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A series LCR circuit is connected to a 220 V variable frequency supply. If L = 20 mH, C = [(800)/(pi^(2))]muF and R = 110 Omega, (a) Find the frequency of the source for which the average power absorbed by the circuit is maximum. (b) Calculate the value of maximum current amplitude. |
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Answer» Solution :Given, `L = 20 mH = 20 XX 10^(-3)H` `C = (800)/(pi^(2)) = muF = (800)/(pi^(2)) xx 10^(-6)F` `R = 110 Omega` (a) `v = (1)/(2pi sqrt(LC))` `= (1)/(2pi sqrt(20 xx 10^(-3) xx (800 xx 10^(-6)))/(pi^(2)))` `= (1)/(2sqrt(16 xx 10^(-6))) = (1)/(8 xx 10^(-3))` `= (1000)/(8) = 125 Hz` (b) `I_(max) = (E)/(Z_(min)) = (E )/(R )` `= (220)/(110) = 2A` |
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