The balance condition for a Wheatstone bridge with arms $P, Q, R, S$ (galvanometer between B and D) is:
A$P + Q = R + S$
B$P/Q = R/S$
C$P/Q = S/R$
D$PQ = RS$
Answer & Solution
Correct answer: C. $P/Q = S/R$
At balance the galvanometer current $I_g = 0$. From the loop equations, $I_1P = I_2S$ and $I_1Q = I_2R$, dividing gives **P/Q = S/R**.
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