Q. A potentiometer wire of length 10 m and resistance 20$\ \Omega$ is connected is series with a 15V battery and an external resistance 40$\ \Omega$. A secondary cell of emf E in the secondary circuit is balanced by 240 cm long the potentiometer wire. The emf E of the cell is
A) 2.4V
B) 1.2V
C) 2.0V
D) 3V
Explanation
(b) Total resistance
Or $R = \ 20 + 40$
$\ \ \ \ \ \ \ \ R = 60\Omega$
Given G=15V
$Current\ \ \ \ I = \frac{V}{R} = \frac{15}{60}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = 0.25A$
$Potential\ gradient\ \ = \frac{V}{l}$
$= \frac{20 \times 0.25}{10} = 0.5Vm^{- 1}$
PD across 240 cm
$E = \ 0.5 \times \ 2.4$
$E = \ 1.2V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.