Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp = 1 Ω as shown in the figure. An external resistance of Re = 2 Ω is connected via the sliding contact. The electric current in the circuit is: (a) 1.35 A (b) 0.9 A (c) 1.0 A (d) 0.3 A

Answer: (c). Each half of the potentiometer is 0.5 Ω; Re is in parallel with the left half. 0.5 ∥ 2 = (0.5 × 2)/(2.5) = 0.4 Ω Total R = 0.4 + 0.5 = 0.9 Ω I = (0.9)/(0.9) = 1.0 A So the answer is (c) 1.0 A.

Sliding contact of a potentiometer is in the middle of the… | JEE Main 2025 (22 Jan, Shift 1)

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Kirchhoff's Laws & Measuring Instruments

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_p = 1\ \Omega$ as shown in the figure. An external resistance of $R_e = 2\ \Omega$ is connected via the sliding contact. The electric current in the circuit is:

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Answer: (c)

Each half of the potentiometer is $0.5\ \Omega$ ; $R_e$ is in parallel with the left half.
$0.5 \parallel 2 = \frac{0.5 \times 2}{2.5} = 0.4\ \Omega$
Total $R = 0.4 + 0.5 = 0.9\ \Omega$
$I = \frac{0.9}{0.9} = 1.0\text{ A}$
So the answer is (c) 1.0 A.

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