Q. When a battery connected across a resistor of 16$\ \Omega$, the voltage across the resistor is 12 V. When the same battery is connected across a resistor of 10$\ \Omega$, voltage across it is 11V. The internal resistance of the battery (in ohm) is
A) $\frac{10}{7}$
B) $\frac{20}{7}$
C) $\frac{25}{7}$
D) $\frac{30}{7}$
Explanation
(b) Here, V<E
∴ $E = V + Ir$
For first case
$E = 12 + \frac{12}{16}r\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
For second case
$E = 11 + \frac{11}{10}r\ \ \ \ \ \ \ \ \ \ \ \ldots\ldots(ii)$
From Eqs. (i) and (ii),
$12 + \frac{12}{16}r = 11 + \frac{11}{10}r$
$\Longrightarrow r = \frac{20}{7}\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.