Superposition: practice problems with step-by-step solutions

Free Superposition practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: 2 voltage sources, voltage + current source, 3 sources.

Free practice problemEasy
R1​ = 8 ΩIR1​R2​ = 2 Ω+−V1​15 VR3​12 ΩA+−V2​8 V

Use superposition to find the current IR1 through R1 in the direction of the arrow.

V1 = 15 VV2 = 8 VR1 = 8 ΩR2 = 2 ΩR3 = 12 Ω
A
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What you practise

  1. Easy2 voltage sources
  2. Mediumvoltage + current source
  3. Hard3 sources

How to solve Superposition problems

  1. Count the independent sources. You will solve one simpler circuit per source.
  2. Keep one source and switch off the others: a voltage source becomes a short circuit (a wire), a current source becomes an open circuit (a gap).
  3. Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
  4. Repeat for every source, then add the contributions with their signs.

Common mistakes

  • Removing a voltage source instead of replacing it with a wire.
  • Adding contributions without checking their directions.
  • Using superposition for power: power is not linear, so find the total current first and then the power.

Key formulas

Superposition
x = x1 + x2 + … (one source at a time)
Switched-off voltage source
short circuit (0 V)
Switched-off current source
open circuit (0 A)
Voltage divider
V2 = V·R2/(R1 + R2)
Current divider
I1 = I·R2/(R1 + R2)

Learn Superposition from scratch

Before this: Series & parallel resistors

1What superposition is

Many circuits have more than one source: two batteries, or a battery and a current source.

Solving them all at once can be hard. Superposition lets you split the problem into simpler pieces.

The idea: each source pushes its own current through the circuit. The real current in a resistor is the sum of what each source causes on its own.

Think of two taps filling the same bath. The water level rises by what the first tap adds plus what the second tap adds.

Self-check What does superposition let you do?
  1. Ignore every source except the largest one
  2. Multiply the effects of the sources
  3. Replace all resistors by one resistor
  4. Solve the circuit once per source and add the results

Answer: Solve the circuit once per source and add the results. You look at one source at a time, find what it causes, and add all the contributions at the end.

2Switching a source off

To look at one source on its own, you switch the other sources off. Switching off means setting the source to zero.

  • A voltage source set to 0 V becomes a plain wire (a short circuit).
  • A current source set to 0 A becomes a gap (an open circuit).
+−0 V0 A

Only the sources are switched off. Every resistor stays exactly where it is.

This works for circuits made of resistors and independent sources. Such circuits are called linear: doubling a source doubles everything it causes.

Self-check How do you switch off a current source?
  1. Remove the resistor next to it
  2. Replace it with a 1 Ω resistor
  3. Replace it with a short circuit (a wire)
  4. Replace it with an open circuit (a gap)

Answer: Replace it with an open circuit (a gap). A switched-off current source carries 0 A. A gap is the only thing that lets no current through, whatever the voltage.

3The tools you need

Each simpler circuit has only one source, so you can solve it with series and parallel resistors.

Rseries = R1 + R2
Rparallel = R1·R2 / (R1 + R2)

The voltage divider is the tool you use most. Two resistors in series share the source voltage in proportion to their size:

V2 = V · R2 / (R1 + R2)

Then Ohm’s law, I = V/R, gives the current in any resistor once you know its voltage.

Self-check A 12 V source feeds 6 Ω and 2 Ω in series. What is the voltage across the 2 Ω resistor?
  1. 3 V
  2. 4 V
  3. 6 V
  4. 9 V

Answer: 3 V. V = 12 · 2/(6 + 2) = 3 V. The smaller resistor gets the smaller share of the voltage.

4The method step by step

  1. Count the independent sources. You will solve one simpler circuit for each of them.
  2. Keep one source and switch off all the others: voltage sources become wires, current sources become gaps.
  3. Solve the simpler circuit for the quantity you need. Keep the same direction (or the same + and −) as in the question.
  4. Repeat for every source.
  5. Add all the contributions with their signs. A contribution in the opposite direction counts as negative.

