Physics - Current Electricity Question with Solution | TestHub

Let the emf of each source be E. When they are connected in series, the current in the circuit $I=\frac{E_{\text{tot }}}{R_{\text{tot }}}=\frac{E+E}{r_1+r_2+R}=\frac{2E}{r_1+r_2+R}$ $\therefore$ potential drop across the cell of internal resistance $r_2,\left(\frac{2E}{r_1+r_2+R}\right)r_2$ Hence, $E-\frac{2E}{\left(r_l+r_2+R\right)}{r}_2=0$ $\begin{array}{rl}& r_1+r_2+R=2r_2\\ \Rightarrow & R=r_2-r_1\end{array}$