Practicing Success
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 ? |
4 Ω 2 Ω 1 Ω 0.5 Ω |
2 Ω |
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$ |