This presentation clearly explains about all theorems with numerical examples including Superposition, Thevenin’s, Norton’s, Reciprocity, Maximum power transfer, Substitution, Tellegen's theorem and example problems.

1. NETWORK ANALYSIS AND SYNTHESIS
B. Tech EE III Semester
Course Code: BTEEC302
Prepared By
Dr. K Hussain
Associate Professor & Head
Dept. of EE, SITCOE

2. Dr. K Hussain
Superposition Theorem

3. Dr. K Hussain 3
Objectives
On completion of this period, you would be able to
• State Superposition theorem
• Explain Superposition theorem
• Apply Superposition theorem to networks

4. Dr. K Hussain 4
Known To Unknown
• Ohm’s law provides the fundamental tool for solving
electrical networks, but it has some limitations.
• It cannot be applied to circuits where more than one source
is acting simultaneously.
• In such cases, we can apply KCL & KVL.

5. Dr. K Hussain 5
If a circuit has two or more independent sources, one way to
determine the value of a specific variable (voltage or current) is
to use nodal or mesh analysis.
Another way is to determine the contribution of each
independent source to the variable and then add them up. The
latter approach is known as the superposition.
The idea of superposition rests on the linearity property.

6. Dr. K Hussain 6
• Even while applying KCL & KVL to the networks
• The number of simultaneous equations to be handled
becomes more when the network is complicated
• This leads to errors in results
• In such cases, we can apply Superposition theorem

7. Dr. K Hussain 7
Superposition Theorem
It states that, “ In a network of linear resistances containing
more than one source of e.m.f. , the current which flows at any
point is the sum of all the currents which would flow at that
point if each source were considered separately & all others
were replaced by their internal resistances”
or
The superposition principle states that the voltage across (or
current through) an element in a linear circuit is the algebraic
sum of the voltages across (or currents through) that element
due to each independent source acting alone.

8. Dr. K Hussain 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 the techniques.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the
contributions due to the independent sources.

9. Dr. K Hussain 9
• Internal resistance of an ideal voltage source is ‘ZERO’ (i.e., a
short can be put)
• Internal resistance of an ideal current source is “INFINITE’ (i.e.,
an open circuit can be put)
• If the given sources are non ideal, replace with their own
resistances

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Example-1
• Use the superposition
theorem to find v in the
circuit of Fig.
Fig.1

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• Since there are two sources, let
• where v1 and v2 are the contributions due to
the 6-V voltage source and the 3-A current source,
respectively.
• To obtain v1, we set the current source to zero, as
shown in Fig.(a).
• Applying KVL to the loop in Fig. (a) gives
• Thus,
• We may also use voltage division to get by writing

12. Dr. K Hussain 12
• To get v2, we set the voltage
source to zero, as in Fig. (b).
Using current division,
• Hence,
• And we find

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Example-2
• Consider the network
shown in Fig.1
• It consists of two sources of
e.m.f. (batteries)
Fig.1

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Application of Superposition Theorem
• Let us apply Superposition
theorem to the circuit shown
in Fig.2
• First consider that battery (B1)
is acting alone in the circuit as
shown in Fig.2
• Therefore, other battery (B2) is
replaced with a short.
Fig.2

15. Dr. K Hussain 15
• As B2 is removed, we can calculate
• The resistance as seen by
the battery as
R = 4 + [( 2 * 1) / (2+1)]
= 4.667 Ω
• Current supplied by battery
as
I = 28 / 4.667
= 6 A
Fig.2

16. Dr. K Hussain 16
• Here direction of actual current flow is considered rather
conventional flow.
• Currents through various
parallel paths can easily be
obtained by applying current
division rule.
• Voltage acting across each
resistor can also be calculated
Fig.3

17. Dr. K Hussain 17
• Now let’s consider that battery B2 is acting
alone
• Replace battery B1 with its
internal resistance. (i.e., a
short)
Fig.4

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• As B1 is removed, we can calculate
• The resistance as seen by
the battery as
R = 1 + ( 2 * 4) / 6
= 2.33 Ω
• Current supplied by battery
is
I = 7 / 2.33
= 3 A
Fig.4

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• This 3 A current divides into two paths as shown in Fig.5
• Current through various
parallel paths can easily be
obtained by applying
current division rule.
• Voltage acting across each
resistor can also be
calculated.
Fig.5

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• Now we have to ‘superimpose’ the results obtained,
i.e., add each value algebraically.
Fig.6 shows voltage acting across various elements, while Fig.7
shows currents.
Fig.7Fig.6

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• Please note that while ‘superimposing’ the results,
• We have to consider the direction
• If currents due to all sources are acting in same direction,
we can simply add
• If they are opposite to each other, we have to subtract
• The same rule applies to voltages acting across the
resistors

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• Analyzing a circuit using superposition has one major disadvantage:
• It may very likely involve more work.
• If the circuit has three independent sources, we may have to
analyze three simpler circuits each providing the contribution
due to the respective individual source.
• However, superposition does help reduce a complex circuit
to simpler circuits through replacement of voltage sources
by short circuits and of current sources by open circuits.

23. Dr. K Hussain 23
Summary
In this period, we have discussed
• Statement of Superposition theorem
• Explanation of Superposition theorem
• Application of Superposition theorem to networks

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Quiz
1. What is the internal resistance of an ideal voltage source?
a) Zero
b) Infinite
c) 10Ω

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Quiz
2. While considering one source, other sources should be,
a) Kept as they are
b) Should be replaced by their internal resistance
c) Can’t say

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Frequently Asked Questions
1. State and explain Superposition theorem as applied
to D.C. circuits.

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Assignment
1. Calculate the current flowing through 2 Ω resistor in the given
circuit by applying Superposition theorem.

