Question

In a potentiometer of wire length $l$, a cell of emf V is balanced at a length $\frac{l}{3}$ from the positive of the wire. For another cell of emf $1.5V$, the balancing length becomes

• A. $\frac{l}{6}$
• B. $\frac{l}{2}$
• C. $\frac{l}{3}$
• D. $\frac{2l}{3}$
• E. $\frac{4l}{3}$

Answer 1

The correct option is B $\frac{l}{2}$

$\frac{E_1}{L_1}=\frac{E_2}{L_2}$

$\frac{V}{l/3}=\frac{1.5V}{x}$

$\therefore x=l/2$