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​ = 20 VV2​ = 12 VIRL​

A linear resistive circuit contains two independent voltage sources, V1 = 20 V and V2 = 12 V. With both sources on, the current through the load resistor is I = −29 mA. With V2 switched off (replaced by a short circuit) and V1 unchanged, the load current is −13 mA. Find the load current I when V1 is switched off and only V2 = 12 V is active.

I = −29 mA (both sources)I = −13 mA (only V1)
I = mA
2.Easy
R1​ = 6 ΩIR1​R2​ = 2 Ω+−V1​6 VR3​5 ΩA+−V2​15 V

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

V1 = 6 VV2 = 15 VR1 = 6 ΩR2 = 2 ΩR3 = 5 Ω
IR1″ = A
3.Medium
R1​ = 10 ΩR3​ = 20 ΩIR3​+−V1​12 VR2​15 ΩAR4​12 ΩBIs​4 A

Use superposition to find the power PR3 absorbed by R3.

V1 = 12 VIs = 4 AR1 = 10 ΩR2 = 15 ΩR3 = 20 ΩR4 = 12 Ω
PR3 = W
4.Medium
R1​ = 12 ΩR3​ = 6 Ω+−V1​6 VR2​20 Ω−+VR2​AR4​12 ΩBIs​1 A

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

V1 = 6 VIs = 1 AR1 = 12 ΩR2 = 20 ΩR3 = 6 ΩR4 = 12 Ω
VR2 = V
5.Hard
R1​ = 12 ΩR2​ = 6 ΩR4​ = 10 Ω+−V1​30 VR3​30 ΩIR3​AIs​5 AB+−V2​?

Find the value of the voltage source V2 (polarity as shown) that makes the current IR3 through R3 equal to zero.

V1 = 30 VIs = 5 AR1 = 12 ΩR2 = 6 ΩR3 = 30 ΩR4 = 10 Ω
V2 = V
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 = −3.5 mA. With V1 doubled and the other two unchanged, I = −3 mA. With the direction of Is reversed and the other two unchanged, I = −6.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = −3.5 mA2·V1: I = −3 mA−Is: I = −6.5 mA
I = mA

Answer key

  1. 1. I = −16 mA
  2. 2. IR1″ = −1.442 A
  3. 3. PR3 = 23.06 W
  4. 4. VR2 = −6.176 V
  5. 5. V2 = 10 V
  6. 6. I = 2 mA

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