Question

The internal resistance of a cell is determined by using a potentiometer. In an experiment, an internal resistance of 100 Ω is used across the given cell. When the key K₂ is closed, the balance length on the potentiometer decreases from 90 cm to 72 cm. Calculate the internal resistance of the cell -

• A. 100 Ω
• B. 75 Ω
• C. 50 Ω
• D. 25 Ω

Answer 1

Answer: (D)

25 Ω

Answer 2

r = R $\left( \frac { \ell _ { 1 } } { \ell _ { 2 } } - 1 \right)$= 100 $\left( \frac { 90 } { 72 } - 1 \right)$ r = 25 Ω