Q. $AB$ is a potentiometer wire of length $100\ cm$ and its resistance is $10\ ohm.$ It is connected in series with a resistance $R = 40\ ohm$ and a battery of e.m.f. $2\ V$ and negligible internal resistance. If a source of unknown e.m.f. $E$ is balanced by $40\ cm$ length of the potentiometer wire, the value of $E$ is
A) $0.8\ V$
B) $1.6\ V$
C) $0.08\ V$
D) $0.16\ V$
Explanation
(d) $E = \frac{e}{(R + R_{h} + r)}\frac{R}{L} \times l = \frac{2}{(10 + 40 + 0)} \times \frac{10}{1} \times 0.4 = 0.16V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.