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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 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. 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. 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 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 When Vin = Vc, (around t=6ms), the capacitor will begin to discharge.the current through the resistor is essentially in line with Vin.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.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.