2
\$\begingroup\$

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."

\$\endgroup\$
2
\$\begingroup\$

Think of this as water in a pipeline. If the capacitor limits the current i.e. like closing a tap in our analogy. Now if the tap is closed water will start to accumulate in the pipeline and pressure(voltage) will start increasing at the point where water flow is cutoff. On the other hand if the capacitor draws current that is we allow the current to flow freely, there will be no accumulation of water and thus increase(change) in pressure (voltage) will be limited. Think of the capacitor as a water tank that stores water. Now the tank takes time to fill. Also a change in pressure does not instantaneously fill the tank. Let the tank have a input and output pipe. If pressure instantaneously increases the tank cannot fill instantaneously so current will flow through output pipe. If pressure increases slowly the tank has time to fill as well. This explains low impedance of capacitor for low frequency (slow change) and high impedance for high frequency (fast change).

This is just a analogy to help visualize the phenomena and is not the exact explanation. Comment if you feel anything is out of place.

New contributor
Dhruv Deshmukh is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
\$\endgroup\$
3
\$\begingroup\$

"Resists" may be an unfortunate choice of word. If I were to describe a capacitor like that I think "reluctant" or "reticent" would be a better choice.

The voltage that develops across a capacitor is the result of charge carriers (electrons typically) building up along the capacitors dielectric.

From Wikipedia: A simple plate capacitor

The build up of charge carriers takes time, and therefore the change in voltage will also take time. Contrast this with a pure resistance where the voltage that develops is dependent on the current flowing through it at any instant in time.

So when the book says the capacitor "resists" changes in voltage, what it is referring to is that any voltage change will take some time depending on how quickly the charge carriers flow in or out of the capacitor.

Hope this helps.

\$\endgroup\$
  • 1
    \$\begingroup\$ How about "react" or "impede"? \$\endgroup\$ – Sredni Vashtar Mar 15 '18 at 2:15
  • 1
    \$\begingroup\$ "The build-up of charge carriers takes time" - Is this why the capacitor is said to have impedance? I mean is this the form of impedance made by the capacitor? \$\endgroup\$ – DigiNin Gravy Mar 15 '18 at 3:10
2
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

The word "resists" here is nothing to do with resistors. It's the plain English meaning of the word. A capacitor opposes changes in voltage.

If you increase the voltage across a capacitor, it responds by drawing current as it charges. In doing so, it will tend to drag down the supply voltage, back towards what it was previously. That's assuming that your voltage source has a non-zero internal resistance.

If you drop the voltage across a capacitor, it releases it's stored charge as current. That tends to keep up the voltage for a short while.

In the example below, Rs is the resistance of the voltage source, assumed to be low, but not zero. All the component values are arbitrary, just for illustration.

schematic

simulate this circuit – Schematic created using CircuitLab


There is a totally different way of looking at it. A capacitor has an impedance given by 1/2πfC, where C is the capacitance, and f the frequency. Notice that as the frequency increases, the impedance drops. Whereas, at DC, the frequency is zero, and the impedance is infinite.

With a bit of hand-waving, we can consider frequency as equivalent to rate-of-change of voltage.

So if we put a capacitor across an unstable DC supply, then once it's charged up, it has no further effect on the DC, as it has infinite impedance to DC. But it effectively shorts high-frequency AC components to ground.

\$\endgroup\$
-1
\$\begingroup\$

Capacitor impedance reduces with rising rate of change in voltage or slew rate dV/dt or rising frequency by increasing current.

This means it resists the rate of change in voltage by absorbing charges with current being the rate of change of charge flow.

So at DC in theory there is open circuit high resistance but for rising AC slew rate, the effective resistance drops inversely with rising dV/dt = I/C

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.