑅 = = = 20Ω
𝑃 500

21
 𝑉 100
𝐼3 = = = 5𝐴
𝑅 20
Current through 4 Ω is also I3i.e, 5A.

Voltage drop across 4 Ω resistor V3 = 13R3 = 5 × 4 = 20 V

Voltage drop across 30 Ω resistor V2 = V + V3 = 100 + 20 = 120 V

𝑉 120
Current through 30 Ω resistor 𝐼2 = 𝑅2 = = 4𝐴
2 30

Current through 5 Q resistor I1 = I2 + I3 = 4 + 5 = 9 A

Voltage drop across 5Ω V, = I1Rl = 9 ×5 = 45 V

Vs = V1+V2=45+120=165V

VS=165V

Power delivered by the source = VSI1

= 165 ×9= 1485 W

Power dissipated in all the resistors

= 𝐼12 𝑅1 + 𝐼22 𝑅2 + 𝐼32 𝑅3 + 𝐼32 𝑅

= 92 × 5 + 42 × 30 + 52 × 4 + 52 × 20

= 1485 W

Power delivered by the source = Power dissipated in all the resistors

= l485W

Problem 15

In the circuit shown in figure, find the value of current through 100 Ω(Nov 2004)

22
Solution:

The current directions are marked as shown in the figure

First, find the equivalent resistance of the circuit.

40 Ω and 100 Ω are connected in series, 40 + 100 = 140 Ω

150×140
150 Ω and 140 Ω are connected in parallel. = 72.41Ω
150+140

120 Ω and 72.41 Ω are connected in series, 120+72.41=192.41 Ω

𝑉 100
𝐼= = = 0.519𝐴
𝑅𝑒𝑞 192.41

Now current through 100 Ω resistor

140 140
𝐼3 = 𝐼 × = 0.519 × = 0.25𝐴
150 + 140 290

23
 Problem 16

Calculate the currents supplied by the batteries in the network shown in figure.(Dec 2003)

Solution

Here 𝐼3 = 𝐼1 + 𝐼2

Applying KVL in the loop ABEFA

−10𝐼2 − 25(𝐼1 + 𝐼2 ) + 90 = 0

−10𝐼2 − 25𝐼1 − 25𝐼2 = −90

25𝐼1 + 35𝐼2 = 90

5𝐼1 + 7𝐼2 = 18 − − − −(1)

Similarly apply KVL in loop CBEDC

−5𝐼1 − 25(𝐼1 + 𝐼2 ) + 125 = 0

−5𝐼1 − 25𝐼1 − 25𝐼2 = −125

30𝐼1 + 25𝐼2 = 125

6𝐼1 + 5𝐼2 = 25 − − − (2)

Using these two equations form matrix equation

5 7 𝐼1 18
[ ][ ] = [ ]
6 5 𝐼2 25

By applying Cramer’s rule, we can find out currents I1 and I2.

5 7
∆= | | = −17
6 5
24
 18 7
∆𝐼1 = | | = −85
25 5
∆𝐼1 −85
𝐼1 = = = 5𝐴
∆ −17
5 18
∆𝐼2 = | | = 17
6 25
∆𝐼2 17
𝐼2 = = = −1𝐴
∆ −17
Here, negative sign indicates, the current direction is opposite to the original direction. It is
shown in figure.

𝐼3 = 𝐼1 − 𝐼2 = 5 − 1 = 4𝐴

The current supplied by the battery 125 V is 5A.

Therefore the 90V battery is getting the charging current of l A.

Problem 17

In the Wheatstone bridge circuit of figure, G is a galvanometer of resistance 10 Ω. Find the

current through it.
Solution:

The currents are marked as in figure, Kirchoff's voltage law is applied to the various loops. There
are three unknowns I1 I2 and I3.

25
Problem 18

What is the different in potential between points X and Y in the circuit shown in figure?

26
 Solution:

The currents in the loops ABCX and DYGF are marked as shown. There can be no current
in the branch CD as otherwise the currents coming out of the positive terminals of the batteries (at
B and D) will not be the currents entering the negative terminals (at A and F).
For loop ABCXA

Mesh Current Method

• In mesh method, KVL is applied to a network.

• Assume that the mesh current direction is clockwise.

27
 The following steps should be implemented to find out the mesh current and branch current.

Step-1: Assume all mesh currents directions are clockwise.

Step-2: If two mesh currents are following through a network element, the actual current in the
circuit element is the algebraic sum of the two currents.

Step-3: Write equation for each mesh in terms of mesh currents by applying KVL.
When writing mesh equations, we assign rise in potential as positive (+) sign and fall in
potential as negative (-) sign.
Step-4: Suppose any value of mesh current becomes negative in the solution, the actual or true
direction of the mesh current is anticlockwise, i.e., opposite to the clockwise direction.
To apply KVL to the given figure,

𝑀𝑒𝑠ℎ 𝑙𝑜𝑜𝑝 − 1: − 𝐼1 𝑅1 − (𝐼1 − 𝐼2 )𝑅2 + 𝑉1 = 0

𝐼1 (𝑅1 + 𝑅2 ) − (𝐼2 )𝑅2 = 𝑉1

𝑀𝑒𝑠ℎ 𝑙𝑜𝑜𝑝 − 2: − 𝐼2 𝑅3 − (𝐼2 − 𝐼1 )𝑅2 − 𝑉2 = 0

𝐼2 (𝑅1 + 𝑅2 ) − (𝐼1 )𝑅2 = −𝑉2

𝑅11 𝑅12 𝐼1 𝑉
[ ] [ ] = [ 1]
𝑅21 𝑅22 𝐼2 𝑉2

𝑅 + 𝑅2 −𝑅2 𝐼 𝑉
[ 1 ] [ 1] = [ 1 ]
−𝑅2 𝑅1 + 𝑅2 𝐼2 −𝑉2

Then solving this matrix, we can find out mesh currents 1, and I2.

