Q. A battery having e.m.f.$\ 5V$ and internal resistance $0.5\ \Omega$ is connected with a resistance of $4.5\ \Omega$ then the voltage at the terminals of battery is
A) $4.5\ V$
B) $4\ V$
C) $0\ V$
D) $2\ V$
Explanation
(a) $i = \frac{E}{R + r} = \frac{5}{4.5 + 0.5} = 1A$
$V = E - ir = 5 - 1 \times 0.5 = 4.5\ Volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.