RG = 60.00Ω, shunt resistance, rs = 0.02Ω Total resistance in the circuit is RG + 3 = 63Ω Hence, I = 3/63 = 0.048 A Resistance of the galvanometer converted to an ammeter is,
RGrs
RG+rs
=
60Ω×0.02Ω
(60+0.02)Ω
= 0.02 Ω Total resistance in the circuit = 0.02 + 3 = 3.02Ω Hence, I = 3/3.02 = 0.99 A