Question

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 $A B$ is $R_0$, then the potential drop $(V)$ across the resistor $R$ is:

• A. $\frac{4 V _0 R }{3 R _0+16 R }$
• B. $\frac{4 V_0 R}{3 R_0+R}$
• C. $\frac{2 V _0 R }{4 R _0+ R }$
• D. $\frac{2 V_0 R}{2 R_0+3 R}$

Answer 1

Answer: (A)

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