I was reading about how capacitors work and two sentences confused me:
"A capacitor’s ability to store energy as a function of voltage (potential difference between the two leads) results in a tendency to try to maintain voltage at a constant level."
"When voltage across a capacitor is increased or decreased, the capacitor “resists” the change by drawing current from or supplying current to the source of the voltage change, in opposition to the change."
If the capacitor draws current from the source when the capacitor's voltage increases, how is this considered a form of resistance by the capacitor? I mean – the way I understood it – for the voltage to remain constant in this case, shouldn't the capacitor not draw any current but instead, limit the flow of current through it? I am really confused in here – I feel like I am missing something very fundamental…
Here is the full paragraph:
"Because capacitors store the potential energy of accumulated electrons in the form of an electric field, they behave quite differently than resistors (which simply dissipate energy in the form of heat) in a circuit. Energy storage in a capacitor is a function of the voltage between the plates, as well as other factors which we will discuss later in this chapter. A capacitor’s ability to store energy as a function of voltage (potential difference between the two leads) results in a tendency to try to maintain the voltage at a constant level. In other words, capacitors tend to resist changes in voltage drop. When the voltage across a capacitor is increased or decreased, the capacitor “resists” the change by drawing current from or supplying current to the source of the voltage change, in opposition to the change."
Best Answer
To see why it's said that a capacitor 'resists', or 'objects to' changes in voltage at its terminals, it's useful to compare its behaviour with a resistor (don't confuse the 'resists', meaning 'tries to stop', with anything to do with the component 'resistor').
If you have 10v across a 1k resistor, then 10mA will flow. If you now try to change the voltage to 20v, ramping it up at 10^6 volts per second, so it takes 10uS to change from 10v to 20v, the current will smoothly increase from 10mA to 20mA in that time.
If you have 10v across a 10uF capacitor, and the voltage has been steady for long enough, then no current flows. If you now try to change the voltage to 20v, ramping it up at 10^6 volts per second, so it takes 10uS to change from 10v to 20v, the current will go to 10A for that 10uS, and back to zero when the voltage is steady again. If you try to change the voltage at 10^7 volts/second, the current pulse will be 100A.
That sort of current amounts to a fairly violent 'objection' to the voltage being changed. If the power supply cannot supply it, then the voltage will not change as fast as expected.