Let the total resistance of the potentiometer wire be $R_0$. Since the contact $C$ is at one-fourth the length from $A$, the resistance of section $AC$ is $R_{AC} = R_0/4$ and the resistance of section $CB$ is $R_{CB} = 3R_0/4$ . The resistor $R$ is connected in parallel with section $AC$. The equivalent resistance ($R_p$) of this parallel combination is calculated as: $R_p = \frac{R \cdot (R_0/4)}{R + R_0/4} = \frac{RR_0}{4R + R_0}$ .

The total resistance of the circuit is the sum of $R_p$ and the series resistance $R_{CB}$: $R_{eq} = R_p + R_{CB} = \frac{RR_0}{4R + R_0} + \frac{3R_0}{4}$.

The potential drop across $R$ is the same as the potential drop across the parallel section $AC$. Using the voltage divider rule (derived from Ohm's Law ): $V = V_0 \times \frac{R_p}{R_{eq}} = V_0 \times \frac{\frac{RR_0}{4R + R_0}}{\frac{RR_0}{4R + R_0} + \frac{3R_0}{4}}$.

Simplifying the expression by clearing the fractions: $V = V_0 \frac{4R}{4R + 3(4R + R_0)} = V_0 \frac{4R}{4R + 12R + 3R_0} = \frac{4V_0R}{16R + 3R_0}$.