Two cells of $e.m.f.$ $s\, E_1$ and $E_2$ and of negligible internal resistances are connected with two variable resistors as shown in Fig. When the galvanometer shows no deflection, the values of the resistances are $P$ and $Q$. What is the value of the ratio $E_2/E_1$ ?
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Potential difference $\mathrm{V}_{\mathrm{P}}$ across $\mathrm{P}$ as determined
from $\mathrm{E}_{1}$ is given by $\mathrm{V}_{\mathrm{P}}=\left(\frac{\mathrm{E}_{1}}{\mathrm{P}+\mathrm{Q}}\right) \mathrm{P}.$
Also, $\mathrm{V}_{\mathrm{P}}=\mathrm{E}_{2}$. Therefore,
$\mathrm{E}_{2}=\mathrm{E}_{1}\left(\frac{\mathrm{P}}{\mathrm{P}+\mathrm{Q}}\right) \Rightarrow \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{P}}{\mathrm{P}+\mathrm{Q}}$
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