A moving coil meter has the following particulars: Numbers of turns, `n=24`, Area of coil, `A=2*0xx10^-3m^2`, magnetic field strength, `B=0*20T`. Resistance of the coil, `R=14Omega`. (a) Indicate a simple way to increase the current sensitivity of the meter by `25%`. (It is not easy to change A or B). (b) If in so doing, the resistance of the coil increases by `7Omega`, is the voltage sensitivity of the modified meter greater or lesser than the original meter?

Answer

(a) n should be increased from 24 to 30 (b) Voltage sensitivity is lesser than original

Explanation

(a) Current sensitivity `I_s=theta/I=(nBA)/(k)`
Since it is easier to change n than k, A or B, therefore the current sensitivity can be increased by increasing n. In order to increase the current sensitivity by 25%, n should be increased by 25%. Therefore new value of n should be
`n^'=n+(25)/(100)n=(125)/(100)n=5/4xx24=30`
(b) New resistance, `R^'=14+7=21Omega`. Voltage
sensitivity `V_s=I_s/R=(nBA)/(kR)=(24BA)/(kxx14)~~(1*7BA)/(k)`
New voltage sensitivity,
`V_s^()'=(n^'BA)/(kR^')=(30BA)/(kxx21)~~1*4(BA)/(k)`
So, `(V_s^()')/(V_s)=(1*4)/(1*7) lt 1` or `V'_s lt V_s`