Two moving coil metres `M_(1) and M_(2)` have the following particular `R_(1) =10Omega, 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 metres. What is the ratio of current sensitivity and voltage sensitivity of `M_(2)" to" M_(1)?`

Answer

(a ) 5/7 (b) 1:1

Explanation

(a) Current sensitivity, `(phi)/(I) = (NBA)/(K)`
Ratio of current Sensitivity `= ((N_1 B_1 A_1)/(K))//((N_2 B_2 A_2)/(K))`
`= (30 xx 0.25 xx 3.6 xx 10^(-3))/(42 xx 0.50 xx 1.8 xx 10^(-3))= 5//7`
(b) Voltage sensitivity, `(phi)/(V)= (NBA)/(k R)`
Ratio of voltage sensitivity = `((N_1 B_1 A_1)/(kR_1))//((N_2 B_2 A_2)/(kR_2))`
`= (30xx 0.25 xx 3.6 xx 10^(-3) x 14)/(42 xx 0.50 xx 1.8 xx 10^(-3) xx 10)=1`