If a shunt `1//10` of the coil resistance is applied to a moving coil galvanometer, its sensitivity becomes

Options

A. `10` fold
B. `11` fold
C. `(1)/(10)` fold
D. `(1)/(11)` fold

Answer

D

Explanation

`(I_(g))/(I) = (S)/(S + G) = ((G //10))/((G//10) + G) = (1)/(11)`
Initially, `alpha_(1) = theta//I_(g)` (i)
Finally, after the shunt is used,
`alpha_(f) = theta//I`(ii)
`:. (alpha_(f))/(alpha_(I)) = (theta//I)/(theta//I_(g)) = (I_(g))/(I) = (1)/(11)`
So current sensitivity becomes `1//11 - fold`.