How can we convert a galvanometer into an ammeter ?

Explanation

Let there be a galvanometer with a coil resistance G. Let it also shows a full deflection for current `I_(g)`. When a small resistance S known as shunt is connected in parallel to the coil then it can be used as an ammeter to measure current up to I, where, `I = ((S + G)/(S))I_(g)`. The shunt reduces overall resistance of an ammeter, which is always desirable. It also increases the maximum magnitude of current which can be measured.