Q. Two sources of equal emf are connected to an 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
A) $R = R_{1}R_{2}/(R_{1} + R_{2})$
B) $R = R_{1}R_{2}/(R_{2} - R_{1})$
C) $R = R_{2} \times (R_{1} + R_{2})/(R_{2} - R_{1})$
D) $R = R_{2} - R_{1}$
Explanation
(d) $i = \frac{2E}{R + R_{1} + R_{2}}$
From cell (2) $E = V + iR_{2} = 0 + iR_{2}$
$\Rightarrow E = \frac{2E}{R + R_{1} + R_{2}} \times R_{2} \Rightarrow R = R_{2} - R_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.