Q. A current $2\ A$ flows through a $2\Omega$ resistor when connected across a battery. The same battery supplies a current $0.5\ A$ when connected across a $9\Omega$ resistor. The internal resistance of the battery is
A) $1\Omega$
B) $0.5\Omega$
C) $1/3\Omega$
D) $1/4\Omega$
Explanation
(c) $2 = \frac{\varepsilon}{2 + r}$
$0.5 = \frac{\varepsilon}{9 + r}$ or $\frac{2}{0.5} = \frac{9 + r}{2 + r}\ \therefore r = \frac{1}{3}\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.