In the adjoining circuit, the battery ${E_1}$ has an $e.m.f.$ of $12\,volt$ and zero internal resistance while the battery $E$ has an $e.m.f.$ of $2\,volt$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ in $ohm$ is
AIEEE 2005, Medium
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For no current through galvanometer, we have
$\left( {\frac{{{E_1}}}{{500 + X}}} \right)X = E$ $ \Rightarrow $ $\left( {\frac{{12}}{{500 + X}}} \right)X = 2$ $ \Rightarrow $ $X = 100 \,\Omega$
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