Q. A new flashlight cell of $e.m.f.\ 1.5\ volt$ gives a current of $15\ amp,$ when connected directly to an ammeter of resistance $0.04\ \Omega.$ The internal resistance of cell is
A) $0.04\ \Omega$
B) $0.06\ \Omega$
C) $0.10\ \Omega$
D) $10\ \Omega$
Explanation
(b) Let the internal resistance of cell be $r,$ then
$i = \frac{E}{R + r} \Rightarrow 15 = \frac{1.5}{0.04 + r} \Rightarrow r = 0.06\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.