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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) =14 omega , N_2 = 42 A_(2) = 1.8 xx 10^(-3) m^2, B_(2) = 0.50 T (The spering constants are identical for the two meters ) Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_2 "and "M_1 |
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Answer» Solution :Current sensitivity , `(theta )/(I) = (BAN)/(C)` Ratio of SECOND to first is`((B_2A_2N_2)/(C_2))/((B_1A_1N_1)/(C_1))= (0.5 xx 1.8 xx 10^(-3) xx 42)/(0.25 xx 3.6 xx 10^3)=7/5 =1.4` b. Voltage sensitivity`(theta)/(V) = (BAN)/(C) xx 1/R` `1.4 xx (R_1)/(R_2) = 1.4 xx (10)/(14) =1` |
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