Q. The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be
A) $\frac{E^{2}}{r}$
B) $\frac{E^{2}}{2r}$
C) $\frac{E^{2}}{3r}$
D) $\frac{E^{2}}{4r}$
Explanation
(d) $i = \frac{R}{r + R}$
$P = i^{2}R$
$\Longrightarrow \ \ P = \frac{E^{2}R}{(r + R)^{2}}$
Power is maximum when r=R
$\therefore\ \ P_{\max\ } = \frac{E^{2}}{4R}\ $
$or\ \ P_{\max\ } = \frac{E^{2}}{4r}\ \ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.