Question 294 of 380

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

A$2.44\mu A$B$2.44mA$C$4.87mA$D$4.87\mu A$

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Answer: (C) $4.87mA$

$\dfrac{x-10}{100}+\dfrac{x-y}{15}+\dfrac{x-0}{10}=0$
$53x-20y=30\ldots \ldots (1)$
$\dfrac{y-10}{60}+\dfrac{y-x}{15}+\dfrac{y-0}{5}=0$
$17y-4x=10\ldots \ldots (2)$
on solving (1)&(2)
$x=0.865$
$y=0.792$
$\Delta V=0.073R=15\Omega$
$\Rightarrow i=4.87mA$