The Superposition Theorem
One source at a time. In any linear circuit with several independent sources, work out what each source does on its own — then simply add the results. Learn the statement, the golden rules for turning sources off, a clear step-by-step method, fully worked examples, and the important power caveat.
Complete Learning Path — Superposition Theorem
From the statement and the linearity idea, to turning sources off, the step-by-step method, worked examples, the power caveat and applications
What is the Superposition Theorem?
The superposition theorem is one of the most useful ideas in circuit analysis. It says that when a linear circuit is driven by more than one independent source, the current or voltage anywhere in that circuit is just the sum of the effects of each source acting on its own.
Instead of solving one complicated circuit with every source active at once, you break the problem into several simpler single-source circuits — each one is easy to solve with basic Ohm’s law and series–parallel rules — and then add up what each source contributed.
Consider a circuit with two batteries V1 and V2 feeding a shared load resistor R3. The current through R3 is caused by both sources together — and that is exactly the kind of problem superposition makes easy.
Independent vs dependent sources
An independent source has a fixed value (a 12 V battery, a 2 A supply). A dependent (controlled) source’s value depends on a voltage or current elsewhere in the circuit. Superposition switches independent sources on and off — dependent ones always stay active.
Statement & the Linearity Principle
Formally, the superposition theorem is stated as:
The whole theorem rests on linearity. A circuit is linear when its response is directly proportional to its source — double the source and you double the response. Mathematically, a linear operation f obeys:
Linear and bilateral only
Superposition works only for circuits built from linear, bilateral elements — resistors, inductors and capacitors. It does not apply to non-linear devices such as diodes and transistors, whose current is not proportional to voltage.
Turning Sources Off — The Two Golden Rules
To find one source’s contribution, you deactivate every other independent source. “Turning off” a source means making its value zero — and that means two different things for the two source types.
Voltage source → short
An ideal voltage source held at 0 V is just a piece of wire. Replace it with a short circuit. (If the source has an internal resistance, that resistance stays in the circuit.)
Current source → open
An ideal current source pushing 0 A carries no current at all — that is an open circuit. Replace it with a gap. (Any parallel internal resistance stays.)
A memory hook
Voltage source off = wire (short). Current source off = cut (open). Zero volts is a wire; zero amps is a cut.
The Step-by-Step Method
Applying superposition is a simple, repeatable recipe. Whatever the circuit, the same five steps get you to the answer.
The procedure in words
- Pick one independent source and leave it active.
- Turn every other independent source off — voltage sources become shorts, current sources become opens (dependent sources stay on).
- Solve the simplified circuit for the branch current or voltage you want. Label it (I′, I″, …).
- Repeat steps 1–3 for every independent source in turn.
- Add all the contributions algebraically, taking care with directions and signs. Currents in the same direction add; opposite directions subtract.
Worked Example 1 — Two Voltage Sources
Find the current through the load R3 in a circuit where V1 = 12 V (through R1 = 6 Ω) and V2 = 6 V (through R2 = 6 Ω) both feed node A, with R3 = 3 Ω from node A to the bottom rail.
Step 1 — V1 acting alone (V2 shorted)
With V2 shorted, R2 sits in parallel with R3: R2 ∥ R3 = (6 × 3)/(6 + 3) = 2 Ω. The voltage at node A is
Step 2 — V2 acting alone (V1 shorted)
By symmetry, R1 ∥ R3 = 2 Ω and
Step 3 — Add the contributions
Check with nodal analysis
Solving directly: (12 − VA)/6 + (6 − VA)/6 = VA/3 gives VA = 4.5 V and IR3 = 4.5/3 = 1.5 A — exactly the superposition result. See nodal analysis for this method.
Worked Example 2 — Voltage Source + Current Source
Superposition really shines when a circuit mixes source types. Find the voltage across the load RL = 5 Ω when a Vs = 10 V source (through R1 = 5 Ω) and a Is = 2 A current source both feed node A.
Vs alone (Is opened)
With the current source open, R1 and RL are in series across Vs. The voltage on RL:
Is alone (Vs shorted)
With Vs shorted, R1 sits in parallel with RL across the current source: R1 ∥ RL = 2.5 Ω.
Superposition with Dependent Sources
A common exam trap: dependent (controlled) sources are never turned off. Only independent sources are deactivated one at a time; every dependent source stays fully active in each sub-circuit.
Because a dependent source’s value tracks a circuit variable (a controlling voltage or current), leaving it active is essential — it re-adjusts itself in each sub-circuit. This keeps the analysis linear while still honouring the control relationship.
