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A coil of inductance 0.50 H and resistance 100 Omega is connected to a 240 V, 50 Hz AC supply. (a) What is the maximum current in the coil ? (b) what is the time lag between the voltage maximum and current maximum ? |
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Answer» SOLUTION : (a) For LR circuit inductance, L = 0.50 H Resistance, R = 100 `Omega` RMS value of voltage , `E_(rms ) = 240 `V Freqeuency, f = 50 Hz angular frequency,` omega = 2 pi` f `I_(0) = (E_(0))/(sqrt(R^(2) + L^(2) omega^(2)))= (sqrt(2) xx E_(rms))/(sqrt(R^(2) + L^(2) (2 pi v)^(2)))` = ` (sqrt(2) xx 240)/(sqrt(10^(4) + (0.5)^(2) xx 4 pi^(2) xx 2500)) ` = 1.82 A (b) RL circuit, instantaneous voltage is ` E = E_(0) cos omega t` Instantaneous current is `I - I_(0) cos (omega t - phi) ` at t = 0 `E = E_(0) ` (maximum voltage ) Att ` = (phi)/(omega)` I = `I_(0)` (maximum current ) THEREFORE, time lag BETWEN maximum voltage and maximum current instant is `(phi)/(omega)` Now tan `phi = (omega L)/(R)` = `(2pi xx 50 xx 0.5)/(100) `1.571 `or"" phi = tan^(-1) ( 1.571) = 57.5 ` time lag = `(phi)/(omega) = (57.5 pi)/(180 xx 2 pi xx 50) = 3.2 ` ms |
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