InterviewSolution
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InterviewSolution
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A series LCR circuit containing a resistance of `120 Omega` has angular resonance frequency `4 xx 10^(5) rad s^(-1)`. At resonance the vlotage across resistance and inductance are 60V and 40 V, repectively, At what frequency, the current in the circuit lags the voltage bu `45^(@)` ?A. `16xx10^(5)"rad "s^(-1)`B. `8xx10^(5)"rad "s^(-1)`C. `4xx10^(5)"rad "s^(-1)`D. `2xx10^(5)"rad "s^(-1)` |
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Answer» Correct Answer - B At resonance as X = 0, `I = (epsilon_(0))/(R )=(60)/(120)=(1)/(2)A=0.5 A` As `V_(L)=IX_(L)=I omega L rArr L = (V_(L))/(I omega)` `therefore L = (40)/((0.5)xx4xx10^(5))=0.2 mH` Also, `omega_(0)=(1)/(sqrt(LC)) therefore C=(1)/(L omega_(0)^(2))` `=(1)/(0.2xx10^(-3)xx(4xx10^(5))^(2))=(1)/(32)mu F` Now in case of series LCR circuit. `therefore tan phi = (X_(L)-X_(C ))/(R )` So current will lag the applied voltage by `45^(@)` if, `tan 45^(@)=(omega L-(1)/(omega C))/(R )` ....(i) or `R = omega L -(1)/(omega C) " " (because tan 45^(@)=1)` or `120=omegaxx2xx10^(4)-(1)/(omega(1//32)xx10^(-6))` (using (i)) `rArr omega^(2)-6xx10^(5)omega-16xx10^(10)=0` `therefore omega = (6xx10^(5)pm sqrt((6xx10^(5))^(2)+64xx10^(10)))/(2)` `therefore omega = (6xx10^(5)+10xx10^(5))/(2)=8xx10^(5)"rad "s^(-1)` |
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