A linear resistive circuit contains two independent voltage sources, V1 = 24 V and V2 = 5 V. With both sources on, the current through the load resistor is I = 33.5 mA. With V2 switched off (replaced by a short circuit) and V1 unchanged, the load current is 17.5 mA. Find the load current I when V1 is raised to 48 V and V2 stays at 5 V.
Superposition worksheet with answers
1.Easy
I = 33.5 mA (both sources)I = 17.5 mA (only V1)
I = mA
2.Easy
Use superposition to find the current IR1 through R1 in the direction of the arrow.
V1 = 15 VV2 = 18 VR1 = 3 ΩR2 = 3 ΩR3 = 2 Ω
IR1 = A
3.Medium
Find the value of the current source Is (direction as shown) that makes the current IR3 through R3 equal to zero.
V1 = 18 VR1 = 20 ΩR2 = 3 ΩR3 = 3 ΩR4 = 10 Ω
Is = A
4.Medium
Use superposition to find the current IR3 through R3 in the direction of the arrow.
V1 = 12 VIs = 4 AR1 = 2 ΩR2 = 4 ΩR3 = 2 ΩR4 = 10 Ω
IR3 = A
5.Hard
Find the value of the voltage source V2 (polarity as shown) that makes the current IR3 through R3 equal to zero.
V1 = 18 VIs = 2.5 AR1 = 20 ΩR2 = 2 ΩR3 = 25 ΩR4 = 10 Ω
V2 = V
6.Hard
Use superposition to find the power PR4 absorbed by R4.
V1 = 36 VIs = 3 AV2 = 6 VR1 = 3 ΩR2 = 20 ΩR3 = 12 ΩR4 = 10 Ω
PR4 = W
Answer key
- 1. I = 51 mA
- 2. IR1 = 1.857 A
- 3. Is = 0.2348 A
- 4. IR3 = 2.4 A
- 5. V2 = 14.2 V
- 6. PR4 = 99.11 W
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