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(A) In a balanced Wheatstone bridge,the condition is $\frac{P}{S} = \frac{Q}{R}$,which implies $PR = QS$. Let the total voltage across the bridge be $

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$R : Q$

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(A) In a balanced Wheatstone bridge,the condition is $\frac{P}{S} = \frac{Q}{R}$,which implies $PR = QS$. Let the total voltage across the bridge be $V$. The power consumed in the upper branch $(P + Q)$ is $P_1 = \frac{V^2}{P + Q}$. The power consumed in the lower branch $(R + S)$ is $P_2 = \frac{V^2}{R + S}$. The ratio of power consumed is $\frac{P_1}{P_2} = \frac{V^2 / (P + Q)}{V^2 / (R + S)} = \frac{R + S}{P + Q}$. From the balanced condition,$S = \frac{PR}{Q}$. Substituting $S$ in the ratio: $\frac{R + \frac{PR}{Q}}{P + Q} = \frac{R(1 + \frac{P}{Q})}{P + Q} = \frac{R(\frac{Q + P}{Q})}{P + Q} = \frac{R}{Q}$. Thus,the ratio is $R : Q$.

Resistances are arranged in a cyclic order to form a balanced Wheatstone bridge as shown in the figure. The ratio of power consumed in the branches $(P + Q)$ and $(R + S)$ is

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