Q. In a potentiometer arrangement, a cell of emf 1.5V gives a balance point at 27cm length of wire. If the cell is replaced by another cell and balance point shifts to 54cm, the emf of the second cell is
A) 3V
B) 1.5V
C) 0.75V
D) 2.25V
Explanation
(a) Given, $E_{1} = 1.5V,\ l_{1} = 27\ cm,\ $
$l_{2} = 54\ cm,\ E_{2} = ?$
$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$
$or\ \ E_{2} = \frac{E_{1}l_{2}}{l_{1}}$
$or\ \ E_{2} = \frac{1.5 \times 54}{27}$
$\ \ \ \ \ \ E_{2} = 3V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.