Draw each simpler circuit before you calculate. Most mistakes come from looking at the original drawing while solving a different circuit.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Find the current I3 through R3 (downwards) in this circuit.

1

Look at the circuit

R1​ = 6 ΩR2​ = 6 ΩR3​3 ΩI3​A+−V1​12 V+−V2​6 V

There are two sources, V1 = 12 V and V2 = 6 V. So we solve two simpler circuits.

2

Only V₁ on

R1​ = 6 ΩR2​ = 6 ΩR3​3 ΩI3​A+−V1​12 V0 V

Replace V2 with a wire. Now R2 runs from node A straight to the bottom wire, so R2 and R3 are in parallel.

R2 ∥ R3 = 6·3 / (6 + 3) = 2 Ω
3

Voltage divider for V₁

V1 sees R1 = 6 Ω in series with the 2 Ω parallel pair. The voltage at A is the share of the 2 Ω part:

VA′ = 12 · 2 / (6 + 2) = 3 V
I3′ = 3 V / 3 Ω = 1 A
4

Only V₂ on

R1​ = 6 ΩR2​ = 6 ΩR3​3 ΩI3​A0 V+−V2​6 V

Now replace V1 with a wire. R1 and R3 are in parallel: 6·3/(6 + 3) = 2 Ω.

VA″ = 6 · 2 / (6 + 2) = 1.5 V
I3″ = 1.5 V / 3 Ω = 0.5 A
5

Add the contributions

Both contributions flow downwards through R3, so both are positive.

I3 = I3′ + I3″ = 1 + 0.5 = 1.5 A
6

Check the answer

With both sources on, VA = 3 + 1.5 = 4.5 V. The currents into A from both sides are (12 − 4.5)/6 = 1.25 A and (6 − 4.5)/6 = 0.25 A.

Together that is 1.5 A, exactly the current that leaves through R3. The answer is right.

Answer

I3 = 1.5 A, flowing downwards.

6Common mistakes

  • Removing a voltage source and leaving a gap. A switched-off voltage source is a wire.
  • Adding contributions without checking their directions. Draw an arrow for every contribution.
  • Calculating power with superposition. Power is not linear: find the total current first, then P = I²·R.
  • Forgetting a source. With three sources you need three simpler circuits.
Self-check In the example, I₃′ = 1 A and I₃″ = 0.5 A in a 3 Ω resistor. What power does R₃ take?
  1. 2.25 W
  2. 3.75 W
  3. 4.5 W
  4. 6.75 W

Answer: 6.75 W. First add the currents: 1.5 A. Then P = 1.5² · 3 = 6.75 W. Adding the two separate powers (3 + 0.75 = 3.75 W) is wrong because power is not linear.

7How exam questions look

  • Easy: two voltage sources and three resistors. Each simpler circuit is a voltage divider.
  • Medium: a voltage source and a current source. Remember that the switched-off current source becomes a gap.
  • Hard: three sources, and the target can be a node voltage or a branch current.

Some questions only give measurements. For example: with both sources on the current is 6.5 mA, and with V2 off it is 4.5 mA. Then V2 alone causes 2 mA.

Because the circuit is linear, doubling V1 doubles its contribution. Use this to predict the current after a source changes.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
R1​ = 10 ΩR2​ = 12 Ω+−V1​18 VR3​2 Ω+−VR3​A+−V2​18 V

Use superposition to find the voltage VR3 across R3 with the polarity shown.

V1 = 18 VV2 = 18 VR1 = 10 ΩR2 = 12 ΩR3 = 2 Ω

Find VR3 (V)

Show hints
  1. This circuit has 2 independent sources, so solve 2 simpler circuits, one per source. Switch the others off: a voltage source becomes a short, a current source becomes an open circuit.
  2. With only V₁ active, R₂ and R₃ end up in parallel. Use a voltage divider to get the voltage at node A.
  3. With only V1 active, VR3′ = 2.634 V. Find the other contribution the same way and add them.
Show the answer and the step-by-step solution

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 10 ΩR2​ = 12 Ω+−V1​18 VR3​2 ΩASHORT

Case 1 · V1 acting alone

V₂ becomes a short, so R₂ now runs from A to ground, in parallel with R₃.

