Department of Electrical Engineering

Academic year: 2024-25, Semester: I

100106: Basic Electrical Engineering

Module 1: DC Circuits

Lecture: Network Theorem Introduction & Superposition Theorem)

Instructor: Dr. Bablesh Kumar Jha

1 Objectives:
• State and explain Superposition theorem with an example.

• Solve circuit problems using Superposition theorem.

2 Background
A major advantage of analyzing circuits using Kirchhoff’s laws as we studied in Lecture#1 is that
we can analyze a circuit without tampering with its original configuration.
A major disadvantage of this approach is that, for a large, complex circuit, tedious computation
is involved.

3 Network Theorem:
The growth in areas of application of electric circuits has led to an evolution from simple to complex
circuits. To handle the complexity, engineers over the years have developed some theorems to
simplify circuit analysis.
When all the theorems and techniques will be known, we will be in a position to apply a par-
ticular theorem or a technique that reduces the time for solving a given problem.

3.1 Superposition Theorem

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 as we discussed in Lecture#3 & 4.
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.

Statement: 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.

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3.1.1 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.

• We consider one independent source at a time while all other independent sources are
turned off. This implies that we replace every voltage source by 0 V (or a short circuit),
and every current source by 0 A (or an open circuit). This way we obtain a simpler and
more manageable circuit.

Note: Dependent sources are left intact because they are controlled by circuit variables.

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 indepen-
dent sources.

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Linearity
Linearity is the property of an element describing a linear relationship between cause and effect.
Although the property applies to many circuit elements, we shall limit its applicability to resistors
in this chapter. The property is a combination of both the homogeneity (scaling) property and the
additivity property.
The homogeneity property requires that if the input (also called the excitation) is multiplied by
a constant, then the output (also called the response) is multiplied by the same constant. For a
resistor, for example, Ohm’s law relates the input i to the output v,

v = iR

If the current is increased by a constant k, then the voltage increases correspondingly by k; that is,

kiR = kv

The additivity property requires that the response to a sum of inputs is the sum of the responses
to each input applied separately. Using the voltage-current relationship of a resistor, if

v1 = i1 R

and,
v2 = i2 R
then applying (i1 + i2 ) gives

v = (i1 + i2 )R = i1 R + i2 R = v1 + v2

We say that a resistor is a linear element because the voltage-current relationship satisfies both the
homogeneity and the additivity properties.
In general, a circuit is linear if it is both additive and homogeneous. A linear circuit consists of
only linear elements, linear dependent sources, and independent sources.

A linear circuit is one whose output is linearly related (or directly proportional) to its input.

v2
Important: since p = i2 R = R
(making it a quadratic function rather than a linear one), the

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 Figure 1: Example-1

Figure 2: Solution of example-1

Figure 3: Solution of example-1

relationship between power and voltage (or current) is nonlinear.

Example-1 Find the current through the 4Ω resistor in Fig. 1.
Solution: When the 40 V source is acting alone (Fig. 2)
By series–parallel reduction technique (Fig. 3),
40
I= = 2.68 A
12 + 2.92
From Fig. 3(a), by current-division rule,
5
I ′ = 2.68 × = 1.12 A(→) = −1.12 A(←)
5+7
When the 8 A source is acting alone (Fig. 4),
By series–parallel reduction technique (Fig. 5),
3
I” = 8 × = 2.8 A(←)
7.53 + 3
By superposition theorem,

I = I ′ + I” = −1.12 + 2.8 = 1.16 A(←)

Example-2 Find the current through the 3 Ω resistor in Fig. 6.

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 Figure 4: Solution of example-1

Figure 5: Solution of example-1

Figure 6: Solution of example-2

Figure 7: Solution of example-2

Solution: When the 5 A source is acting alone (Fig. 7)

By series–parallel reduction technique (Fig. 8),
15
I′ = 5 × = 3.75 A(↓)
15 + 2 + 3
When the 20 V source is acting alone (Fig. 9)
By series–parallel reduction technique (Fig. 10),
20
I= =6A
3.33

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 Figure 8: Solution of example-2

Figure 9: Solution of example-2

Figure 10: Solution of example-2

From Fig. 10(a), by current-division rule,

4
I” = 6 × = 1 A(↑) = −1 A(↓)
20 + 4
By superposition theorem,
I = I ′ + I” = 3.75 − 1 = 2.75 A(↓)
Example-3 Find the current Iy in Fig. 11.

Figure 11: Example-3

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 Figure 12: Solution of example-3

Solution When the 120 V source is acting alone (Fig. 12),

Applying KVL to the mesh,
120 − 4Iy′ − 10Iy′ − 8Iy′ = 0
Iy′ = 5.45 A(→)
When the 12 A source is acting alone (Fig. 13).

Figure 13: Solution of example-3

From Fig. 13,

Iy ” = I1 (1)
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,

I2 − I1 = 12 (2)

Applying KVL to the outer path of the supermesh,

−4I1 − 10Iy ” − 8I2 = 0

−4I1 − 10I1 − 8I2 = 0

14I1 + 8I2 = 0 (3)
Solving Eqs (2) and (2),
I1 = −4.36 A; I2 = 7.64 A
Iy ” = I1 = −4.36 A(→)

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 Figure 14: Solution of example-3

When the 40 V source is acting alone (Fig. 14)

Applying KVL to the mesh,

−4Iy ”′ − 10Iy ”′ − 8Iy ”′ − 40 = 0

40
Iy ”′ = − = −1.82 A(→)
22
By superposition theorem,

Iy = Iy′ + Iy ” + Iy ”′ = 5.45 − 4.36 − 1.82 = −0.73 A(→)

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 Problems for Practice
Question-1: Find the current through the 4 Ω resistor in Fig. 15.

Figure 15: Practice problem-1

Answer: 4.29 A(↓)

Question-2 Find the current through the 5 Ω resistor in Fig. 16.

Figure 16: Practice problem-2

Answer: 1.14 A(←)

Question-3 Find the current I1 in Fig. 17

Figure 17: Practice problem-3

Answer: 0.68 A(↑)

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If you have any doubts or queries, please do not hesitate to contact me.

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Your success as an engineer will be directly proportional to your ability to communicate!

-Charles K. Alexander