That there is a 90° phase shift where "current leads the voltage" in a "pure capacitance" circuit such as in schematic below, is easy to understand. It also fits 100% with the hydraulic analogy of it, an elastic rubber sheet. In both hydraulic and electric circuits, the current will flow when the voltage changes, and not at the peaks. It is the derivative of the voltage.
So, I am capable of understanding that.
But going from pure capacitance to RC filters, I read that:
"In the low-pass case, the output of the filter lags the input (negative phase shift); in the high-pass case the output leads the input (positive phase shift). " (source)
This to me makes no sense, and I do not know where to start to make sense of it. There must be a very easy way to make sense of it, and so I figured I would ask. I understand high pass and low pass filters effect on the amplitude, e.g., why they are actual filters, and I understand them at a circuit level too. But, not why they would have reverse phase shift.
The phase shift curves used in examples seem to mix what parameters they show, but they all show the same thing. Some use Vs, voltage source, others Vc, the voltage at the capacitor, but those are equivalent in the context in my understanding (the capacitor reaches its fully charged state when Vs is at its peak. ) To simplify, I will use Vs, and then Io, the current measured at the output. In the low pass RC filter, and the "pure capacitance" circuit, the effect is the same. But, in the high pass RC filter, the measured current is in the reverse direction. But it must still be flowing from capacitor plate closest to Vs, and away from Vs, when Vs starts to decrease from its peak.
Based on the circuits for low pass and high pass RC filter, I do not understand why current measured at output is reverse in one case.
Based on my assumption that the current must be reversed since that is what it seems to measure, in a "pure inductance" circuit, the current is the opposite to "pure capacitance" because the inductor will oppose the current as the voltage is increasing towards positive peak, and then it will oppose decreasing the current. This makes the current “lag” behind the voltage. If a "high pass RC filter" has an output phase shift similar to an inductor, what is causing this? My source for the claim that it has is still this article.
Adding another image to explore why the current is reversed, I drew a little output circuit onto the schematic in the image above, for both low pass and high pass RC filters. And, specified a point t where a measurement is made. Unless I misunderstand something, it must be measuring the reverse current direction in either filter. Starting to get it now.
My interpretation is that Vc and Vs will cause currents to flow in opposite directions along output as shown in the circuits below.
This seems to be equivalent to these circuits, that measure opposite voltages at the output.
Edit: I finally understood, and posted an answer. The reason it was not directly intuitive to me was that the causes for the phase shift are entirely different in the low pass filter.