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​ = 15 VIRL​

A linear resistive circuit contains two independent voltage sources, V1 = 12 V and V2 = 15 V. With both sources on, the current through the load resistor is I = 6 mA. With V2 switched off (replaced by a short circuit) and V1 unchanged, the load current is 18 mA. Find the load current I when V1 is switched off and only V2 = 15 V is active.

I = 6 mA (both sources)I = 18 mA (only V1)
I = mA
2.Easy
R1​ = 12 ΩR2​ = 10 Ω+−V1​12 VR3​12 ΩIR3​A+−V2​20 V

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

V1 = 12 VV2 = 20 VR1 = 12 ΩR2 = 10 ΩR3 = 12 Ω
IR3 = A
3.Medium
R1​ = 2 ΩR3​ = 12 Ω+−V1​24 VR2​3 Ω+−VR2​AR4​15 ΩBIs​3 A

Use superposition to find the voltage VR2 across R2 with the polarity shown.

V1 = 24 VIs = 3 AR1 = 2 ΩR2 = 3 ΩR3 = 12 ΩR4 = 15 Ω
VR2 = V
4.Medium
R1​ = 20 ΩR3​ = 15 Ω+−V1​24 VR2​3 ΩAR4​20 ΩIR4​BIs​3 A

Use superposition to find the power PR4 absorbed by R4.

V1 = 24 VIs = 3 AR1 = 20 ΩR2 = 3 ΩR3 = 15 ΩR4 = 20 Ω
PR4 = W
5.Hard
R1​ = 20 ΩR2​ = 25 ΩR4​ = 30 ΩIR4​+−V1​24 VR3​25 ΩAIs​3 AB+−V2​36 V

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

V1 = 24 VIs = 3 AV2 = 36 VR1 = 20 ΩR2 = 25 ΩR3 = 25 ΩR4 = 30 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 = 9.5 mA. With V1 doubled and the other two unchanged, I = 16.5 mA. With the direction of Is reversed and the other two unchanged, I = 2.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 9.5 mA2·V1: I = 16.5 mA−Is: I = 2.5 mA
I = mA

Answer key

  1. 1. I = −12 mA
  2. 2. IR3 = 0.9375 A
  3. 3. VR2 = 15.7 V
  4. 4. PR4 = 34.92 W
  5. 5. IR4 = −1.296 A
  6. 6. I = 10.5 mA

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