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Circuit Analysis: AC, Linear Circuits and Transients

In the circuit in the figure, E = 12 V, R1 = 4 Ω, R2 = 12 Ω and J = 2 A. Which of the following correctly gives the magnitude and direction of the current in R1?

DC circuit with a voltage source and a current source + E R1 A J R2
DC circuit with a voltage source and a current source · View in diagrams
  1. A0.75 A from node A toward E
  2. B0.75 A from E toward node A
  3. C2.25 A from E toward node A
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Answer: A. 0.75 A from node A toward E

Using the superposition theorem, apply the sources one at a time. With E alone (J open), 12/(4 + 12) = 0.75 A flows through R1 and R2 in series, from E toward node A. With J alone (E shorted), the 2 A of J divides between R1 and R2, and 2 × 12/(4 + 12) = 1.5 A flows in R1 from node A toward E. Adding them with their directions gives 1.5 − 0.75 = 0.75 A from node A toward E, a direction in which E absorbs power (is charged). The node voltage VA = 15 V confirms this: (15 − 12)/4 = 0.75 A.

  • A. Correct. Superposition gives 0.75 A from node A toward E.
  • B. The magnitude is right but the direction is reversed; this is the direction of the current due to E alone. The 1.5 A due to J is larger, so the total flows the other way.
  • C. This adds the two currents as 0.75 + 1.5 = 2.25 A without considering their directions.
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