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.

Explanation

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.