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In an LR circuit with an AC source (\$V=V_0\cos(\omega t)\$) we have that $$I(t)=\frac{V_0}{\sqrt{R^2+(\omega L)^2}}cos(\omega t -\phi)$$ where \$\phi=\frac{\omega L}{R}\$. I find this formula rather intuitive. 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. Similarly, if the frequency \$\omega\$ is high, then there will always be a very high change in magnetic flux through the inductor and thus a large back emf will produced meaning the currents amplitude will ultimately be reduced. But now if we hook up a simple RC circuit to the same AC source, we get the solution $$I(t)=\frac{v_0}{\sqrt{R^2+(\frac{1}{\omega c})^2}}cos(\omega t+\phi)$$ where \$\phi = \frac{1}{R\omega C}\$. But now I do not find this formula intuitive at all. Why should the currents amplitude approach a maximum as we increase the frequency? In the case of the LR circuit, the current goes to zero and the reason is rather intuitive (It's because of the increasingly rapid changes in magnetic flux which generate an increasingly strong back emf) except for an RC circuit, I can't think of any reason why the amplitude approaches a maximum. Also, why would the currents amplitude approach a maximum value if the capacitance is extremely high but approach zero if the capacitance is extremely low? And finally why is there a phase difference at all for a RC circuit? The phase delay in the LR circuit can easily be chalked up to the back emf but for an RC circuit there doesn't seem to be any immediately apparent reason for the phase difference.

Basically, I can't seem to come up with any reasonable explanation as to why the capacitive reactance (\$\frac{1}{\omega c}\$ ) takes the form that it does. Any help on this would be most appreciated!

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    \$\begingroup\$ Try to read this electronics.stackexchange.com/questions/287394/… \$\endgroup\$ – G36 Sep 1 '20 at 16:52
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    \$\begingroup\$ And about inductor electronics.stackexchange.com/questions/470171/… if you are interested. \$\endgroup\$ – G36 Sep 1 '20 at 17:02
  • \$\begingroup\$ Salah the Goat! How's your "calculus thinking?" Any much good? Or is it obscure to you? \$\endgroup\$ – jonk Sep 1 '20 at 18:10
  • \$\begingroup\$ (By the way it's hard not to think of Zlateh instead of Salah. But that's another story, I suppose.) \$\endgroup\$ – jonk Sep 1 '20 at 18:17
  • \$\begingroup\$ I'd considered an approach, but deleted it upon re-reading you. You want a more physical explanation. The electronic explanation is easy: moving small bits of charges around fast, which you have to do at higher frequencies, means you need a lot of current to achieve it. But I think what you want will be found in the dipole approximation, rotational inertia, and its restoring torque magnitudes. (Where the speed of light isn't important to the space-time distance between the plates, anyway.) I'll need to think more to see if I can get a back-EMF equivalent concept, though. \$\endgroup\$ – jonk Sep 2 '20 at 1:08
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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 clarification

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)
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I think you need to consider these basic relationships for a capacitor: -

$$Q = CV \hspace{1cm}\text{(charge = capacitance}\cdot\text{voltage)}$$

$$\dfrac{dQ}{dt} = I\hspace{1cm}\text{(rate of change of charge = current)}$$

$$\text{Therefore,}\hspace{1cm}I = C\cdot \dfrac{dV}{dt}$$

So, in the last formula, current is greater if the rate of change of voltage rises and, current is greater (for the same rate of change of voltage) when capacitance increases.

The last formula above directly leads to the AC reactance of a capacitor being \$\frac{1}{\omega C}\$.

If a sinewave voltage is applied to a capacitor then current will be maximum when the rate of change of voltage is maximum hence, for AC circuits you should be able to visualize that current and voltage waveforms are displaced by a phase angle of 90°.

For an inductor we have: -

$$V = L\cdot \dfrac{di}{dt}$$

So, do you see the similarities and shouldn't you expect a phase angle to be produced in both scenarios?

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  • \$\begingroup\$ Okay this definitely helps a lot. I'm still not at 100% though. So in the case of a capacitor connected to an AC voltage source with absolutely no resistance present (a perfectly ideal circuit assumed), the voltage across the capacitor should always equal the voltage across the source (that is, the voltage provided by the source). Now when the source voltage is at its zero point, it has its max rate of change magnitude \$|\frac{dV}{dt}|\$ and similarly the capacitor will have its maximum \$|\frac{dV}{dt}|\$ . Thus the current should only reach a max when V_source and V_cap are at zero right? \$\endgroup\$ – SalahTheGoat Sep 1 '20 at 15:54
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    \$\begingroup\$ @SalahTheGoat yes, for a sinewave applied voltage, the rate of change of voltage is greatest as the voltage passes through zero hence, this is when the maximum current occurs. Given that the current is also a sinewave, it is shifted forward by 90° from the voltage wave form (i.e. said to lead voltage by 90°) \$\endgroup\$ – Andy aka Sep 1 '20 at 16:06

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