Question

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 as below for this frequency.

The figure shows a series LCR circuit connected to a variable frequency $230V$ source. $L=5.0H$, $C=80$ $\mu F$, $R=40\Omega$.

Answer 1

$1/Z=\sqrt{\frac{1}{R^2}+(1/\omega L-\omega C{)}^2}$

At resonance, $\omega =\sqrt{1/LC}=50rad/s$

at $\omega =50$, $Z$ is maximum and current is minimum.

$I_L=V/\omega L=0.92A$

$I_c=\omega CV=0.92A$

$I_R=V/R=5.75A$