How can DC charge a capacitor?

I read that a capacitor shows infinite resistance towards DC. While solving questions we use DC to charge the capacitor and use a constant potential difference and current. This contradicts that DC cannot charge a capacitor. Is there any clear idea on how DC cannot charge a capacitor?

The problem is sloppy terminology. You're using "DC" to mean two different things:

1. A circuit in which none of the voltages or currents change over time. This is a mathematical ideal.

2. A voltage or current source that produces a constant voltage, such as a battery.

In a DC circuit (meaning #1), a capacitor acts like an open circuit. No current flows through it. If your circuit has a charging capacitor, it's not a DC circuit, because the capacitor voltage and current are changing over time.

But a DC voltage or current source (meaning #2) can definitely charge a capacitor. Connecting that source to the capacitor changes the circuit. If the circuit changes, it's not a DC circuit anymore (meaning #1)

A circuit that contains only DC sources (meaning #2) and passive components (resistors, capacitors, and inductors) will eventually (asymptotically) become a DC circuit (meaning #1).

I read that a capacitor shows infinite resistance towards DC

That is the steady state behavior, as opposed to transient behavior.

Is there any clear idea on how DC cannot charge a capacitor?

DC can charge a capacitor.

Suppose you have a capacitor with zero volts between its terminals, and suppose you connect its terminals to the terminals of a battery. For a brief interval of time, a current will flow, and the voltage of the capacitor will rise until it reaches the voltage of the battery. What happens during that interval of time is called "transient behavior." What happens after that interval (no current flow, no voltage change) is the "steady state."

The time scale—the duration of the transient—could be nanoseconds, or it could be many seconds, or it could be anywhere in between. It all depends on the size of the capacitor, the voltage of the battery, the resistance of the battery and the wires, and on what you deem to be "significant" current or "significant" change in voltage.

See @JRE's answer for more detail including plots of voltage and current vs. time during the transient.

DC can charge a capacitor. It cannot pass through a capacitor.

Have a look at this circuit:

simulate this circuit – Schematic created using CircuitLab

This is the current flowing into the capacitor:

You have a DC source. Current flows into the capacitor until the capacitor is charged. Once the capacitor is charged, no current flows through the circuit.

1. DC can flow into the empty capacitor.
2. No DC flows through the charged capacitor.

Alternatively, look at this circuit:

simulate this circuit

Again, current flows into the capacitor until it is charged. Thereafter, no current flows.

• But is there any special role for AC in charging a Capacitor?AC can only uncharge the capacitor am I right here? If the capacitor is fully charged and again I'm providing charge to it then what would happen? (Mostly It would act like a open circuit) and I'm confused that can Ac charge the capacitor even after the capacitor is fully charged. (I'm lacking the concept somewhere). Is it possible to explain this? Commented Aug 2 at 18:34
• AC can pass through a capacitor by repeatedly charging and discharging the capacitor.
– JRE
Commented Aug 2 at 18:35
• Electrons cannot pass through the dielectric but for every electron brought to a cathode one electron is expelled from the anode. So the device is conducting. AC "pass through" because the capacitor is always charging or discharging. Commented Aug 3 at 16:10
• The two circuits are the same; only the capacitor and the resistor have been swapped, which is not relevant here. Commented Aug 4 at 6:24
• The plots and schematic are of different things. The plots are of a switched DC source turning on at t=0 and the schematic is just a DC source. There is an implicit assumption that the source and cap. start at 0 V. Commented Aug 5 at 17:13

An ideal capacitor appears initially to have zero resistance. It's like filling a bucket with water. By the time it's full no more can be added. AC means the bucket is being emptied and filled repeatedly

DC is the only way to charge a capacitor, at least in the long term. In the following circuit, I pass a constant DC current through a capacitor, and plot the voltage across the capacitor over time:

simulate this circuit – Schematic created using CircuitLab

As you can see, capacitor voltage just keeps on rising, and as long as that current keeps flowing, it will continue to rise until it explodes from dielectric breakdown.

If I replace the DC current source with AC, the changing current direction alternately charges and discharges the capacitor, and the capacitor's voltage averages out to be zero, in the long term:

simulate this circuit

If current direction frequently changes (that's what we call AC), the capacitor doesn't have time to charge very far before the current changes direction, causing the capacitor to discharge again. In this way, voltage fluctuations across the capacitor diminish in amplitude as AC current frequency increases.

By definition, the impedance $$\Z\$$ (which is the AC equivalent of DC resistance) of something is the ratio of the voltage $$\v\$$ across it, to current $$\i\$$ through it. That is Ohm's law for AC:

$$Z =\frac{v}{i}$$

A lower impedance would have a lower voltage across it, for some given current. Even if the current amplitude remains the same, the voltage fluctuations across the capacitor decrease with frequency, and the conclusion is that its impedance is therefore lower at higher frequency.

