Question

When a resistance of 12 Ω is connected across a cell, its terminal potential difference is balanced by 120 cm of a potentiometer wire. When a resistance of 18 Ω is connected across the same cell, the balancing length is 150 cm. Find the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.

Answer 1

Data : Part I : R = 12 Ω, l₂ = 120 cm; 

Part II : R = 18 Ω, l₂ = 150 cm

From Eqs, (1) and (2), we get,

This is the balancing length when the cell is in open circuit.

∴ r = 12 \(\left(\cfrac{I_1-120}{120}\right)\) = \(\cfrac{300-120}{10}\) = \(\cfrac{180}{10}\) = 18 Ω

This is the internal resistance of the cell a unit