Q. The electromotive force of a primary cell is $2\ volt.$ When it is short-circuited it gives a current of $4\ ampere.$ Its internal resistance in $ohm$ is
A) 0.5
B) 5.0
C) 2.0
D) 8.0
Explanation
(a) $E = V + ir$
After short-circuiting, $V = 0 \Rightarrow r = \frac{E}{i} = \frac{2}{4} = 0.5\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.