Question

The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance of 10 Ω, the balancing length is reduced by 20 cm. Find the internal resistance of the cell.

Answer 1

Data :R = 10 Ω, l₁ = 120 cm, l₂ = 120 – 20 = 100 cm
r = R(\(\frac{l_{1}-l_{2}}{l_{2}}\))
= 10 (\(\frac{120-100}{100}\))
= 2 Ω
The internal resistance of the cell is 2 Ω.