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Superposition Theorem: Statement, Steps and Solved Examples
Learn the superposition theorem in simple words: statement, why it works, how to replace voltage and current sources, step-by-step method, limitations, and 5 fully solved numerical problems for BEE and GATE EE.
⚡ In 30 seconds
- In a linear circuit, the response to all sources together = the sum of the responses to each source acting alone.
- To keep one source, kill the others: voltage source → short circuit, current source → open circuit.
- Dependent sources are never killed.
- Use it for current and voltage only, never directly for power.
Statement of Superposition Theorem
Many circuits have more than one source, for example two batteries feeding the same load. The superposition theorem lets us study the effect of each source separately and then simply add the results. Each separate circuit has only one source, so it can often be solved with series–parallel reduction alone.
📖 Statement
In a linear, bilateral network containing two or more independent sources, the current through (or voltage across) any element is the algebraic sum of the currents (or voltages) produced by each independent source acting alone, with all the other independent sources replaced by their internal resistances.
∑ In symbols
\[ I = I' + I'' + I''' + \cdots \]Here \(I'\) is the current due to the first source alone, \(I''\) due to the second source alone, and so on. “Algebraic sum” means we add with signs: a current in the opposite direction is subtracted.
📝 Before you start
You should be comfortable with Ohm’s law, Kirchhoff’s laws (KCL and KVL) and series–parallel combination of resistors. For practice on loop equations, see KVL solved examples.
Why It Works: Linearity
A circuit made of resistors, inductors, capacitors and independent sources is linear. Ohm’s law \(V = IR\) is a straight-line relation: double the voltage and the current doubles. A linear system has two properties:
Homogeneity (scaling)
If an input \(x\) gives output \(y\), then input \(kx\) gives output \(ky\).
Additivity
If \(x_1\) gives \(y_1\) and \(x_2\) gives \(y_2\), then \(x_1 + x_2\) gives \(y_1 + y_2\).
The superposition theorem is just the additivity property applied to circuits. Each source is one input; the branch current is the output. Because the response to each source does not depend on the others, we may find them one by one and add them.
💡 Simple picture
Two people push a box on a smooth floor. The final speed of the box equals the speed that the first push alone would give plus the speed that the second push alone would give. This works because the motion is linear in force. A diode or a saturated iron core is not linear, so this simple adding does not work there.
How to Replace (Kill) Sources
When one source acts alone, every other independent source is “switched off”. Switching off a source means setting its value to zero, and keeping only its internal resistance in the circuit:
| Source | Set to zero means | Replace by |
|---|---|---|
| Ideal voltage source | \(V = 0\) | Short circuit (a plain wire) |
| Ideal current source | \(I = 0\) | Open circuit (remove it) |
| Practical voltage source | \(V = 0\) | Its internal resistance (in series) |
| Practical current source | \(I = 0\) | Its internal resistance (in parallel) |
| Dependent (controlled) source | — | Keep it active; it is not killed |
⚠️ Common mistake
Students often open a voltage source or short a current source. Remember: a zero-volt source has no voltage across it, which is a wire; a zero-ampere source passes no current, which is a break in the circuit.
Steps to Apply the Superposition Theorem
- Mark the branch whose current or voltage you need, with an assumed direction.
- Keep one independent source and replace all other independent sources by their internal resistances (short or open as in the table).
- Solve this one-source circuit for the required current (series–parallel reduction, current divider rule, or KVL/KCL). Note its direction.
- Repeat for every independent source.
- Add all the results algebraically, taking direction into account. A result opposite to your assumed direction is negative.
A circuit with \(n\) independent sources needs \(n\) separate one-source circuits.
Explanation with a Two-Source Circuit
Consider the circuit in Figure 1. Two voltage sources \(V_1\) and \(V_2\) feed a load resistor \(R_L\) through \(R_1\) and \(R_2\). We want the current \(I\) through \(R_L\).
Step 1: V₁ acting alone
Replace \(V_2\) by a short circuit (an ideal voltage source has zero internal resistance). Now \(R_2\) and \(R_L\) are in parallel, and that combination is in series with \(R_1\). Let the load current be \(I'\).
Step 2: V₂ acting alone
Now replace \(V_1\) by a short circuit. \(R_1\) and \(R_L\) are in parallel, in series with \(R_2\). Let the load current be \(I''\).
