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. $\frac{I_{1} R_{2}-I_{2} R_{1}}{I_{1}-I_{2}}$

contexto.120

3 years ago

Correct answer is C.

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