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InterviewSolution
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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`. |
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Answer» 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. |
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