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The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 Ω$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

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$2.44 μ A$

$4.87 μ A$

$2.44 m A$

$4.87 m A$

answer is D.

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Detailed Solution

$\frac{10 - x}{60} + \frac{0 - x}{5} + \frac{y - x}{15} = 0$ and $\frac{10 - y}{100} + \frac{0 - y}{10} + \frac{x - y}{15} = 0$

$x = 0.79 V, y = 0.865$

$∴ i_{P B} = (\frac{y - x}{15}) = 4.87 \times 10^{- 3} A$

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