Physics - Current Electricity Question with Solution | TestHub

Key Idea : This problem is based on the application of potentiometer in which we find the internal resistance of a cell. 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 $\frac{E}{V}=\frac{l_1}{l_2}$ where $l_1$ 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 $V=IR]$ $\therefore \frac{R+r}{R}=\frac{l_1}{l_2}$ $\begin{array}{rl}1+\frac{r}{R}& =\frac{110}{100}\\ \frac{r}{R}& =\frac{10}{100}\\ r& =\frac{1}{10}\times 10=1\Omega \end{array}$ $r=\frac{1}{10}\times 10=1\Omega$