A Wheatstone bridge is balanced. Find the rat | vedclass

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(B) In a balanced Wheatstone bridge,the potential difference across the bridge is the same for both branches if connected in parallel. However,the sta

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$R : P$

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(B) In a balanced Wheatstone bridge,the potential difference across the bridge is the same for both branches if connected in parallel. However,the standard configuration shows the branches $(P, Q)$ and $(R, S)$ are in series with each other in parallel arms. Let the total voltage across the parallel branches be $V$. The power dissipated in a resistor is given by $P_{diss} = \frac{V^2}{R_{eq}}$. For the branch $(P+Q)$,the power is $P_1 = \frac{V^2}{P+Q}$. For the branch $(R+S)$,the power is $P_2 = \frac{V^2}{R+S}$. The ratio is $\frac{P_1}{P_2} = \frac{V^2/(P+Q)}{V^2/(R+S)} = \frac{R+S}{P+Q}$. Since the bridge is balanced,$\frac{P}{R} = \frac{Q}{S}$,which implies $PS = QR$,or $\frac{S}{Q} = \frac{R}{P}$. Also,from the balanced condition $\frac{P}{Q} = \frac{R}{S}$,we can write $S = \frac{RQ}{P}$. Substituting $S$ into the ratio: $\frac{R + RQ/P}{P + Q} = \frac{R(1 + Q/P)}{P + Q} = \frac{R(P+Q)/P}{P+Q} = \frac{R}{P}$. Thus,the ratio is $R:P$.

$A$ Wheatstone bridge is balanced. Find the ratio of electric power dissipated in $(P + Q)$ to $(R + S)$.

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