A Wheatstone bridge is balanced. Find the rat | vedclass

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(B) In a balanced Wheatstone bridge,the potential difference across the central resistor is zero,so no current flows through it. The circuit consists

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$R : P$

Vedclass · Answer

(B) In a balanced Wheatstone bridge,the potential difference across the central resistor is zero,so no current flows through it. The circuit consists of two parallel branches: one with resistors $P$ and $Q$ in series,and the other with resistors $R$ and $S$ in series.The total resistance of the first branch is $(P + Q)$ and the total resistance of the second branch is $(R + S)$.Since both branches are connected across the same potential difference $V$,the power dissipated in each branch is given by $P_{power} = \frac{V^2}{R_{eq}}$.Therefore,the ratio of power dissipated is:$\frac{	ext{Power in } (P+Q)}{	ext{Power in } (R+S)} = \frac{\frac{V^2}{P+Q}}{\frac{V^2}{R+S}} = \frac{R+S}{P+Q}$For a balanced Wheatstone bridge,the condition is $\frac{P}{R} = \frac{Q}{S}$,which implies $PS = QR$,or $\frac{S}{Q} = \frac{R}{P}$.We can rewrite the ratio as:$\frac{R+S}{P+Q} = \frac{R(1 + S/R)}{P(1 + Q/P)}$Since $\frac{S}{R} = \frac{Q}{P}$,the terms in the brackets cancel out:$\frac{R(1 + Q/P)}{P(1 + Q/P)} = \frac{R}{P}$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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