Step-5: The branch current can be easily found out by using the mesh currents I1 and I2.

Problem 19

Solve the mesh and branch currents shown in figure.

28
 Solution:

First assign mesh currents I1, and I2 to meshes PQSP and QRSQ respectively. It is shown in
figure.

29
30
Problem 20

Use mesh analysis to determine the three mesh currents in the circuit of figure shown in below.
(AU/Mech - May 2012)

The three required mesh currents are assigned as shown in figure. Applying KVL, we can get
three mesh equations.

31
32
 Problem 21

Determine the value of current through the branch DC of the network shown below in figure,
when the current through the branch BD is zero.

(AU, Coimbatore - Dec. 2010), (AU/Mech - June 2005)

Solution:

We have marked in AB as I, and AD as I2. Since there is no current in BD, I, lows through BC
and I2 through DC.

33
Problem 22
In the circuit given in figure, obtain the load current and power delivered to the load.

(AU, Coimbatore/EEE - Dec 2010), (AU/Mech - Dec 2005)

Solution:

34
 Applying KVL, we can get three mesh equations.

Loop 1

Loop 2

Loop 3

Using these three equation and form matrix equation.

By applying Cramer's rule, we can find out current through load (15 Ω). Mesh current I3
flows through 15 Ωresistor

35
 Current through load resistor (15 Ω) is 2 A
Power delivered to the load = I2R = 22 x 15=60 Watts
Problem 23
Determine the current supplied by each battery in the circuit shown in figure using mesh analysis.

(AU/CSE - Dec 2006)

Solution:
The three loop currents are shown in figure.
Loop 1

Loop 2

36
Loop 3

Using these three equations form matrix equation.

Here we have to solve the mesh currents I,, I2 and I3 by applying Cramer's rule.

37
 Current supplied by battery E1 is I1 =2.55 A
Current supplied by battery E2 is I1-I2 = 2.55 - 1.822 = 0.728 A
Current supplied by battery E3 is I2 = 1.822 A
Current supplied by battery E4 is I3 + I2 = 3.135 + 1.822 =4.957 A
Current supplied by battery E5 is I3 = 3.135 A

Nodal Voltage Method

• In nodal method, KCL is applied to a network.

• Node is defined as junction (or) joining point of two or more component terminals.
• In nodal method, select one node as a reference-node, with respect to which the voltages
at all other nodes are measured. Thus the reference node acts as ground or common for
the circuits.
Step-1: First, convert all the voltage sources to current sources.

Step-2: The conductances of all branches connected to node 1 are added and denoted by G11. It
is called the self conductance of node.
Step-3: All the conductances connected to nodes 1 and 2 are added and denoted by G12. It is
called mutual conductance of node 1 and 2. This is written with negative sign. If no
conductance is connected between node 1 and 2 then
G12 = 0, G12 = G21.
Step-4: I1 denotes the value of the current source current to node 1 and is written on the right
hand side of the equation. The sign I1 is positive if it is flowing towards node I1 otherwise it is
negative. If no current source is connected to node 1, then I1 = 0.

38
 𝑉1 𝑉2
𝐼1 = 𝐼𝑆𝐶1 = ; 𝐼2 = 𝐼𝑆𝐶2 =
𝑅1 𝑅3

This circuit consists of two nodes 1 and 2 and common (ref.) node

Conductance at node 1:
1 1 1 1
𝐺11 = + + ; 𝐺12 = − = 𝐺21
𝑅1 𝑅4 𝑅2 𝑅2
Conductance at node 2:
1 1 1
𝐺22 = + +
𝑅2 𝑅3 𝑅5
Then form the matrix model,
𝐺11 𝐺12 𝑉1 𝐼
[ ] [ ] = [ 1]
𝐺21 𝐺22 𝑉2 𝐼2

1 1 1 1
+ + −
𝑅1 𝑅4 𝑅2 𝑅2 𝑉 𝐼
[ 1] = [ 1]
1 1 1 1 𝑉2 𝐼2
− + +
[ 𝑅2 𝑅2 𝑅3 𝑅5 ]

Then solving this matrix, we can find out nodal voltages v1 and v2.

Problem 24
Using nodal analysis, determine the current in the 20 Ω resistor.

Solution:
Convert all the sources into their equivalent current sources.
10
𝐼1 = = 1𝐴;
10
39
 20
𝐼2 = = 2𝐴;
10
Then apply KCL to node V1
𝑉1 𝑉1 𝑉1
2+1= + +
10 20 10
1 1 1
3 = 𝑉1 ( + + )
10 20 10
0.25𝑉1 = 3
𝑉1 = 12𝑉
𝑉1
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑇ℎ𝑟𝑜𝑢𝑔ℎ 20Ω = = 0.6𝐴
20
Problem 25
Using nodal analysis, obtain the currents flowing in all the resistors of the circuit shown in figure.
(June 2007)

Solution:
Convert all the voltage sources into their equivalent current sources.

Now, the circuit becomes as shown in figure.

40
 Then apply KCL to node V1

Problem 26

Calculate the voltage across the 15 Ω resistor in the network shown in figure below using
nodal analysis.

(AU/ECE - May 2005)

Solution:

Convert all the voltage sources into their equivalent current sources.

41
Then the circuit becomes

At node 1, the current is 5A.

At node 2, the current is - 8 A
Node equations in matrix form by inspection method

1 1 1 1
+ + − 𝑉 5
[20 10 15 15 ] [ 1] = [ ]
1 1 1 1 𝑉2 −8
− + +
15 10 10 15

0.216 −0.066 𝑉1 5
[ ][ ] = [ ]
−0.066 0.216 𝑉2 −8

Problem 27

Using Nodal method find current through 8 Ω resistor.

