To determine the internal resistance of a cell by using a potentiometer, the null point is at 1 m when shunted by 3 Ω resistance and at a length 1.5 m, when cell is shunted by 6 Ω resistance The internal resistance of the cell is (a) 1 Ω (b) 4 Ω (c) 2 Ω (d) 6 Ω

Answer: (d). l = l₀R/R+r: 1 = l₀3/3+r, 1.5 = l₀6/6+r. Dividing: 1.5 = (2(3+r))/(6+r) ⇒ 9+1.5r = 6+2r r = 6 Ω So the answer is (d) 6 Ω.

To determine the internal resistance of a cell by using a… | MHT CET 2021 (23 Sep, Shift 2)

Q59MHT CET 2021 · 23 SEP · SHIFT 2Medium

Potentiometer

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1\text{ m}$ when shunted by $3\ \Omega$ resistance and at a length $1.5\text{ m}$ , when cell is shunted by $6\ \Omega$ resistance The internal resistance of the cell is

Show answer and solution

Answer: (d)

$l = l_0\dfrac{R}{R+r}$ : $1 = l_0\dfrac{3}{3+r}$ , $1.5 = l_0\dfrac{6}{6+r}$ .
Dividing: $1.5 = \dfrac{2(3+r)}{6+r} \Rightarrow 9+1.5r = 6+2r$
$r = 6\ \Omega$
So the answer is (d) $6\ \Omega$ .

Report a mistake

Potentiometer question from MHT CET 2021 (23 Sep, Shift 2)

More questions on Potentiometer