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
R1​ = 12 ΩR2​ = 12 Ω+−V1​6 VR3​5 ΩIR3​A+−V2​6 V

Use superposition to find the current IR3 through R3 in the direction of the arrow.

V1 = 6 VV2 = 6 VR1 = 12 ΩR2 = 12 ΩR3 = 5 Ω
IR3 = A
2.Easy
R1​ = 8 ΩR2​ = 2 ΩIR2​+−V1​6 VR3​4 ΩA+−V2​15 V

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

V1 = 6 VV2 = 15 VR1 = 8 ΩR2 = 2 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 = 5 V and Is = 6 mA, I = 2.5 mA; with Vs = 12 V and Is = 6 mA, I = 3.9 mA. Find the load current I when Vs = 10 V and Is = 1 mA.

5 V, 6 mA → 2.5 mA12 V, 6 mA → 3.9 mAVs = 10 V, Is = 1 mA
I = mA
4.Medium
R1​ = 6 ΩR3​ = 4 Ω+−V1​10 VR2​4 ΩAR4​6 Ω+−VR4​BIs​4 A

Use superposition to find the voltage VR4 across R4 with the polarity shown.

V1 = 10 VIs = 4 AR1 = 6 ΩR2 = 4 ΩR3 = 4 ΩR4 = 6 Ω
VR4 = V
5.Hard
linear circuitV1​V2​Is​IRL​

A linear resistive circuit contains three independent sources: V1, V2 and Is. The current I through a load is measured three times. With all sources at their normal values, I = 12.5 mA. With V1 doubled and the other two unchanged, I = 21.5 mA. With the direction of Is reversed and the other two unchanged, I = −0.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 12.5 mA2·V1: I = 21.5 mA−Is: I = −0.5 mA
I = mA
6.Hard
R1​ = 6 ΩR2​ = 12 Ω+−VR2​R4​ = 30 Ω+−V1​36 VR3​25 ΩAIs​5 AB−+V2​12 V

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

V1 = 36 VIs = 5 AV2 = 12 VR1 = 6 ΩR2 = 12 ΩR3 = 25 ΩR4 = 30 Ω
VR2 = V

Answer key

  1. 1. IR3 = 0.5455 A
  2. 2. IR2 = −2.786 A
  3. 3. I = 2.25 mA
  4. 4. VR4 = 14.32 V
  5. 5. I = 15.5 mA
  6. 6. VR2 = −27.92 V

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