A linear resistive circuit contains two independent voltage sources, V1 = 9 V and V2 = 24 V. With both sources on, the current through the load resistor is I = 16.5 mA. With V2 switched off (replaced by a short circuit) and V1 unchanged, the load current is 9.5 mA. Find the load current I when V1 is switched off and only V2 = 24 V is active.
Superposition worksheet with answers
1.Easy
I = 16.5 mA (both sources)I = 9.5 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 = 15 VR1 = 4 ΩR2 = 8 ΩR3 = 6 Ω
IR1 = A
3.Medium
Use superposition to find the power PR4 absorbed by R4.
V1 = 30 VIs = 1.5 AR1 = 2 ΩR2 = 4 ΩR3 = 5 ΩR4 = 3 Ω
PR4 = W
4.Medium
A linear resistive circuit contains one voltage source Vs and one current source Is. The current I through a load resistor is measured twice: with Vs = 5 V and Is = 2 mA, I = 1.75 mA; with Vs = 15 V and Is = 1 mA, I = 4 mA. Find the load current I when Vs = 10 V and Is = 1 mA.
5 V, 2 mA → 1.75 mA15 V, 1 mA → 4 mAVs = 10 V, Is = 1 mA
I = mA
5.Hard
Find the value of the voltage source V2 (polarity as shown) that makes the current IR3 through R3 equal to zero.
V1 = 24 VIs = 2 AR1 = 30 ΩR2 = 20 ΩR3 = 10 ΩR4 = 25 Ω
V2 = V
6.Hard
Use superposition to find the current IR2 through R2 in the direction of the arrow.
V1 = 12 VIs = 3 AV2 = 36 VR1 = 20 ΩR2 = 3 ΩR3 = 25 ΩR4 = 20 Ω
IR2 = A
Answer key
- 1. I = 7 mA
- 2. IR1 = 1.154 A
- 3. PR4 = 3.797 W
- 4. I = 2.75 mA
- 5. V2 = 14 V
- 6. IR2 = −2.619 A
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