A potentiometer wire has length of 5 m and resistance of 16 Ω. The driving cell has an e.m.f. of 5 V and an internal resistance of 4 Ω. When the two cells of e.m.f.s 1.3 V and 1.1 V are connected so as to assist each other and then oppose each other, the balancing lengths are respectively (a) 3 m, 0.25 m (b) 0.25 m, 3 m (c) 2.5 m, 0.3 m (d) 0.3 m, 2.5 m

Answer: (a). Wire current I = 5/16+4 = 0.25 A; p.d. across wire = 0.25 × 16 = 4 V. Potential gradient k = 4/5 = 0.8 V/m. Assisting: l₁ = (1.3+1.1)/(0.8) = 3 m. Opposing: l₂ = (1.3-1.1)/(0.8) = 0.25 m. So the answer is (a) 3 m, 0.25 m.

A potentiometer wire has length of 5 m and resistance of 16 Ω. The… | MHT CET 2023 (12 May, Shift 1)

Q46MHT CET 2023 · 12 MAY · SHIFT 1Difficult

Potentiometer

A potentiometer wire has length of $5\text{ m}$ and resistance of $16\ \Omega$ . The driving cell has an e.m.f. of $5\text{ V}$ and an internal resistance of $4\ \Omega$ . When the two cells of e.m.f.s $1.3\text{ V}$ and $1.1\text{ V}$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively

Show answer and solution

Answer: (a)

Wire current $I = \frac{5}{16+4} = 0.25\ \text{A}$ ; p.d. across wire $= 0.25 \times 16 = 4\ \text{V}$ .
Potential gradient $k = \frac{4}{5} = 0.8\ \text{V/m}$ .
Assisting: $l_1 = \frac{1.3+1.1}{0.8} = 3\ \text{m}$ .
Opposing: $l_2 = \frac{1.3-1.1}{0.8} = 0.25\ \text{m}$ .
So the answer is (a) 3 m, 0.25 m.

Report a mistake

Potentiometer question from MHT CET 2023 (12 May, Shift 1)

More questions on Potentiometer