Q. Two sources of equal E.M.F. are connected to external resistance $R$ . The internal resistances of the two sources are $R_{1}$ and $R_{2}$ $\left(R_{2} > R_{1}\right)$ . If the potential difference across the source having internal resistance $R_{2}$ is zero, then
NTA AbhyasNTA Abhyas 2022
Solution:
Given,
The potential difference across the cell is,
$E-IR_{2}=0$
Current flowing through the circuit is given by,
$I=\frac{2 E}{R_{1} + R_{2} + R}$
$\therefore $ $E=IR_{2}$
$\Rightarrow E=\frac{2 E R_{2}}{R_{1} + R_{2} + R}$
$\Rightarrow R_{1}+R_{2}+R=2R_{2}$
$\Rightarrow R=R_{2}-R_{1}$
The potential difference across the cell is,
$E-IR_{2}=0$
Current flowing through the circuit is given by,
$I=\frac{2 E}{R_{1} + R_{2} + R}$
$\therefore $ $E=IR_{2}$
$\Rightarrow E=\frac{2 E R_{2}}{R_{1} + R_{2} + R}$
$\Rightarrow R_{1}+R_{2}+R=2R_{2}$
$\Rightarrow R=R_{2}-R_{1}$