42
 (AU, Chennai / Mech - May 2009)
Solution:
Convert all the voltage sources into its equivalent current sources.
4
𝐼1 = = 0.8𝐴
5
6
𝐼2 = = 0.5𝐴
12
Now, the circuit becomes

At node 1, the current is 0.8 A

At node 2, the current is 0.5 A
Node equations in matrix form by inspection method.

43
 Problem 28

Use nodal analysis to determine the voltage across 5Q resistance and the current in the 12 V
source. (AU, Chennai / Mech - Dec 2009)

Solution:

First voltage source convert into current source.

44
 12
𝐼= = 3𝐴
4
Now the circuit becomes

At node A, the current is 3 - 9 = - 6A

At node C, the current is 9 A
At node B, the current is zero
Node equations in matrix form by inspection method.

Solve VB and Vc by Cramer's rule.

= 1 (0.355 - 0.04) + 0.5 (- 0.25 - 0.05) - 0.2.5 (0.1+ 0.1775) = 0.0956

45
 = 1 (6.39) + 0.5 (-4.5)-6 (0.1 + 0.1775) = 2.475

Problem 29

Describe Kirchhoff’s laws. For the circuit shown in the figure below, determine the current through
6Ω resistor and the power supplied by the current source.
(A.U Chennai,June 2010)

Solution:

Kirchhoff’s laws:
Step 1:

46
 3Ω and 6 Ω are in parallel and hence
3 × 6 18
R𝐴 = = = 2Ω
3+6 9
This circuit can be redrawn as

Step 2:
3Ω and 2 Ω are in series and hence
RA = 3Ω +2Ω = 5Ω

21 × 2
I5 = = 6𝐴
2+5

6×3
𝐼6 = = 2𝐴
6+3
Problem 30

Find the current through 8Ω resistor.

47
Solution:

Writing the kirchoff’s voltage equation for loops, we get

(i)Loop ABGHA
-5I1-15(I1-I2) + 4=0

20I1-15I2=4 (1)
(ii)Loop BCFGB

-10I2-8(I2+I3)+15(I1-I2)=0

15I1-10I2-8I2-15I2-8I3=0

15I1-33I2-8I3=0 (2)
(iii)Loop CDEFC
12I3-6=8(I2+I3) = 0

8I2+20I3-6 (3)

Solving 1,2,&3 we get,

I1=0.22A
I2=0.03A
I3=0.29A

Here current flowing through the 8Ω resistor is assumed to be the sum of I2 and I3 .so,

I8Ω =I2+I3=0.32A

48
Problem 31

Find the current in the 8Ω resistor in the following circuit using Kirchhoff’s laws.

(May 2013)
Solution:

In Loop ABEFA,Apply KVL,

20I1-8I2=0 --------------- (1)
In Loop BCDEB,Apply KVL,
20(I1+I2)+8I2=40
20I1+28I2=40---------------(2)
Solving eqn (1) & (2)
I2 =1.11A
Therefore current flow through 8Ω resistor =1.11A
Problem 32

Calculate (i) equivalent resistance across the terminal of the supply

(ii) Total current supplied by the source
(iii) Power delivered to 16Ω resistor for the circuit shown below.
(May 2015)

49
Solution:
(a) Equivalent resistance across the terminal of the supply

Step: i) 4Ω, 16Ω & 4Ω are in series,

Req=4 + 16 + 4 = 24Ω

Step: ii) 12Ω & 24Ω are in parallel

12 × 24
𝑅𝑒𝑞 = = 8𝛺
12 + 24

Step: iii) 6Ω, 8 Ω & 6Ω are in series,

Req= 6 + 8 + 6 = 20 Ω

50
 Step: iv) 12 Ω & 20Ω are in Parallel
12 × 20
𝑅𝑒𝑞 = = 7.5𝛺
12 + 20

Step: v) 8 Ω, 7.5Ω & 8Ω are in series,

Req= 8 + 7.5 + 8 = 23.5Ω

Therefore equivalent resistance across the terminal of the supply is 23.5Ω

(b) Total current supplied by the source

𝑉 100
𝐼= = = 4.26𝐴
𝑅 23.5

(c)Power delivered to 16Ω resistor

P = 𝐼2 R = 4.262 × 23.5 = 426.47 W

51
 Problem 33
Determine the current, power in the 4Ω resistance of the circuit. (May/June 2014 &
Dec2015)

Solution:

First convert the current source into voltage source

Loop-1
12=2I1+10(I1+ I2) +1(I1+ I3)
12=2I1+ 10I1+ 10I2+I1+ I3
12=13I1+ 10I2+ I3 ------- (1)
Loop-2
10=2I2+10(I2+ I1) +3(I2-I3)
10=2I2+10I2+10I1+3I2-3I3
10=10I1+15I2-3I3 -----(2)
Loop-3
24=4I3+3(I3-I2) +1(I3+I1)
24=4I3+3I3-3I2+I3+I1
24=I1-3I2+8I3 -------(3)
On solving the above equations, we get
I1=-1 A
I2=2.13A
I3=3.39A

The current in the 4Ω resistance = 3.39A

52
 The Power in the 4Ω resistance=(I3)2R
= (3.93)2×4
= 61.7796 W

Problem 34
For the given circuit, determine the current in 5Ω resistor(Nov/Dec 2016)

