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This may be a dumb/begginer question, but I'm having trouble to understand what exactly happens when we connect a real capacitor directly with a battery.

In my understanding, theoretically, when an uncharged capacitor is connected directly to a battery of, let's say, 9 volts, instantly the capacitor will be charged and its voltage will also become 9V. This will happen because there is no resistance between the capacitor and the battery, so the variation of current by time will be infinite. Obviously, this is true when talking about ideal components and non-realistic circuits.

I thought that doing it in real life would cause sparks, damaged components, explosions, or whatever. However, I saw some videos and people usually do connect batteries directly with capacitors. Also, the current that flows from the battery to the capacitor is somehow of low magnitude, since it takes some considerable time to make the capacitor have the same voltage as the battery.

I would like to know why this happens, thanks.

This is an example of the circuit I talked about:

enter image description here

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Both the battery and the capacitor have an internal resistance.

Your capacitor looks a bit like this on the inside:

schematic

simulate this circuit – Schematic created using CircuitLab

Of course, I don't know your capacitor, so I don't know the exact internal resistance, but 3Ohm will be a close enough approximation.

The same happens in your battery, so in fact you are doing this:

schematic

simulate this circuit

So now for a tiny amount of time the current will be maximum, but it is only about 0.9A

Of course when you put a capacitor onto a battery like that, you will not make great contact, so there will be some extra resistance there as well, so it might even be 0.7A.

The reason it now takes time, is that when the capacitor charges, the voltage across the resistors decreases, so the current decreases as well, so the voltage on the capacitor will increase more slowly, and so on and so on, so it will actually approach the battery voltage slower and slower.

The larger the resistors or the capacitors the more time it will take.

The moment it is at 67% can be calculated by R * C.

So in the example that is: t(67%) = R * C = 10 * 220u = 2.2ms.

But if the capacitor is 22000uF (= 22mF) then the RC-time, as it is called, will be 220ms, or 0.22s for it to charge with a total resistance of 10Ohm. But with a capacitor that size it might also have a slightly higher resistance, so that'll make it even slower.

And then it's only at 67%. The next 30% will take much more time.

EDIT: Note; increased the 9V-bat resistance as per Nick's comment.

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    \$\begingroup\$ 9V batteries in particular have very high internal resistance. \$\endgroup\$ – Nick Johnson Jun 19 '15 at 8:45
  • \$\begingroup\$ @NickJohnson I thought so, but couldn't remember for sure right then, I'll up the value in the images and adjust the current. \$\endgroup\$ – Asmyldof Jun 19 '15 at 11:06
  • \$\begingroup\$ @NickJohnson And that's an interesting side-effect of how they're usually constructed: 6x 1.5V cells in series. Though, even then it's only ~1.7 Ω, compare ~0.25 Ω for AAA and ~0.12 Ω for AA. \$\endgroup\$ – Bob Jun 19 '15 at 12:13
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Real batteries and capacitors have an internal resistance which will act to reduce the current charging the capacitor. This will prevent the death and destruction you were expecting. :-)

In any case, it is hard to see a spark produced with 9 volts...

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    \$\begingroup\$ In electronics school we burnt up a 750V multimeter with a 9V battery when showing how transformers work.(It was quite a shock, quite literally) \$\endgroup\$ – akaltar Jun 19 '15 at 9:33
  • \$\begingroup\$ "it is hard to see a spark produced with 9 volts..." - well, you can easily burn ("burn" as in "open fire & flame") a low-ohm carbon resistor with 9V battery; also, with 9V PSU (not battery, Rint too high - but a PSU with < 1 ohm Rint usually "works") you can easily create sparks by "lightly" (i.e. for short time and only with a tip of the connector) shorting the PSU terminals with (obviously thickly insulated) wire. I've done this twice, by accident - thinly insulated wire with permanent short got its insulation completely melted; another wire (loose) created a fountain of sparks. #DTTAH \$\endgroup\$ – vaxquis Jun 19 '15 at 11:05
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In addition to Asmyldofs helpful answer, it is worth noting that even if all the conductors were superconducting with zero resistance, the initial current would not be infinite and the current would decay to zero.

Why not infinite current? As there is a loop of current the circuit will have some inductance. So the current will initially rise at a rate of Vbatt/L. The voltage across the capacitor will shoot past Vbatt to nearly twice that value and then reverse, giving a damped sinusoid centered at Vbatt.

Why damped? We are generating a time-varying magnetic field. That is how to make an electromagnetic (radio) wave. The power in the radiated field will cause the oscillation in current to die out.

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As you say, only in "theory," can we obtain "ideal" results.
Using realistic power sources and capacitors, one obtains non-ideal results. This is because real components have "additional" resistance, inductance, and capacitance.
Although one can never obtain ideal results, by keeping the "additional" components as small as possible, we can obtain results "close" to ideal.
In your particular case, the reason there were no "dramatic effects," is that the battery and the capacitor have internal resistance. Therefore, the capacitor will not instantly charge up to the battery voltage. It will "slowly" charge up at the "normal" rate specified by the product of Rint and the capacitance C.
To sum up, the reason why capacitors take time to charge up is - internal resistance.

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