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Current from source in ladder circuit

Short answer

A short circuit across the 1 Ω resistor bypasses it and the last branch. The circuit simplifies to 2 Ω in series with the parallel combination of 2 Ω and 2 Ω (effective 1 Ω). Total resistance is 3 Ω, so source current I = 3 V / 3 Ω = 1 A.

Find current from the source in the following circuit diagram. [Figure: 3 V source with 1 Ω in parallel; three 2 Ω series resistors along the top; a 1 Ω resistor and a 5 Ω resistor in the shunt branches.]

Reading the figure

A 3 V source feeds a ladder: series 2 Ω, then node N₁ with a 2 Ω shunt, series 2 Ω to node N₂, then series 2 Ω and 5 Ω. The 1 Ω shunt at N₂ has a wire connected across it (from its top to the bottom rail), so N₂ is shorted to the bottom rail. The 1 Ω carries no current, and the last 2 Ω + 5 Ω branch is also at zero potential difference.

Reduction

  • From N₁ to the bottom rail: the 2 Ω shunt in parallel with the 2 Ω series leg leading to the short at N₂. 2 ∥ 2 = 1 Ω
  • Total resistance seen by the source = 2 Ω (first series) + 1 Ω = 3 Ω

Source current

I = 3/3 = 1 A

Answer: Source current = 1 A

Assumption: the loop of wire drawn round the 1 Ω is a short circuit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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