R2 ∥ R3 = 12·2/(12+2) = 1.714 Ω
VA = V1·(R2∥R3)/(R1+R2∥R3) = 18·1.714/(10+1.714) = 2.634 V
VR3′ = VA − 0 = 2.634 − 0 = 2.634 V
R1​ = 10 ΩR2​ = 12 ΩSHORTR3​2 ΩA+−V2​18 V

Case 2 · V2 acting alone

V₁ becomes a short, so R₁ now runs from A to ground, in parallel with R₃.

R1 ∥ R3 = 10·2/(10+2) = 1.667 Ω
VA = V2·(R1∥R3)/(R2+R1∥R3) = 18·1.667/(12+1.667) = 2.195 V
VR3″ = VA − 0 = 2.195 − 0 = 2.195 V

Add the contributions

VR3 = VR3′ + VR3″
VR3 = 2.634 + 2.195 = 4.829 V

A negative result means the real polarity is opposite to the one marked.

Example 2Medium
R1​ = 15 ΩR3​ = 8 Ω+−V1​6 VR2​6 ΩAR4​12 ΩIR4​BIs​2 A

Use superposition to find the power PR4 absorbed by R4.

V1 = 6 VIs = 2 AR1 = 15 ΩR2 = 6 ΩR3 = 8 ΩR4 = 12 Ω

Find PR4 (W)

Show hints
  1. This circuit has 2 independent sources, so solve 2 simpler circuits, one per source. Switch the others off: a voltage source becomes a short, a current source becomes an open circuit.
  2. Find the current IR4 through R4 by superposition, then square it: P = I²R.
  3. With only V1 active, IR4′ = 0.07059 A.
Show the answer and the step-by-step solution

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 15 ΩR3​ = 8 Ω+−V1​6 VR2​6 ΩAR4​12 ΩBOPEN

Case 1 · V1 acting alone

Iₛ becomes an open circuit, so R₃ and R₄ form one series branch from A to ground, in parallel with R₂.

R3+R4 = 8+12 = 20 Ω
R2 ∥ (R3+R4) = 6·20/(6+20) = 4.615 Ω
VA = 6·4.615/(15+4.615) = 1.412 V
VB = VA·R4/(R3+R4) = 1.412·12/20 = 0.8471 V
IR4′ = (VB − 0)/R4 = (0.8471 − 0)/12 = 0.07059 A
R1​ = 15 ΩR3​ = 8 ΩSHORTR2​6 ΩAR4​12 ΩBIs​2 A

Case 2 · Is acting alone

V₁ becomes a short, so R₁ is in parallel with R₂. Iₛ splits at B between R₄ and the path through R₃.

R1 ∥ R2 = 15·6/(15+6) = 4.286 Ω
Req = R4 ∥ (R3+R1∥R2) = 12 ∥ 12.29 = 6.071 Ω
VB = Is·Req = 2·6.071 = 12.14 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 12.14·4.286/12.29 = 4.235 V
IR4″ = (VB − 0)/R4 = (12.14 − 0)/12 = 1.012 A

Add the current contributions

IR4 = IR4′ + IR4″ = 0.07059 + 1.012 = 1.082 A

Power in the resistor

Power depends on the square of the current, so it does not obey superposition: add the currents first, then square.

PR4 = IR4²·R4 = 1.082²·12 = 14.06 W

Answer

PR4 = 14.06 W

Adding the powers of the separate cases would give 12.34 W, which is wrong.

Example 3Hard
R1​ = 10 ΩR2​ = 4 Ω+−VR2​R4​ = 25 Ω+−V1​12 VR3​8 ΩAIs​1 AB+−V2​15 V

Use superposition to find the voltage VR2 across R2 with the polarity shown.

