Question

Two moving coil meters `M_1` and `M_2` have the following particulars:
`R_1=10Omega`, `N_1=30`, `A_1=3*6xx10^-3m^2`, `B_1=0*25T`,
`R_2=14Omega`, `N_2=42`, `A_2=1*8xx10^-3m^2`, `B_2=0*50T`
(The spring constants are identical for the two metres). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of `M_2` and `M_1`.

Answer 1

For moving coil meter `M_(1)`:
Resistance, `R_(1) = 10 Omega`
Number of turns, `N_(1) = 30`
Area of cross-section, `A_(1) = 3.6 xx 10^(-3)m^(2)`
Magnetic field strength, `B_(1) = 0.25T`
Spring constant `K_(1) = K`
For moving coil meter `M_(2)`:
Resistance, `R_(2) = 14 Omega`
Number of turns, `N_(2) = 42`
Area of cross-section, `A_(2) = 1.8 xx 10^(-3)m^(2)`
Magnetic field strength, `B_(2) = 0.50T`
Spring constant, `K_(2) = K`
(a) Current sensitivity of `M_(1)` is given as:
`I_(s1) = (N_(1)B_(1)A_(1))/(K_(1))`
And, current sensitivity of `M_(2)` is gives as:
`I_(s2) = (N_(2)B_(2)A_(2))/(K_(2))`
`(I_(s2))/(I_(s1)) = (N_(2)B_(2)A_(2)K_(1))/(K_(2)N_(1)B_(1)A_(1))`
`:.` Ratio
`=(42 xx 0.5 xx 1.8 xx 10^(-3)xxK)/(K xx 30 xx 0.25 xx 3.6 xx 10^(-3)) = 1.4`
Hence, the ratio of current sensitivity of `M_(2)` to `M_(1)` is 1.4.
(b) Voltage senstivity for `M_(2)` is given as:
`V_(s2) = (N_(2)B_(2)A_(2))/(K_(2)R_(2))`
And, voltage sensitivity for `M_(1)` is given as:
`V_(s1) = (N_(1)B_(1)A_(1))/(K_(1))`
`(V_(s2))/(V_(s1)) = (N_(2)B_(2)A_(2)K_(1)R_(1))/(K_(2)R_(2)N_(1)B_(1)A_(1))`
`:.` Ratio
`= (42 xx 0.5 xx 1.8 xx 10^(-3)xx 10 xx K)/(Kxx 14 xx 30 xx 0.25 xx 3.6 xx 10^(-3)) =1`
Hence, the ratio of voltage sensitivity of `M_(2)` to `M_(2)` to 1.