= Solution
In the <parallel LC circuit> both components have the same voltage <phasor>. Adding their current <phasors> by <Kirchhoff's current law> gives the total <electrical admittance>:
$$
\frac1Z=\frac1{j\omega L}+j\omega C
=j\left(\omega C-\frac1{\omega L}\right).
$$
Thus
$$
\boxed{Z=\frac{j\omega L}{1-\omega^2LC},\qquad
|Z|=\frac{\omega L}{|1-\omega^2LC|},\qquad
\arg Z=\begin{cases}+\pi/2,&\omega<1/\sqrt{LC},\\-\pi/2,&\omega>1/\sqrt{LC}.\end{cases}}
$$
At the ideal parallel <resonance>, $\omega_0=1/\sqrt{LC}$, the two branch current <phasors> cancel. The net <electrical admittance> is zero and the magnitude of the <complex impedance> tends to infinity: the parallel combination behaves as an open circuit at that <frequency>. Its <complex argument> is undefined exactly at the pole. Finite component losses replace the infinite peak by a large finite <electrical impedance>. This conclusion concerns a voltage-driven sinusoidal response; an ideal resonant current drive does not have a bounded sinusoidal steady-state voltage.
Back to article page
