What would the output waveform be in each of these cases?  In the second picture, the source is a step function from 0 to 1V.

I feel like I solved them, but I am not convinced that my understanding is air-tight. Any help is appreciated.

Edit: Adding my analysis here of both circuits.

1. Initially, there is 1V on the top-plate of the capacitor, but since the capacitor prevents any flow of DC current, there is no current through the resistor and its bottom plate is at 0V. Hence Vc, initial is 1V. Now when a pulse of 1->2V is applied, the voltage across the capacitor cannot change instantly, so the capacitor acts as a short, except, with its top-plate at 2V, the bottom plate can only go to 1V (to keep Vc at 1V). So the output waveform jumps to 1V as a response to the impulse portion of the step (1->2V), but decays to 0V.

2. The bottom plate has 1V on it. Initially there is no current in the circuit so the top plate has 0V on it. When the voltage source(step) jumps to 1V, the voltage across the capacitor cannot charge instantaneously, so it tries to change the charge on its top plate by charging through the R, and reaches 1V in 5 time constants.

Please let me know if my analysis is incorrect, any inputs appreciated.

• It would help in discerning if your "understanding is air-tight" if you would tell us what your understanding actually is.
– jonk
Nov 30 '21 at 20:35
• Apologies for not adding that my analysis. In the first case, there is 1V on the top-plate of the capacitor, but since the capacitor prevents any flow of DC current, there is no current through the resistor and its bottom plate is at 0V. Hence Vc, initial is 1V. Now when a pulse of 1->2V is applied, the voltage across the capacitor cannot change instantly, so the capacitor acts as a short, except, with its top-plate at 2V, the bottom plate can only go to 1V (to keep Vc at 1V). So the output waveform jumps to 1V as a response to the impulse portion of the step (1->2V), but decays to 0V. Nov 30 '21 at 20:53