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Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements L,C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency. |
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Answer» SOLUTION :As they are CORRECTED in the parallel combination `(1)/(Z)=(1)/(R )+((1)/(X_(L))-(1)/(X_(C )))` As the reactance `(X_(C )-X_(L))` is perpendicular to the ohmic resistance R, therefore we can write as `(1)/(Z)=sqrt((1)/(R^(2))+((1)/(X_(L))-(11)/(X_(C )))^(2))` `= sqrt((1)/(R^(2))+((1)/(omega L)-omega C)^(2))` At resonance, `omega = omega r = (1)/(sqrt(LC))` That means `(1)/(Z)` = minimum and thus Z = maximum. As Z is maximum, current will be minimum. current through INDUCTOR `I_("RMS")L=(V_("rms"))/(X_(L))=(230)/(50xx5)` `I_("rms")L=(230)/(250)=0.92 A` current through capacitor `I_("rms")C=(V_("rms"))/(X_(C ))=(V_("rms")xx omega C)/(1)` `= 230xx50xx80xx10^(-6)` `= 0.92A`. Current throught resistor `I_("rms")R=(I_("rms"))/(R )=(230)/(40)=5.75 A` |
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