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 $rms$ current value in each branch of the circuit given below.
Figure shows a series $LCR$ circuit connected to a variable frequency $230\; V$ source. $L=5.0\; H, C=80\; \mu F, R=40\; \Omega$.

(N/A) An inductor $(L)$,a capacitor $(C)$,and a resistor $(R)$ are connected in parallel with each other in a circuit.
Given:
$L = 5.0\; H$
$C = 80\; \mu F = 80 \times 10^{-6}\; F$
$R = 40\; \Omega$
$V = 230\; V$
The impedance $(Z)$ of the given parallel $LCR$ circuit is given by:
$\frac{1}{Z} = \sqrt{\frac{1}{R^2} + \left(\frac{1}{\omega L} - \omega C\right)^2}$
At resonance,the reactive part is zero,i.e.,$\frac{1}{\omega L} - \omega C = 0$.
Therefore,the resonant angular frequency is:
$\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{0.02} = 50\; rad/s$.
At this frequency,the term $(\frac{1}{\omega L} - \omega C)^2$ becomes zero,making $\frac{1}{Z}$ minimum,which means the impedance $Z$ is maximum. Consequently,the total current $I = \frac{V}{Z}$ is minimum.
The $rms$ current in each branch is:
$1$. Current through inductor $(I_L)$:
$I_L = \frac{V}{\omega L} = \frac{230}{50 \times 5.0} = \frac{230}{250} = 0.92\; A$.
$2$. Current through capacitor $(I_C)$:
$I_C = V \omega C = 230 \times 50 \times 80 \times 10^{-6} = 230 \times 4000 \times 10^{-6} = 0.92\; A$.
$3$. Current through resistor $(I_R)$:
$I_R = \frac{V}{R} = \frac{230}{40} = 5.75\; A$.