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Thevenin’s Theorem

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In the previous lecture, we have discussed
• Statement and explanation of Superposition Theorem .
• Solved problems on Superposition Theorem.
Recap

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On completion of this lecture, you would be able to
• State Thevenin’s theorem
• Explain Thevenin’s theorem
• Apply Thevenin’s theorem to circuits
Objectives

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• Many laws and theorems are available for solving the networks
• Some of them are, Kirchhoff’s laws, Loop Current method,
nodal analysis, Superposition theorem, etc.,
• Thevenin’s Theorem is one of such theorems
Known To Unknown

32. Dr. K Hussain 32
It states that,
“A given network when viewed from its any two terminals (load terminals),
can be replaced by a single voltage source in series with a single
resistance.” That single voltage source is called Thevenin’s voltage source
(VTh) and Resistance is called Thevenin’s resistance (RTh)
OR
“Thevenin’s theorem states that a linear two-terminal circuit can be
replaced by an equivalent circuit consisting of a voltage source VTh in series
with a resistor RTh, where VTh is the open-circuit voltage at the terminals
and RTh is the input or equivalent resistance at the terminals when the
independent sources are turned off”
Thevenin’s Theorem

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THEVENIN’S EQUIVALENT CIRCUITGIVEN CIRCUIT
Thevenin’s Circuit
Fig.1

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• To calculate the Thevenin’s voltage or open circuit
voltage, the load resistance has to be removed
• Then find the voltage appearing across the open
circuited terminals
V Th = V OC
Rules To Find Thevenin’s Voltage

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• Replace the voltage source (battery) with its internal
resistance
• Calculate the resistance of the network as viewed by
the open circuited load terminals. (i.e., view inwards
from terminals A & B)
R Th = R i (internal resistance)
Rules To Find Thevenin’s Resistance

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Fig.2
Thevenin’s Equivalent Circuit
• The given circuit is replaced by the Thevenin’s equivalent
circuit

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• Current, I L = V TH / ( R TH + R L)
• Voltage, V L = I L * R L
• Power consumed, P L = V L * I L
Parameters at load terminals

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A circuit with a load: (a) original circuit, (b) Thevenin equivalent.
In general,

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Find the Thevenin equivalent circuit of the circuit shown in Fig,
to the left of the terminals a-b. Then find the current through and
an RL= 6, 16 and 36 ohms.
Solution:
We find RTh by turning off the 32-V voltage source
(replacing it with a short circuit) and the 2-A
current source (replacing it with open circuit).
The circuit becomes what is shown in Fig (a).
Thus,
Example

40. Dr. K Hussain 40
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,
Alternatively, it is even easier to use nodal analysis. We ignore the 1ohm resistor
since no current flows through it. At the top node, KCL gives
as obtained before.
We could also use source transformation to find The Thevenin equivalent circuit
is shown in Fig.
The current through RL is
When RL=6,
When RL=16,
When RL=36,

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Q. Using Thevenin’s theorem, find the equivalent circuit to the left of
the terminals in the circuit of Fig. Then find I.
Practice Problem

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In this lecture, we have discussed
• Statement and explanation of Thevenin’s theorem
• Thevenin’s equivalent circuit for a given circuit with an example.
Summary

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1. VTh stands for
Ans: Thevenin’s voltage
2 . The formula to calculate IL =
Ans: VTH/(RTH+RL)
Quiz

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3 . While calculating, R Th, one has to
a) leave sources as they are
b) replace sources with their internal resistances
c) short circuit the sources
d) open circuit the sources
Quiz

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1. Find the current flowing through the load resistance
(15 Ω) by applying Thevenin’s theorem.
Frequently Asked Questions
RL

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Norton’s Theorem
Dr. K Hussain

47. 47
Recap
In the previous lecture, we have discussed
• Thevenin’s theorem
• Thevenin’s equivalent circuit of a given circuit
47
Dr. K Hussain

48. 48
Objectives
On completion of this lecture, you would be able to
• State Norton’s theorem
• Explain Norton’s theorem
• Apply Norton’s theorem to a given circuit
Dr. K Hussain

49. 49
Known To Unknown
We have already discussed that
• Many laws and theorems are available for solving networks
• Some of them are, Kirchhoff’s laws, Maxwell’s Loop Current
theorems, Superposition theorem etc.,
• Norton’s Theorem is also one of such theorems
Dr. K Hussain

50. 50
Norton’s Theorem
• Statement:
• “Any two terminal active network when viewed from its load terminals,
can be replaced by a single current source in parallel with a single
resistance.”
• That single current source is called Norton’s current source (IN)
And Resistance is called Norton’s resistance (RN)
OR
• Norton’s theorem states that a linear two-terminal circuit can be
replaced by an equivalent circuit consisting of a current source IN in
parallel with a resistor RN, where IN is the short-circuit current through
the terminals and RN is the input or equivalent resistance at the
terminals when the independent sources are turned off.
Dr. K Hussain

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Norton’s Equivalent Circuit
Norton’s equivalent circuitTwo terminal circuit
Fig.1 Fig.2
Dr. K Hussain

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To Find IN :
• Remove the load resistance & put a short there
• Then find the current flowing through the short circuited
terminals
• I N = I SC
Dr. K Hussain

53. 53
To Find RN :
• Replace the voltage source (battery) with its internal
resistance
• Calculate the resistance of the circuit as viewed by the open
circuited load terminals. (i.e. terminals A & B as shown in
Fig.1)
• R N = R i (internal resistance)
Dr. K Hussain

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Replace the given circuit with its Norton’s equivalent
circuit as shown in Fig.3
Fig.3
Dr. K Hussain

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At load terminals,
• Current I L = [ I N / ( R N + R L)] * R N
• Voltage V L = IL * R L
• Power consumed =P L = V L * I L
Dr. K Hussain

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Example 1: Using Norton’s theorem, calculate current
through 15 Ω resistor in the circuit shown in Fig.4.
Fig.4
Ω
Ω Ω
Dr. K Hussain

57. 57
Solution
Step 1: To find out IN:
• Remove the load resistance
• Short circuit the load
terminals as shown in Fig.5
• Find the current flowing
through the short
• 12 Ω resistor is in parallel
with the short as shown in
Fig.5 Fig.5
Ω
Ω
Dr. K Hussain

