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​ = 4 ΩIR2​+−V1​18 VR3​6 ΩA+−V2​8 V

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

V1 = 18 VV2 = 8 VR1 = 4 ΩR2 = 4 ΩR3 = 6 Ω
IR2 = A
2.Easy
linear circuitV1​ = 9 VV2​ = 10 VIRL​

A linear resistive circuit contains two independent voltage sources, V1 = 9 V and V2 = 10 V. With both sources on, the current through the load resistor is I = −4.5 mA. With V2 switched off (replaced by a short circuit) and V1 unchanged, the load current is −3 mA. Find the load current I when V1 is raised to 18 V and V2 stays at 10 V.

I = −4.5 mA (both sources)I = −3 mA (only V1)
I = mA
3.Medium
R1​ = 8 ΩR3​ = 5 Ω+−V1​30 VR2​12 Ω+−VR2​AR4​5 ΩBIs​1.5 A

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

V1 = 30 VIs = 1.5 AR1 = 8 ΩR2 = 12 ΩR3 = 5 ΩR4 = 5 Ω
VR2 = V
4.Medium
R1​ = 5 ΩR3​ = 3 ΩIR3​+−V1​9 VR2​15 ΩAR4​4 ΩBIs​1.5 A

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

V1 = 9 VIs = 1.5 AR1 = 5 ΩR2 = 15 ΩR3 = 3 ΩR4 = 4 Ω
IR3 = A
5.Hard
R1​ = 2 ΩR2​ = 10 ΩR4​ = 12 Ω+−V1​15 VR3​25 ΩIR3​AIs​2.5 AB−+V2​15 V

Use superposition to find the power PR3 absorbed by R3.

V1 = 15 VIs = 2.5 AV2 = 15 VR1 = 2 ΩR2 = 10 ΩR3 = 25 ΩR4 = 12 Ω
PR3 = W
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 = 17.5 mA. With V1 doubled and the other two unchanged, I = 26.5 mA. With the direction of Is reversed and the other two unchanged, I = 11.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 17.5 mA2·V1: I = 26.5 mA−Is: I = 11.5 mA
I = mA

Answer key

  1. 1. IR2 = 0.4375 A
  2. 2. I = −7.5 mA
  3. 3. VR2 = 9.73 V
  4. 4. IR3 = 0.06977 A
  5. 5. PR3 = 3.472 W
  6. 6. I = 12 mA

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