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

Let Z be the effective impedance of the parallel LCR circuit
then , `(1)/(Z) = sqrt((1)/(R^(2)) + ( omega C - (1)/(omega L ) )^(2) ) `
Which is minimum at `omega = omega_(0)= (1)/(sqrt(LC))`
Therefore Z is maximum at `omega= omega_(0)` and the total current amplitdue is minimum

Given, rms value of voltage , `E_(rms) = 230` V
rms value of current in the branch containing resistor is
`I_(R) = (E_(rms))/(R) = (230)/(40) `= 5.75 A
rms value of current in the branch containing inductor is
` I_(L) = (E_(rms))/( omega_(0)L) = (230)/(50 x 5) ` = 0.92 A
rm.s value of current in the branch containing capacitor is
`I_(C) = E_(rms) xx omega_(0 C)`
= ` 230 xx 50 xx 80 xx 10^(-6) ` = 0.92 A