The Superposition
Theorem: Taming
Multiple Sources
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Anup Mondal Swipe >
Electrical Engineer
1
What is Superposition
Theorem?
In any linear circuit with multiple
independent sources, the response (voltage or
current) across any element is the algebraic
sum of the responses caused by each source
acting alone.
2
Why Use It?
Simplifies analysis of circuits with multiple
sources.
Helps understand individual source
contributions.
Works for both voltage and current sources.
3
Core Principle
⦿ Consider one source at a time.
⦿ Turn OFF (deactivate) other independent
sources:
➢ Voltage sources → replace with short circuit.
➢ Current sources → replace with open circuit.
⦿ Then add up all individual effects.
4
Steps to Apply Superposition
Theorem
1️⃣ Select one independent source, turn OFF
others.
2️⃣ Calculate current/voltage contribution
from that source.
3️⃣ Repeat for all sources.
4️⃣ Add all contributions algebraically →
Final result.
5
Example Application
Find the current flowing through 20 Ω using the
superposition theorem.
» Step 1: First, let us find the current flowing through a
circuit by considering only the 20 V voltage source. The
current source can be open-circuited hence, the modified
circuit diagram is shown in the following figure.
6
» Step 2: The nodal voltage V1 can be determined using
the nodal analysis method.
The nodal equation at node 1 is written as follows:
The current flowing through the 20 Ω resistor can be
found using the following equation:
𝑉1
I1=10+20
Substituting the value of the V 1 in the above equation, we
get I1 = 0.4 A
Therefore, the current flowing through the 20 Ω resistor to
due 20 V voltage source is 0.4 A.
7
» Step 3: Now let us find out the current flowing through
the 20 Ω resistor considering only the 4 A current source.
We eliminate the 20 V voltage source by short-circuiting
it. The modified circuit, therefore, is given as follows:
In the above circuit, the resistors 5 Ω and 10 Ω are parallel
to each other, and this parallel combination of resistors is
in series with the 10 Ω resistor. Therefore, the equivalent
resistance will be:
5𝑥10 40
RAB=5+10 + 10 = 3 Ω
8
Now, the simplified circuit is shown as follows:
The current flowing through the 20 Ω resistor can be
determined using the current division principle.
𝑅1
I2=Is 𝑅1+𝑅2
Substituting the values, we get
I2= 1.6 A
Therefore, the current flowing through the circuit when
only 4 A current source is 1.6 A.
» Step 4: The summation of currents I1 and I2 will give us
the current flowing through the 20 Ω resistor.
Mathematically, this is represented as follows:
I = I1 + I2
Substituting the values of I1 and I2 in the above equation,
we get
I = 0.4+1.6 = 2 A
Therefore, the current flowing through the resistor is 2 A.
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Key Limitations
Only valid for linear circuits.
Cannot be applied directly for power
(since power ∝ I² or V²).
More useful for voltages & currents, not
nonlinear components.
10
Practical Applications
⦿ Multi-source DC & AC circuits.
⦿ Signal analysis in communication
systems.
⦿ Complex power system network studies.
Key Takeaways:
Combine with Thevenin/Norton
Theorems to simplify further.
Superposition shows contributions;
Thevenin/Norton simplify structure.
Superposition Theorem = “Divide &
Conquer” for circuits.
Master it to analyze multi-source networks
with confidence.
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Anup Mondal
Electrical Engineer
