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A coil of inductance 0.50 H and resistance 100 ohm is connected to 240 V, 50 Hz a.c. supply. (a) What is the peak current in the coil ? (b) What is the time lag between the peak voltage and the peak current ? |
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Answer» Solution :Here `L = 0.50 H, R = 100` ohm, `E_(rms) = 240 V, v = 50 Hz` Step I : We know `E_(0) = sqrt(2) E_(rms) = 1.414 XX 240 = 339.4 V` Impedance of LR circuit is `Z_(L) = sqrt(R^(2) + omega^(2)L^(2)) = sqrt(R^(2) + 4PI^(2)v^(2)L^(2))` `= sqrt((100)^(2) + 4 xx (3.14)^(2) xx (50)^(2) xx (0.50)^(2)) = 186.1` ohm Peak current, `I_(0) = (E_(0))/(Z_(L)) = (339.4)/(186.1) = 1.82 A` In LR circuit the phase difference between current and voltage `phi` is given by `tan phi = (0.50 xx2 xx 3.14 xx 50)/(100) = 1.57` `phi = tan^(-1)(1.57) = 57^(@)30. = 57.5^(@)` `= (57.5 xx pi)/(180) = 0.3194 pi` radians Here `E = E_(0) sin omega t , E_(0) = E_(rms) sqrt(2)` `therefore` Time LAG between `E_(0)` and `I_(0)` is given by `t = (phi)/(omega) = (phi)/(2pi v) = (0.3194 pi)/(2pi xx 50)` `= 0.003194 s = 3.194 xx 10^(-3)s`. |
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