Physics - Current Electricity Question with Solution | TestHub

In potentiometer experiment in which we find internal resistance of a cell, let $E$ be the emf of the cell and $V$ the terminal potential difference, then $$\frac{E}{V}=\frac{\mathrm{h}}{l_2}$$ where $\mathrm{l}$ and $l_2$ are lengths of potentiometer wire with and without short circuited through a resistance. Since, $\frac{E}{V}=\frac{R+r}{R}$ $[\because E=I(R+r)$ and $V=IR]$ $\therefore \frac{R+r}{R}=\frac{4}{l_2}$ or $1+\frac{r}{R}=\frac{110}{100}$ Or $\frac{r}{R}=\frac{10}{100}$ or $r=\frac{1}{10}\times 10=1\Omega$