Step 3: Add the two results
Both \(I'\) and \(I''\) flow downward through \(R_L\) in this circuit, so they add:
∑ Result
\[ I = I' + I'' \]If one source were reversed, its current through \(R_L\) would flow upward and would be subtracted. The individual currents can also be found with KVL (mesh equations) instead of series–parallel reduction; the final answer is the same.
Limitations of Superposition Theorem
- Not for power. Power \(P = I^2R\) depends on the square of current, which is not linear. So \(P \neq I'^2R + I''^2R\). Find the total current first, then the power (see Q5).
- Only for linear circuits. It cannot be used with diodes, transistors in switching mode, or saturated magnetic cores.
- Only independent sources are switched off. Dependent sources stay in every sub-circuit.
- Long for many sources. With three or more sources, nodal or mesh analysis is usually quicker.
🎯 GATE tip
Superposition is very useful when a circuit has sources of different frequencies (for example, a DC source and an AC source together). Each source is analysed separately at its own frequency and the time-domain responses are added. Phasors of different frequencies must never be added directly.
Superposition Theorem Solved Problems
Try each question first, then tap Show solution. Currents marked ↓ flow from node A to node B.
Using the superposition theorem, find the current \(I_3\) through the 2 Ω resistor.
Show solution
Step 1: 20 V source alone (12 V shorted)
The 4 Ω (R₂) and 2 Ω are in parallel, in series with R₁ = 4 Ω:
\[ \begin{aligned} R_2 \parallel R_3 &= \frac{4 \times 2}{4 + 2} = \frac{4}{3}\ \Omega \\ R_{T} &= 4 + \frac{4}{3} = \frac{16}{3}\ \Omega \\ I_{T} &= \frac{20}{16/3} = 3.75\ \text{A} \end{aligned} \]By the current divider rule:
\[ I_3' = 3.75 \times \frac{4}{4 + 2} = 2.5\ \text{A}\ (\downarrow) \]Step 2: 12 V source alone (20 V shorted)
\[ \begin{aligned} R_1 \parallel R_3 &= \frac{4}{3}\ \Omega, \quad R_{T} = \frac{16}{3}\ \Omega \\ I_{T} &= \frac{12}{16/3} = 2.25\ \text{A} \\ I_3'' &= 2.25 \times \frac{4}{4 + 2} = 1.5\ \text{A}\ (\downarrow) \end{aligned} \]Step 3: Add
\[ I_3 = I_3' + I_3'' = 2.5 + 1.5 = 4\ \text{A}\ (\downarrow) \]Check with nodal analysis: \(V_A = \dfrac{20/4 + 12/4}{1/4 + 1/4 + 1/2} = 8\ \text{V}\), so \(I_3 = 8/2 = 4\ \text{A}\). ✅
The circuit of Q1 is changed: the 12 V source is now reversed. Find the current \(I_3\) in the 2 Ω resistor and the current \(I_2\) in R₂ (from A towards the 12 V source).
Show solution
Step 1: 20 V source alone
This is the same as in Q1: \(I_T = 3.75\ \text{A}\) enters node A and divides between R₃ and R₂:
\[ \begin{aligned} I_3' &= 3.75 \times \frac{4}{6} = 2.5\ \text{A}\ (\downarrow) \\ I_2' &= 3.75 \times \frac{2}{6} = 1.25\ \text{A}\ (\text{A} \rightarrow \text{E}_2) \end{aligned} \]Step 2: reversed 12 V source alone
The source current is still 2.25 A, but now it leaves the + terminal at the bottom, goes to node B, rises through R₃ and R₁ to node A, and returns through R₂:
\[ \begin{aligned} I_3'' &= 2.25 \times \frac{4}{6} = 1.5\ \text{A}\ (\uparrow) \\ &= -1.5\ \text{A}\ (\downarrow) \\ I_2'' &= 2.25\ \text{A}\ (\text{A} \rightarrow \text{E}_2) \end{aligned} \]Step 3: Add with signs
\[ \begin{aligned} I_3 &= 2.5 + (-1.5) = 1\ \text{A}\ (\downarrow) \\ I_2 &= 1.25 + 2.25 = 3.5\ \text{A}\ (\text{A} \rightarrow \text{E}_2) \end{aligned} \]KCL check at A: current from R₁ \(= (20 - V_A)/4\) with \(V_A = 2\ \text{V}\) gives 4.5 A, and \(1 + 3.5 = 4.5\ \text{A}\). ✅
Find the current through the 8 Ω resistor using superposition.