Mesh equation

8 − 1(𝑖1 − 𝑖2 ) − 2(𝑖1 − 𝑖3 ) = 0

8 − 𝑖1 + 𝑖2 − 2𝑖1 + 2𝑖3 = 0

−3𝑖1 + 𝑖2 + 2𝑖3 = 8 − − − (1)

ii) Mesh equation

10 − 3𝑖2 − 2(𝑖2 − 𝑖3 ) − 1(𝑖2 − 𝑖1 ) = 0

10 − 3𝑖2 − 2𝑖2 + 2𝑖3 − 𝑖2 + 𝑖1 = 0

𝑖1 − 6𝑖2 + 2𝑖3 = −10 − − − (2)

Mesh equation

12 − 5𝑖3 − 2(𝑖3 − 𝑖1 ) − 2(𝑖3 − 𝑖2 ) = 0

2𝑖1 + 2𝑖2 − 9𝑖3 = −12 − − − (3)

−3 1 2 𝑖1 8
(1 −6 2 ) (𝑖2 ) = [−10]
2 2 −9 𝑖3 −12

=-109

Solving above 3 equation

I1=-0.936

53
I2=2.037

I3=1.578

Current across 5Ω resistor is =1.578A

Problem 35

Find I in the circuit shown by superposition theorem. (May’09)

Solution: If 10V is an active:

6×4 24
R AB = = = 1.6Ω
6 + 4 + 5 15
6 × 5 30
RB = = = 2Ω
15 15
20
RC = = 1.33Ω
15

54
 2 + 1.6 = 3.6Ω
3 + 1.33 Ω = 4.33Ω
R eq = 1 + 1.96 + 1.33
= 4.29 Ω
3.6 × 4.33
3.6 parallel with 4.38: =
3.6 + 4.33
15.588
= = 1.96Ω
7.93

10
𝐼𝐿′ = = 2.33A
4.29
If 6V is an active:
Loop – 1:
−I1 − 2(I1 − I2 ) − 4(I1 − I3 ) = 0
−I1 − 2I1 + 2I2 − 4I1 + 4I3 = 0
−7I1 + 2I2 + 4I3 = 0
7I1 − 2I2 − 4I3 = 0 − − − − − −(1)

Loop – 2:
−2(I2 − I1 ) − 3I2 + 6 − 6(I2 − I3 ) = 0
−2I2 + 2I1 − 3I2 − 6I2 + 6I3 = −6
55
 2I1 − 11I2 + 6I3 = −6 − − − − − (2)
Loop – 3:
−6(I3 − I2 ) − 6 − 5I3 − 4(I3 − I1 ) = 0
−6I3 + 6I2 − 6−5I3 − 4I3 + 4I1 = 0
4I1 + 6I2 − 15I3 = 6 − − − − − −(3)
I1 = −0.02 A (Current flow in opp. direction)
I2 = 0.40 A
I3 = −0.24 A (Current flow in opp. direction)
By applying superposition theorem the current through:
I = 𝐼𝐿′ + 𝐼𝐿′′
= 2.23 + 0.02
I = 2.25 A

Problem 36

Find the current through 5 Ω resistor using superposition theorem, in the circuit shown in
figure. (Nov’14)

Solution:
Converting all the current source to voltage source

Step 1:
-4I1 – 2 (I1+I2) +32=0
6I1 + 2 I2=32 - - - - - - - - - - - -- - - -- - - - - - - (1)
-5I2 -10I2 – 2 (I2+I1) +32=0
2I1 + 17 I2=32- - - - - - - - - - - -- - - -- - - - - - - (2)
Therefore I2’=1.306A

56
Step 2:

-1.33I2 -10I2 – 5I2+45=0

I2’’ = 2.755A

Step 3:

-1.33I2 -5I2 – 10I2+40=0

I2’’’ = 2.449A
Hence I2 = I’2 + I2’’ + I2’’’
= 1.306 + 2.755 + 2.449 = 6.51A

Problem 37
Solve for current in 5 Ω resistor by principle of superposition theorem. (Apr’03)

Solution:
Step: 1

R T = 10 + 5 parallel with 21
5 × 21
= 10 + = 14.04 Ω
5 + 21
57
 R T = 14.04 Ω
20
IT = = 1.425 A
14.04
21
I1 = 1.425 × = 1.151 A
26
Step – 2: Allow 1A source to act. 20 V is removed.

10 × 5
= = 3.33 A
15

1×1
I1 = = 0.041A
1 + 20 + 3.3
10
III = 0.041 × = 0.027 A
15

Step – 3: I = I1 + III = 1.151 + 0.027 = 1.178 AI = 1.178 A

Problem 38
Calculate the current in the 4Ω resistor of figure using superposition.(May’14)

Solution: Converting all the current source to voltage source

58
 Step 1: I4Ω due to 10Vsource

4 ×4
𝑅𝑒𝑞 = 3 + =5Ω
4+4
𝑉 10
𝐼𝑇1 = = = 2𝐴
𝑅𝑒𝑞 5
𝐼𝑇1 × 4 2 × 4
𝐼4Ω (10V) = = = 1A
4+4 4+4

Step 2: I4Ω due to 2V source

3 ×4
𝑅𝑒𝑞 = 4 + = 5.714Ω
3+4
𝑉 2
𝐼𝑇2 = = = 0.35𝐴
𝑅𝑒𝑞 5.714
𝐼𝑇2 × 3 0.35 × 3
𝐼4Ω (2V) = = = 0.15A
4+3 7

Therefore I4Ω = 1+ 0.15 = 1.15A

Problem 39
Compute the current in the 23 Ω resistor of the following figure shown below by applying
the super position principle.(Apr’15)
59
 Solution:
Step 1: Considering 200V voltage source alone,

Now, 4 Ω is in series with 23 Ω

∴ 𝑎𝑡𝐵, 4 + 23 = 27Ω

Now 27 Ω is in parallel with 27 Ω

1 1
∴ 𝑎𝑡𝐵, + = 13.5Ω
27 27

∴ 𝑅𝑒𝑞 = 47 + 13.5 = 60.5 Ω

𝑉 200
𝐼= = = 3.31 𝐴
𝑅𝑒𝑞 60.5
60
Applying current division technique at node A
27
𝐼𝐿 = 3.31 × = 1.66 𝐴
27 + 4 + 23

