Question

To determine the internal resistance of a cell by using potentiometer, the null point is at 1 m when cell is shunted by 3 Ω resistance and at a length 1.5 m when cell is shunted by 6 Ω resistance. The internal resistance of the cell is ______.

Options

• 6 Ω

• 4 Ω

• 2 Ω

• 3 Ω

Answer 1

To determine the internal resistance of a cell by using potentiometer, the null point is at 1 m when cell is shunted by 3 Ω resistance and at a length 1.5 m when cell is shunted by 6 Ω resistance. The internal resistance of the cell is 6 Ω.

Explanation:

We know r = `"R"((l_1 - l_2)/l_2)`

`therefore "r" 3 ((l_1 - 1)/1) = 3(l_1 - 1)`

`= 6((l_1 - 1.5)/1.5) = 6((2l_1 - 3)/3)`

`therefore 3(l_1 - 1) = (6(2l_1 - 3))/3`

`3l_1 - 3 = 4l_1 - 6`

`l_1 = 3`

∴ r = 3(3 - 1) = 6 Ω