Hi!

Ever since Veritasium’s video on current flow I have been re-examining and questioning everything i “know”. I was looking into Low Pass filters - I have always applied the formulas for the 3dB cutoff point I need, maybe used a simulator, got the result and called it a day! I am comfortable with the math if need be or to use a phasor diagram. However, I was thinking about the current flow and I fear I am missing a fundamental concept

let’s take this circuit

and running it through the simulator, we can see this

From t=0

- Vin increases in the positive direction, as such current starts to flow, and the voltage across the capacitor increases as charge builds on the plates.
- as long as Vin > Vc (Vc = Vout), the voltage on the capacitor will continue to rise, albeit after the peak of Vin this will slow
- the current through the resistor is essentially in line with Vin
- When Vin = Vc, (around t=6ms), the capacitor will begin to discharge. This will mean the current will reverse - and indeed, at 6ms Ir goes negative - this implies that despite Vin still being positive and pushing current into the circuit, it is being opposed by the voltage on the capacitor, as the voltage on the capacitor is higher it is
*forcing*current back into the voltage source

my question here is, what is the real world implications of this? I have been thinking about various voltage sources, like an AC generator, and how that would handle current coming back in when it’s still generating a positive voltage and trying to drive current. This lead me to think about smoothing capacitors in rectifiers - however in this case, there are diodes that prevent reverse currents and so the smoothing capacitor will discharge into the load - RC filters don’t appear to do this.

I pulled out my Art of Electronics to see what they had to say on the topic, but didn’t find much on the implications of the current flow. I can follow the math and can draw the waveforms I see - however thinking about the actual current flow here has got me stumped i must admit!

I have seen examples where the RC filter is driving an op-amp; an op-amp having an extremely high (ideally infinite) input impedance, won’t accept current in, and so in this case again, the discharge current from the capacitor in the filter must go back toward the source.

When the capacitor discharges, for a period in time whilst Vin is still positive, we have two voltage sources in parallel with different voltages which is generally not considered a good idea. The resistor in between will create a voltage drop of IR - if we look at the point 7.8ms, the current through R1 is -1A, with a 30 ohm resistor we would expect to see -30v across R1 and indeed Vin - Vout is - 30v. However the current must still flow *somewhere* and as far as I can see it can only go back into the source

I had thought that as there was no load, it doesn’t really make sense - sure you can construct the circuit without a load (and again, the only place the discharge current can flow is back to the source; this would also be the same case as connecting an op-amp with high input impedance), but to be sure I can add a load; the voltage across is still Vout = Vc, but we can see a current through that follows Vc/Rload - which makes sense at face value, but again, not if I think about where the currents are flowing!

As the capacitor is discharging, we know there must be a current flowing! but *where* does it flow to? without a load, it can only go one way; with a load there is a discharge path, but then i’d expect the current through R1 to change accordingly…

I have been thinking back and forth on this for some time! I am really, really sure I have missed a fundamental concept here - I can follow the math to get the answer, but conceptually it seems to be messing with me!