Show solution
Step 1: 10 V source alone (3 A source → open circuit)
Step 2: 3 A source alone (10 V source → short circuit)
The 3 A entering node A divides between the 2 Ω path and the 8 Ω resistor:
\[ I'' = 3 \times \frac{2}{2 + 8} = 0.6\ \text{A}\ (\downarrow) \]Step 3: Add
\[ I = I' + I'' = 1 + 0.6 = 1.6\ \text{A}\ (\downarrow) \]Check: nodal equation \(\dfrac{V_A - 10}{2} + \dfrac{V_A}{8} = 3\) gives \(V_A = 12.8\ \text{V}\), and \(12.8/8 = 1.6\ \text{A}\). ✅
Three batteries with series resistors are connected in parallel between nodes A and B. Find the current \(I_3\) through the 1 Ω resistor (taken upward, from B to A).
Show solution
Step 1: 8 V alone (9 V and 4 V shorted)
\[ \begin{aligned} R_T &= 2 + (1 \parallel 2) = 2 + \tfrac{2}{3} = \tfrac{8}{3}\ \Omega \\ I_T &= \frac{8}{8/3} = 3\ \text{A} \\ I_3' &= 3 \times \frac{2}{1 + 2} = 2\ \text{A}\ (\downarrow) \\ &= -2\ \text{A}\ (\uparrow) \end{aligned} \]Step 2: 9 V alone (8 V and 4 V shorted)
\[ \begin{aligned} R_T &= 1 + (2 \parallel 2) = 2\ \Omega \\ I_3'' &= \frac{9}{2} = 4.5\ \text{A}\ (\uparrow) \end{aligned} \]Step 3: 4 V alone (8 V and 9 V shorted)
\[ \begin{aligned} R_T &= 2 + (2 \parallel 1) = \tfrac{8}{3}\ \Omega \\ I_T &= \frac{4}{8/3} = 1.5\ \text{A} \\ I_3''' &= 1.5 \times \frac{2}{1 + 2} = 1\ \text{A}\ (\downarrow) \\ &= -1\ \text{A}\ (\uparrow) \end{aligned} \]Step 4: Add
\[ I_3 = -2 + 4.5 - 1 = 1.5\ \text{A}\ (\uparrow) \]Check with Millman’s theorem: \(V_{AB} = \dfrac{8/2 + 9/1 + 4/2}{1/2 + 1 + 1/2} = 7.5\ \text{V}\), so \(I_3 = (9 - 7.5)/1 = 1.5\ \text{A}\). ✅
In a linear circuit with two sources, source 1 alone sends 1 A and source 2 alone sends 2 A through a 10 Ω resistor, both in the same direction. Find the actual power in the resistor. Is it equal to the sum of the powers due to each source?
Show solution
Adding the separate powers gives
\[ \begin{aligned} P' + P'' &= 1^2 \times 10 + 2^2 \times 10 \\ &= 10 + 40 = 50\ \text{W} \neq 90\ \text{W} \end{aligned} \]The difference \(2I'I''R = 40\ \text{W}\) is the cross term that superposition of powers misses.
Summary
| Point | Remember |
|---|---|
| Applies to | Linear, bilateral circuits with two or more independent sources |
| Finds | Currents and voltages (not power) |
| Voltage source off | Short circuit (or its internal resistance) |
| Current source off | Open circuit (or its internal resistance) |
| Dependent source | Always kept active |
| Final step | Add individual responses with their signs |
Frequently Asked Questions
What is the superposition theorem?
In a linear, bilateral network with two or more independent sources, the current or voltage in any element equals the algebraic sum of the currents or voltages produced by each independent source acting alone, with the other independent sources replaced by their internal resistances.
How are voltage and current sources replaced in superposition?
An ideal voltage source is replaced by a short circuit and an ideal current source by an open circuit. Practical sources are replaced by their internal resistances. Dependent sources are kept active.
Why can’t superposition be used to find power?
Power is proportional to the square of current (P = I²R), which is not linear. The sum of powers from each source misses the cross term 2I′I″R, so find the total current first and then the power.
Is superposition theorem applicable to AC circuits?
Yes, for linear AC circuits using phasors. If the sources have different frequencies, each is analysed separately and the time-domain responses are added.
Are dependent sources switched off in superposition theorem?
No. Only independent sources are switched off. Dependent (controlled) sources remain in the circuit for every sub-circuit.
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