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​ = 4 ΩR2​ = 3 Ω+−V1​8 VR3​6 ΩIR3​A+−V2​6 V

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

V1 = 8 VV2 = 6 VR1 = 4 ΩR2 = 3 ΩR3 = 6 Ω
IR3′ = A
2.Easy
R1​ = 6 ΩIR1​+−V1​8 VR2​12 ΩAIs​2 A

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

V1 = 8 VIs = 2 AR1 = 6 ΩR2 = 12 Ω
IR1 = A
3.Medium
R1​ = 15 ΩR3​ = 2 ΩIR3​+−V1​12 VR2​12 ΩAR4​5 ΩBIs​4 A

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

V1 = 12 VIs = 4 AR1 = 15 ΩR2 = 12 ΩR3 = 2 ΩR4 = 5 Ω
IR3 = A
4.Medium
R1​ = 10 ΩR3​ = 8 Ω+−V1​15 VR2​2 ΩIR2​AR4​2 ΩBIs​3 A

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

V1 = 15 VIs = 3 AR1 = 10 ΩR2 = 2 ΩR3 = 8 ΩR4 = 2 Ω
IR2 = A
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 = 8 mA. With V1 doubled and the other two unchanged, I = 15 mA. With the direction of Is reversed and the other two unchanged, I = −12 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 8 mA2·V1: I = 15 mA−Is: I = −12 mA
I = mA
6.Hard
R1​ = 10 ΩR2​ = 3 ΩR4​ = 15 Ω+−V1​15 VR3​4 ΩIR3​AIs​2 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 = 2 AR1 = 10 ΩR2 = 3 ΩR3 = 4 ΩR4 = 15 Ω
V2 = V

Answer key

  1. 1. IR3′ = 0.4444 A
  2. 2. IR1 = −0.8889 A
  3. 3. IR3 = 1.854 A
  4. 4. IR2 = 0.6429 A
  5. 5. I = 17 mA
  6. 6. V2 = 3 V

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