Question

A resistance of R $\Omega$ draws current from a potentiometer as shown in the figure. The potentiometer has a total resistance $R_0\Omega$. 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.

Answer 1

As the slide is in middle of the potentiometer only half of its resistance $\left(R_0/2\right)$ will be in parallel with the resistance R. Hence, the total resistance $\left(R_1\right)$ will be given by the following expression:

$\frac{1}{R_1}=\frac{1}{R}+\frac{1}{\left(R_0/2\right)}$

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

The total resistance between $A$ and $C$ is $R_1+R_0/2$.

$\therefore$ 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 $R_1$.

$V_1=I{R}_1=\left(\frac{2V}{2R_1+R_0}\right)\times R_1$

Substituting for $R_1$, we have

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

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

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