Le's fit that principle into the context of your question, which is about signals, and more importantly potentials. To set the scene, consider a resistor potential divider, below left:

simulate this circuit

The total voltage across the resistors in series is 12V, but that voltage is shared by both resistors. Each will develop a fraction of the total in proportion to its resistance. If either of the resistances increases, so will its share of the total available. That can be seen above right, in which I have increased R1 to 1.5kΩ. Consequently, its share of the voltage has risen from 3V to 4V, and the other resistor's voltage has had to reduce, so that the total is always 12V.

In the context of AC, we refer to "impedance", which is the dynamic equivalent of resistance at DC. A resistor's impedance is that same as its resistance, since that quantity does not vary with frequency. However, as we established before, a capacitor's impedance does vary with frequency, falling as frequency increases.

If we replace one of the resistors with a capacitor, the ratio of impedances will now be frequency dependent. Here I impose a small 3V amplitude AC signal, added to a constant 12V DC offset, across a series connected capacitor and resistor. That is, the input signal $$\V_{IN}=+12V\pm 3V\$$:

simulate this circuit

At DC (0Hz freqeuncy), the capacitor will have infinite impedance, and will adopt all of the available DC voltage across it. This is shown on voltmeter VM1.

At higher frequencies, though, capacitor impedance will drop to where it is much smaller than R1, leaving R1 to adopt those higher frequency components of the signal. In other words, R1 will have no DC across it, but all the AC fluctuations, which is $$\V_{OUT}\$$ shown in orange below. The input $$\V_{IN}\$$ is blue:

Since the output in this example is a copy of the input, but with the 12V DC offset removed, that is the reason why some say that capacitors block DC.

I would remind you once again, though, that strictly speaking, capacitors don't block DC current, and given a constant DC current they would charge until they die. It would be more accurate to say that they have a "frequency dependent resistance" (impedance), which, in the context of an alternating current through them, causes them to develop smaller and smaller AC voltage across them as current frequency increases.

• So, a current-supplied capacitor is a ticking time bomb waiting to happen, isn't it :-)? Commented Aug 4 at 12:27
• @Circuitfantasist I think that the real ticking time bomb is the ideal current source required to do that! Commented Aug 4 at 12:46
• Ah... The capacitor only sets the moment when the 'bomb' goes off. But it also provides a large current at the moment of discharge, like a flash. Commented Aug 4 at 14:05

But DC current can charge a capacitor.

Just like a water hose with steady flow of water can fill up a bucket.

AC is just a water hose that both fills the bucket (charges the cap) and then sucks the bucket empty (discharges the cap).

• Nice analogy... Commented Aug 3 at 19:59

"infinite resistance" is a way of speaking. If you solve the equations for a discharged capacitor, you get that under a constant voltage the charge does increase and tends asymptotically to a constant, while the current decreases exponentially. Thus, the instantaneous resistance grows exponentially over time.

If the time constant is small compared to your time scale, then it is legit to consider the resistance infinite.

A capacitor can be functionally represented as a voltage source (a rechargeable battery) connected in reverse polarity to the input voltage, creating the charging current. In its extreme states of being fully discharged and fully charged, it behaves like a short circuit and an open circuit, respectively. The "passage" of alternating current through a capacitor is an illusion. In fact, current does not pass through it because its plates are separated by an insulator. It causes an accumulation of charge on one of its plates and a removal of charge from the other plate, which creates the illusion of current flow.

Real capacitor

The Time-Domain Simulation below shows us how the voltage across the capacitor changes over time. At the start, the voltage is 0 V. After 0.6 ms, it reaches 5 V (half the input voltage V). Finally, it gets very close to the 10 V input voltage.

simulate this circuit – Schematic created using CircuitLab

Simulated capacitor

By modeling the capacitor as a rechargeable battery, we can disregard time and employ a more straightforward yet powerful DC Live Simulation for analysis (hovering the mouse over the circuit).

"Short circuit"

At the beginning, the capacitor is uncharged. Its voltage is zero and it behaves as a source with zero voltage (short circuit).

simulate this circuit

So we can think of it as a piece of wire.

simulate this circuit

Charging voltage source

As the capacitor charges, it behaves as a variable voltage source. Its voltage (exponentially) increases from 0 V to the 10 V input voltage, while the current gradually decreases.

simulate this circuit

"Open circuit"

I read that a capacitor shows infinite resistance towards DC.

As time passes, the capacitor voltage gets closer and closer to the supply voltage. Eventually, the current flow becomes negligible.

simulate this circuit

Once fully charged, the capacitor behaves as if it had infinite resistance, acting as an open circuit.

simulate this circuit

The intriguing aspect is that this "open circuit" can be seen as having an equivalent voltage source equal to the input, rather than infinite resistance. This equivalent voltage source effectively behaves as a "virtual infinite resistance" . Such a behavior is commonly referred to as "bootstrapping" in circuits.

