Electrical impedance I really just wanted to confirm if my understanding about electrical impedance in an AC circuit is correct.
So essentially the equation of impedance that $(\text{mod }Z)^2 = R^2 + (\omega L-\frac{1}{\omega C})^2$ means that at $\omega_0$, the natural frequency or at resonance, $(\text{mod }Z)^2$ is minimized, hence implying that the current is maximized due to $V=IZ$. So does this mean that even though the rate of flow of charge is maximized (current) the amplitude of the charge isn't actually maximized because of the amplitude curve and the fact that max amplitude doesn't occur at $\omega =\omega_0$ when there is damping (resistance, $R$,  in this case). And we can understand this intuitively by the fact that increasing current means more charge flows but there are also more collisions and so there will become a point after which the resistance is too high and the charge actually decreases, even if current increases ?

 A: The maximum charge on the capacitor depends on the current and the frequency.  If the frequency goes up, the current will have less time on each cycle to charge the capacitor. This suggests the the maximum charge may occur at a frequency less than the resonant frequency. (The rate of change of the current is small near the peak of the resonance curve.)  To find the frequency for maximum charge, solve rhe voltage equation  for the charge as a function of frequency, and take the derivative to maximize that.
A: According to the added picture, the AC source has a frequency different from $\omega_0$. That is why the formula for $A(\omega)$ depends on $L$ and $C$.
Note that if $\omega = \omega_0$:$$A(\omega) = \frac{\frac{V_0}{L}}{\sqrt{\gamma^2\omega^2}} = \frac{V_0}{\omega R}$$
Differentiating q with respect to time we get the current, and its amplitude in this case ($\omega = \omega_0)$ is only function of the source and the ohmic resistance $R$:
$I = \omega \frac{V_0}{\omega R} e^{i(\omega t - \delta)} = \frac{V_0}{R}e^{i(\omega t - \delta)}$
