Question

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

Answer 1

Given: A resistance of $R$ draws the current form of the potentiometer. The potentiometer wire, AB, has a total resistance of $R_0$. A voltage $V$ is supplied to the perpendicular

To derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire.

While the slide is in the middle of the potentiometer only half of its resistance, $\frac{R_0}{2}$ will be between the points A and B. Hence, the total resistance between A and B, say. $R_1$ will be given by the following expression:

$\frac{1}{R_1}=\frac{1}{R}+\frac{1}{\frac{R_0}{2}}⟹R_1=\frac{R{R}_0}{R_0+2R}$

The total resistance between A and C will be sum of the resistance between A and B and B and C, i.e., $R_1+\frac{R_0}{2}$

Therefore, the current flowing 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 the product of current of $I$ 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

$V_1=\left(\frac{2V}{2\frac{R{R}_0}{R_0+2R}+R_0}\right)\times \frac{R{R}_0}{R_0+2R}⟹V_1=\frac{2VR}{2R+R_0+2R}⟹V_1=\frac{2VR}{R_0+4R}$