58. 58
• So, 12 Ω resistor will also
be short circuited
• Therefore the circuit will be
as shown in Fig.6
• IN = 24 / (3+1)
= 6 A
Fig.6
Ω
Dr. K Hussain

59. 59
Step 2: To find out RN :
• Replace the battery with its
internal resistance
• Open circuit load terminals as
shown in Fig.7
• Calculate the resistance of the
circuit as viewed from the open
circuited load terminals
Fig.7
Dr. K Hussain
Ω

60. 60
• When we look into the circuit
as shown in Fig.7, from the
open circuited load terminals
• 3Ω and 1Ω resistors are in
series
• The combination is in parallel
with 12 Ω
• Therefore,
RN = 4 * 12 / (4+12)
= 48 / 16 = 3 Ω
Fig.7
Dr. K Hussain

61. 61
• Replace the given
network with Norton’s
equivalent circuit as
shown in Fig.8
• I L = I N * R N
(RN + RL)
= 6 * [3 / (3 + 15)]
= 18 / 18
= 1 A Fig.8
Ω
Dr. K Hussain
Ω

62. 62
Summary
In this period, we have discussed
• Statement of Norton’s theorem
• Explanation of Norton’s theorem
• Application of Norton’s theorem to a given circuit
Dr. K Hussain

63. 63
Quiz
1. What is the internal resistance for an ideal current
source?
a) Infinite
b) Zero
c) Depends on other factors of the circuit
d) None
Dr. K Hussain

64. 64
2. Norton’s equivalent circuit will consist of
a) A current source in parallel with resistance
b) A voltage source in series with resistance
c) None
Quiz
Dr. K Hussain

65. 65
Frequently Asked Questions
1. State & explain Norton’s theorem.
2. Explain the steps to obtain the Norton’s equivalent circuit
for a given network.
Dr. K Hussain

66. 66
1. Determine current supplied by the battery, using Norton’s
theorem.
Assignment
Dr. K Hussain

67. Source Transformation
• Source transformation is another tool for
simplifying circuits.
• A source transformation 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.
Dr. K Hussain 67
Fig. Transformation of dependent sources

68. Dr. K Hussain 68
Source transformation also applies to dependent sources, provided
we carefully handle the dependent variable.
• It is easy to show that they are indeed equivalent.
• If the sources are turned off, the equivalent resistance
at terminals a-b in both circuits is R.
• Also, when terminals are short-circuited, the short-circuit
current flowing from a to b is isc=vs/R in the circuit on the left-hand
side and isc=is for the circuit on the right-hand side.
• Thus, vs/R=is in order for the two circuits to be equivalent.
• Hence, source transformation requires that
Fig. Transformation of dependent sources

69. 69
• The Thevenin and Norton equivalent circuits are
related by a source transformation.
• This is essentially source transformation. For this reason,
source transformation is often called Thevenin-Norton
transformation.
Dr. K Hussain

70. Dr. K Hussain 70
Example 2. Find the Norton equivalent circuit of the
circuit in Fig. at terminals a-b.
Solution:
We find RN in the same way we find RTh,
in the Thevenin equivalent circuit.
Set the independent sources equal
to zero.
This leads to the circuit in Fig. (a),
from which we find RN .
Thus,
Fig.

71. Dr. K Hussain 71
To find IN, we short-circuit terminals a and b,
as shown in Fig.(b).
We ignore the 5-ohm resistor because it has been
short-circuited.
Applying mesh analysis, we obtain
From these equations, we obtain

72. Dr. K Hussain 72
Alternatively, we may determine IN from VTh
.
We obtain VTh as the open-circuit voltage across terminals a and
b in Fig (c).
Using mesh analysis, we obtain
as obtained previously.
This also serves to confirm Eq. that
RTh=(voc/isc)=(4/1)=4 ohm
Thus, the Norton equivalent circuit is as
shown in Fig. d.
Hence
Fig (d)

73. 73
Reciprocity Theorem
Dr. K Hussain

74. 74
Recap
In the previous lecture, we have discussed
• Thevenin’s theorem
• Thevenin’s equivalent circuit of a given circuit
74
Dr. K Hussain

75. 75
Objectives
On completion of this lecture, you would be able to
• State Norton’s theorem
• Explain Norton’s theorem
• Apply Norton’s theorem to a given circuit
Dr. K Hussain

76. 76
Known To Unknown
We have already discussed that
• Many laws and theorems are available for solving networks
• Some of them are, Kirchhoff’s laws, Maxwell’s Loop Current
theorems, Superposition theorem etc.,
• Norton’s Theorem is also one of such theorems
Dr. K Hussain

77. 77
Norton’s Theorem
• Statement:
• “Any two terminal active network when viewed from its load terminals,
can be replaced by a single current source in parallel with a single
resistance.”
• That single current source is called Norton’s current source (IN)
And Resistance is called Norton’s resistance (RN)
OR
• Reciprocity Theorem states that – In any branch of a network or circuit,
the current due to a single source of voltage (V) in the network is equal
to the current through that branch in which the source was originally
placed when the source is again put in the branch in which the current
was originally obtained. This theorem is used in the bilateral linear
network which consists of bilateral components.
Dr. K Hussain

78. Dr. K Hussain 78
This theorem is used for solving many DC and AC network which have many applications
in electromagnetism electronics.
These circuits do not have any time-varying element.
Explanation of Reciprocity Theorem

79. It is clear from the figure above that the voltage source and
current sources are interchanged for solving the network with
the help of Reciprocity Theorem.
• The limitation of this theorem is that it is applicable only to
single-source networks and not in the multi-source network.
The network where reciprocity theorem is applied should be
linear and consist of resistors, inductors, capacitors and
coupled circuits. The circuit should not have any time-varying
elements.
Dr. K Hussain 79

