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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)? |
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Answer» 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` |
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