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Answer: 2. circuit (2)

Key idea: A Wheatstone bridge is balanced when no current flows through G, i.e. \(\frac{P}{Q}=\frac{R}{S}\).

The top arms are 10 \(\Omega\) and 10 \(\Omega\), so the ratio is 1. The bottom arms are 15 \(\Omega\) and the 5 \(\Omega\) resistor in series with the diode D, so balance needs the 5 \(\Omega\)-plus-diode arm to equal 15 \(\Omega\).

The official key gives circuit (2). With ideal diodes none of the four circuits gives exactly this balance, so we cannot derive the key from the figure.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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