# voltage in a capacitor

#### Rufinus

Joined Apr 29, 2020
27
Hello. I am learning electronics and I have 3 questions.

1) How does behave the voltage in a capacitor when I connect it to a DC power supply, for example 10V? Is instantly 10V or is raising until is charged?

2) If I connect a capacitor to a AC power supply, at for example 60hz, the capacitor charge and discharge constantly. Is it possible that if I disconnect the capacitor will be without charge because in that exactly moment it was in the discharge moment?

Thank you

Kind regards

#### OBW0549

Joined Mar 2, 2015
3,454
Hello. I am learning electronics and I have 3 questions.
I only count two questions.

1) How does behave the voltage in a capacitor when I connect it to a DC power supply, for example 10V? Is instantly 10V or is raising until is charged?
You can't change the voltage across a capacitor instantaneously, as it would require an infinite amount of current. Depending on the size of the capacitor and the current capability of the power supply, the capacitor will charge quickly or not so quickly.

2) If I connect a capacitor to a AC power supply, at for example 60hz, the capacitor charge and discharge constantly. Is it possible that if I disconnect the capacitor will be without charge because in that exactly moment it was in the discharge moment?
Yes. It could also have a voltage across it, if disconnected at the right moment.

#### crutschow

Joined Mar 14, 2008
25,126
1) How does behave the voltage in a capacitor when I connect it to a DC power supply, for example 10V? Is instantly 10V or is raising until is charged?
The capacitor voltage will rise at a rate determined by the current the supply can provide and the size of the capacitor per the formula t = (C*V) / i, where t is time, C is capacitance, V is the capacitor voltage and i is the charging current.
2) If I connect a capacitor to a AC power supply, at for example 60hz, the capacitor charge and discharge constantly. Is it possible that if I disconnect the capacitor will be without charge because in that exactly moment it was in the discharge moment?
It can have zero volts across it (zero charge) if disconnected exactly at the zero crossing.

#### WBahn

Joined Mar 31, 2012
25,918
Hello. I am learning electronics and I have 3 questions.

1) How does behave the voltage in a capacitor when I connect it to a DC power supply, for example 10V? Is instantly 10V or is raising until is charged?

2) If I connect a capacitor to a AC power supply, at for example 60hz, the capacitor charge and discharge constantly. Is it possible that if I disconnect the capacitor will be without charge because in that exactly moment it was in the discharge moment?

Thank you

Kind regards
The voltage across a capacitor cannot change instantaneously. Any circuit that would require such a change is not a viable model of a realizable circuit. In your case the model must reflect the resistances associated with the power supply, the capacitor, and the interconnecting wires. Those will establish the maximum current that can flow (assuming the power supply can actually supply that much to begin with) which, in turn, will establish the maximum rate at which the capacitor's voltage can change.

When you disconnect a capacitor, it will retain whatever charge it had at the moment it was disconnected (at least initially) and have the corresponding voltage across it. Thus the voltage across the capacitor after you disconnect it from a 120 Vrms 60 Hz power supply could be anywhere from about -170 V to +170 V, and yes 0 V is a possibility, but it is quite unlikely that you will hit exactly 0 V. Also, the voltage across the capacitor can change somewhat even after it is disconnected due to things like dielectric absorption, but that is probably well beyond the scope of what you need to know at this stage.

#### dl324

Joined Mar 30, 2015
10,987
1) How does behave the voltage in a capacitor when I connect it to a DC power supply, for example 10V? Is instantly 10V or is raising until is charged?
The voltage will rise exponentially at a rate determined by the resistance limiting current.

$$V_C = V_{initial}(1-e^{\frac{-t}{RC}})$$

It's considered fully charged after 5 time constants.

EDIT: If you don't have an external series resistor, you'd use the internal resistance of the power source.

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#### Rufinus

Joined Apr 29, 2020
27
Thank you for your answers. All of you have been very clear. This is a great forum,

After posting, I have been experimenting with a large capacitor (330v 60 uf) and with a small transformer that gives me 6 volts. Sometimes one plate was positive and sometimes was negative, sometimes I measured 4, 6 or even 9 volts. So yes, now everything makes sense.

Only to be sure I have understood. In the diagram, when the switch is open, de voltmeter shows 9V. When I close the switch, the voltmeter shows a lower voltage and start rising until the capacitor charge. That´s right?

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#### dl324

Joined Mar 30, 2015
10,987
After posting, I have been experimenting with a large capacitor (330v 60 uf) and with a small transformer that gives me 6 volts.
You shouldn't do that unless the capacitor is unpolarlzed. A polarized electrolytic capacitor can explode when reverse biased. I've had it happen to me when some assemblers installed them backwards.

#### crutschow

Joined Mar 14, 2008
25,126
EDIT: If you don't have an external series resistor, you'd use the internal resistance of the power source.
If the source has a current limit circuit, then the charging voltge will be basically linear rather than exponential.

#### dl324

Joined Mar 30, 2015
10,987
If the source has a current limit circuit, then the charging voltge will be basically linear rather than exponential.
That information wasn't provided, so we would assume a constant voltage.

#### Rufinus

Joined Apr 29, 2020
27
Yes, I am familiarised with electrolytic capacitors an explosions. The capacitor is unporalized

#### dl324

Joined Mar 30, 2015
10,987
Yes, I am familiarised with electrolytic capacitors an explosions. The capacitor is unporalized
That's good. I was concerned after you described your qualifications in your first post.

#### Rufinus

Joined Apr 29, 2020
27
Yes yes, I know, but when I was a child I loved playing with electronics. I used to etch PCB´s and that things, but now I am more interested in the theory, the physics behind electicity.

And anybody can confirm if in the diagram, when the switch is open, de voltmeter shows 9V. When I close the switch, the voltmeter shows a lower voltage and start rising until the capacitor charge. That´s right?

