Superposition Theorem
Superposition theorem is based on the concept of linearity between the response and
excitation of an electrical circuit. It states that the response in a particular branch of a linear
circuit when multiple independent sources are acting at the same time is equivalent to the
sum of the responses due to each independent source acting at a time.
In this method, we will consider only one independent source at a time. So, we have to
eliminate the remaining independent sources from the circuit. We can eliminate the voltage
sources by shorting their two terminals and similarly, the current sources by opening their
two terminals.
Therefore, we need to find the response in a particular branch ‘n’ times if there are ‘n’
independent sources. The response in a particular branch could be either current flowing
through that branch or voltage across that branch.
Procedure of Superposition Theorem
Follow these steps in order to find the response in a particular branch using superposition
theorem.
Step 1 − Find the response in a particular branch by considering one independent source
and eliminating the remaining independent sources present in the network.
Step 2 − Repeat Step 1 for all independent sources present in the network.
Step 3 − Add all the responses in order to get the overall response in a particular branch
when all independent sources are present in the network.
Example
Find the current flowing through 20 Ω resistor of the following circuit using superposition
theorem.
Step 1 − Let us find the current flowing through 20 Ω resistor by considering only 20 V
voltage source. In this case, we can eliminate the 4 A current source by making open
circuit of it. The modified circuit diagram is shown in the following figure.
There is only one principal node except Ground in the above circuit. So, we can use nodal
analysis method. The node voltage V1 is labelled in the following figure. Here, V1 is the
voltage from node 1 with respect to ground.
The nodal equation at node 1 is
Substitute the value of V1 in the above equation.
Therefore, the current flowing through 20 Ω resistor is 0.4 A, when only 20 V voltage source
is considered.
Step 2 − Let us find the current flowing through 20 Ω resistor by considering only 4 A
current source. In this case, we can eliminate the 20 V voltage source by making short-
circuit of it. The modified circuit diagram is shown in the following figure.
In the above circuit, there are three resistors to the left of terminals A & B. We can replace
these resistors with a single equivalent resistor. Here, 5 Ω & 10 Ω resistors are connected
in parallel and the entire combination is in series with 10 Ω resistor.
The equivalent resistance to the left of terminals A & B will be
The simplified circuit diagram is shown in the following figure.
We can find the current flowing through 20 Ω resistor, by using current division principle.
Substitute IS = 4A, R1 = 40/3Ω and R2 = 20Ω in the above equation.
Therefore, the current flowing through 20 Ω resistor is 1.6 A, when only 4 A current source
is considered.
Step 3 − We will get the current flowing through 20 Ω resistor of the given circuit by doing
the addition of two currents that we got in step 1 and step 2. Mathematically, it can be
written as
I = I1 + I2
Substitute, the values of I1 and I2 in the above equation.
I = 0.4 + 1.6 = 2A
Therefore, the current flowing through 20 Ω resistor of given circuit is 2 A.
Note − We can’t apply superposition theorem directly in order to find the amount
of power delivered to any resistor that is present in a linear circuit, just by doing the addition
of powers delivered to that resistor due to each independent source. Rather, we can
calculate either total current flowing through or voltage across that resistor by using
superposition theorem and from that, we can calculate the amount of power delivered to
that resistor using I2R or V2/R.
Questions
1. What kinds of circuits can be solved by superposition?
2. What are the steps to solve a circuit using the superposition theorem?
3. How are different kinds of sources treated when solving by superposition?
4. How do I use superposition to solve a circuit?
