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Keeping the source of frequency equal to the resonating frequency of the series LCR circuit, if the three elements L, C and R in are arranged in parallel , show that the total current in the parallel LCR circuit is a minimum at this frequency. Obtain the r.m.s. value of current in each brach of the circuit for the elements and source specified in for this frequency. |
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Answer» If Z is the effecitve impedance of the parallel LCR circuits shown in Fig then in vector form, `(1)/(vec Z) = (1)/(vec R) + ((1)/(vec X_(L)) - (1)/(vec X_(C )))` As reactance `(X_(C ) - X_(L))` is perpendicular to ohmic resistance R, therefore, in magnitude, we may write `(1)/(Z) = sqrt((1)/(R^(2)) + ((1)/(omega L) - omega C)^(2))` At `omega = omega_(r) = (1)/(sqrt(LC)) , (1)/(Z)` = minimum. `:.` Z is maximum. `:.` Total current will be minimum. Through L, `I_(v) = (E_(v))/(omega L) = (230)/(50 xx 5) = 0.92 A` Through C, `I_(v) = (E_(v))/(1 // omega C) = omega C E_(v) = 5 xx (80 xx 10^(-6)) xx 230 = 0.92 A` Through R, `I_(v) = (E_(v))/(R ) = (230)/(40) = 5.75 A` As currents in L and C are equal, but `180^(@)` out of phase, therefore, they cancel out. `:.` Total current `= I_(v) =` current through `R = 5.75 A` |
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