Range of an ammeter is increases by n times. What is the change in its resistance ?

Explanation

The shunt that is requied to current a galvanometer of range `I_(G)` and resistance G into an ammeter of range In is,
`S=(I_(G))/(I-I_(G))*G=(1)/((I)/(I_(G))-1)*G=(G)/(n-1) ["given" (I)/(I_(G))=n]`
Now, if R be the equivalent resistance of the parallel combination of G and S, then,
`(1)/(R)=(1)/(G)+(1)/(S)=(1)/(G)+(n-1)/(G)=(n)/(G)`
`:. R=(G)/(n)`
So, if range of the ammeter is n times, then its reistance decreases from G to `(G)/(n)`.