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NEET PHYSICSCURRENT ELECTRICITYHard

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

A

4V₀R / (3R₀ + 16R)

B

4V₀R / (3R₀ + R)

C

2V₀R / (4R₀ + R)

D

2V₀R / (2R₀ + 3R)

Step-by-Step Solution

Let the total resistance of the potentiometer wire be R0R_0. Since the contact CC is at one-fourth the length from AA, the resistance of section ACAC is RAC=R0/4R_{AC} = R_0/4 and the resistance of section CBCB is RCB=3R0/4R_{CB} = 3R_0/4 . The resistor RR is connected in parallel with section ACAC. The equivalent resistance (RpR_p) of this parallel combination is calculated as: Rp=R(R0/4)R+R0/4=RR04R+R0R_p = \frac{R \cdot (R_0/4)}{R + R_0/4} = \frac{RR_0}{4R + R_0} .

The total resistance of the circuit is the sum of RpR_p and the series resistance RCBR_{CB}: Req=Rp+RCB=RR04R+R0+3R04R_{eq} = R_p + R_{CB} = \frac{RR_0}{4R + R_0} + \frac{3R_0}{4}.

The potential drop across RR is the same as the potential drop across the parallel section ACAC. Using the voltage divider rule (derived from Ohm's Law ): V=V0×RpReq=V0×RR04R+R0RR04R+R0+3R04V = V_0 \times \frac{R_p}{R_{eq}} = V_0 \times \frac{\frac{RR_0}{4R + R_0}}{\frac{RR_0}{4R + R_0} + \frac{3R_0}{4}}.

Simplifying the expression by clearing the fractions: V=V04R4R+3(4R+R0)=V04R4R+12R+3R0=4V0R16R+3R0V = V_0 \frac{4R}{4R + 3(4R + R_0)} = V_0 \frac{4R}{4R + 12R + 3R_0} = \frac{4V_0R}{16R + 3R_0}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from CURRENT ELECTRICITY. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSCURRENT ELECTRICITYslidingcontactfourthlengthpotentiometer

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