80. Steps for Solving a Network Utilizing Reciprocity Theorem
• Step 1 – Firstly, select the branches between which reciprocity
has to be established.
• Step 2 – The current in the branch is obtained using any
conventional network analysis method.
• Step 3 – The voltage source is interchanged between the
branch which is selected.
• Step 4 – The current in the branch where the voltage source
was existing earlier is calculated.
• Step 5 – Now, it is seen that the current obtained in the
previous connection, i.e., in step 2 and the current which is
calculated when the source is interchanged, i.e., in step 4 are
identical to each other.
Dr. K Hussain 80

81. 81
Norton’s Equivalent Circuit
Norton’s equivalent circuitTwo terminal circuit
Fig.1 Fig.2
Dr. K Hussain

82. 82
To Find IN :
• Remove the load resistance & put a short there
• Then find the current flowing through the short circuited
terminals
• I N = I SC
Dr. K Hussain

83. 83
To Find RN :
• Replace the voltage source (battery) with its internal
resistance
• Calculate the resistance of the circuit as viewed by the open
circuited load terminals. (i.e. terminals A & B as shown in
Fig.1)
• R N = R i (internal resistance)
Dr. K Hussain

84. 84
Replace the given circuit with its Norton’s equivalent circuit as
shown in Fig.3
Fig.3
Dr. K Hussain

85. 85
At load terminals,
• Current I L = [ I N / ( R N + R L)] * R N
• Voltage V L = IL * R L
• Power consumed =P L = V L * I L
Dr. K Hussain

86. 86
Example 1: Using Norton’s theorem, calculate current through 15 Ω
resistor in the circuit shown in Fig.4.
Fig.4
Ω
Ω Ω
Dr. K Hussain

87. 87
Solution
Step 1: To find out IN:
• Remove the load resistance
• Short circuit the load
terminals as shown in Fig.5
• Find the current flowing
through the short
• 12 Ω resistor is in parallel
with the short as shown in
Fig.5 Fig.5
Ω
Ω
Dr. K Hussain

88. 88
• So, 12 Ω resistor will also
be short circuited
• Therefore the circuit will be
as shown in Fig.6
• IN = 24 / (3+1)
= 6 A
Fig.6
Ω
Dr. K Hussain

89. 89
Step 2: To find out RN :
• Replace the battery with its
internal resistance
• Open circuit load terminals as
shown in Fig.7
• Calculate the resistance of the
circuit as viewed from the open
circuited load terminals
Fig.7
Dr. K Hussain
Ω

90. 90
• When we look into the circuit
as shown in Fig.7, from the
open circuited load terminals
• 3Ω and 1Ω resistors are in
series
• The combination is in parallel
with 12 Ω
• Therefore,
RN = 4 * 12 / (4+12)
= 48 / 16 = 3 Ω
Fig.7
Dr. K Hussain

91. 91
• Replace the given
network with Norton’s
equivalent circuit as
shown in Fig.8
• I L = I N * R N
(RN + RL)
= 6 * [3 / (3 + 15)]
= 18 / 18
= 1 A Fig.8
Ω
Dr. K Hussain
Ω

92. Source Transformation
• Source transformation is another tool for simplifying
circuits.
• A source transformation 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.
Dr. K Hussain 92
Fig. Transformation of dependent sources

93. Dr. K Hussain 93
Source transformation also applies to dependent sources, provided
we carefully handle the dependent variable.
• It is easy to show that they are indeed equivalent. If the sources are
turned off, the equivalent resistance at terminals a-b in both circuits is
R.
• Also, when terminals are short-circuited, the short-circuit current
flowing from a to b is isc=vs/R in the circuit on the left-hand side and
isc=is for the circuit on the right-hand side.
• Thus, vs/R=is in order for the two circuits to be equivalent.
• Hence, source transformation requires that
Fig. Transformation of dependent sources

94. 94
• The Thevenin and Norton equivalent circuits are related by a
source transformation.
• This is essentially source transformation. For this reason,
source transformation is often called Thevenin-Norton
transformation.
Dr. K Hussain

95. Dr. K Hussain 95
Example 2. Find the Norton equivalent circuit of the circuit in Fig. at terminals a-b.
Solution:
We find RN in the same way we find RTh,
in the Thevenin equivalent circuit.
Set the independent sources equal to zero.
This leads to the circuit in Fig. (a), from which
we find RN .
Thus,
To find IN, we short-circuit terminals a and b,
as shown in Fig.(b).
We ignore the 5-ohm resistor because it has been
short-circuited.
Applying mesh analysis, we obtain
From these equations, we obtain

96. Dr. K Hussain 96
Alternatively, we may determine IN from VTh
. We
obtain VTh as the open-circuit voltage across
terminals a and b in Fig (c).
Using mesh analysis, we obtain
as obtained previously.
This also serves to confirm Eq. that
RTh=(voc/isc)=(4/1)=4 ohm
Thus, the Norton equivalent circuit is as
shown in Fig. d.
Hence

97. 97
Summary
In this period, we have discussed
• Statement of Norton’s theorem
• Explanation of Norton’s theorem
• Application of Norton’s theorem to a given circuit
Dr. K Hussain

98. 98
Quiz
1. What is the internal resistance for an ideal current
source?
a) Infinite
b) Zero
c) Depends on other factors of the circuit
d) None
Dr. K Hussain

99. 99
2. Norton’s equivalent circuit will consist of
a) A current source in parallel with resistance
b) A voltage source in series with resistance
c) None
Quiz
Dr. K Hussain

100. 100
Frequently Asked Questions
1. State & explain Norton’s theorem.
2. Explain the steps to obtain the Norton’s equivalent circuit
for a given network.
Dr. K Hussain

101. 101
1. Determine current supplied by the battery, using Norton’s
theorem.
Assignment
Dr. K Hussain

102. 102
Norton’s Theorem
Dr. K Hussain

103. 103
Recap
In the previous lecture, we have discussed
• Thevenin’s theorem
• Thevenin’s equivalent circuit of a given circuit
103
Dr. K Hussain

104. 104
Objectives
On completion of this lecture, you would be able to
• State Norton’s theorem
• Explain Norton’s theorem
• Apply Norton’s theorem to a given circuit
Dr. K Hussain

