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 $10\, V$ is maintained across $AC.$
JEE MAIN 2021, Diffcult
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$\frac{x-10}{100}+\frac{x-y}{15}+\frac{x-0}{10}=0$
$53 x-20 y=30 \ldots \ldots(1)$
$\frac{y-10}{60}+\frac{y-x}{15}+\frac{y-0}{5}=0$
$17 y -4 x =10$
on solving $(1)\,\&\,(2)$
$x=0.865$
$y =0.792$
$\Delta V =0.073 R =15\, \Omega$
$i =4.87 \,mA$
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