Question

A battery of $V$ voltage is connected across the potentiometer wire $AC$ of total resistance $R_0$ as shown. Calculate the potential across the resistance $R$ if the sliding contact point $B$ is exactly at the middle of the potentiometer wire.

• A. $\frac{2VR}{4R_0+R}$
• B. $\frac{4V{R}_0}{2R_0+R}$
• C. $\frac{2VR}{R_0+4R}$
• D. $\frac{4V{R}_0}{R_0+2R}$

Answer 1

The correct option is A $\frac{2VR}{4R_0+R}$

When the slide is in the middle of the potentiometer only half of its total resistance

Therefore, $\frac{R_0}{2}$ ($R\propto length$) will be between $A$ and $B$ say $R_1$ will be given by

$\frac{1}{R_0}=\frac{1}{R}+\frac{1}{R_0}$

$R_1=\frac{R_{0}R}{R_0+2R}$

Total resistance between $A$ and $C$ will be the sum of the resistance between $A$ & $B$ and $B$ & $C$, given by

$R_1+\frac{R_0}{2}$

The current flowing throughout the potentiometer is

$I=\frac{V}{R_1+\frac{R_0}{2}}=\frac{2V}{2R_1+R_0}$

The voltage $V_1$ taken from potentiometer is given by

$V_1=I{R}_1=\frac{2V}{2\left(\frac{R_{0}R}{R_0+2R}\right)+R_0}\times \left(\frac{R_{0}R}{R_0+2R}\right)$

$V_1=\frac{2VR}{2R+R_0+2R}=\frac{2VR}{R_0+4R}$

Hence, Option $A$ is the correct answer. which is $\frac{2VR}{4R_0+R}$