Question

A potentiometer is used to measure the potential difference between A and B, the null point is obtained at 0.9 m. Now the potential difference between A and C is measured, the null point is obtained at 0.3 m. The ratio `E_2/E_1` is (E₁ > E₂) ______

 

Options

• 3 : 1

• 3 : 2

• 2 : 3

• 1 : 3

Answer 1

A potentiometer is used to measure the potential difference between A and B, the null point is obtained at 0.9 m. Now the potential difference between A and C is measured, the null point is obtained at 0.3 m. The ratio `E_2/E_1` is (E₁ > E₂) 2 : 3.

Explanation:

`E_{ab}/E_{ac} = L_{ab}/L_{ac} = E_1/(E_1 - E_2) = 0.9/0.3 = 3`

`E_1 = 3E_1 - 3E_2`

`2E_1 = 3E_2`

`E_1/E_2 = 3/2`

∴ `E_2/E_1 = 2/3`