When a resistance ‘ R_{1}{ }^{\prime} is connected across the terminal of a cell of e.m.f. ‘ \mathrm{E}_{1} ‘ the current is ‘ \mathrm{I}_{1} ‘. When the resistance is changed to \mathrm{R}_{2} ‘ the current is ‘ I_{2} ‘. The internal resistance of the cell is
A. \frac{I_{1} R_{2}+I_{2} R_{1}}{I_{1}+I_{2}}
B. \frac{I_{2} R_{2}+I_{1} R_{1}}{I_{1}+I_{2}}
C. \frac{I_{2} R_{2}-I_{1} R_{1}}{I_{1}-I_{2}}
D. \frac{I_{1} R_{2}-I_{2} R_{1}}{I_{1}-I_{2}}
When a resistance ‘ R_{1}{ }^{\prime} is connected across the terminal of a cell of e.m.f. ‘ \mathrm{E}_{1} ‘ the current is ‘ \mathrm{I}_{1} ‘. When the resistance is changed to \mathrm{R}_{2} ‘ the current is ‘ I_{2} ‘. The internal resistance of the cell is
A. \frac{I_{1} R_{2}+I_{2} R_{1}}{I_{1}+I_{2}}
B. \frac{I_{2} R_{2}+I_{1} R_{1}}{I_{1}+I_{2}}
C. \frac{I_{2} R_{2}-I_{1} R_{1}}{I_{1}-I_{2}}
D. \frac{I_{1} R_{2}-I_{2} R_{1}}{I_{1}-I_{2}}

Correct answer is C.

\begin{array}{l} I_{1}=\frac{E}{r+R_{1}} \\ I_{2}=\frac{E}{r+R_{2}} \\ \frac{I_{1}}{I_{2}}=\frac{r+R_{2}}{r+R_{1}} \\ I_{1} r+I_{1} R_{1}=I_{2} r+I_{2} R_{2} \\ r\left(I_{1}-I_{2}\right)=I_{2} R_{2}-I_{1} R_{1} \\ r=\frac{I_{2} R_{2}-I_{1} R_{1}}{I_{1}-I_{2}} \end{array}