I am confused with this! How does a capacitor block DC?

  • I have seen many circuits using capacitors powered by a DC supply. So, if capacitor blocks DC, why should it be used in such circuits?
  • Also, the voltage rating is mentioned as a DC value on the capacitor. What does it signify?
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    \$\begingroup\$ If you apply a direct current source to a capacitor, it will pass DC just fine. (The voltage will increase until the cap explodes, of course...) \$\endgroup\$
    – endolith
    Aug 16, 2011 at 16:10
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    \$\begingroup\$ If you apply DC voltage to a capacitor it is not at all blocked at first. Eventually, the capacitor gets charged and puts out its ow n DC. At that point no current flows through it. \$\endgroup\$ Oct 22, 2018 at 18:25

9 Answers 9


I think it would help to understand how a capacitor blocks DC (direct current) while allowing AC (alternating current).

Let's start with the simplest source of DC, a battery:

just a battery

When this battery is being used to power something, electrons are drawn into the + side of the battery, and pushed out the - side.

Let's attach some wires to the battery:

a battery with wires attached

There still isn't a complete circuit here (the wires don't go anywhere), so there is no current flow.

But that doesn't mean that there wasn't any current flow. You see, the atoms in the copper wire metal are made up of a nuclei of the copper atoms, surrounded by their electrons. It can be helpful to think of the copper wire as positive copper ions, with electrons floating around:

illustration of copper ions with electrons

Note: I use the symbol e- to represent an electron

In a metal it is very easy to push the electrons around. In our case we have a battery attached. It is able to actually suck some electrons out of the wire:

movement of an electron from the wire

The wire attached to the positive side of the battery has electrons sucked out of it. Those electrons are then pushed out the negative side of the battery into the wire attached to the negative side.

It's important to note that the battery can't remove all the electrons. The electrons are generally attracted to the positive ions they leave behind; so it's hard to remove all the electrons.

In the end our red wire will have a slight positive charge (cause it's missing electrons), and the black wire will have a slight negative charge (cause it has extra electrons).

flow of current due to charge in the wires

So when you first connect the battery to these wires, only a little bit of current will flow. The battery isn't able to move very many electrons, so the current flows very briefly, and then stops.

If you disconnected the battery, flipped it around, and reconnected it: electrons in the black wire would be sucked into the battery and pushed into the red wire. Once again there would only a tiny amount of current flow, and then it would stop.

The problem with just using two wires is that we don't have very many electrons to push around. What we need is a large store of electrons to play with - a large hunk of metal. That's what a capacitor is: a large chunk of metal attached to the ends of each wire.

With this large chunk of metal, there are a lot more electrons we can easily push around. Now the "positive" side can have a lot more electrons sucked out of it, and the "negative" side can have a lot more electrons pushed into it:

illustration of more charge on a larger surface

So if you apply an alternating current source to a capacitor, some of that current will be allowed to flow, but after a while it will run out of electrons to push around, and the flow will stop. This is fortunate for the AC source, since it then reverses, and current is allowed to flow once more.

But why is a capacitor rated in DC volts

A capacitor isn't just two hunks of metal. Another design feature of the capacitor is that it uses two hunks of metal very close to each other (imagine a layer of wax paper sandwiched between two sheets of tin foil).

The reason they use "tin foil" separated by "waxed paper" is because they want the negative electrons to be very close to the positive "holes" they left behind. This causes the electrons to be attracted to the positive "holes":

charge attraction between capacitor plates

Because the electrons are negative, and the "holes" are positive, the electrons are attracted to the holes. This causes the electrons to actually stay there. You can now remove the battery and the capacitor will actually hold that charge.

This is why a capacitor can store a charge; electrons being attracted to the holes they left behind.

But that waxed paper isn't a perfect insulator; it's going to allow some leakage. But the real problem comes if you have too many electrons piled up. The electric field between the two "plates" of the capacitor can actually get so intense that it causes a breakdown of the waxed paper, permanently damaging the capacitor:

capacitor plate breakdown

In reality a capacitor isn't made of tin foil and waxed paper (anymore); they use better materials. But there is still a point, a "voltage", where the insulator between the two parallel plates breaks down, destroying the device. This is the capacitor's rated maximum DC voltage.

