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A resistance of 600 Omega an inductor of 0.4 H and a capacitor of 0.01 muFare connected in series to an AC source of variable frequency . Find the frequency of AC source for which current in the circuit is maximum. Also calculate the bandwidth and quality factor for the circuit . |
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Answer» Solution :The current in the circuit is maximum at resonant frequency. At resonant` X_(L) = X_(C) ` `omega_(r)L = (1)/(omega_(r)C)omega_(r) = (1)/(sqrt(LC))` `omega_(r) = (1)/(sqrt(0.4 xx 0.01 xx 10^(-6)))` ` = 1.6 xx 10^(4)` Hz Quality FACTOR `Q = (1)/(R) sqrt((L)/(C)) = (1)/(600) sqrt((0.4)/(0.01 xx 10^(-6)))` ` = 0.01 xx 10^(3)` As `Q = ("Resonant frequency ")/("Bandwidth")` `THEREFORE "Bandwidth " = ("Resonant FREQUNCY" )/(Q ) ` = ` (16 xx 10^(3))/(0.01 xx 10^(3))` = 1600 Hz |
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