Considering 20A source alone,

Now 27 Ω is in parallel with 47 Ω

1 1
+ = 17.15Ω
27 47
Applying current division technique at node B
4 + 17.15
𝐼𝐿 = 20 × = 9.58 𝐴
4 + 17.15 + 23
Current through 23 Ω resistor due to both sources
𝐼𝐿 = 1.66 + 9.58 = 11.24 𝐴

Problem 40
For the circuit shown in figure, using Thevenin’s theorem, find the current in the 10Ω resistor.
(May’14)

Solution:
Step 1: To calculate Vth – let us disconnecting RL

10
𝐼1 = = 2𝐴
2+3
Voltage drop across 12 Ω resistor 𝑉1 = 3 × 2 = 6𝑉
4
𝐼2 = = 1𝐴
3+1
61
 Voltage drop across 3 Ω resistor𝑉2 = 1 × 3 = 3𝑉
∴ 𝑉𝐴𝐵 = 𝑉1 − 𝑉2 = 6 − 3 = 3𝑉
Step 2: To find Rth

2×3
= 1.2𝐴
2+3
3×1
= 0.75𝐴
3+1
Therefore Rth = 1.2+0.75 = 1.95Ω
Step 3:The equivalent circuit will be

Problem 41
A Wheatstone bridge PQRS has the following details, PQ =1000Ω, QR = 100 Ω, RS = 450 Ω, SP
= 5000 Ω. A galvanometer of resistance 500Ω is connected between Q and S. A 4.5 Volt battery
of negligible resistance is connected between P and R with P positive. Find the magnitude and
direction of current through the galvanometer (Apr’03).
Solution:

Step – 1: To calculateVth - Disconnect galvanometer between Q and S

4.5 × 1000
V1000 = = 4.09 V
1000 + 100
4.5 × 5000
V5000 = = 4.128 V
5000 + 450

62
 VQS = Vth = V1000 − V5000
= 4.09 − 4.128 = −0.0375 V = 0.0375 V

Step – 2: To calculateR th , from the circuit in step 2, kill the source (short circuit)

1000 × 100 5000 × 450

R th = R QS = + = 90.9 + 413 = 504 Ω
1100 5450

Step 3: Thevenin’s equivalent circuit between Q and S

Vth 0.0375
IL = Ig = =
R th + R L 504 + 500
= 37.4 × 10−6 A = 37.4 μA

Problem 42
Find the thevenin’s equivalent circuit at (a, b) (May’09)

63
 Solution: Apply delta – star transformation.

R AB R CA 4 × 10
RA = = ⟹ 1.67 Ω
R AB + R BC + R CA 4 + 10 + 10
R AB R BC 4 × 10
RB = = ⟹ 1.67 Ω
R AB + R BC + R CA 4 + 10 + 10
R CA R BC 10 × 10
RC = = ⟹ 4.17 Ω
R AB + R BC + R CA 4 + 10 + 10

The circuit reduced in series connection

4.17 + 5 = 9.167 Ωand1.67 + 6 = 7.67 Ω

R th = (1.67 parallel with 9.167) + 7.67

= 9.08 Ω
To find Vth :

Apply KVL in loop (l)

50 – 1.67 I − 9.167I = 0
50 = 10.833 I
50
I= = 4.62 A
10.833
Vth = I × R = 4.62 × 9.167
64
 = 42.35 V
Thevenin’ s equivalent:

Problem 43
Determine the Thevenin’s equivalent across AB.(Jun’09)

Solution: To find Rth:

10 × 5 50
R th = 10 parallel with 5 = = = 3.33 Ω
10 + 5 15

To find Vth :

50 − 10I − 5I − 25 = 0
−15I = −25
25
I= = 1.67 A
15
∴ 50 − 10I = 0
50 − 10(1.67) = 0
I = 33.3 A

65
 Problem 44
Calculate I in 2 Ω resistor using Thevenin’s theorem.(Nov’10)

Solution: To find R th

1×1 1
R th = = = 0.5 Ω
1+1 2
To find Vth :

2−I−I= 0
−2I = −2
I=1A
VAB = Vth = 1 × 2
= 2 Volts
Vth
∴ IL =
R th + R c
2 2
= = = 0.8 A
0.5 + 2 2.5

Problem 45
Find the current through the 5 Ω resistor by Thevenin’s theorem. (Nov’10)

Solution: To find R th

66
R th = (2 parallel with 3) + 1
2×3
= +1
2+3
6
= + 1 = 2.2 Ω
5

To findVth :

10 − 2I − 3I = 0
−5I = −10
I = 2A

Vth = I × R = 2 × 3 = 6
Vth = 6 Volts
To findIL :
Vth 6
IL = =
R L + R th 5 + 2.2
6
= = 0.83 A
7.2
Problem 46
Find the current I, through the 20Ω resistor shown in figure using Thevenin’s theorem.
(Nov’14).

