Superposition Theorem
Superposition Theorem – Statement, Explanation and Solved Numerical Problems
The Superposition Theorem is an important network theorem used in electrical circuit analysis. It is especially useful for analyzing linear circuits that contain more than one independent source. This theorem helps simplify complex circuits by allowing us to consider the effect of one source at a time.
This tutorial is designed for BEE, Diploma, Engineering students, and JEE aspirants. The explanation is beginner-friendly, exam-oriented, and focuses on developing a clear understanding of both theory and numerical problem solving.
Before learning the Superposition Theorem and solving numerical problems, you should have a good understanding of the following basic concepts:
📌Statement: In any linear, bilateral network containing atleast two energy sources, the cur rent flowing through a particular branch is the algebraic sum of the currents flow ing through that branch when each source is considered separately and remaining sources are replaced by their respective internal resistances.
While considering the effect of one source, all other independent sources must be replaced by their internal resistances. This simplifies the circuit and makes analysis easier.
How to Deactivate Sources in Superposition Theorem
While applying the Superposition Theorem, sources are deactivated as follows:
- Ideal voltage source → replaced by a short circuit
- Ideal current source → replaced by an open circuit
Dependent sources are not deactivated and remain active during the analysis.
Physical Meaning of Superposition Theorem
The Superposition Theorem is based on the principle of linearity. In a linear circuit, the response produced by multiple sources acting together is equal to the sum of responses produced by each source acting independently.
This principle does not apply to power calculations directly, because power is a nonlinear quantity. Therefore, superposition is used to find voltages or currents, not power.
Steps to Apply Superposition Theorem
Follow these systematic steps while solving numerical problems using the Superposition Theorem:
- Identify all independent sources in the circuit
- Choose one source and deactivate all others
- Analyze the circuit to find current or voltage due to that source
- Repeat the process for each independent source
- Add all individual responses algebraically
Explanation:
Consider the circuit shown in the above figure. We need to estimate the current \( I \) through \( R_L \) using the Superposition Theorem.
Consider voltage source \( V_1 \) acting alone. Remove the other voltage source i.e. replace it by its internal resistance. As the internal resistance of an ideal voltage source is zero, it is replaced by a short circuit. The current through \( R_L \) is \( I' \). (Refer below circuit)
Now consider the second voltage source \( V_2 \) acting alone. Replace the first voltage source with a short circuit. The current through \( R_L \) is \( I'' \). (Refer below circuit)
\( I' \) and \( I'' \) can be calculated by using Kirchhoff’s Voltage Law.
Therefore, the current flowing through \( R_L \) when both the voltage sources are acting is:
\[ I = I' + I'' \]
Before learning numerical problems based on the Superposition Theorem, you should be familiar with the application of Kirchhoff’s Voltage Law (KVL). If you have not studied this concept yet, please visit the Kirchhoff’s Laws: Solved examples page first, learn the basics, and then continue with this tutorial.
Superposition Theorem: Solved Example Problems
The following solved examples demonstrate how the Superposition Theorem is applied step by step. Carefully observe the method and sign convention used in each case.
Q.1 Find the current flowing through the 20 Ξ© resistance.
Consider Only 4 V source active
π Mesh–1 Equation:
\[ 4 - 10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = -4 \]
π Mesh–2 Equation:
\[ -20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = 0 \]
\[ I_1 = 0.28~\text{A} \] \[ I_2 = 0.04~\text{A} \] \[ I'_{20\Omega} = 0.04~\text{A} \;(\rightarrow) \]
Consider Only 8 V source active
π Mesh–1 Equation:
\[ -10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = 0 \]
π Mesh–2 Equation:
\[ 8 - 20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = -8 \]
\[ I_1 = 0.08~\text{A} \] \[ I_2 = 0.24~\text{A} \] \[ I''_{20\Omega} = 0.24~\text{A} \;(\rightarrow) \]
\[ I_{20\Omega} =I'_{20 \Omega} + I''_{20 \Omega} \] \[ I_{20\Omega} = 0.04 + 0.24 \] \[ I_{20\Omega} = \boxed{0.28~\text{A}} \;(\rightarrow) \]
Q.2 Find the current flowing through the 10 Ξ© resistance.
