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 R0, then the potential drop (V) across the resistor R is:

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 R0, then the potential drop (V) across the resistor R is: (A) (4 V 0 R /3 R 0+16 R ) (B) (4 V0 R/3 R0+R) (C) (2 V 0 R /4 R 0+ R ) (D) (2 V0 R/2 R0+3 R)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9bbad9d8c0bab7e1513cd1089dddb6ef-.png" /> In series, potential divides in direct ratio of resistance, So, VA C=(RA C/RA C+RC B) V0 =(( RR 0/4 R + R 0)/( RR 0)4 R + R 0+(3 R 0)4 × V 0=(4 RV 0/16 R +3 R 0)

Q. The sliding contact $C$ is at one fourth of the length of the potentiometer wire $(A B)$ 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:

$\frac{4 V _{0} R}{3 R _{0} + 16 R}$

$\frac{4 V _{0} R}{3 R _{0} + R}$

$\frac{2 V _{0} R}{4 R _{0} + R}$

$\frac{2 V _{0} R}{2 R _{0} + 3 R}$

In series, potential divides in direct ratio of resistance,
So, $V_{A C} = \frac{R _{A C}}{R _{A C} + R _{CB}} V_{0}$
$= \frac{\frac{R R _{0}}{4 R + R _{0}}}{\frac{R R _{0}}{4 R + R _{0}} + \frac{3 R _{0}}{4}} \times V_{0} = \frac{4 R V _{0}}{16 R + 3 R _{0}}$

Q. 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 R0​, then the potential drop (V) across the resistor R is:

3R0​+16R4V0​R​

3R0​+R4V0​R​

4R0​+R2V0​R​

2R0​+3R2V0​R​