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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 spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity, and (b) voltaage sensitivity of M_2 and M_1. |
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Answer» Solution :(a) We know current sensitivity (C.S.) of a GALVONOMETER `= (phi)/(I) = (NAB)/(k)` `:. ((C.S)_(M_2))/((C.S)_(M_1)) = (N_2 A_2B_2)/(N_1 A_1 B_1) = (42 xx 1.8 xx 10^(-3) xx 0.5)/(30 xx 3.6 xx 10^(-3) xx 0.25) = 1.4 = 1.4 : 1`. (B) VOLTAGE sensitivity (V.S) of galvanometer is `(phi)/(V) = (N A B)/(k) cdot 1/R` `:. ((V.S)_(M_2))/((V.S)_(M_1)) = (N_2 A_2 B_2 R_1)/(N_1 A_1 B_1 R_2) = (42 xx 1.8 xx 10^(-3) xx 0.5 xx 10)/(30 xx 3.6 xx 10^(-3) xx 0.25 xx 14) = 1 = 1 : 1`. |
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