In the adjoining circuit, the battery $E_1$ has an emf of $12\, volt$ and zero internal resistance while the battery $E_2$ has an $emf$ of $2\, volt$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ (in $ohm$ ) is
Medium
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Current across $(500 \Omega) \rightarrow$
$I=\frac{E_{1}-E_{2}}{500}$ ............$(1)$
$\mathrm{E}_{2}=\mathrm{IX}$ .........$(2)$
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