105. 105
Known To Unknown
We have already discussed that
• Many laws and theorems are available for solving networks
• Some of them are, Kirchhoff’s laws, Maxwell’s Loop Current
theorems, Superposition theorem etc.,
• Norton’s Theorem is also one of such theorems
Dr. K Hussain

106. 106
Norton’s Theorem
• Statement:
• “Any two terminal active network when viewed from its load terminals,
can be replaced by a single current source in parallel with a single
resistance.”
• That single current source is called Norton’s current source (IN)
And Resistance is called Norton’s resistance (RN)
OR
• Substitution Theorem states that the voltage across any branch or the
current through that branch of a network being known, the branch can
be replaced by the combination of various elements that will make the
same voltage and current through that branch. In other words,
the Substitution Theorem says that for branch equivalence, the
terminal voltage and current must be the same.
Dr. K Hussain

107. Dr. K Hussain 107
The concept of the theorem is based on the substitution of one element from another
element. This theorem gives intuition on the behaviour of the circuit. It also helps in
proving several other theorems.
But the substitution theorem cannot use for solving the theorem which has more than
two sources which are neither connected in series nor parallel.
Explanation of Reciprocity Theorem
Simply one can say that the Substitution Theorem is the replacement of one element
with another equivalent element. In a network, if any element is substituted or
replaced by a voltage or current source whose voltage and current across or through
that element will remain unchanged as the previous network.

108. Dr. K Hussain 108
Let us understand the theorem with the help of the circuit diagram shown below:
The current I, is flowing through the circuit, which is divided into current I1 flowing
through the resistance R1 and the current I2 flowing through the resistance R2. V1, V2 and
V3 are the voltage drop across the resistance R1, R2 and R3 respectively.

109. Dr. K Hussain 109
Now if the resistance R3 is substituted by the voltage source V3 as shown in the circuit
diagram below:
In the circuit diagram shown below the resistance, R3 is replaced by the current flowing
through that element, i.e. I1

110. Dr. K Hussain 110
In both the cases shown above if the element is substituted by the voltage source or the
current source, then also, the initial conditions of the circuit does not alter.
This means that the voltage across the resistance and current flowing through the
resistance unaltered even if they are substituted by other sources.
Steps for Solving Network Using Substitution Theorem

111. Dr. K Hussain 111
Steps for Solving Network Using Substitution Theorem

112. It is clear from the figure above that the voltage source and
current sources are interchanged for solving the network with
the help of Reciprocity Theorem.
• The limitation of this theorem is that it is applicable only to
single-source networks and not in the multi-source network.
The network where reciprocity theorem is applied should be
linear and consist of resistors, inductors, capacitors and
coupled circuits. The circuit should not have any time-varying
elements.
Dr. K Hussain 112

113. Steps for Solving a Network Utilizing Reciprocity Theorem
• Step 1 – Firstly, select the branches between which reciprocity
has to be established.
• Step 2 – The current in the branch is obtained using any
conventional network analysis method.
• Step 3 – The voltage source is interchanged between the
branch which is selected.
• Step 4 – The current in the branch where the voltage source
was existing earlier is calculated.
• Step 5 – Now, it is seen that the current obtained in the
previous connection, i.e., in step 2 and the current which is
calculated when the source is interchanged, i.e., in step 4 are
identical to each other.
Dr. K Hussain 113

114. 114
Norton’s Equivalent Circuit
Norton’s equivalent circuitTwo terminal circuit
Fig.1 Fig.2
Dr. K Hussain

115. 115
To Find IN :
• Remove the load resistance & put a short there
• Then find the current flowing through the short circuited
terminals
• I N = I SC
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116. 116
To Find RN :
• Replace the voltage source (battery) with its internal
resistance
• Calculate the resistance of the circuit as viewed by the open
circuited load terminals. (i.e. terminals A & B as shown in
Fig.1)
• R N = R i (internal resistance)
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117. 117
Replace the given circuit with its Norton’s equivalent circuit as
shown in Fig.3
Fig.3
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118. 118
At load terminals,
• Current I L = [ I N / ( R N + R L)] * R N
• Voltage V L = IL * R L
• Power consumed =P L = V L * I L
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119. 119
Example 1: Using Norton’s theorem, calculate current through 15 Ω
resistor in the circuit shown in Fig.4.
Fig.4
Ω
Ω Ω
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120. 120
Solution
Step 1: To find out IN:
• Remove the load resistance
• Short circuit the load
terminals as shown in Fig.5
• Find the current flowing
through the short
• 12 Ω resistor is in parallel
with the short as shown in
Fig.5 Fig.5
Ω
Ω
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121. 121
• So, 12 Ω resistor will also
be short circuited
• Therefore the circuit will be
as shown in Fig.6
• IN = 24 / (3+1)
= 6 A
Fig.6
Ω
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122. 122
Step 2: To find out RN :
• Replace the battery with its
internal resistance
• Open circuit load terminals as
shown in Fig.7
• Calculate the resistance of the
circuit as viewed from the open
circuited load terminals
Fig.7
Dr. K Hussain
Ω

123. 123
• When we look into the circuit
as shown in Fig.7, from the
open circuited load terminals
• 3Ω and 1Ω resistors are in
series
• The combination is in parallel
with 12 Ω
• Therefore,
RN = 4 * 12 / (4+12)
= 48 / 16 = 3 Ω
Fig.7
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124. 124
• Replace the given
network with Norton’s
equivalent circuit as
shown in Fig.8
• I L = I N * R N
(RN + RL)
= 6 * [3 / (3 + 15)]
= 18 / 18
= 1 A Fig.8
Ω
Dr. K Hussain
Ω

125. Source Transformation
• Source transformation is another tool for simplifying
circuits.
• A source transformation 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.
Dr. K Hussain 125
Fig. Transformation of dependent sources

