![]() This finally answers the first part of your question (Why is C discharged?) - It is not discharged, it really is charged we're just not looking at the upper plate, but at the lower plate connected to the resistor, gradually being pulled low through R. ![]() We can imagine how the upper plate of C remains at V in, the lower plate becomes charged towards 0 V, and finally, there is no voltage left across the resistor, between the lower plate and 0 V. Here's your original circuit again, with some symbols we will need for the explanation, the assumption that we have no load, and the equations showing V out for C on top and R at the bottom. This almost answers your first question about the decrease in output voltage already we just have to turn this configuration upside down again. Also, we can see how the voltage remaining across the resistor on top of the capacitor becomes less the more we charge the capacitor: V R = V in - V C. We can see how C is charged according to the RC time constant and according to the magnitude of the input voltage step from 0 V to V in. This is the picture you usually see for a capacitor being charged through a resistor, so it may be worth the effort: We take the freedom to start with changed positions of R and C note that I in = I C = I R, so we really are allowed to do this (KCL). ![]() Long story short: For a low-to-high transition of your input signal, your capacitor isn't discharged, it is charged, and it remains charged until the high-to-low transition occurs.
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