Circuit Analysis: AC, Linear Circuits and Transients
In the circuit in the figure, E = 12 V, R1 = 4 Ω, R2 = 12 Ω and the current source is J = 2 A (flowing into node A). Which of the following is the voltage VA of node A with respect to ground?
- A6 V
- B9 V
- C15 V
Show answer and explanation
Answer: C. 15 V
Applying Kirchhoff's current law at node A and setting the currents flowing out through R1 and R2 equal to the incoming current J gives (VA − 12)/4 + VA/12 = 2. Multiplying both sides by 12 gives 3VA − 36 + VA = 24, so VA = 60/4 = 15 V. Checking with the superposition theorem, E alone contributes 12 × 12/(4 + 12) = 9 V and J alone contributes 2 A × 3 Ω (R1 and R2 in parallel) = 6 V, for a total of 15 V. The node voltage is higher than the source voltage E because the current pushed in by J flows through R1 toward E.
- A. This is the contribution of the current source J alone (with E shorted); the contribution of E is missing.
- B. This is the voltage-divider result for the voltage source E alone (with J open); the contribution of J is missing.
- C. Correct. The node equation gives VA = 15 V.
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