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Current in 5Ω using Thevenin theorem

Short answer

Thevenin voltage is −2 V and Thevenin resistance is 3 Ω. The current through the 5 Ω resistor is 2/(3+5) = 0.25 A, flowing from terminal d to c.

Find out the current through 5 ohm resistor connected across the terminal c and d in the network shown below using the Thevenin's theorem. [Figure: 1 A current source on the left with 2 Ω shunt; 5 Ω between terminals c and d; 2 Ω shunt after d; 2 Ω in series with an 8 V source on the right.]

Reading the figure

Bottom rail = ground. The 1 A source (arrow up) feeds node c. c has a 2 Ω shunt. The 5 Ω is connected between c and d. d has a 2 Ω shunt, and a branch with 2 Ω in series with the 8 V battery (+ on top, towards the 2 Ω).

Step 1: Remove the 5 Ω; find V_th across c-d

  • V_c = 1 A × 2 Ω = 2 V
  • Node d: the 8 V source with 2 Ω and the 2 Ω shunt form a divider: V_d = 8 × 2/(2 + 2) = 4 V

V_th = V_c − V_d = 2 − 4 = −2 V (d is 2 V higher than c)

Step 2: Thevenin resistance (sources off)

1 A opened, 8 V shorted:

  • At c: 2 Ω
  • At d: 2 Ω ∥ 2 Ω = 1 Ω

R_th = 2 + 1 = 3 Ω

  c o--[3 Ω]--(2 V)--[5 Ω]--o d      (polarity: d is positive)

Step 3: Current in the 5 Ω

I = |V_th|/(R_th + 5) = 2/(3 + 5) = 0.25 A, flowing from d to c through the 5 Ω.

Answer: I₅Ω = 0.25 A (from d to c)

✓ Answer checked by Claude Opus 5.5 (high effort)

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