Q. A cell of emf $E$ and internal resistance $r$ supplies currents for the same time $t$ through external resistance $R_{1} =$100 $\Omega\ and\ R_{2} = 40\ \Omega$ separately. If the heat developed in both the cases in the same, then the internal resistance of the cell is given by
A) 28.6$\ \Omega$
B) 70$\ \Omega$
C) 63.3$\ \Omega$
D) 140$\ \Omega$
Explanation
(c) Current drawn from the cell in resistance $R_{1}$ will be $I = E$/$(R_{1} + r)$
Therefore, heat produced in $R_{1}\ ie,$
$H_{1} = \frac{E^{2}R_{1}t}{\left( R_{1} + r \right)^{2}}$
Heat produced in $R_{2}\ ie,\ H_{2} = \frac{E^{2}R_{2}t}{\left( R_{2} + r \right)^{2}}$
As per question $H_{1} = H_{2}$
or $\frac{E^{2}R_{1}t}{\left( R_{1} + r \right)^{2}} =$ $\frac{E^{2}R_{2}t}{\left( R_{2} + r \right)^{2}}$
On solving we get;
$r = \sqrt{R_{1}R_{2}}$
$= \sqrt{100 \times 40}$=63.25$\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.