A galvanometer of resistance 100 ohm gives full deflection for a current 10^-5 A. The value of shunt requiredmto convert it into an ammeter of range 1 A is:-

Q20. A galvanometer of resistance 100$\mathrm{\Omega }$ gives full deffection for a current 10^–5 A. The value of shunt required to convert it into a ammeter of range
1 ampere, is:-
(1) 1 Ω
(2) 10⁻³ Ω 
(3) 1⁻⁵ Ω
(4) 100 Ω

Galvanometer is a very sensitive instrument therefore it cannot measure heavy currents. In order to convert a Galvanometer into an Ammeter, a very low shunt resistance is connected in parallel to Galvanometer. Value of shunt is so adjusted that most of the current passes through the shunt.

If resistance of galvanometer is R_g and it gives full-scale deflection when current I_g is passed through it. Then,

V = I_gR_g

V = 10^-5 A × 100 Ω = 10^-3 V

Let a shunt of resistance (R_s) is connected in parallel to galvanometer. If total current through the circuit is I.

I = 1A = I_s + I_g

V = I_sR_s = (I - I_g)R_s

(I - I_g)R_s = I_gR_g

(1 - 10^-5)R_s = 10^-3 

R_s ≈ 0.001 Ω =  10^-3 Ω

Regards