Q. A 50V battery is connected across a $10\ \Omega$ resistor and a current of 4.5 A flows. The internal resistance of the battery is
A) $10\ \Omega$
B) $0.5\ \Omega$
C) $1.1\ \Omega$
D) $5\ \Omega$
Explanation
(c) $E - V = ir\ or\ r = (E - V)i$
and $V = iR = 4.5 \times 10 = 45V$
$\therefore\ r = \frac{(50 - 45)}{4.5} = \frac{5}{4.5}$
$= 1.10\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.