Question

A resistance of $R\Omega$ draws current from a potentiometer. Potentiometer has a total resistance $R_0\Omega$ as shown in figure. A voltage $V$ is supplied to the potentiometer. Derive an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.

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

Answer 1

Answer: (A)

$\frac{1}{R_1}=\frac{1}{R}+\frac{1}{\left(R_0/2\right)}⟹R_1=\frac{R_{0}R}{R_0+2R}$
The total resistance between $A$ and $C$ will be sum of resistance between $A$ and $B$ and $B$ and $C$, i.e., $R_1+R_0/2$.
So, the current flowing through the potentiometer will be
$I=\frac{V}{R_1+R_0/2}=\frac{2V}{2R_1+R_0}$
The voltage $V_1$ taken from the potentiometer will be the product of current $I$ and resistance $R_1$
$V_1=I{R}_1=\left(\frac{2V}{2R_1+R_0}\right)\times R_1=\frac{2VR}{R_0+4R}$