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​ = 12 VV2​ = 9 VIRL​

A linear resistive circuit contains two independent voltage sources, V1 = 12 V and V2 = 9 V. With both sources on, the current through the load resistor is I = 2 mA. With V2 switched off (replaced by a short circuit) and V1 unchanged, the load current is −4 mA. Find the load current I when V1 is raised to 24 V and V2 stays at 9 V.

I = 2 mA (both sources)I = −4 mA (only V1)
I = mA
2.Easy
R1​ = 5 ΩR2​ = 5 Ω+−V1​6 VR3​6 ΩIR3​A+−V2​12 V

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

V1 = 6 VV2 = 12 VR1 = 5 ΩR2 = 5 ΩR3 = 6 Ω
IR3 = A
3.Medium
R1​ = 3 ΩR3​ = 10 ΩIR3​+−V1​15 VR2​6 ΩAR4​5 ΩBIs​?

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

V1 = 15 VR1 = 3 ΩR2 = 6 ΩR3 = 10 ΩR4 = 5 Ω
Is = A
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 = 12 V and Is = 10 mA, I = 4.2 mA; with Vs = 12 V and Is = 6 mA, I = 5.4 mA. Find the load current I when Vs = 5 V and Is = 6 mA.

12 V, 10 mA → 4.2 mA12 V, 6 mA → 5.4 mAVs = 5 V, Is = 6 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 = 30.5 mA. With V1 doubled and the other two unchanged, I = 42.5 mA. With the direction of Is reversed and the other two unchanged, I = 13.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 30.5 mA2·V1: I = 42.5 mA−Is: I = 13.5 mA
I = mA
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 = 14 mA. With V1 doubled and the other two unchanged, I = 9 mA. With the direction of Is reversed and the other two unchanged, I = −2 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 14 mA2·V1: I = 9 mA−Is: I = −2 mA
I = mA

Answer key

  1. 1. I = −2 mA
  2. 2. IR3 = 1.059 A
  3. 3. Is = 2 A
  4. 4. I = 1.2 mA
  5. 5. I = 20.5 mA
  6. 6. I = 3 mA

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