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. $\frac{2 V R}{R_{0}+4 R}$
• B. $\frac{2 V R}{R_{0}}$
• C. $\frac{4 V R}{R_{0}+2 R}$
• D. $\frac{V R}{R_{0}+2 R}$

Answer 1

Answer: (A)

$\frac{1}{R_{1}}=\frac{1}{R}+\frac{1}{\left(R_{0} / 2\right)} \Longrightarrow R_{1}=\frac{R_{0} R}{R_{0}+2 R}$
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
$I=\frac{V}{R_{1}+R_{0} / 2}=\frac{2 V}{2 R_{1}+R_{0}}$
The voltage $V_{1}$ taken from the potentiometer will be the product of current $I$ and resistance $R_{1}$
$V_{1}=I R_{1}=\left(\frac{2 V}{2 R_{1}+R_{0}}\right) \times R_{1}=\frac{2 V R}{R_{0}+4 R}$