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Circuit Analysis: AC, Linear Circuits and Transients

In the circuit in the figure, E = 12 V, R1 = 4 Ω and J = 2 A. R2 is removed. Which of the following is the Thevenin equivalent circuit (equivalent voltage Vth and equivalent resistance Rth) seen from the two terminals to which R2 was connected (node A and ground)?

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. AVth = 12 V, Rth = 4 Ω
  2. BVth = 20 V, Rth = 4 Ω
  3. CVth = 20 V, Rth = 3 Ω
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Answer: B. Vth = 20 V, Rth = 4 Ω

Vth is the voltage with the terminals open; all 2 A of J then flows through R1 toward the positive terminal of E, so Vth = E + J R1 = 12 + 2 × 4 = 20 V. Rth is the resistance seen from the terminals with the voltage source shorted and the current source opened, which is just R1 = 4 Ω. Connecting R2 = 12 Ω to this equivalent gives VA = 20 × 12/(4 + 12) = 15 V, matching the result of nodal analysis. As a Norton equivalent, it is a current source equal to the current with the terminals shorted, E/R1 + J = 5 A, in parallel with 4 Ω.

  • A. This forgets the contribution of the current source J and takes only E as the open-circuit voltage.
  • B. Correct. Vth = 20 V and Rth = 4 Ω.
  • C. Rth = 3 Ω is R1 in parallel with R2; it includes the load R2, which is removed when finding the Thevenin equivalent.
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  1. 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?
  2. 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?
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