A Wheatstone bridge is initially at room temp | vedclass

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(C) Let the total voltage be $V = 40 	ext{ V}$.Initially, all resistances are $R$. The bridge is balanced, so $V_a = V_b = V/2 = 20 	ext{ V}$.After

Vedclass · Accepted Answer

$0.95$

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(C) Let the total voltage be $V = 40 	ext{ V}$.Initially, all resistances are $R$. The bridge is balanced, so $V_a = V_b = V/2 = 20 	ext{ V}$.After heating, $R_3$ becomes $R' = R + 0.1R = 1.1R$.The bridge now consists of two parallel branches: one with $(R_1 + R_2)$ and one with $(R_3 + R_4)$.However, the nodes $a$ and $b$ are the midpoints of these branches.Voltage at node $a$ (between $R_1$ and $R_2$): $V_a = V 	imes \frac{R_2}{R_1 + R_2} = 40 	imes \frac{R}{R + R} = 20 	ext{ V}$.Voltage at node $b$ (between $R_3$ and $R_4$): $V_b = V 	imes \frac{R_4}{R_3 + R_4} = 40 	imes \frac{R}{1.1R + R} = 40 	imes \frac{R}{2.1R} = \frac{40}{2.1} \approx 19.0476 	ext{ V}$.The potential difference is $V_a - V_b = 20 - 19.0476 = 0.9524 	ext{ V}$.Rounding to two decimal places, we get $0.95 	ext{ V}$.

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