Question

A resistance of $R$ drawn current from a potentiometer . The potentiometer wire, $AB$ has a total resistance of $R_0$ A voltage V is supplied to the potentiometer . Derive an expansion for the voltage across $R$ when the sliding contact is in the middle of potentiometer wire.

Answer 1

Given, $Resistance=R,Potentiometer=AB,Voltage=V$

When the slides are in the middle of the potentiometer only half of its total resistance i;e $\frac{R_0}{2}$ will between A and B,

So, $\frac{1}{R}=\frac{1}{R}+\frac{1}{\frac{R_0}{2}}\Rightarrow R_1=\frac{R_{0}R}{R_0+2R}$

The total resistance between the A and B is will be the sum of the resistance between A and C or C and B,

$R_1+\frac{R_{)}}{2}$

The current is following through the potentiometer will be:

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

The voltage $V_1$taken from the potentiometer will be product of the current $I$ and the resistance $R_1$,

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

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