In a potentiometer, a cell is balanced at 240 cm. On shunting the cell with 2 Ω resistance, the balancing comes at 120 cm. What is the internal resistance of the cell ?
Answer & explanation
Correct answer: option 2
Let EMF of the Cell is E and internal resistance r. Potential gradient of the potentiometer wire is $V_0 per cm$.
$ V_0 \times 240 = E$
When cell is shunted by a resistance R, then potential difference across cell is $ V = E - Ir = \frac{ER}{r+R} = V_0\times 120$
Dividing the two Equations $\frac{r+R}{R}=2$
$ r = R = 2\Omega$