# How is voltage applied across a capacitor changing when I connect it to a DC power source?

Current entering/leaving a capacitor is proportional to the change in voltage applied across its plates.

Lets say I have a capacitor in series with a DC power supply, a resistor, and an open switch. Once I close the switch, the voltage across the capacitor is equal to the voltage supplied by the DC source, but this voltage remains constant since DC power supplies are obviously constant.

This is all under the assumption that all components are ideal. I know that for current to enter the capacitor, the voltage across it has to be changing, which is what happens in the real world. Where does this change in voltage come from?

• The voltage across the capacitor/resistor series combination is constant and equal to the power supply voltage. The voltage across the capacitor starts at zero and rises exponentially to the supply voltage when it's fully charged. – Chu Dec 6 '15 at 23:25
• But the current entering a capacitor is proportional to the change in voltage applied across its plates, right? If the voltage applied across its plates is constant, how can current enter it? – Sam D20 Dec 6 '15 at 23:28
• The only explanation I can think of is having the applied voltage across its plates be the sum of the DC source and the reverse voltage given off by the capacitor. Then I could see a changing dv/dt. – Sam D20 Dec 6 '15 at 23:29
• According to your description, there's a resistor (R) in series with the capacitor (C), so the capacitor is not connected directly across the power supply. The voltage across C is $\small (V_S -RI)$, where $\small V_s$ is the supply voltage, $\small I$ is the current flowing through R and C. – Chu Dec 6 '15 at 23:38
• Ah, that is precisely the answer I was looking for. Thank you very much. – Sam D20 Dec 6 '15 at 23:43