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    \$\begingroup\$ A good explanation but it doesn't answer the OP's question in a direct fashion: With AC, you have an instantaneous variation in the voltage. At every point of the AC wave, the voltage is varying and when you have a capacitor in the ckt, this change/variation in voltage can be transmitted across the dielectric to the other side/plat via an electric field of varying intensity. Therefore current flows in the circuit even though the dielectric is an insulator to the flow of electrons. \$\endgroup\$
    – user41067
    Oct 15, 2015 at 8:27
  • \$\begingroup\$ with DC, you have a fixed/non-varying electric field across the dielectric of some field strength/intensity and the dielectric polarizes to oppose this field so the electrons at the other end of the wire are not under the influence of any electric field and therefore don't move barring the usual/routine movement of electrons in a metal. \$\endgroup\$
    – user41067
    Oct 15, 2015 at 8:30
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    \$\begingroup\$ you need to think of this in terms of waves with electrons/molecular polarization acting as a means/medium for waves. \$\endgroup\$
    – user41067
    Oct 15, 2015 at 9:45
  • \$\begingroup\$ After reading this answer, Can i conclude that capacitors do not allow current to flow in DC when connected in series? \$\endgroup\$
    – Fennekin
    Apr 16, 2017 at 7:43
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    \$\begingroup\$ @Fennekin Capacitors do not allow DC current to flow whether alone, or connected in series, or connected in parallel. But, again, that's in the steady state. There will still be an initial rush of some current; whether they are connected in series, parallel, or alone. \$\endgroup\$
    – Ian Boyd
    Apr 16, 2017 at 15:14

Let me see if I can add one more perspective to the other 3 answers.

Capacitors act like a short at high frequencies and an open at low frequencies.

So here are two cases:

Capacitor in series with signal

enter image description here

In this situation, AC is able to get through, but DC is blocked. This is commonly called a coupling capacitor.

Capacitor in parallel with signal

enter image description here

In this situation, DC is able to get through, but AC is shorted to ground causing it to be blocked. This is commonly called a decoupling capacitor.

What is AC?

I have used the terms "High Freq" and "Low Freq" rather loosely as they don't really have any numbers associated with them. I did this because what is considered low and high depends on what is going on in the rest of the circuit. If you want to learn more about this you can read about low-pass filters on Wikipedia or some of our RC filter questions.

Voltage Rating

The voltage that you see with capacitors is the maximum voltage you can safely apply to the capacitor before you start to run the risk of the capacitor physically breaking down. Sometimes this happens as an explosion, sometimes fire, or sometimes just gets hot.

  • \$\begingroup\$ Kellen, I appreciate your use of pictures, but I'm missing an answer to the question how the cap blocks DC. You just say that it does. \$\endgroup\$
    – stevenvh
    Aug 17, 2011 at 6:26
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    \$\begingroup\$ @Stevenvh I felt the the confusion that the OP had was not about the physics of how it blocks DC, but rather why it is used if it does block DC. Plus I figured your answer did a pretty good of explaining it at a more physical level and didn't think I could explain that part any better than you. \$\endgroup\$
    – Kellenjb
    Aug 17, 2011 at 11:54

The explanation is in the fact that opposite charges attract each other. A capacitor is a compact construction of 2 conducting plates separated by a very thin insulator. If you put DC on it one side will be positively charged and the other side negatively. Both charges attract each other but can't pass the insulating barrier. There's no current flow. So that's end of story for DC.
For AC it's different. One side will successively be positively and negatively charged, and attract negative and positive charges resp. So changes on one side of the barrier provoke changes on the other side, so that it appears that the charges cross the barrier, and that current effectively flows through the capacitor.

A charged capacitor is always DC charged, i.e. one side has the positive charges and the other side the negative. These charges are a storage for electrical energy, which is necessary in many circuits.

The maximum voltage is determined by the insulating barrier. Above a certain voltage it will breakdown and create a short circuit. That can happen under DC but also under AC.


A simple way of thinking about it is that a series capacitor blocks DC, while a parallel capacitor helps maintain a steady voltage.

This is really two applications of the same behavior - a capacitor reacts to try to keep the voltage across itself constant. In the series case, it's quite happy to remove a steady voltage difference, but any abrupt change in one side will be passed through to the other to keep the voltage difference constant. In the parallel case, any abrupt change in voltage will be reacted to.

  • \$\begingroup\$ nice simple answer, kudos \$\endgroup\$
    – vicatcu
    Aug 17, 2011 at 17:38

This is not a very technical answer, but it's a graphical explanation that I find very funny and simple:

enter image description here

  • 2
    \$\begingroup\$ Yes, nice, but you'll get in trouble if you actually try to explain that AC path! :-) \$\endgroup\$
    – stevenvh
    May 15, 2012 at 15:07
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    \$\begingroup\$ @stevenvh yeah, of course I know it's a bit stupid, but I've always found it clever :) \$\endgroup\$
    – clabacchio
    May 15, 2012 at 15:16
  • \$\begingroup\$ Its a weird answer :D :D :D :D :D \$\endgroup\$
    – perilbrain
    Sep 15, 2012 at 3:55
  • \$\begingroup\$ Actually it helped me lot to understand clearly what is a capacitor. Thank u! \$\endgroup\$
    – Mr Bonjour
    Dec 9, 2013 at 15:01
  • \$\begingroup\$ Not an answer, just a (bad) joke. \$\endgroup\$ Jun 25, 2023 at 20:41

The amount of charge that develops across the plates of a capacitor with a given voltage across its terminals is governed by the formula:

\$ Q = C \times V \$ (charge = capacitance * voltage)

Differentiating both sides (current is the time derivative of charge), gives:

\$ I = C \times \dfrac {dV}{dt} \$ (current = capacitance * the rate of change in voltage)

DC voltage is the same as saying \$ \dfrac{dV}{dt} = 0 \$.