Solution:
Remove the 20Ω resistor

67
 To find Rth:

2 Ω is parallel to 5 Ω,
5×2
= = 1.428 Ω
5+2

a. Ω is in series with 10 Ω,
1.428 + 10 = 11.428 Ω
11.428 Ω is parallel to 1 Ω,
11.428 × 1
= = 0.919 Ω
11.428 + 1

To find Voc:

−10𝐼1 + 10𝐼2 = −20

10𝐼1 − 10𝐼2 = 20 − − − − − − − − − − − − − − − − − (1)

𝐼2 − 9 + 1.428[𝐼2 − 𝐼3 ] + 10[𝐼2 + 𝐼1 ] = 0
10𝐼1 + 12.428𝐼2 − 1.428𝐼3 = 9 − − − − − − − − − −(2)
68
 −10 + 1.428[𝐼3 − 𝐼2 ] = 0
1.428𝐼2 − 1.428𝐼3 = −10 − − − − − − − − − − − − − −(3)

∴ 𝐼2 = −0.047 𝐴

Problem 47
Find the Thevenin’s equivalent of the network shown in fig. (Nov’15)

To find Vth: Converting the current sources into voltage sources,

In matrix form
3+2 −2 𝐼 10 − 10
[ ] [ 1] = [ ]
−2 2 + 1 + 2 2 𝐼 10 + 5 + 10
5 −2 𝐼1 0
[ ][ ] = [ ]
−2 5 𝐼2 25
5 −2
∆=[ ] = 25 − 4 = 21
−2 5
5 0
∆ 𝐼2 = [ ] = 125 − 0 = 125
−2 25

∆ 𝐼2 125
𝐼2 = = = 5.95 𝐴
∆ 21
∴ 𝑉𝑂𝐶 = 𝑉𝑡ℎ = 𝑉𝐴𝐵 = 𝑉2Ω − 10

= (5.95 × 2) − 10 = 1.9 𝑉
To find Rth:

69
 3 Ω is parallel to 2 Ω,
3×2
= = 1.2 Ω
3+2
1.2 Ω is in series with 1 Ω,
= 1.2 + 1 = 2.1 Ω
2.1 Ω is parallel to 2 Ω,
2.1 × 2
= = 1.02 Ω
2.1 + 2

Thevenin’s equivalent:

Statement of Norton’s theorem:

Statement:
Any linear active network with output terminals A, B as shown in figure can be replaced by a
single current source. Isc (=IN) in parallel with a single impedance.
ZTH (= Zn ) = R TH (= R N )

Problem 48
Using Norton’s theorem find current through 6Ω resistance shown in figure. (May’06)

70
 Solution:
To find R TH .
5 × 10
5Ω and 10Ω are in parallel R TH = = 3.33 Ω
5 + 10
To find IN :

No current in 10Ω resistance because, it is short circuit.

20
∴ IN = = 4A
5
Norton’s Equivalent Circuit:

4 × 3.33
IL = = 1.427 A
3.33 + 6
Current through 6 Ω = 1.427 A
Problem 49
Evaluate the current I. flowing through the 1 ohm resistance by applying Norton’s theorem.

Solution:
Let remove the load resistor 1Ω and short circuit it.

71
 Convert the voltage sources into equivalent current sources.

IN = 1 + 1.5 + 1 = 3.5 A
To calculateRth:

2×2
R TH = =1Ω
2+2
Norton’s Equivalent Circuit:

IN R TH 3.5 × 1
IL = = = 1.75 A
R TH + R L 1+1

Problem 50
Find the voltage across 12 Ω resistor by using Norton’s theorem.(Apr’11)

Solution:
To find R TH :

72
 12 × 4 48
12 ∥l 4 = = = 3Ω
12 + 4 16

2×3 6
2 ∥l 3 = = = 1.2 Ω
2+3 5
1.2 + 4 = 5.2 Ω
To find ISC ;

Loop – 1:
4 − 4I1 − 2(I1 − I2 ) = 0
4 − 4I1 − 2I1 + 2I2 = 0
4 − 6I1 + 2I2 = 0
6I1 − 2I2 = 4 − − − − − (1)
Loop – 2:
−4 − 4I2 − 2(I3 − I1 ) = 0
−4I2 − 2I2 + 2I1 = 4
2I1 − 6I2 = 4 − − − (2)
By solving, (1) and (2)
I1 = 0.5 A
I2 = −0.5 A

I12 = 0.5 − (−0.5)

= 0.5 + 0.5
I12 = 1Ω
5.2
IL = 1 ×
5.2 + 12
5.2
IL = 1 ×
17.2
IL = 0.302 A
VL = IL × R L
73
 = 0.302 × 12
VL = 3.624 Volts

Problem 51
Obtain the Norton’s model and find the maximum power that can be transferred to the 100 Ω
load resistance, in the circuit shown in figure.[May’16]

Solution:

To find Rth:
Let us remove the load resistor RL= 100 Ω.

And short circuit the voltage sources,

Since 220 Ω is in parallel with 470 Ω,

220 × 470
= 220 + 470 = 149.82 Ω

74
 And 330 Ω is in parallel with 200 Ω,
330 × 200
= 330 +200 = 124.52 Ω

Since 149.85 Ω is in series with 124.52 Ω,

149.85 + 124.52 = 274.38 Ω
Rth = 274.38 Ω

To find IN:

By Inspection method:
220 + 470 −470 0 𝐼1 10
[ −470 470 + 330 −330 ] [𝐼2 ] = [ 0 ]
0 −330 200 + 330 𝐼3 −5
690 −470 0 𝐼1 10
[−470 800 −330] [𝐼2 ] = [ 0 ]
0 −330 530 𝐼3 −5

690 −470 0 𝐼1
∆ = [−470 800 −330] [𝐼2 ]
0 −330 530 𝐼3

∆ = 690[[800 × 530] − [−330 × −330]] − [−470][[−470 × 530] − [−330 × 0]]

+ 0[[−470 × −330] − [800 × 0]]
∆ = 690 [424000 − 108900] − [−470][[−249100] − 0] + 0
∆ = 690 [315100] − 117077000
∆ = 217419000 − 117077000
∆ = 100342000

