The current sensitivity of a moving coil galvanometer increases by `20%` when its resistance is increased by a factor 2. Calculate by what factor does the voltage sensitivity change?

Explanation

Current sensitivity `(alpha)/I=(nBA)/k`……….i
Voltage sensitivity `(alpha)/V=(nBA)/(kR)`………ii
Resistance of a galvanometer increase when n and A are changed
Given `R'=2R`
Then `n=n'` and `A=A'`
New current sensitivity
`(alpha')/(I')=(n'A'B)/k`.....iii
New voltage sensitivity
`(alpha')/V=(alpha')/(I'R')=(n'A'B)/(2kR)`.....iv
Since `(alpha')/(I')=120/100 (alpha)/I`........v
From i and iii `,(n'A'B)/R=(n'A'B)/k=(nAB)/k 120/100,n'A'=6/5nA`
Using equation (iv)
`(alpha')/V=6/5(nAB)/(2kR), (alpha')/V=(3nAB)/(5kR), (alpha')/V=3/5(alpha)/V`
Thus voltage sensitivity decreases by a factor of `3/5`