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Circuit Analysis: AC, Linear Circuits and Transients

In the circuit in the figure, E1 = 14 V, E2 = 7 V, R1 = 2 Ω, R2 = 1 Ω and R3 = 4 Ω. When the circuit is solved by mesh analysis with the mesh currents I1 and I2 in the directions shown in the figure, which of the following is the magnitude of the current in R3?

DC circuit with two meshes + E1 R1 R2 + E2 R3 I1 I2
DC circuit with two meshes · View in diagrams
  1. A1 A
  2. B2 A
  3. C3 A
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Answer: B. 2 A

For the left mesh, E1 = (R1 + R3)I1 − R3 I2 gives 6I1 − 4I2 = 14; for the right mesh, E2 opposes I2, so −E2 = (R2 + R3)I2 − R3 I1 gives −4I1 + 5I2 = −7. Solving these simultaneous equations gives I1 = 3 A and I2 = 1 A. The two mesh currents flow through R3 in opposite directions, so its current is I1 − I2 = 2 A (downward). The voltage of the node above R3, 4 × 2 = 8 V, also confirms the current in R1, (14 − 8)/2 = 3 A, and in R2, (8 − 7)/1 = 1 A.

  • A. 1 A is the mesh current I2; in R3 it overlaps with I1 flowing in the opposite direction.
  • B. Correct. I1 − I2 = 3 − 1 = 2 A.
  • C. 3 A is the mesh current I1, the current in R1.
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