InterviewSolution
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InterviewSolution
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An inductance of `(200)/(pi) mH` a capacitance of `(10^(-3))/(pi)` and a resistance of `10 Omega` are connected in series with an `AC` source of `220 V, 50Hz`. The phase angle of the circuit is |
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Answer» Here, `L = (200)/(pi) mH = (200 xx 10^(-3))/(pi) H = (0.2)/(pi) H` `C = (10^(-3))/(pi) F, R = 10 Omega , E_(v) = 220 V, n = 50 Hz` `X_(L) = omega L = 2 pi n L = 2 pi xx 50 xx (0.2)/(pi) = 20 Omega` `X_(C )= (1)/(omega C) = (1)/(2 pi n C) = (pi)/(2 pi xx 50 xx 10^(-3)) = 10 Omega` `tan phi = ((X_(L) - X_(C )))/(R ) = (20 - 10)/(10) = 1 , phi = (pi)/(4)` |
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