A galvanometer has a current sensitivity of 1 mA per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of $500\Omega$  and cell of internal resistance $1\Omega$ . It gives a defection of 5 division for shunt of 5 ohm and 20 division for shunt of 25 ohm. The emf of cell is

$$\begin{gathered} \mathrm{Here}, \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}+\frac{\mathrm{GS}}{\mathrm{G}+\mathrm{S}}}\mathrm{and}{\mathrm{I}}_{\mathrm{g}}=\frac{\mathrm{IS}}{\mathrm{G}+\mathrm{S}} \\ {\mathrm{I}}_{\mathrm{g}}=\frac{\mathrm{E}}{(\mathrm{R}+\mathrm{r})+\frac{\mathrm{GS}}{(\mathrm{G}+\mathrm{S})}}\times \frac{\mathrm{S}}{(\mathrm{G}+\mathrm{S})} \\ \therefore {\mathrm{I}}_{\mathrm{g}}=\frac{\mathrm{ES}}{(\mathrm{R}+\mathrm{r})(\mathrm{G}+\mathrm{S})+\mathrm{GS}} \\ \mathrm{For} \mathrm{S} = 5 \mathrm{ohm}, {\mathrm{I}}_{\mathrm{g}}=5\times {10}^{-3}\mathrm{A} \\ \mathrm{Hence}, 5\times {10}^{-3}=\frac{\mathrm{E}\times 5}{501(\mathrm{G}+5)+5\mathrm{G}} \dots .\left(\mathrm{i}\right) \\ \mathrm{For} \mathrm{S} = 25 \mathrm{ohm}, \mathrm{Ig} = 20 \mathrm{x} {10}^{-3} \mathrm{A} \\ \mathrm{And} 20\times {10}^{-3}=\frac{\mathrm{E}\times 25}{501(\mathrm{G}+25)+25\mathrm{G}} \dots \left(\mathrm{ii}\right) \\ \mathrm{Dividing} \mathrm{and} \mathrm{solving}, \mathrm{G}=88.2\mathrm{\Omega } \\ \mathrm{From} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \mathrm{E}={10}^{-3}[501(88.2+5)+5\times 88.2]=47.1 \mathrm{Volt} \end{gathered}$$