So you want some intuiton about the impedance of the capacitor.

We will need first an intuiton on the mechanism of current flow on a capacitor. You can watch this great and short video, that will give you more intuition than I could ever give by just writing.

Once we have some intuition about the flow of curent in a capacitor, lets talk about the impedance. We can build the intuiton this way:

• The capacitor has a certain charge \$+Q\$ and \$-Q\$ in each plate.
• There is a force on the carges on each of the plate, exerted by the carges on the other plate.
• The higher the carge \$Q\$, the higher this force
• This force opposes the flow of current

So, up to this point, we have some intuition on the realtionship between the carge \$Q\$ on each plate and the flow of current. We know that the higher the carge \$Q\$, there will be more opposition to the flow of current. We can summarize this as

• More charge \$Q\$ -> capacitor presents a higher resistance \$R\$

So, what ís the relationship between the frequency \$w\$ of an ideal AC source and the carge \$Q\$ on each pĺate of the capacitor?

• The frequency \$w\$ tells us how long is the period of charge/discharge of the capacitor
• Higher \$w\$, shorter charge/discharg times. Lower \$w\$, longer charge/discharge times
• A shorter charge/discharge time means less charge \$Q\$ will be built up on the plates. A longer charge/discharge time means more charge \$Q\$ will be built up.

So, the higher the AC source frequency \$w\$, less charge \$Q\$ will be built up. Lower frequency \$w\$ will mean more charge \$Q\$. We know from our previous discussion that more (less) charge \$Q\$ means more (less) resistance \$R\$.

So, we can conclude

• The lower the frequency \$w\$ -> The higher the resistance \$R\$
• The higher the frequency \$w\$ -> The lower the resistance \$R\$

The exact functional relationship is given by \$X_C(w)=\frac{1}{wC}\$. You can refer to any textbook for the details of the calculus.

Important aclaration

The formulas you gave are only valid for a permanent AC state. This means, those solutions are valid for an AC source of fixed amplitude \$A\$ and frequency $w$. They are not valid for any change in this parameters.

If any of this parameters change, you have to analyze the transient response of your system. Here we are talking about the permanent response. It is an important difference.

You said

If the inductance L is high then the system has a strong 'ability' to fight back against changes in the current thereby reducing the currents amplitude.

This is stament is true, but the equation you gave for the current in a RL circuit is not representing the physicial situation you are talking about. You are talking about a transient. The equation is describing a permanent state.

If you have taken a course involving differential equations, you have seen this: a differential equation has a solution that is the sum of two functions: the transient and the permanent

• The transient solution does not depend on the input. It depends on your system.
• The permanent solution depends on your input and your system, and is a linear combination of the input (for linear systems)