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?

  • \$\begingroup\$ 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. \$\endgroup\$ – Chu Dec 6 '15 at 23:25
  • \$\begingroup\$ 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? \$\endgroup\$ – Sam D20 Dec 6 '15 at 23:28
  • \$\begingroup\$ 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. \$\endgroup\$ – Sam D20 Dec 6 '15 at 23:29
  • \$\begingroup\$ 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. \$\endgroup\$ – Chu Dec 6 '15 at 23:38
  • \$\begingroup\$ Ah, that is precisely the answer I was looking for. Thank you very much. \$\endgroup\$ – Sam D20 Dec 6 '15 at 23:43

The basic relationship in a capacitor is that the voltage is proportional to the charge on the "+" plate. However, we need to know how current and voltage are related. To derive that relationship you need to realize that the current flowing into the capacitor is the rate of charge flow into the capacitor. Here's the situation. We'll start with a capacitor with a time-varying voltage, v(t), defined across the capacitor, and a time-varying current, i(t), flowing into the capacitor. The current, i(t), flows into the "+" terminal taking the "+" terminal using the voltage polarity definition. Using this definition we have:

ic(t) = C dvc(t)/dt

This relationship is the fundamental relationship between current and voltage in a capacitor. It is not a simple proportional relationship like we found for a resistor. The derivative of voltage that appears in the expression for current means that we have to deal with calculus and differential equations here - whether we want to or not.


The voltage across the series combination of capacitor and resistor is equal to the DC source. What changes over time is the proportion across each. At the instant the switch closes it's all across the resistor, because the cap approximates a short circuit. After infinite time the cap voltage equals the DC source voltage. That means there is no voltage across the resistor and thus no current flow through it.


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