Two cells having unknown e.m.f.s $\mathrm{E}_{1}$ and $\mathrm{E}_{2}\left(\mathrm{E}_{1}\mathrm{E}_{2}\right)$ are connected in potentiometer circuit so as to assist each other. The null point obtained is at 490 $\mathrm{cm}$ from the higher potential end. When cell $\mathrm{E}_{2}$ is connected…

Question

Two cells having unknown e.m.f.s $\mathrm{E}_{1}$ and $\mathrm{E}_{2}\left(\mathrm{E}_{1}>\mathrm{E}_{2}\right)$ are connected in potentiometer circuit so as to assist each other. The null point obtained is at 490 $\mathrm{cm}$ from the higher potential end. When cell $\mathrm{E}_{2}$ is connected so as to oppose cell $\mathrm{E}_{1}$, Ithe null point is obtained at $90 \mathrm{~cm}$ from the same end. The ratio of the e.m.f.s of two cells $\left(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\right)$ is

Aicharya · Accepted Answer

$\begin{aligned} \frac{E_{1}}{E_{2}} &=\frac{\ell_{1}+\ell_{2}}{\ell_{1}-\ell_{2}} \ &=\frac{490+90}{490-90}=\frac{580}{400}=1.45 \end{aligned}$

Two cells having unknown e.m.f.s $\mathrm{E}_{1}$ and…

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