InterviewSolution
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InterviewSolution
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In a series `LCR` circuit with an `ac` source of `50 V,R=300 Omega`,frequency `v=50/piHz`.The average electric field energy, stored in the capacitor and average magnetic energy stored in the coil are `25 mJ` and `5 mJ` respectively.The `RMS` current in the circuit is `0.10 A`.Then find |
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Answer» Correct Answer - (a)`C=20 muF` , (b)`1 H` , (c )`35.36 V` Av. Electric field energy =`(1/2CV_(rms)^(2))=25xx10^(-3) J` `therefore 1/2xxc.I_(rms)^(2)xx1/(2pi^(2)v^(2)c^(2))=25xx10^(-3)J thereforeC=20 muF` Av.magnetic energy `(1/2Li_(rms)^(2))=5xx10^(-3)` `therefore L=(2xx5xx10^(-3))/(0.10)^(2) rArrL=1` henry `V_(R)=I_(rms)R , V_(C)=I_(rms)X_(C) , V_(L)=I_(rms)xxomegal.` `=0.10xx300 , =(0.10)xx1/(2pi(50/pi)xx20xx10^(-6)) ,=0.10xx2pixx50/pi`(1) =`V_(R)=30 V , V_(C)=50 V , V_(L)=10 V` `rms` voltage of source `E_(rms)=50 V` (given in the equation) |
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