V1 = 12 VIs = 1 AV2 = 15 VR1 = 10 ΩR2 = 4 ΩR3 = 8 ΩR4 = 25 Ω

Find VR2 (V)

Show hints
  1. This circuit has 3 independent sources, so solve 3 simpler circuits, one per source. Switch the others off: a voltage source becomes a short, a current source becomes an open circuit.
  2. Start with V₁ alone: short V₂, open Iₛ. Seen from A, R₃ is in parallel with R₂+R₄.
  3. With only V1 active, VR2′ = 0.6379 V. Find the other contributions the same way and add them.
Show the answer and the step-by-step solution

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 10 ΩR2​ = 4 ΩR4​ = 25 Ω+−V1​12 VR3​8 ΩAOPENBSHORT

Case 1 · V1 acting alone

V₂ becomes a short and Iₛ an open circuit. Seen from A, R₃ is in parallel with the series pair R₂+R₄.

R3 ∥ (R2+R4) = 8 ∥ 29 = 6.27 Ω
VA = 12·6.27/(10+6.27) = 4.625 V
VB = VA·R4/(R2+R4) = 4.625·25/29 = 3.987 V
VR2′ = VA − VB = 4.625 − 3.987 = 0.6379 V
R1​ = 10 ΩR2​ = 4 ΩR4​ = 25 ΩSHORTR3​8 ΩAIs​1 ABSHORT

Case 2 · Is acting alone

Both voltage sources become shorts. Iₛ splits at B between R₄ (now to ground) and R₂ in series with R₁ ∥ R₃.

R1 ∥ R3 = 4.444 Ω, so the left path is R2+4.444 = 8.444 Ω
Req = R4 ∥ 8.444 = 6.312 Ω
VB = Is·Req = 1·6.312 = 6.312 V
VA = VB·4.444/8.444 = 3.322 V
VR2″ = VA − VB = 3.322 − 6.312 = −2.99 V
R1​ = 10 ΩR2​ = 4 ΩR4​ = 25 ΩSHORTR3​8 ΩAOPENB+−V2​15 V

Case 3 · V2 acting alone

V₁ becomes a short and Iₛ an open circuit. R₁ ∥ R₃ hangs from A, in series with R₂ back to B.

R1 ∥ R3 = 10·8/(10+8) = 4.444 Ω
VB = V2·(R2+R1∥R3)/(R4+R2+R1∥R3) = 15·8.444/33.44 = 3.787 V
VA = VB·(R1∥R3)/(R2+R1∥R3) = 3.787·4.444/8.444 = 1.993 V
VR2‴ = VA − VB = 1.993 − 3.787 = −1.794 V

Add the contributions

VR2 = VR2′ + VR2″ + VR2‴
VR2 = 0.6379 + (−2.99) + (−1.794) = −4.146 V

A negative result means the real polarity is opposite to the one marked.

Questions and answers

What is Superposition?

Many circuits have more than one source: two batteries, or a battery and a current source.

Solving them all at once can be hard. Superposition lets you split the problem into simpler pieces.

How do I solve Superposition problems step by step?
  1. Count the independent sources. You will solve one simpler circuit per source.
  2. Keep one source and switch off the others: a voltage source becomes a short circuit (a wire), a current source becomes an open circuit (a gap).
  3. Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
  4. Repeat for every source, then add the contributions with their signs.
What are common mistakes in Superposition?
  • Removing a voltage source instead of replacing it with a wire.
  • Adding contributions without checking their directions.
  • Using superposition for power: power is not linear, so find the total current first and then the power.
What do the Superposition exercises cover?
  • Easy: 2 voltage sources
  • Medium: voltage + current source
  • Hard: 3 sources

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Series & parallel resistors

Courses with this topic

Circuit Analysis 1FE Electrical and Computer exam practice

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Printable worksheet

Superposition worksheet with answers
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