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The sliding contact C is at one fourth of the length of the potentiometer wire (AB) from A as shown in the circuit diagram. If the resistance of the wire AB is $R_0$, then the potential drop (V) across the resistor R is
Detailed Solution
Portion AC has resistance $\frac{R_0}{4}$, in parallel with R; portion CB ($\frac{3R_0}{4}$) is in series. $R_{eq}=\frac{\frac{R_0}{4}R}{\frac{R_0}{4}+R}+\frac{3R_0}{4}=\frac{R_0(16R+3R_0)}{4(R_0+4R)}$. With $I=\frac{V_0}{R_{eq}}$, $V_R=I\times\frac{\frac{R_0}{4}R}{\frac{R_0}{4}+R}=\frac{4V_0R}{16R+3R_0}$.