So a capacitor allows no current to flow "through" it for DC voltage (i.e. it blocks DC).

The voltage across the plates of a capacitor must also change in a continuous manner, so capacitors have the effect of "holding up" a voltage once they are charged to it, until that voltage can be discharged through a resistance. A very common use for capacitors is therefore stabilize rail voltages and decouple rails from ground.

The voltage rating is how much voltage you can apply across the plates before the electro-static forces break down the material properties of the dielectric material between the plates rendering it broken as a capacitor :).


My answer to such questions is always "water". Water flowing through pipes is a surprisingly accurate analogy for current flowing through wires. Current is how much water flows through a pipe. Voltage difference becomes the difference in water pressure. The pipes are supposed to lie flat, so that gravity plays no role.

In such an analogy, a battery is a water pump, and a capacitor is a rubber membrane which completely blocks the pipe. DC is water flowing constantly in one direction through a pipe. AC is water flowing back and forth all the time.

With this in mind, it should be obvious that a capacitor blocks DC: since the membrane can only stretch so far, water can't just keep on flowing in the same direction. There will be some flow while the membrane stretches (i.e. the capacitor charges), but at one point it becomes stretched enough to completely balance the water pressure, thus blocking any further flow.

It also becomes obvious that a capacitor won't block AC completely, but it does depend on the membrane properties. If the membrane is sufficiently stretchy (high capacitance), it will pose no challenge to water flowing back and forth quickly. If the membrane is really rather stiff (e.g. a thin sheet of plastic), this corresponds to low capacitance, and if the water flows back and forth slowly, such flow will be blocked, but very high frequency oscillations will still make it through.

This analogy has been so exceptionally useful to me that I really wonder why it isn't used more widely.

  • \$\begingroup\$ A friend helped me understand why this analogy is not used more widely: apparently he has as little intuition for water flow in pipes as he does for current flow in wires! \$\endgroup\$ Jun 5, 2014 at 22:59

First off, a capacitor blocks DC and is a lower impedance to AC, while an inductor tends to block AC yet pass DC very easily. By "blocking", we mean than it offers a high impedance to the signal we're talking about.

First, though, we need to define a few terms to explain this. You know what resistance is, right? Resistance is the opposition to current flow that results in the burning of power, measured in watts. It does not matter if the current is AC or DC, the power dissipated by a perfect resistor is the same amount for either.

So resistance is one kind of "impedance" to current flow. There are 2 others - "inductive reactance", and "capacitive reactance". Both are also measured in ohms, like resistance, but both are different in that, for one thing, they vary with frequency, and for another, they don't actually consume power like a resistance does. So all together, there are 3 kinds of impedance - resistive, inductive, and capacitive.

The amount of blocking or impedance of inductors in ohms can be determined by:

$$X_L = 2\pi fL$$

Where 2pi is approximately 6.28, f is the frequency (AC, obviously) of a signal, L is the inductance measured in henries, and where "X sub L" is the inductive reactance in ohms.

Inductive reactance is the impedance of a component due to inductance; it is a kind of resistance, but does not actually burn power in watts like a resistor does, and since "f" for frequency needs to be supplied, the value of it varies with frequency for a given inductor.

Notice that as the frequency goes up, so does the impedance ( AC resistance) in ohms. And notice that if the frequency equals zero, then so does the impedance - a frequency of zero means DC, so inductors have virtually no resistance to DC current flow. And as the frequency goes up, so does the impedance.

Capacitors are the opposite- the formula for capacitive reactance is

$$X_C = \frac{1}{2\pi fC}$$

Here, C is the capacitance of the cap in farads, "2pi" and "f" are the same as above, and "X-sub-C" is the capacitive reactance in ohms. Notice that here, the reactance is "one divided by" the frequency and the capacitance - this results in values of impedance that go down with frequency and capacitance. So if the frequency is high, the impedance will be low, and if the frequency is near zero, which is DC, the impedance will be nearly infinite - in other words, capacitors block DC, but pass AC, and the higher the frequency of the AC signal, the less the impedance to it.


I'll go for the shortest-answer qualitative-take-away approach:

A capacitor across DC rails is there, in effect, to short any AC signals that might otherwise get onto the supply rails, so the amount of AC across your DC circuit is reduced.

The voltage rating on a cap is the maximum voltage (sum of DC and any AC present!) that the cap should see. Exceed this voltage and the cap will fail.


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