Superposition worksheet with answers

Free printable Superposition worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
linear circuitV1​ = 10 VV2​ = 15 VIRL​

A linear resistive circuit contains two independent voltage sources, V1 = 10 V and V2 = 15 V. With both sources on, the current through the load resistor is I = 23 mA. With V2 switched off (replaced by a short circuit) and V1 unchanged, the load current is 15.5 mA. Find the load current I when V1 is switched off and only V2 = 15 V is active.

I = 23 mA (both sources)I = 15.5 mA (only V1)
I = mA
2.Easy
R1​ = 2 ΩR2​ = 8 ΩIR2​+−V1​20 VR3​4 ΩA+−V2​10 V

Use superposition to find the current IR2 through R2 in the direction of the arrow.

V1 = 20 VV2 = 10 VR1 = 2 ΩR2 = 8 ΩR3 = 4 Ω
IR2 = A
3.Medium
linear circuitVs​Is​IRL​

A linear resistive circuit contains one voltage source Vs and one current source Is. The current I through a load resistor is measured twice: with Vs = 12 V and Is = 6 mA, I = 3 mA; with Vs = 20 V and Is = 2 mA, I = 9 mA. Find the load current I when Vs = 4 V and Is = 10 mA.

12 V, 6 mA → 3 mA20 V, 2 mA → 9 mAVs = 4 V, Is = 10 mA
I = mA
4.Medium
R1​ = 10 ΩR3​ = 4 ΩIR3​+−V1​12 VR2​6 ΩAR4​10 ΩBIs​?

Find the value of the current source Is (direction as shown) that makes the current IR3 through R3 equal to zero.

V1 = 12 VR1 = 10 ΩR2 = 6 ΩR3 = 4 ΩR4 = 10 Ω
Is = A
5.Hard
R1​ = 10 ΩIR1​R2​ = 25 ΩR4​ = 12 Ω+−V1​10 VR3​12 ΩAIs​0.5 AB−+V2​15 V

Use superposition to find the power PR1 absorbed by R1.

V1 = 10 VIs = 0.5 AV2 = 15 VR1 = 10 ΩR2 = 25 ΩR3 = 12 ΩR4 = 12 Ω
PR1 = W
6.Hard
R1​ = 15 ΩR2​ = 12 ΩR4​ = 2 Ω+−V1​6 VR3​5 ΩIR3​AIs​4 AB+−V2​?

Find the value of the voltage source V2 (polarity as shown) that makes the current IR3 through R3 equal to zero.

V1 = 6 VIs = 4 AR1 = 15 ΩR2 = 12 ΩR3 = 5 ΩR4 = 2 Ω
V2 = V

Answer key

  1. 1. I = 7.5 mA
  2. 2. IR2 = 0.3571 A
  3. 3. I = −3 mA
  4. 4. Is = 0.45 A
  5. 5. PR1 = 4.099 W
  6. 6. V2 = 2.4 V

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