Question

In the adjoining potentiometer circuit, the resistance of uniform cross section the potentiometer wire AB of Length 1 m is 10$\Omega$. When the variable resistance R = 10 $\Omega$, the balance point is obtained for length l as shown. If the variable resistance is doubled, the new balance length is

• A. $l$
• B. $1.5l$
• C. $2l$
• D. $\frac{l}{2}$

Answer 1

Answer: (B)

$1.5l$

Let E be the voltage of main source. Current through the circuit in both the conditions is,

$I_1=\frac{E}{R+R}$ and

$I_2=\frac{E}{2R+R}$

In both the conditions the voltage across points A and J will remains constant.

$V_{AJ}=\frac{E}{R+R}\times R_{l_1}=\frac{E}{2R+R}\cdot R_{l_2}$

$\frac{R_{l_1}}{2R}=\frac{R_{l_2}}{3R}$

$\therefore \frac{R_{l_2}}{R_{l_1}}=\frac{3}{2}$