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​ = 8 ΩR2​ = 2 Ω+−V1​20 VR3​3 ΩIR3​A+−V2​10 V

Use superposition to find the contribution IR3′ of V1 alone to the current IR3 through R3 in the direction of the arrow.

V1 = 20 VV2 = 10 VR1 = 8 ΩR2 = 2 ΩR3 = 3 Ω
IR3′ = A
2.Easy
R1​ = 4 ΩR2​ = 12 Ω+−V1​15 VR3​6 ΩIR3​A+−V2​18 V

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

V1 = 15 VV2 = 18 VR1 = 4 ΩR2 = 12 ΩR3 = 6 Ω
IR3 = A
3.Medium
R1​ = 6 ΩR3​ = 20 Ω+−V1​10 VR2​20 ΩAR4​3 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 = 20 ΩR3 = 20 ΩR4 = 3 Ω
VR4 = V
4.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 = 4 V and Is = 2 mA, I = 1.6 mA; with Vs = 10 V and Is = 2 mA, I = 3.1 mA. Find the load current I when Vs = 12 V and Is = 4 mA.

4 V, 2 mA → 1.6 mA10 V, 2 mA → 3.1 mAVs = 12 V, Is = 4 mA
I = mA
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 = 18.5 mA. With V1 doubled and the other two unchanged, I = 30.5 mA. With the direction of Is reversed and the other two unchanged, I = 15.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 18.5 mA2·V1: I = 30.5 mA−Is: I = 15.5 mA
I = mA
6.Hard
R1​ = 8 ΩR2​ = 2 ΩR4​ = 2 Ω+−V1​15 VR3​15 ΩIR3​AIs​6 AB+−V2​?

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

V1 = 15 VIs = 6 AR1 = 8 ΩR2 = 2 ΩR3 = 15 ΩR4 = 2 Ω
V2 = V

Answer key

  1. 1. IR3′ = 0.8696 A
  2. 2. IR3 = 1.75 A
  3. 3. VR4 = 11.53 V
  4. 4. I = 4.2 mA
  5. 5. I = 13.5 mA
  6. 6. V2 = 4.5 V

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