Explain : "Increasing the current sensitivity may not necessarily increase the voltage sensitivity".

Explanation

1. Current sensitivity is given by,
`phi/I=(NAB)/k" "...(1)`
2. If N `to` 2N that is we double the number of turns, then
`phi/I=(2phi)/I`
3. Thus, the current sensitivity doubles. However the resistance of the galvanometer is likely to double since it is proportional the length of the wire.
4. Voltage sensitivity,
`phi/V=((NAB)/k)(1/V)=((NAB)/k)1/R`
If `Nto2NandRto2R` then, voltage sensitivity does not change.
`phi/Vto phi/V`
5. So, it is not necessary to change in voltage sensitivity by changing current sensitivity.