A moving coil meter has the following particulars. Number of turns, `N=20`, Area of the coil, `A=2*0xx10^-3m^2`, Magnetic field strength, `B=0*20T`, Resistance of the coil, `R=12Omega`.
(a) Indicate a simple way to increase the current sensitivity of the meter by 20% (It is not easy to change A or B).
(b) If in so doing, the resistance of the coil increases to `18Omega`, is the voltage sensitivity of the modified meter greater or lesser than the original meter?

Answer

(a) Increasing the turns by `20%`
(b) Lesser

Explanation

(i) Current sensitivity, `I_s=theta/I=(nBA)/(k)`
As, it it not easy to charge A or B, hence sensitivity can be increased either by increasing n or by decreasing k. As it is easier to change n, than k hence current sensitivity can be increased by increasing n, `I_s` can be increased by 20% if n is increased by 20% i.e. by increasing n from 20 to 25.
(ii) Voltage sensitivity, `V_s=theta/V=(nBA)/(kR)`
Voltage sensitivity of modified meter will be
`V_s^()'=(n^'BA)/(kR^')`
Hence `(V_s^()')/(V_s)=(n^')/(R^')R/n=25/18xx12/20=5/6=0*83lt1`
or `V'_s lt V_s`
Therefore, the voltage sensitivity of the modified meter is lesser than the original meter.