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
NEET 2022, Medium
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In series, potential divides in direct ratio of resistance,
So, $V_{A C}=\frac{R_{A C}}{R_{A C}+R_{C B}} V_0$
$=\frac{\frac{ RR _0}{4 R + R _0}}{\frac{ RR _0}{4 R + R _0}+\frac{3 R _0}{4}} \times V _0=\frac{4 RV _0}{16 R +3 R _0}$
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