$A$ resistance of $R \; \Omega$ draws current from a potentiometer. The potentiometer has a total resistance $R_{0} \; \Omega$ (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.

(N/A) When the sliding contact $B$ is in the middle of the potentiometer,the resistance $R_{0}$ is divided into two equal parts,each of $R_{0}/2$. The resistance $R$ is connected in parallel with the lower half of the potentiometer (between points $A$ and $B$).
The equivalent resistance $R_{1}$ between points $A$ and $B$ is given by:
$\frac{1}{R_{1}} = \frac{1}{R} + \frac{1}{R_{0}/2} = \frac{1}{R} + \frac{2}{R_{0}} = \frac{R_{0} + 2R}{R \cdot R_{0}}$
$R_{1} = \frac{R \cdot R_{0}}{R_{0} + 2R}$
The total resistance of the circuit between points $A$ and $C$ is the sum of the equivalent resistance $R_{1}$ and the remaining part of the potentiometer resistance $(R_{0}/2)$:
$R_{total} = R_{1} + \frac{R_{0}}{2}$
The total current $I$ flowing from the source $V$ is:
$I = \frac{V}{R_{total}} = \frac{V}{R_{1} + R_{0}/2} = \frac{2V}{2R_{1} + R_{0}}$
The voltage $V_{1}$ across the resistance $R$ is the same as the voltage across the parallel combination $R_{1}$:
$V_{1} = I \cdot R_{1} = \left( \frac{2V}{2R_{1} + R_{0}} \right) \cdot R_{1}$
Substituting $R_{1} = \frac{R \cdot R_{0}}{R_{0} + 2R}$ into the expression for $V_{1}$:
$V_{1} = \frac{2V \cdot \left( \frac{R \cdot R_{0}}{R_{0} + 2R} \right)}{2 \left( \frac{R \cdot R_{0}}{R_{0} + 2R} \right) + R_{0}} = \frac{2V \cdot R \cdot R_{0}}{2R \cdot R_{0} + R_{0}(R_{0} + 2R)} = \frac{2V \cdot R \cdot R_{0}}{2R \cdot R_{0} + R_{0}^{2} + 2R \cdot R_{0}} = \frac{2V \cdot R \cdot R_{0}}{R_{0}^{2} + 4R \cdot R_{0}}$
Dividing numerator and denominator by $R_{0}$:
$V_{1} = \frac{2VR}{R_{0} + 4R}$