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Suppose I have a capacitor, and I have charged it to say 35V. Now at this moment, I disconnect the battery, so currently it's charged to 35V. Now to this charged capacitor I add a 12V battery (replacing that 35volts battery), what will happen? Will the capacitor push charges into the battery? Because the capacitor has higher voltage level than the battery.

This question came to my mind because when we apply AC in a pure capacitor,at a point, the capacitor gains the voltage V_max. But then the supply voltage starts to decrease, this means the capacitor should have larger voltage than the source after crossing V_max, so what exactly happens after the source crosses V_max?, for 0V to V_max the source pushes charges from one plate to other, now on the other hand does the capacitor pushes charge through the battery during V_max to 0?

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    \$\begingroup\$ An ideal capacitor in parallel with a AC voltage source is basically a short circuit. \$\endgroup\$ Dec 24, 2021 at 12:27
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    \$\begingroup\$ If you follow your train of thought with capacitors in AC circuits you'll discover why they cause a phase-shift between the current & voltage... \$\endgroup\$
    – brhans
    Dec 24, 2021 at 14:29
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    \$\begingroup\$ @Unimportant No, an ideal capacitor is not a short circuit to an ac voltage. It is a non-zero impedance with a value inversely proportional to the ac frequency. A short circuit allows infinite current with no voltage drop, but a capacitor has a finite current and non-zero voltage drop. \$\endgroup\$ Dec 24, 2021 at 18:12
  • \$\begingroup\$ @Unimportant: Reactance is 1/(2pi * f * C), so AC voltage across a capacitor only drives infinite current through it at infinite frequency or capacitance. \$\endgroup\$ Dec 24, 2021 at 22:14

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Your intuition is in the correct direction.

If you have ideal components, and connect a capacitor to a V source that is a different value, you will get an infinite current (for an infinitesimal time) -- normal circuit analysis isn't possible in this case.

However, imagine that the capacitor and V source are initially at the same value (say 0), and the V source is a sinewave. As the V rises, current will flow into the capacitor. As the V peaks, the capacitor voltage rate of change slows and it is 0 at the top of the voltage peak. Then as the voltage source falls, current flows out of the capacitor back into the voltage source. This behaviour continues as the V source becomes negative.

Remember that the current in a capacitor is proportional to the derivative of the capacitor voltage. This is the behaviour described above -- when the V source is changing rapidly, there is a capacitor current; when it is not changing (at the + and - peaks), there is no capacitor current.

The net result if you plotted voltage and capacitor current vs. time is that the capacitor current would follow the slope of the V source. If the V source is a sinusoid, the capacitor waveform would look like it is lagging this (a sinusoid 90 ° delayed).

Also (over an integer number of cycles), there is no net current in the capacitor (the + and - currents cancel out). Equally there is no net current in the voltage source. This means that no overall power is dissipated; it just shuttles back and forth between the capacitor and the voltage source.

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If you connect an ideal capacitor directly to an ideal voltage source with a different voltage then you have an invalid circuit. Our definition of "parallel" for circuit analysis says that the elements in parallel must have the same voltage, always. So the scenarios you describe are nonsensical for purposes of circuit analysis using ideal elements.

To make sense of your scenarios you must add an element between the voltage source and the capacitor, such as an ideal resistor. Then we can analyze the circuit in a meaningful way. A current will flow through the resistor. The voltage across the resistor will be the difference between the capacitor and source voltage. The resistor obeys Ohm's Law; the current through it is proportional to the voltage across it.

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  • \$\begingroup\$ Using your definiton of an invalid circuit, you may never connect an uncharged ideal capacitor directly to an ideal voltage source. \$\endgroup\$
    – Uwe
    Dec 24, 2021 at 17:44
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    \$\begingroup\$ @Uwe That is correct. Doing so causes infinite current to flow for zero time. Such a circuit can not be analyzed by normal circuit analysis techniques. If you check for conservation of energy you find that some energy mysteriously disappears. \$\endgroup\$ Dec 24, 2021 at 18:06
  • \$\begingroup\$ @ElliotAlderson The energy does not have to get turned into heat. We have superconductors which have zero resistance. But while we can remove resistance from a circuit, we still have inductance. The circuit will ring at the frequency matched by the parasitic inductor and capacitor value \$\endgroup\$
    – Ferrybig
    Dec 25, 2021 at 22:42
  • \$\begingroup\$ @Ferrybig So where does the energy go? I'm not talking about power, but energy when the circuit has reached a dc final state. You say we have inductance, but the circuit suggested by the OP does not include an ideal inductor. You can't change the rules in the middle of the game. \$\endgroup\$ Dec 25, 2021 at 23:43
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Suppose I have a capacitor, and I have charged it to say 35V. [....] Now to this charged capacitor I add a 12V battery [....], what will happen? Will the capacitor push charges into the battery? Because the capacitor has higher voltage level than the battery.

It's exactly the same scenario as: -

  • Charging the capacitor initially to 23 volts
  • Shorting it out

Now, can you answer your own question here?

This question came to my mind because when we apply AC in a pure capacitor, at a point, the capacitor gains the voltage V_max. But then the supply voltage starts to decrease, this means the capacitor should have larger voltage than the source

No, that's impossible; if the capacitor is connected to the source, it will always have the same voltage as the source.

so what exactly happens after the source crosses V_max?

The capacitor voltage is forced to follow the source voltage. No smoke and mirrors; it has to or infinite current is flowing and that's not really a practical circuit or one that is analysable. Sure the capacitor tries to fight any change in source voltage as per the standard equation for a capacitor: -

$$Q = CV \hspace{1cm}\text{hence}\hspace{1cm} \dfrac{dq}{dt} = C\dfrac{dv}{dt} = I$$

So, capacitor current is proportional to rate of change of applied voltage. The faster you change the voltage the higher the current is. This is a fundamental rule in EE about capacitors.

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