Watch out
If you accidentally deactivate a dependent source, your answer will be wrong. When only dependent sources remain (no independent source active), that particular sub-circuit contributes zero on its own — but the dependent source still influences every other sub-circuit.
The Power Caveat — Power Does NOT Superpose
This is the single most-missed point about superposition: you cannot add the powers produced by individual sources. Power is not a linear quantity.
Power depends on the square of current (or voltage): P = I²R. Squaring is non-linear. If the total current is I = I′ + I″, then
With the numbers from Example 1 (I′ = 1 A, I″ = 0.5 A, R3 = 3 Ω): adding powers gives 3 + 0.75 = 3.75 W, but the true power is (1.5)²×3 = 6.75 W. Always superpose currents or voltages first, then compute power.
Advantages & Limitations
Limitations
- Works only for linear, bilateral circuits.
- Cannot be used for power directly (square-law).
- Gets tedious with many sources — one sub-circuit each.
- Dependent sources cannot be deactivated.
Applications
Superposition is not just an exam exercise — it is a working tool for analysing and designing real linear systems.
Amplifier biasing
Separate the DC bias from the AC signal, source by source.
Signal analysis
Add the responses to many input signals in a linear system.
Multi-source networks
Solve circuits with several batteries or supplies cleanly.
Filter & AC design
Handle each frequency or source independently, then combine.
Other theorems
The reasoning behind Thevenin’s and Norton’s theorems.
Fault & sensitivity
See how a single source or change affects the whole network.
Key Terms — Glossary
| Term | Meaning |
|---|---|
| Superposition theorem | Total response = sum of responses from each independent source acting alone. |
| Linear circuit | A circuit whose response is directly proportional to its source (resistors, L, C). |
| Independent source | A source with a fixed value that does not depend on the circuit. |
| Dependent source | A controlled source whose value depends on a circuit voltage or current; never turned off. |
| Deactivate / turn off | Set a source to zero: voltage source → short circuit, current source → open circuit. |
| Short circuit | A zero-resistance wire (0 V); what an ideal voltage source becomes when off. |
| Open circuit | A break with no current (0 A); what an ideal current source becomes when off. |
| Contribution | The current or voltage produced in a branch by one source alone (I′, I″…). |
| Algebraic sum | Adding contributions with correct signs/directions. |
| Cross term | The 2·I′·I″·R part of power lost if powers are added — why power is not superposable. |
Frequently Asked Questions
Quick, expert answers to the questions people ask most about the superposition theorem.
What is the superposition theorem in simple words?
In a linear circuit with several independent sources, you find the effect of each source on its own (with the other sources turned off) and then add all those effects together. The sum gives the true current or voltage when every source is acting at once.
How do you turn off a source in the superposition theorem?
An ideal independent voltage source is turned off by replacing it with a short circuit (a plain wire, 0 V). An ideal independent current source is turned off by replacing it with an open circuit (a gap, 0 A). Any internal source resistance is left in place.
Why does the superposition theorem only work for linear circuits?
Superposition depends on linearity, where the response is proportional to the source so that f(a + b) = f(a) + f(b). Non-linear elements such as diodes and transistors do not obey this rule, so their responses cannot simply be added.
Can the superposition theorem be used to find power?
No. Power depends on the square of current or voltage, which is non-linear, so the powers from individual sources do not add up. First superpose the currents or voltages, and only then compute the power from the total value.
Are dependent sources turned off in superposition?
No. Only independent sources are deactivated one at a time. Dependent (controlled) sources depend on a circuit variable, so they must stay active in every sub-circuit while you switch the independent sources on and off.
What are the steps of the superposition theorem?
Keep one independent source active and turn the rest off (voltage sources to shorts, current sources to opens); solve the simplified circuit for the wanted current or voltage; repeat for every independent source; then add all the individual contributions algebraically, respecting their directions or signs.
What are the limitations of the superposition theorem?
It applies only to linear, bilateral circuits, it cannot be used directly for power, and it becomes tedious when a circuit has many sources because each source needs its own sub-circuit. It also cannot deactivate dependent sources.
Where is the superposition theorem used?
It is used to analyse linear circuits driven by several sources, to separate DC and AC contributions in amplifier biasing, in signal and filter analysis, and as the reasoning behind other network theorems such as Thevenin’s and Norton’s.
Conclusion & Key Takeaways
The superposition theorem turns a tangled multi-source circuit into a set of easy single-source problems — solve each, then add. It is one of the most powerful shortcuts in linear circuit analysis.
One at a time
Analyse each source alone.
V → short
Turn off a voltage source with a wire.
I → open
Turn off a current source with a gap.
Add up
Sum the contributions algebraically.
Linear only
No diodes, no transistors.
Not for power
Superpose V or I first, then find P.