devarshi-dt-logo

Question:

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of (i) infinity (ii) 9.5Ω, the balancing lengths, on the potentiometer wire are found to be 3m and 2.85 m, respectively. The value of internal resistance of the cell is?

0.95Ω

0.75Ω

0.25Ω

0.5Ω

Solution:

Let l1 be the balancing length when R = ∞, and l2 be the balancing length when R = 9.5Ω.
Given:
l1 = 3m
l2 = 2.85m
R = 9.5Ω
The internal resistance of the cell, r, is given by the formula:
r = (l1 - l2)/l2 * R
r = (3 - 2.85)/2.85 * 9.5
r = 0.15/2.85 * 9.5
r = 0.5Ω