126. Dr. K Hussain 126
Source transformation also applies to dependent sources, provided
we carefully handle the dependent variable.
• It is easy to show that they are indeed equivalent. If the sources are
turned off, the equivalent resistance at terminals a-b in both circuits is
R.
• Also, when terminals are short-circuited, the short-circuit current
flowing from a to b is isc=vs/R in the circuit on the left-hand side and
isc=is for the circuit on the right-hand side.
• Thus, vs/R=is in order for the two circuits to be equivalent.
• Hence, source transformation requires that
Fig. Transformation of dependent sources

127. 127
• The Thevenin and Norton equivalent circuits are related by a
source transformation.
• This is essentially source transformation. For this reason,
source transformation is often called Thevenin-Norton
transformation.
Dr. K Hussain

128. Dr. K Hussain 128
Example 2. Find the Norton equivalent circuit of the circuit in Fig. at terminals a-b.
Solution:
We find RN in the same way we find RTh,
in the Thevenin equivalent circuit.
Set the independent sources equal to zero.
This leads to the circuit in Fig. (a), from which
we find RN .
Thus,
To find IN, we short-circuit terminals a and b,
as shown in Fig.(b).
We ignore the 5-ohm resistor because it has been
short-circuited.
Applying mesh analysis, we obtain
From these equations, we obtain

129. Dr. K Hussain 129
Alternatively, we may determine IN from VTh
. We
obtain VTh as the open-circuit voltage across
terminals a and b in Fig (c).
Using mesh analysis, we obtain
as obtained previously.
This also serves to confirm Eq. that
RTh=(voc/isc)=(4/1)=4 ohm
Thus, the Norton equivalent circuit is as
shown in Fig. d.
Hence

130. 130
Summary
In this period, we have discussed
• Statement of Norton’s theorem
• Explanation of Norton’s theorem
• Application of Norton’s theorem to a given circuit
Dr. K Hussain

131. 131
Quiz
1. What is the internal resistance for an ideal current
source?
a) Infinite
b) Zero
c) Depends on other factors of the circuit
d) None
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132. 132
2. Norton’s equivalent circuit will consist of
a) A current source in parallel with resistance
b) A voltage source in series with resistance
c) None
Quiz
Dr. K Hussain

133. 133
Frequently Asked Questions
1. State & explain Norton’s theorem.
2. Explain the steps to obtain the Norton’s equivalent circuit
for a given network.
Dr. K Hussain

134. 134
1. Determine current supplied by the battery, using Norton’s
theorem.
Assignment
Dr. K Hussain

135. Maximum Power Transfer Theorem
Dr. K Hussain 135

136. Dr. K Hussain 136
Recap
In the previous Lecture, we have discussed
• Statement of Norton’s theorem
• Explanation of Norton’s theorem
• Application of Norton’s theorem to a given circuit

137. Dr. K Hussain 137
Objectives
On completion of this lecture, you will be
able to
• State Maximum power transfer theorem
• Derive the expression for Maximum power transferred to load
terminals
• Apply Maximum power Transfer theorem to a given circuit

138. Dr. K Hussain 138
Known To Unknown
• The expression for power is
P = V * I (or) I2R (or) V2/R
• In a resistor, power will be consumed in the form of heat

139. Dr. K Hussain 139
Maximum Power Transfer Theorem
• As applied to D.C. networks, it may stated as
• “ A resistive load will abstract max. power from a
network, when load resistance is equal to resistance of
the network as viewed from the output terminals, with
all sources of e.m.f removed leaving behind their internal
resistances.”
Maximum power is transferred to the load when the load resistance equals the
Thevenin resistance as seen from the load (RL RTh).
OR

140. Let
• RL - Load resistance
• r - internal resistance
of the battery
• Ri - total resistance
of the circuit, as seen
from the load terminals
(A and B)
• R i = R2 ║(r + R1)
• R L will abstract max. power from the network when
R L = R i
Fig.1
Dr. K Hussain 140

141. • Proof:
Circuit current I = E / (R L + R i )
Power consumed by the load is P L
P L = I2 RL
= E2 RL / (RL+Ri)2
For P L to be maximum, (dPL /dR L)= 0 ----------eq(1)
[RL*2*(RL+ Ri)]-(RL+ Ri)2*1]/[(RL+ Ri)2]=0
2RL-(RL+ Ri)=0
Dr. K Hussain 141

142. Dr. K Hussain 142
• Differentiating eq(1), and simplifying we get
2 RL = RL + Ri
OR
RL = Ri
Hence Max. power is
P L = I2 RL
= E2 RL / (RL+Ri)2
=E2/4RL

143. Dr. K Hussain 143
• The voltage across the load is half the open circuit
voltage at the terminals A & B
• Overall efficiency of a network supplying max. power to
load terminals is 50 percent only
• Half of the power is wasted in the network
OR

144. Dr. K Hussain 144
• The application of this theorem to power transmission and
distribution networks is limited, because in their case, the goal is
high efficiency.
• This theorem is applicable and widely used in Communication
networks, because in their case, the goal is either to receive or
transmit maximum power.
The source and load are said to be matched when RL= RTh.

145. Dr. K Hussain 145
Example: Find the value of for maximum power transfer in the circuit
of Fig. Find the maximum power.
Solution:

146. Dr. K Hussain 146
Summary
In this lecture, we have discussed
• Max power transfer theorem
• Expression of maximum power transferred to load terminals
• Application of Maximum power Transfer theorem

147. Dr. K Hussain 147
Quiz
1. Max. power transfer theorem is used in
a) Communication networks
b) Transmission lines
c) Distribution lines
d) All the above

148. Quiz
2. If max. power is delivered to load what is the
efficiency of the system?
a) 100 %
b) 50%
c) 85%
d) 0%
148
Dr. K Hussain 148

149. Dr. K Hussain 149
Frequently Asked Questions
1. State and explain maximum power transfer theorem.
2. Derive an expression for maximum power received by the load
terminals and state where this theorem is more useful.

