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A series `LCR` circuit with `L=0.125//pi H,C=500//pi nF,R=23Omega` is connected to a `230 V` variable frequency supply. (a)What is the source frequency for which current amplitudes is maximum? Obtain this maximum value. (b)What is the source frequency for which average power absorbed by the circuit is maximum ?Obtain the value of this maximum power.(c )For what reactangle of the circuit, the power transferred to the circuit is half the power at resonance?What is the current amplitude at this reactangle ? (d)if `omega` is the angular frequency at which the power consumed in the circuit is half the power at resonance, write an expression for `omega` (e)What is the `Q`-factor (Quality factor) of the given circuit? |
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Answer» Correct Answer - (a)`2000 Hz,10sqrt2 A` , (b)`2000 Hz, 2300` watt , ( c)`23 Omega , 10 A` , (d)`0.125/pi omega-(1xx10^(9))/(omega(500)/(pi))+-23` , (e)`500//23` Given that `L=0.125//pi , C=500/pixx10` `R=23 , V_(rms)=230` volt (a)`f_(R)=1/(2pi) 1/sqrt(CL)=2000 Hz , I_(rms)(max)=230/R=10` Amp. (b)at Resonance `f_(R)=2000 , P_(max)=I_(rms)^(2) R=(10)xx23=2300` watt ( c)`P_(max)/2=I_(rms)^(2) R , (I_(rms)^(2) R)/ 2 , =I_(rms)^(2) R=I_(rms)=I_(rms)/sqrt2` `V_(rms)/sqrt(R^(2)+x^(2))=I_(rms)/sqrt2=2300/sqrt((23)^(2)+x^(2))=10/sqrt2 x=23 Omega`. `I_(0)=I_(rms)xxsqrt2=10` Amp. (d)`omegaL-1/(omegac)=+23 Omega rArr omega =?` (e)`Q=(omega_(r)L)/R=(2xxpixx2000xx0.125/pi)/23=50/23` |
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