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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 Exercisefor this frequency. |
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Answer» Solution :Effective impedance of the parallel LCR circuit is given by `=(1)/(Z)=sqrt((1)/(R^(2))+(omegaC-(1)/(OMEGAL))^(2))` which is minimum at `omega=omega_(0)=(1)/(sqrt(LC))` Therefore, |Z| is maximum at `omega = omega_(0)` , and the total current amplitude is minimum. In R branch, `I_("RMS")=5.75A` In L branch, `L_("Lrms")=0.92A` In C branch, `I_("Crms")=0.92A` Note: total current `I_("rms") = 5.75 A` , since the currents in L and C branch are `180^(@)` out of phase and add to zero at EVERY INSTANT of the cycle. |
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