Question

A galvanometer of resistance 100 Ω is connected to a battery of 2V, with a resistance of 1900 Ω in series. The deflection obtained is 30 divisions. To reduce this deflection by 10 divisions, the additional resistance required to be connected in series is ______ 

Options

• 1500 Ω

• 500 Ω

• 2000 Ω

• 1000 Ω

Answer 1

A galvanometer of resistance 100 Ω is connected to a battery of 2V, with a resistance of 1900 Ω in series. The deflection obtained is 30 divisions. To reduce this deflection by 10 divisions, the additional resistance required to be connected in series is 1000 Ω.

Explanation:

Total resistance in the circuit is R = 1900 + 100 = 2000 Ω

∴ Current `I_1 = 2/2000 = 10^-3 A`

The deflection is decreased from 30 to 20 divisions

∴ `I_2 = 2/3I_1 = 2/3 xx 10^-3 A`

If total resistance now is R' then

`R^' = V/I_2 = (2 xx 3)/(2 xx 10^-3) = 3 xx 10^3  Omega = 3000  Omega`

∴ Additional resistance = R' - R = 3000 - 2000 = 1000 Ω