AC + DC charging

But is there any special role for AC in charging a Capacitor? I'm confused that can Ac charge the capacitor even after the capacitor is fully charged.

The AC voltage is simply added to the DC voltage. Imagine we have a 1 V AC signal riding on a 10 V DC level (offset).

simulate this circuit

The graph below shows how the capacitor charges exponentially to the DC voltage and then wiggles around it.

How current "passes" through a capacitor

I read that a capacitor shows infinite resistance towards DC... How can DC charge a capacitor?

Indeed, how can something (current) pass through a barrier (insulator)? Clearly, there is some trick here that needs to be unraveled. It can be observed everywhere in life where there is accumulation.

Conceptual diagram

Imagine two tanks that are half-filled with something (water, air, oil, food, money, information, current...), and we continue filling the first one. So we have the notion that the "something" is disappearing.

simulate this circuit

Concurrently, we are draining the second container of its contents. So we have the notion that the "something" is appearing.

simulate this circuit

Now, imagine placing both tanks close together inside a 'black box' (dotted line). This creates the illusion that the substance is passing through the box...

simulate this circuit

... and we imagine that inside there is simply a "pipe" connecting the input to the output.

simulate this circuit

Electrical diagram

Let's now transfer our general observations to electrical circuits. The electric current accumulates charge on the left plate of the capacitor...

simulate this circuit

.. and removes charge from the right capacitor plate.

simulate this circuit

This makes it seem as if the current is flowing directly through the insulating material...

simulate this circuit

... and we imagine that inside there is simply a "piece of wire" connecting the input to the output.

simulate this circuit

This answer is very similar to that of JRE (which I upvoted, by the way) but I'm trying to be a little more qualitative and descriptive.

Disclaimer: This is a classical description using concept of electron flow.

First, no insulator is absolutely perfect. So in the real world, capacitor resistance can never be truly infinite. It can, however, be so large that it is practically infinite. In introductory studies, it's common to idealize capacitors and state that the dielectric has infinite resistance.

Second, it's true that a capacitor will show virtually infinite resistance to DC current. However, that doesn't mean that a capacitor can't charge. Charging, resistance to DC current through the capacitor itself, and current flow in the circuit, are different things. Keep reading for an explanation why.

Third, wires or conductors in a circuit have some finite resistance, but it is usually small enough to be considered zero for short wire runs and macroscopic circuits. That's why all points along an uninterrupted conductor are considered "the same point electrically."

Lastly, a capacitor acts like an "open" in a circuit. It's often taught that no current can flow in a circuit with an open. That is an oversimplification. A more precise statement might be that constant, sustained current can't flow in a circuit with an open. (I think the biggest cause of confusion in introductory teaching about capacitors is failing to point out and correct that oversimplification.) Because electrons repel each other, some small amount of current will flow, from a high concentration of electrons to a low concentration, along an uninterrupted length of conductor, until charge is equally distributed along the length of that conductor. If there is an open in a DC circuit, once charge is equally distributed, then current will cease.

Putting all this together: When you attach a capacitor across a DC voltage source, it takes a finite amount of time for electrons to move from the negative side of the DC voltage source to the negative side of the capacitor. And it will take the same amount of time for electrons on the positive side of the capacitor to be drawn towards the positive pole of the DC voltage source. No current will flow through the capacitor. But for a small time, a tiny amount of current will flow though the conductors on either side of the capacitor—in order to supply or remove electrons from the capacitor plates—until each plate of the capacitor reaches the same electrical potential as the pole of the DC source it is attached to.

And, of course, capacitors are actually designed with rolled up plates inside, to increase the surface area, to deliberately increase the amount of electrons it takes to equalize potential, in order to increase the time it takes for the plates on either side of the capacitor to reach the same electrical potential as each pole of the DC voltage source. (The electric field between the two plates of the capacitor plays an important role in the operation of capacitors, but being a technician and not an engineer, I am sadly ill-equipped to understand those details.)

In summary:

1. "Charging" a capacitor means to accumulate a surplus or deficiency of electrons on the plates of the capacitor, not to pass current through the capacitor.
2. Some tiny amount of current needs to flow through the conductors of a circuit in order to supply these electrons to or from the plates of a capacitor.
3. Once the voltage across the plates of the capacitor equals the voltage across the DC voltage source, current through the conductors stops since there is no longer a potential difference between each plate of the capacitor and the pole of the DC source it is connected to.
4. No current ever flows through an (ideal) capacitor. If it did, that would actually bleed off the potential difference between the plates and defeat the purpose of having a capacitor to begin with.

There is an even more verbose explanation at this Stack Exchange answer: If AC current can flow through a capacitor, why can't it flow through an open circuit?