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A coil of inductance 0.50 H and resistance 100 Omegais connected to a 240 V, 50 Hz a.c. supply. (a) What is the maximum current in the coil ? (b) What is the time lag between the voltage maximum and the current maximum ? |
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Answer» Solution :Here L = 0.5 H, R `=100 Omega, V_(rms) = 240 V` and v=50 Hz `THEREFORE omega = 2pi v= 2 xx 3.14 xx 50 = 314 s^(-1)`. (a) `therefore` Impedance `Z = sqrt(R^(2) + L^(2) omega^(2)) = sqrt((100)^(2) + (0.5)^(2).(3.14)^(2)) = 186 Omega` `therefore I_(m) = V_(m)/Z = (sqrt(2) V_(rms))/Z = (sqrt(2) xx 240)/186 = 1.82 A` (b) If the current lags behind the voltage in phase by `phi`, then `phi = tan^(-1) (L omega)/R = tan^(-1) ((0.5 xx 314)/100) = tan^(-1) (1.5700) = 57.5^(@) = (57 xx pi )/180` rad `therefore` TIME lag between voltage maximum and current maximum `t= pi/(2pi) , T = phi/(2piV) = (57.5 xx pi)/(180 xx 2pi xx 50) = 3.2 xx 10^(-3)` s or 3.2 rms. |
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