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A series LCR circuit is made by taking `R = 100 Omega, L = (2)/(pi)` and `C = (100)/(pi) mu F`. The series combination is connected across an a.c. source of 220 V, 50 Hz. Calculate impedance of the circuit and peak value of current flowing n the circuit. What is power factor of the circuit ? Compare it with one at resonance frequency. |
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Answer» Here, `R = 100 Omega`, `L = (2)/(pi) H, C = (100)/(pi) xx 10^(-6) F`, `E_(v) = 220 V, v = 50 Hz, Z = ? I_(0) = ? Cos phi = ?` `X_(L) = omega L = 2 pi L = 2 pi xx 50 xx (2)/(pi) = 200 Omega` `X_(C ) = (1)/(omega C) = (1)/(2 pi v xx (100)/(pi) xx 10^(-6)) = 100 Omega` `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` `= sqrt(100^(2) + (200 - 100)^(2)) = 141.4 Omega` `I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(Z) = (sqrt2 xx 200)/(141.4) = 2.2 A` Power factor `= cos phi = (R )/(Z) = (100)/(100 sqrt2) = (1)/(sqrt 2)` At resonance, `Z = R , cos phi = (R )/(R ) = 1` |
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