Q. When a resistance of $2ohm$ is connected across the terminals of a cell, the current is $0.5\ ampere.$ When the resistance is increased to $5\ ohm,$ the current is $0.25\ ampere.$ The internal resistance of the cell is
A) $0.5\ ohm$
B) $1.0\ ohm$
C) $1.5\ ohm$
D) $2.0\ ohm$
Explanation
(b) Let the e.m.f. of cell be $E$ and internal resistance be $r$
Then $0.5 = \frac{E}{(r + 2)}$ and $0.25 = \frac{E}{(r + 5)}$
On dividing, $2 = \frac{5 + r}{2 + r} \Rightarrow r = 1\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.