To find the Norton’s Current IN, the current flowing across it is I2.
690 10 0 𝐼1
∆2 = [−470 0 −330] [𝐼2 ]
0 −5 530 𝐼3
∆2 = 690[[0 × 530] − [−330 × −5]] − 10 [[−470 × 530] − [−330 × 0]]
+ 0[[−470 × −5] − [0 × 0]]
∆2 = 690 [0 − 1650] − 10 [[−249100] − 0] + 0
∆2 = −1138500 + 2491000
∆2 = 1352500

75
 ∆2 1352500
𝐼2 = = = 0.0135 𝐴
∆ 100342000
𝐼𝑁 = 𝐼2 = 0.0135 𝐴

Norton’s equivalent circuit:

To find the maximum power across 100 Ω,

𝐼𝑁 × 𝑅𝑡ℎ
𝐼𝐿 =
𝑅𝑡ℎ + 𝑅𝐿
0.0135 × 274.38
𝐼𝐿 =
274.38 + 100
𝐼𝐿 = 9.89 × 10−3 𝐴

Maximum power = IL2 × R L

Maximum power = [9.89 × 10−3 ]2 × 274.38
= 0.00978 𝑊
Problem 52
Obtain the Thevenin and Norton equivalent circuits for the active network shown below.
(Apr’15)

Thevenin’s equivalent circuit:

By using mesh analysis,

To find Vth
− 3 𝐼 − 6𝐼 + 10 + 20 = 0
−9𝐼 + 30 = 0

76
 −30
𝐼= = 3.33 𝐴
−9
𝑉𝑡ℎ = 10 + 6 𝐼 = 10 + (6 × 3.33) = 29.98𝑉
To find Rth

1 1
𝑅𝑒𝑞 = + = 2Ω
3 6
∴ 𝑅𝑡ℎ = 2 + 3 = 5Ω
Thevenin equivalent circuit:

Norton equivalent circuit:

By using mesh analysis,

Loop 1:
− 3 𝐼1 − 6(𝐼1 − 𝐼2 ) + 10 + 20 = 0
− 9 𝐼1 + 6𝐼2 + 30 = 0
9 𝐼1 − 6𝐼2 = 30 … … … … … … … … … … … … … … … … … . (1)
Loop 2:
3 𝐼2 + 6(𝐼2 − 𝐼1 ) − 10 = 0
− 6 𝐼1 + 9𝐼2 − 10 = 0
− 6 𝐼1 + 9𝐼2 = 10 … … … … … … … … … … … … … … … … … . (2)
Solving (1) and (2),
I1 = 7.33A
I2 = Isc= 6A

Norton equivalent circuit

77
 Maximum power transfer theorem:
In DC circuits maximum power is transferred from a source to the load when a load
resistance is made equal to the resistance of the network. As we would from the load terminals
with load removed and all the sources replaced by the internal resistance.

Problem 53
Find the value of RL for maximum power delivered to it:

Solution:
According to maximum power transfer theorem R L must be equal to the resistance between
A and B with R L disconnected and all the sources killed (=R th ). The corresponding circuit is as
follow.

5 × 5 25
5 Ω is in parallelwith 5Ω = = = 2.5 Ω
5 + 5 10
R th = R AB = (10 + 2.5) Ω parallel with 20 Ω

78
 12.5 × 20
=
12.5 + 20
R L = R th = 7.7 Ω

Problem 54
Determine the value of resistance that may be connected across A and B so that maximum power
is transferred from the circuit to the resistance. Also, estimate the maximum power transferred to
the resistance shown in fig.(Nov’15)

To find Vth:
10Ω is in series with 4 Ω,
= 10 + 4 = 14Ω
14 Ω is parallel to 8 Ω,
14 × 8
= = 5.09 Ω
14 + 8
5.09Ω is in series with 2Ω,
= 5.09 + 2 = 7.09Ω
20
𝐼2 Ω = = 2.82 𝐴
7.09
8
𝐼10 Ω = 𝐼2 Ω ×
8 + 4 + 10
8
= 2.82 × = 1.025 𝐴
8 + 4 + 10

∴ 𝑉𝐴𝐵 = 𝑉𝑡ℎ = 5 + 𝑉10 Ω

= 5 + (𝐼10 Ω × 10)
= 5 + (1.025 × 10) = 15.25 𝑉
To find Rth:

2 Ω is parallel to 8 Ω,
2×8
= = 1.6 Ω
2+8
1.6Ω is in series with 4Ω,
= 1.6 + 4 = 5.6Ω
5.6Ω is parallel to 10 Ω,

79
 5.6 × 10
= = 3.59 Ω
5.6 + 10

Thevenin’s equivalent circuit,

Maximum power transferred:

2
𝑉𝑂𝐶
𝑃𝑚𝑎𝑥 =
4𝑅𝑡ℎ
15.252
𝑃𝑚𝑎𝑥 =
4 × 3.59
𝑃𝑚𝑎𝑥 = 16.19 𝑊

Problem 55
Find the value of R L which maximum power transferred toR L and hence the maximum power
transferred to R L figure.(May’13)

Solution:
To findR th :
1 Ω is parallel to 3 Ω
1×3 3
= = 0.75 Ω
3+1 4
0.75 Ω is in series with 2Ω

0.75 + 2 + 0.5 = 3.25 Ω

R th = 3.25 Ω
To find Vth :
−0.5I1 + 12 − 2I1 − 1I1 = 0
−0.5I1 − 2I1 − 1I1 = −12
−3.5I1 = −12
12
I1 = = 3.42 A
3.5
Vth = 3.42 × 1 = 3.42 Volts
80
Thevenin’s equivalent circuit is,

VOC2 (3.42)2
Pmax = =
4R th 4 × 3.25
11.696
=
13
= 0.899 Watts
Problem 56
Calculate the value of R L so that maximum power is transferred from battery. (Apr’09)

Solution:
Thevenin’s resistance (Rth): Remove the load resistor and Short circuit the voltage source.

