Q. A battery of emf 2V and internal resistance 0.1$\ \Omega$ is being charged by a current of 5A. the potential difference between the terminals of the battery is
A) 2.5$\ \Omega$
B) 1.5$\ \Omega$
C) 0.5$\ \Omega$
D) 1$\ \Omega$
Explanation
(a) When r is internal resistance of the battery and $i\ $the charging current, then
$\ \ \ \ \ V = E + ir$
$\ \ \ \ V = 2 + 5 \times 0.1 = 2.5V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.