Q. If the potential difference across the internal resistance $r_{1}$ is equal to the emf E of the battery, then
A) $R = r_{1} + r_{2}$
B) $R = \frac{r_{1}}{r_{2}}$
C) $R = r_{1} - r_{2}$
D) $R = \frac{r_{2}}{r_{1}}$
Explanation
(c) The total emf $= E\ + E\ = 2E$
Total resistance =$R + r_{1} + r_{2}$
∴ Current flowing through the circuit
$i = \frac{2E}{R + r_{1} + r_{2}}$
According to question $E = ir_{1}$
$\Longrightarrow \ \ \ i = \frac{E}{r_{1}}$
$\therefore\ \ \frac{E}{r_{1}} = \frac{2E}{R + r_{1} + r_{2}}$
$\ \ \ \ \ \ R = r_{1} - r_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.