What is the value of the shunt that passes `10%` of the main current through a galvenomenter of `99 Omega`?

Explanation

As in figure `R_(g)I_(g)=(I-I_(g))S``rArr99xx(I)/(10)=(I-(I)/(10))xxS`
`rArrS=11Omega`
For calculation it is simply a resistance

Resistance of ammeter
`R_(A)=(R_(G)-S)/(R_(G)+S)`

for `SltltR_(G)rArrR_(A)=S`