Two moving coil meters `M_1` and `M_2` have the following particulars :
`R_1 = 10 Omega , N_1 = 30`
`A_1 = 3.6 xx 10^(-3) m^2 , B_1 = 0.25 T`
`R_2 = 14Omega , N_2 = 42`
`A_2 = 1.8 xx 10^(-3) m^2 , B_2 = 0.50 T`
(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of `M_2` and `M_1` .

Explanation

When current carrying coil in a galvanometer comes under equilibrium, external torque and restoring torque become equal in magnitude. Hence, `BINAsin90^(@)=kphi`
`(because"Angle "theta" between "vecAandvecB" is "90^(@))`
`BINA=kphi" "...(1)`
(Where k = torsional constant `phi` = angular deflection of coil in the equilibrium condition)
From equation (1),
`phi/I=(BNA)/k" "...(2)`
Here ratio `phi/I` gives amount of deflection in galvanometer per unit current passing through it. It is called current sensitivity, shown by symbol `S_(i)`.
Thus, `S_(i)=phi/I=(BNA)/k" "...(3)`
Here, k is same for both the galvanometer,
And so, `S_(i)propBNA`
`therefore((S_(i))_(2))/((S_(i))_(1))=(B_(2)N_(2)A_(2))/(B_(1)N_(1)A_(1))`
= `((0.5)(42)(1.8xx10^(-3)))/((0.25)(30)(3.6xx10^(-3)))`
= 1.4
Deflection produced in the galvanometer per unit voltage applied across it, is called its voltage sensitivity, shown by symbol `S_(V)`.
Thus,
`S_(V)=phi/V=phi/(IR)=S_(i)/R" "...(4)`
`therefore((S_(V))_(2))/((S_(V))_(1))=((S_(i))_(2))/((S_(i))_(1))xxR_(1)/R_(2)`
= `1.4xx10/14`
= 1