Initially current in the circuit is
I=505=0.1
and the current in circuit when ammeter is connected,is 1 of initial current , i.e., I′=1001×0⋅1=0.099A
To increase the range of galvanometer, let resistance rs
is connected parallel to it and along with 50Ω is connected in series to ammeter. So, equivalent resistance of circuit is:
Req=50+100+rS100rS
Now, From equation V=I′Req
5=0.099×(50+100+rs100rs)⇒50+100+rs100rs=0.0995⇒50+100+rs100rs=50.5050⇒100+rs100rs=50.5050⇒rs=0.5Ω