- The voltage drop across a resistor in series with a capacitor should be in my opinion only the potential difference between the plate of the capacitor connected to the same wire with the resistor and thep late
For simplicity assume that there is a +4V in the positive terminal of the battery and -4V in the negative one and let A be the capacitor plate connected to the positive terminal and B the capacitor plate connected to the negative terminal.
During the first (milisecond) The capacitor is uncharged so all the voltage drop is across the resistor: 8V,
In my opinion it should be only 4V:
The red wire going out from the positive terminal,To the resistor and the capacitor is physically disconnected from the negative terminal.
And the only potential difference would be the difference between the capacitor's plate A potential and the positive terminal potential,During charging ( first miliseconds) it should be : +4V - 0V = 4V potential difference and not 8V
I know this is wrong but i don't know why.
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