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​ = 15 VV2​ = 24 VIRL​

A linear resistive circuit contains two independent voltage sources, V1 = 15 V and V2 = 24 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 5 mA. Find the load current I when V1 is switched off and only V2 = 24 V is active.

I = −2 mA (both sources)I = 5 mA (only V1)
I = mA
2.Easy
R1​ = 5 ΩIR1​R2​ = 2 Ω+−V1​20 VR3​10 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 = 20 VV2 = 8 VR1 = 5 ΩR2 = 2 ΩR3 = 10 Ω
IR1″ = A
3.Medium
R1​ = 3 ΩR3​ = 2 ΩIR3​+−V1​30 VR2​6 ΩAR4​4 ΩBIs​?

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

V1 = 30 VR1 = 3 ΩR2 = 6 ΩR3 = 2 ΩR4 = 4 Ω
Is = A
4.Medium
R1​ = 3 ΩR3​ = 15 ΩIR3​+−V1​18 VR2​10 ΩAR4​15 ΩBIs​1.5 A

Use superposition to find the power PR3 absorbed by R3.

V1 = 18 VIs = 1.5 AR1 = 3 ΩR2 = 10 ΩR3 = 15 ΩR4 = 15 Ω
PR3 = W
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 = 10 mA. With V1 doubled and the other two unchanged, I = 10.5 mA. With the direction of Is reversed and the other two unchanged, I = 9 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 10 mA2·V1: I = 10.5 mA−Is: I = 9 mA
I = mA
6.Hard
R1​ = 12 ΩR2​ = 10 ΩR4​ = 15 Ω+−V1​9 VR3​4 Ω+−VR3​AIs​2 AB+−V2​6 V

Use superposition to find the voltage VR3 across R3 with the polarity shown.

V1 = 9 VIs = 2 AV2 = 6 VR1 = 12 ΩR2 = 10 ΩR3 = 4 ΩR4 = 15 Ω
VR3 = V

Answer key

  1. 1. I = −7 mA
  2. 2. IR1″ = −1 A
  3. 3. Is = 5 A
  4. 4. PR3 = 1.076 W
  5. 5. I = 1 mA
  6. 6. VR3 = 5.866 V

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