InterviewSolution
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InterviewSolution
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Figure here, shows a series L-C-R circuit connected to a variable frequency 230 V source. L = 5.0H, C = `80 muF` and r =`40 Omega` (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. show that the potential drop across the L-C combination is zero at the resonating frequency. |
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Answer» a) If `omega_(0)` be the resonant angular frequency = source frequency at resonance, then `omega_(0) = 1/(sqrt(LC)` = `1/sqrt(5 xx 80 xx 10^(-6))= 50 rads^(-1)` `f_(0) = omega_(0)/(2pi) = (50)/(2 xx 3.141) = 7.96 Hz` b) At resonance, Z=R = `40Omega` `I_(rms) = V_(rms)/Z= 230/40 = 5.75 A` and `I_(0) = V_(0)/Z = (sqrt(2)V_(rms))/Z` `(sqrt(2) xx 230)/(40) = 8.13 A` c) Tjhe rms potential drop across R is given by `V_(rms) = I_(rms) R= 5.75 xx 40 = 230 V` The rms potential drop across L is given by `V_(Lrms) = I_(rms) X_(L) = I_(rms)(omega_(0)L)` `=5.75 xx 50 xx 5 = 1437.5 V` The rms potential drop across C is given by `V_(rms) = I_(rms)X_(C)=I_(rms) xx 1/(omega_(0)C) = 5.75 xx 1/(50 xx 80 xx 10^(-6))` =1437.5 V The rms potential drop across L-C is given by `V_(LC) = V_(L) - V_(C) = 1437.5 - 1437.5=0` |
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