In the adjoining circuit, the battery $E_1$ has an $e.m.f.$ of $12\, volts$ and zero internal resistance, while the battery $E_2$ has an $e.m.f.$ of $2\, volts$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ in $ohms$ is
Medium
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$i=\frac{10}{500}$
$2=\mathrm{i} \times \mathrm{x}$
$x=100$
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