150. 150
Tellegen’s Theorem
Dr. K Hussain

151. 151
Recap
In the previous lecture, we have discussed
• Norton’s theorem and source transformation
151
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152. 152
Objectives
On completion of this lecture, you will be able to
• State Tellegen’s theorem
• Explain Tellegen’s theorem
• Apply Tellegen’s theorem to a given circuit
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153. 153
Known To Unknown
We have already discussed that
• Many laws and theorems are available for solving networks
• Some of them are, Kirchhoff’s laws, Maxwell’s Loop Current
theorems, Superposition theorem, Thevini’s and Norton’s
theorem, etc.,
• Tellegen’s Theorem is also one of such theorems
Dr. K Hussain

154. 154
Tellegen’s Theorem
• Statement:
• “Tellegen’s Theorem states that the summation of
power delivered is zero for each branch of any electrical
network at any instant of time”.
OR
• Tellegen’s Theorem can also be stated in another word
as, in any linear, nonlinear, passive, active, time-variant or
time-invariant network the summation of power
(instantaneous or complex power of sources) is zero.
Dr. K Hussain

155. 155
Tellegen’s Theorem
• Tellegen’s theorem is independent of the network elements. Thus, it is
applicable for any lump system that has linear, active, passive and time-
variant elements. Also, the theorem is convenient for the network
which follows Kirchoff’s current law and Kirchoff’s voltage law.
• It is mainly applicable for designing the filters in signal processing.
• It is also used in complex operating systems for regulating stability. It is
mostly used in the chemical and biological system and for finding the
dynamic behaviour of the physical network.
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156. 156
Explanation of Tellegen’s Theorem
• Tellegen’s Theorem can also be stated in another word as, in any linear,
nonlinear, passive, active, time-variant or time-invariant network the
summation of power (instantaneous or complex power of sources) is
zero.
• Thus, for the Kth branch, this theorem states that:
• Where,
n is the number of branches; vK is the voltage in the branch
iK is the current flowing through the branch.
Dr. K Hussain

157. Dr. K Hussain 157
Equation (1) shows the Kth branch through current
vK is the voltage drop in branch K and is given as:
We have
Let,
Where vp and vq are the respective node voltage at p and q nodes.

158. Dr. K Hussain 158
Also
Obviously
Summing the above two equations (2) and (3), we get
Such equations can be written for every branch of the network.
Assuming n branches, the equation will be:

159. Dr. K Hussain 159
However, according to the Kirchhoff’s current law (KCL), the
algebraic sum of currents at each node is equal to zero.
Therefore,
Thus, from the above equation (4) finally, we obtain
Thus, it has been observed that the sum of power delivered to a
closed network is zero.
This proves the Tellegen’s theorem and also proves the
conservation of power in any electrical network.

160. • It is also evident that the sum of power delivered to the network by
an independent source is equal to the sum of power absorbed by all
passive elements of the network.
• Note:
• It depends on voltage and current product of an element but not on
the type of element.
• While verifying Tellegen’s theorem do not disturb original network.
Dr. K Hussain 160

161. Dr. K Hussain 161
Step 2 – Find the corresponding branch currents using
conventional analysis methods.
Step 1 – In order to justify this theorem in an electrical
network, the first step is to find the branch voltage drops.
Steps for Solving Networks Using Tellegen’s Theorem
The following steps are given below to solve any electrical
network by Tellegen’s theorem:
Step 3 – Tellegen’s theorem can then be justified by summing
the products of all branch voltages and currents.

162. • Now, if the set of voltages and currents is taken, corresponding
the two different instants of time, t1 and t2, then Tellegen’s
theorem is also applicable where we get the equation as shown
below:
Dr. K Hussain 162
• For example, if a network having some branches “b” then:

163. Dr. K Hussain 163
Example:
Verify the Tellegen’s theorem for the given circuit.
Solution:
If current flows from + to – then treat it as
power absorption.
If current flow from – to + then treat it as
power delivering.
∴ P10V = V. I = 10 × 1 = 10 watt (Pabsorbed)
P2A = V. I = 10 × 2 = 20 watt (Pdelivered)
P10Ω = I2. R = 1 × 10 = 10 watt (Pabsorbed)
∴ Pdelivered = Pabsorbed = 20 watt
Hence Tellegen’s theorem is verified.

164. Dr. K Hussain 164
For this network, we will assume a set of branch
voltages satisfy the Kirchhoff voltage law and a set
of branch current satisfy Kirchhoff current law at
each node.
We will then show that these arbitrary
assumed voltages and currents satisfy
the equation.
And it is the condition of Tellegen’s
theorem.
In the network shown in the figure,
let v1, v2 and v3 be 7, 2 and 3 volts respectively.
Applying Kirchhoff Voltage Law around loop ABCDEA.
We see that v4 = 2 volt is required. Around loop CDFC, v5 is required
to be 3 volt and around loop DFED, v6 is required to be 2.
Example:

165. Dr. K Hussain 165
We next apply Kirchhoff’s Current Law successively
to nodes B, C and D.
At node B let ii = 5 A, then it is required that i2 = – 5 A.
At node C let i3 = 3 A and then i5 is required to be – 8.
At node D assume i4 to be 4 then i6 is required to be – 9.
Carrying out the operation of equation,
We get,
Hence Tellegen theorem is verified.

166. Dr. K Hussain 166
Application of Tellegen’s Theorem
• It is used in the digital signal processing system for designing
filters.
• In the area of the biological and chemical process.
• In topology and structure of reaction network analysis.
• The theorem is used in chemical plants and oil industries to
determine the stability of any complex systems.

