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​ = 3 ΩIR1​R2​ = 2 Ω+−V1​12 VR3​5 ΩA+−V2​8 V

Use superposition to find the contribution IR1″ of V2 alone to the current IR1 through R1 in the direction of the arrow.

V1 = 12 VV2 = 8 VR1 = 3 ΩR2 = 2 ΩR3 = 5 Ω
IR1″ = A
2.Easy
linear circuitV1​ = 20 VV2​ = 24 VIRL​

A linear resistive circuit contains two independent voltage sources, V1 = 20 V and V2 = 24 V. With both sources on, the current through the load resistor is I = 22 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 raised to 40 V and V2 stays at 24 V.

I = 22 mA (both sources)I = 15.5 mA (only V1)
I = mA
3.Medium
R1​ = 8 ΩR3​ = 2 ΩIR3​+−V1​12 VR2​2 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 = 8 ΩR2 = 2 ΩR3 = 2 ΩR4 = 10 Ω
Is = A
4.Medium
R1​ = 10 ΩR3​ = 2 ΩIR3​+−V1​15 VR2​20 ΩAR4​6 ΩBIs​2 A

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

V1 = 15 VIs = 2 AR1 = 10 ΩR2 = 20 ΩR3 = 2 ΩR4 = 6 Ω
IR3 = A
5.Hard
R1​ = 8 ΩR2​ = 8 ΩR4​ = 3 ΩIR4​+−V1​10 VR3​12 ΩAIs​3 AB−+V2​9 V

Use superposition to find the current IR4 through R4 in the direction of the arrow.

V1 = 10 VIs = 3 AV2 = 9 VR1 = 8 ΩR2 = 8 ΩR3 = 12 ΩR4 = 3 Ω
IR4 = A
6.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 = 28.5 mA. With V1 doubled and the other two unchanged, I = 40.5 mA. With the direction of Is reversed and the other two unchanged, I = 9.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 28.5 mA2·V1: I = 40.5 mA−Is: I = 9.5 mA
I = mA

Answer key

  1. 1. IR1″ = −1.29 A
  2. 2. I = 37.5 mA
  3. 3. Is = 0.24 A
  4. 4. IR3 = −0.1364 A
  5. 5. IR4 = −1.481 A
  6. 6. I = 21.5 mA

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