Consider Only 4 V source active
π Mesh–1 Equation:
\[ -4 - 10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = 4 \]
π Mesh–2 Equation:
\[ -20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = 0 \]
\[ I_1 = -0.28~\text{A} \] \[ I_2 = -0.04~\text{A} \] \[ I'_{20\Omega} = -0.04~\text{A} \;(\rightarrow) \]
Consider Only 8 V source active
π Mesh–1 Equation:
\[ -10I_1 - 5(I_1 - I_2) = 0 \]
\[ -15I_1 + 5I_2 = 0 \]
π Mesh–2 Equation:
\[ 8 - 20I_2 - 10I_2 - 5(I_2 - I_1) = 0 \]
\[ 5I_1 - 35I_2 = -8 \]
\[ I_1 = 0.08~\text{A} \] \[ I_2 = 0.24~\text{A} \] \[ I''_{20\Omega} = 0.24~\text{A} \;(\rightarrow) \]
\[ I_{20\Omega} =I'_{20 \Omega} + I''_{20 \Omega} \] \[ I_{20\Omega} = -0.04 + 0.24 \] \[ I_{20\Omega} = \boxed{0.2~\text{A}} \;(\rightarrow) \]
Q.3 Find the current flowing through the 5 Ξ© resistance.
Consider Only 50 V source active
π Mesh–1 Equation:
\[ 50 - 2I_1 - 4(I_1 - I_2) = 0 \]
\[ -6I_1 + 4I_2 = -50 \]
π Mesh–2 Equation:
\[ -3I_2 -5I_2 - 4(I_2-I_1) = 0 \]
\[ 4I_1 -12I_2 = 0 \]
\[ I_1 = 10.7142~\text{A} \] \[ I_2 = 3.5714~\text{A} \] \[ I'_{5\Omega} = 3.5714~\text{A} \;(\downarrow) \]
Consider Only 100 V source active
π Mesh–1 Equation:
\[ -2I_1 - 4(I_1 - I_2) = 0 \] \[ -6I_1 + 4I_2 = 0 \]
π Mesh–2 Equation:
\[ -3I_2 -5I_2 - 4(I_2-I_1) -100= 0 \] \[ 4I_1 -12I_2 = 100 \]
\[ I_1 = -7.1428~\text{A}\] \[ I_2 = -10.7142~\text{A}\]
\[ I''_{5 \Omega} = -10.7142~\text{A} (\downarrow) \]
\[ I_{5\Omega} =I'_{5 \Omega} + I''_{5 \Omega} \] \[ I_{20\Omega} = 3.5714 + (-10.7142) \] \[ I_{20\Omega} = \boxed{-7.1428~\text{A}} \;(\downarrow) \] \[ I_{20\Omega} = \boxed{7.1428~\text{A}} \;(\uparrow) \]
Q.4 Find the current flowing through the 1 Ξ© resistance.
Consider Only 16 V source active
π Mesh–1 Equation:
\[ 16 -4I_1 -4(I_1 - I_2) = 0 \] \[ -8I_1 + 4I_2 = -16 \]
π Mesh–2 Equation:
\[ -2I_2 -1I_2 -4(I_2 - I_1) = 0 \] \[ 4I_1-7I_2=0 \]
\[ I_1 = 2.8~\text{A}\] \[ I_2 = 1.6~\text{A}\]
\[ I'_{2 \Omega} = 1.6~\text{A} (\rightarrow) \]
Consider Only 12 V source active
π Mesh–1 Equation:
\[ -4I_1 -4(I_1 - I_2)-12 = 0 \] \[ -8I_1 + 4I_2 = 12 \]
π Mesh–2 Equation:
\[ -2I_2 -1I_2 +12 -4(I_2 - I_1) = 0 \] \[ 4I_1-7I_2=-12 \]
\[ I_1 = -0.9~\text{A}\] \[ I_2 = 1.2~\text{A}\]
\[ I''_{2 \Omega} = 1.2~\text{A} (\rightarrow) \]
Consider Only 20 V source active
π Mesh–1 Equation:
\[ -4I_1 -4(I_1 - I_2) = 0 \] \[ -8I_1 + 4I_2 = 0 \]
π Mesh–2 Equation:
\[ -2I_2 -1I_2 -20 -4(I_2 - I_1) = 0 \] \[ 4I_1-7I_2=20 \]
.\[ I_1 = -2~\text{A}\] \[ I_2 = -4~\text{A}\]
\[ I'''_{2 \Omega} = -4~\text{A} (\rightarrow) \]
\[ I_{2\Omega} = I'_{2\Omega}+I''_{2\Omega}+I'''_{2\Omega} \] \[ I_{2\Omega} =1.6 +1.2+(-4) \] \[ I_{2\Omega} =-1.2 (\rightarrow) \] \[ I_{2\Omega} =1.2 (\leftarrow) \]
Summary
The Superposition Theorem is a powerful tool for analyzing linear electrical circuits containing multiple sources. By breaking a complex problem into simpler parts, it makes circuit analysis systematic and manageable. Regular practice of numerical problems will help students apply this theorem confidently in examinations.
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