167. THANK YOU
Dr. K Hussain 167

168. 168
Substitution Theorem
Dr. K Hussain

169. 169
Recap
In the previous lecture, we have discussed
• Tellegen’s theorem
• Explanation of Tellegen’s theorem
• Application of Tellegen’s theorem to the circuit
169
Dr. K Hussain

170. 170
Objectives
On completion of this lecture, you will be able to
• State Substitution theorem
• Explain Substitution theorem
• Apply Substitution theorem to a given circuit
Dr. K Hussain

171. 171
Known To Unknown
We have already discussed that
• Many laws and theorems are available for solving networks
• Some of them are, Kirchhoff’s laws, Maxwell’s Loop Current
theorems, Superposition theorem, Thevini’s and Norton’s
theorem, etc.,
• Substitution Theorem is also one of such theorems
Dr. K Hussain

172. Dr. K Hussain 172
Substitution theorem states that if an element in a network is
replaced by a voltage source whose voltage at any instant of time
is equals to the voltage across the element in the previous
network then the initial condition in the rest of the network will
be unaltered.
Substitution Theorem
Alternately if an element in a network is replaced by a current
source whose current at any instant of time is equal to the current
through the element in the previous network then the initial
condition in the rest of the network will be unaltered.
or

173. Dr. K Hussain 173
Explanation of Substitution Theorem:
The current I, is flowing through the circuit, which is divided into
current I1 flowing through the resistance R1 and the current
I2 flowing through the resistance R2. V1, V2 and V3 are the voltage
drop across the resistance R1, R2 and R3 respectively.

174. Dr. K Hussain 174
Now if the resistance R3 is substituted by the voltage source V3 as
shown in the circuit diagram below:
In the circuit diagram shown below the resistance, R3 is replaced
by the current flowing through that element, i.e. I1

175. Dr. K Hussain 175
In both the cases shown above if the element is substituted by the
voltage source or the current source, then also, the initial
conditions of the circuit does not alter.
This means that the voltage across the resistance and current
flowing through the resistance unaltered even if they are
substituted by other sources.

176. Dr. K Hussain 176
Example
Let us take a circuit as shown in fig – d.
For more efficient and clear understanding let us go
through a simple practical example:
As per voltage division rule voltage across
3Ω and 2Ω resistance are

177. Dr. K Hussain 177
If we replace the 3Ω resistance with a voltage source of 6 V as
shown in fig – e, then
According to Ohm’s law the voltage across 2Ω
resistance and current through the circuit is
Alternately if we replace 3Ω resistance with a current source of 2A as shown
in fig – f, then
Voltage across 2Ω is V2Ω = 10 – 3× 2 = 4 V
and
voltage across 2A current source is V2A = 10 – 4 = 6 V
We can see the voltage across 2Ω resistance and current through the circuit is
unaltered
i.e., all initial condition of the circuit is intact.

178. Dr. K Hussain 178
THANK YOU

179. Dr. K Hussain 179
Reciprocity Theorem

180. Dr. K Hussain 180
the reciprocity theorem states that if an emf E in one branch
of a reciprocal network produces a current I in another, then
if the emf E is moved from the first to the second branch, it
will cause the same current in the first branch, where the emf
has been replaced by a short circuit.
Reciprocity theorem states that if we consider two loops A
and B of a reciprocal network N, and if an ideal voltage
source, E, in loop A, produces a current I in loop B, then an
identical source in loop B will produce the same current I in
loop A.

181. Dr. K Hussain 181
Explanation of Reciprocity Theorem
The voltage source and the ammeter used in this
theorem must be ideal. That means the internal
resistance of both the voltage source and
ammetermust be zero.
The reciprocal circuit may be a simple or complex network. But
every complex reciprocal passive network can be simplified into
a simple network.
The ratio of V and I is called the transfer resistance.
As per reciprocity theorem, in a linear passive network, supply
voltage V and output current I are mutually transferable.

182. Dr. K Hussain 182
The theorem can easily be understood by this
following example.

183. Dr. K Hussain 183
In a network if we interchange the position of response
and excitation then the ratio of response to excitation
is constant.
Then, (IL/VS) = (IS/VL)

184. Dr. K Hussain 184
Find the value of I in the given network?
Example:
Solution:
By reciprocity theorem
5 = − I
10 30
I = - 15A

185. Dr. K Hussain 185
Steps for Solving Network Using Substitution Theorem
Step 1 – First obtain the concerned branch voltage and the current flowing through the
branch given by Vxy and Ixy as shown below in figure A
Step 2 – The branch may be substituted by an independent voltage source as shown
above in figure B

186. Dr. K Hussain 186
Step 3 – Similarly the branch may be replaced by
an independent current source as shown below
in the circuit diagram C.
Step 4 – Hence it is seen that the voltage drop and the
current flowing through the circuit A shown above will be
same if it is substituted by any independent voltage or
current source shown in the figure B and C.

187. • It is clear from the figure above that the voltage source and current
sources are interchanged for solving the network with the help of
Reciprocity Theorem.
• The limitation of this theorem is that it is applicable only to
single-source networks and not in the multi-source network.
Dr. K Hussain 187
• The network where reciprocity theorem is applied should be
linear and consist of resistors, inductors, capacitors and
coupled circuits. The circuit should not have any time-varying
elements.

188. Steps for Solving a Network Utilizing Reciprocity Theorem
• Step 1 – Firstly, select the branches between which reciprocity
has to be established.
• Step 2 – The current in the branch is obtained using any
conventional network analysis method.
• Step 3 – The voltage source is interchanged between the
branch which is selected.
• Step 4 – The current in the branch where the voltage source
was existing earlier is calculated.
• Step 5 – Now, it is seen that the current obtained in the
previous connection, i.e., in step 2 and the current which is
calculated when the source is interchanged, i.e., in step 4 are
identical to each other.
Dr. K Hussain 188

189. 189
Summary
In this period, we have discussed
• Statement of Norton’s theorem
• Explanation of Norton’s theorem
• Application of Norton’s theorem to a given circuit
Dr. K Hussain

190. 190
Quiz
1. What is the internal resistance for an ideal current
source?
a) Infinite
b) Zero
c) Depends on other factors of the circuit
d) None
Dr. K Hussain

191. 191
2. Norton’s equivalent circuit will consist of
a) A current source in parallel with resistance
b) A voltage source in series with resistance
c) None
Quiz
Dr. K Hussain

192. 192
Frequently Asked Questions
1. State & explain Norton’s theorem.
2. Explain the steps to obtain the Norton’s equivalent circuit
for a given network.
Dr. K Hussain

193. 193
1. Determine current supplied by the battery, using Norton’s
theorem.
Assignment
Dr. K Hussain