10 × 10
R th = = 5Ω
10 + 10

Open circuit voltage Voc:Remove the load resistor.

50 × 10
Voc = VAB =
50 + 10
= 25V

81
 The maximum power is transferred to the load resistor only when the load resistance equals
the thevenin’s resistance.

Voc2 252
Maximum power Pmax = = = 31.25
4R th 4 × 5
Pmax = 31.25 W
Problem 57
(a) Find the value of R for maximum power to R.

(b) Determine the maximum power to R for the given network.(Nov’13)

Solution:
CalculateR th
 Short circuit the voltage source.
 Remove the load resistor.

5×5
R th = + 5 = 7.5Ω
5+5

Calculate Voc :Remove the load resistor R.

20 × 5
Voc = = 10 V
5+5

Thevenin’s Equivalent Circuit:

The maximum power is transferred to the load resistor only when the load resistance equals
the Thevenin’s resistance.

82
 R L = R th = 7.5Ω

Calculation of Pmax :
Voc2
Pmax =
4R th
(10)2
=
4 × (7.5)
= 3.33W

Problem 58
In figure IRabsorbs maximum power. Find the value of R and the maximum power. (Apr’05)

Solution: Calculate R th :

Short circuit the voltage source: Remove the load resistor

6 × 5 30
R th = = = 2.72 Ω
6 + 5 11

Calculate Voc :
Remove the load resistor R.

36 × 5
Voc = VPQ =
5+6
= 16.36 V

The maximum power is transferred to the load resistor only when the load resistance equals
the Thevenin’s resistance.
R L = R th = 2.72 Ω

83
 Voc2
Pmax =
4R th
(16.36)2
=
4 × 2.72
= 24.6 W

Problem 59
Find the load resistance for maximum power across it as shown in figure. What is the maximum
power?(Aug’07)

Solution:

To find R th :

4Ωand 12Ωare in parallel,

4 × 12
= 3Ω
4 + 12
3Ωis in series with 12Ω,
12 + 3 = 15Ω
15Ωand 30 Ωare in parallel,
15 × 30
= 10Ω
15 + 30

R th = 10 + 5
R th = 15Ω

To find Voc :
By inspection method,

16 −12 I1 120
[ ][ ] = [ ]
−12 54 I2 0

16 −12
∆=[ ]
−12 54
= 864 − 144
∆ = 720
16 120
∆I2 = [ ]
−12 0
= 0 + 1440

84
 ∆I2 = 1440

∆I2
I2 =
∆
1440
=
720
I2 = 2A

Voltage across AB,

Voc = 30 × 2
Voc = 60V.
For maximum power transfer,
R L = R th = 15Ω
Voc2
Maximum power Pmax =
4R th
(60)2
=
4 × 15
Pmax = 60 W
Problem 60
Determine the load resistance to receive maximum power from the source and also find the
power. Deliver to the load on the circuit shown in figure. (Aug’10)

Solution:
To find R th :

5 × 15 10 × 20
R th = +
5 + 15 10 + 20
= 3.75 + 6.66
R th = 10.41 Ω

To find Voc :

85
Voltage at point A is
50 × 15
VA = = 37.5 V
15 + 5
Voltage at point B is
50 × 20
VB = = 33.33 V
20 + 10

Voc = VAB = VA − VB
= 37.5 − 33.33
Voc = 4.17 V

Thevenin’s equivalent circuit:

According to maximum power transfer.R L = R th = 10.41 Ω

Voc2
Maximum power Pmax =
4R th
(4.17)2
= = 0.4176 W
4 × 10.41
Pmax = 0.4176 W

Problem 61
In the circuit of figure find the value of R for maximum power transfer. Also, calculate the
maximum power.(May’14)

Step 1: Converting all the voltage source to current source

86
 Step 2: Calculate Rth- Remove the load resistor and short circuit the voltage source

15 × 10
=
15 + 10
𝑅𝑡ℎ = 6Ω

Step 3: Calculate Voc – Remove the load resistor

Loop 1:

15 × 10
= = 6Ω
15 + 10
𝑉1 = 6 × 0.8
= 4.8 𝑉

Loop 2:

15 + 10 = 25 Ω
𝑉2 = 25 × 2 = 50 𝑉

Therefore V= V1 + V2
= 4.8 + 50
V = 54.8V

87
 Thevenin’s equivalent circuit:

RL = Rth = 6Ω

Step 4:
Calculate
2
𝑉𝑂𝐶
𝑃𝑚𝑎𝑥 =
4𝑅𝑡ℎ
54.82
=
4 ×6
𝑃𝑚𝑎𝑥 = 125 𝑊

Problem 62
Find the value of RL in figure for maximum power to RL and calculate the maximum power. [
Nov’16]

To find Rth:

Let the current source 6 A be open circuited and voltage source be short circuited.

88
3 Ωand 10 Ωare in parallel,
3 × 10
= 2.307 Ω
3 + 10
2.307 Ωand 2 Ωare in parallel,
2.307 × 2
= 1.069 Ω
2.307 + 2
Rth = 1.069 Ω
To find Vth:

Converting the E1 voltage source to Current source I1,

68
I1 = = 22.6 A
3

6 A and 22.6 A are in parallel,

6 + 22.6 = 28.6 A
10 Ωand 3 Ωare in parallel,
3 × 10
= 2.307 Ω
3 + 10

89
Converting the Current source toVoltage source,
Vth = I x R
= 28.6 x 1.069
Vth= 30.57 V

2
𝑉𝑡ℎ
𝑃𝑚𝑎𝑥 =
4𝑅𝑡ℎ
[30.57]2
=
4 × 1.069

𝑃𝑚𝑎𝑥 = 218.85 𝑊

90