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#### Tonyr1084

Joined Sep 24, 2015
4,719
Voltmeter shows 9V. When I close the switch, the voltmeter shows a lower voltage and start rising until the capacitor charge. That´s right?
Yes. That's because - assuming the capacitor is completely discharged - when the voltage is applied the capacitor will appear as a dead short. As much current as the battery can deliver will be driven into the capacitor. As its voltage rises its resistance rises. Since DC does not pass through a capacitor, it appears as a resistor of sorts. As the cap reaches full charge the resistance appears as an infinity of ohms. In other words, it appears as an open circuit to the battery.

Depending on each microsecond, as the cap charges current drops as resistance rises. Ohm's law at play. A given voltage and an assumed resistance of near zero will give the highest current. Let's look at it mathematically:

A 9V battery and a nearly discharged capacitor. Assume the battery can deliver 200 mA. The cap appears as a near short. Let's assume it's one ohm. Ohm's law predicts that at 9 volts and 1 ohm the current should be 9 amps. But the battery can't deliver that much current. So you can actually watch the voltage rise. Next lets look at the cap at 50% capacity; And for sake of argument let's also assume 1KΩ is = to infinity. At 50% capacity, that resistance would be 500Ω. At 9V and 500Ω, the current would be: 9V ÷ 500 = 0.018 amps (18 mA). Of course, assuming a 1KΩ resistance is equivalent to infinity is absolutely wrong. I just used that to help make sense of how the current is changing as the voltage in the cap rises.

One last factor that will determine how fast a capacitor charges up - - - it's Farad rating. Assuming the same example above, a 1 micro farad (1 µF) will take X microseconds to charge. A 100 µF cap will take 100 times longer to reach the same full charge. So size matters. That's why caps are used as part of a filter. A higher frequency will pass through a 1 µF cap whereas a 100µF cap will pass a much lower frequency. I'm no expert on frequencies, so I can't give more than that. I just know that when using a tweeter speaker in a stereo system, use of a capacitor of a given value will block all lower frequency audio and only pass the frequencies that are equal to or higher than the capacitor's rating.

#### Tonyr1084

Joined Sep 24, 2015
4,719
Edit alert - I had to edit some of my numbers in post #13. I had somethings backwards. Hey! It happens.

#### MrChips

Joined Oct 2, 2009
21,350
A capacitor cannot reach the charge voltage instantly because of:

1) Internal resistance of the battery
2) Charge capacity of the battery
3) Resistance and inductance of all connecting wires
4) Internal resistance and inductance of the capacitor.

#### dl324

Joined Mar 30, 2015
10,987
when the switch is open, de voltmeter shows 9V. When I close the switch, the voltmeter shows a lower voltage and start rising until the capacitor charge. That´s right?
It depends.

If the battery was fresh and the capacitor was small enough, the battery voltage wouldn't drop enough for you to measure.

If you used a power supply, a voltage drop might happen, but it might be so brief that you wouldn't be able to measure it with a voltmeter. A more likely scenario is that you'd see the capacitor voltage increase exponentially with the time constant being determined by the capacitor value and the internal resistance of the power source.

If you want to verify the "physics", you should use a resistor to limit the charging current and choose values that give you a time constant large enough for you to take measurements.

#### Rufinus

Joined Apr 29, 2020
27
Ahhh ok, now I understand. I didn´t realise that the capacitor when is discharged acts like a short. Thank you again for your answers. Finally the third question!

I have seen in youtube (I leave the link at the end) a guy making tiny arcs with a flyback transformer. (I only watch it, I don’t play with high voltage), and later he puts in parallel a homemade HV capacitor and the arcs get stronger but not continuous. What I think is happening is that the capacitor charges, and when reaches the voltage to start the arc discharge all its charge making a more power arc. That´s right?

And writing this I wonder one thing more. One difference between a battery and a capacitor is that the capacitor can gives all its charge much quicker, isn´t it? I have been reading about the capacitive time constant and the charging and discharging current of a capacitor and how decrease, but how can I calculate how much current gives a capacitor if I short its contacts. For example an electrolytic cap 40V and 5.000uF

Thank you very much

#### Tonyr1084

Joined Sep 24, 2015
4,719
the capacitor charges, and when reaches the voltage to start the arc discharge all its charge making a more power arc. That´s right?
Yes.
the capacitor can gives all its charge much quicker, isn´t it?
Yes.

The capacitor is drawing current away from the arc. When the voltage is high enough the arc can strike. At that time the capacitor is now giving back its stored energy. When the capacitor drops below the arc voltage level it again starts drawing the charge. Excellent demonstration. Thanks for the video.

#### Chris65536

Joined Nov 11, 2019
178
And anybody can confirm if in the diagram, when the switch is open, de voltmeter shows 9V. When I close the switch, the voltmeter shows a lower voltage and start rising until the capacitor charge. That´s right?
One thing that is missing from your circuit diagram, but which always exists in the real world, is a series resistor. As drawn, the resistor would be part of the battery, in series with its voltage source. This is what limits the battery's maximum current. The wires connecting everything would have a small resistance too. This allows you do the math, without dividing by zero.

#### crutschow

Joined Mar 14, 2008
25,126
As its voltage rises its resistance rises. Since DC does not pass through a capacitor, it appears as a resistor of sorts.
I'm sorry, but using "resistance" as a measure of the capacitors slowing charge rate as its voltage rises, is not an appropriate use of that word, and likely confusing to a newbie.
The only resistance in play here is the series resistance of the source voltage.
It's the voltage difference between the source and capacitor across that resistance that slows the charge rate as the capacitor voltage rises, which causes the exponential rise in the capacitor voltage with time.