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Why can't capacitors be connected in parallel with ideal voltage sources? In various questions involving it we were asked to ignore the capacitors, by our teachers without valid explanations.

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    \$\begingroup\$ Why not ask your teacher. Go on don't be shy. \$\endgroup\$ – Andy aka Jul 24 '18 at 15:36
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    \$\begingroup\$ What do you think would happen at the instant a capacitor is connected across an ideal voltage source; and subsequently? \$\endgroup\$ – Chu Jul 24 '18 at 15:53
  • \$\begingroup\$ @Chu Capacitors allow AC to pass through it, so it should charge and discharge the capacitors with each cycle. \$\endgroup\$ – mepkn Jul 24 '18 at 16:03
  • \$\begingroup\$ Could you give ALL the relevant information, not drip-feed in response to comments. \$\endgroup\$ – Chu Jul 24 '18 at 16:54
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    \$\begingroup\$ Don't even ask your teacher. Analyze the circuit both ways, with and without the capacitor. Did adding the capacitor change the voltage or current of any element in the circuit other than the source? \$\endgroup\$ – The Photon Jul 24 '18 at 18:43
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The problem is that you can not connect an ideal voltage source of a given voltage in parallel with an ideal capacitor that has some initial voltage different from the source voltage. Once these two are connected, our definitions of "ideal voltage source" and "in parallel" demand that the voltage across the capacitor instantaneously changes.

Now, since $$i_C = C\frac{dV}{dt}$$ we have a finite change of voltage in zero time, which requires an undefined/infinite amount of current. Therefore, such a connection is not valid within our definitions of ideal circuits.

In the real world, all wires are also resistors and there are no ideal voltage sources, so what you see is a very large current flowing through the capacitor until it reaches the voltage of the source.

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In DC power sources, you will see large capacitors in parallel with the output used to filter the DC voltage output. In an "ideal" DC voltage source (like a fully charged car battery), putting capacitors in parallel with the battery terminals will initially change the total circuit current until the capacitor is fully charged wherein the current drawn by the capacitor is negligible. After this initial charge, the capacitor "may be ignored" when computing circuit voltage drops and current flows. The capacitor will have very little current flow and the voltage would be equal to the DC source so can be "ignored".

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... if you connect a charged capacitor, specifically, one charged to a certain voltage different from the voltage source, you are creating a short circuit between two nodes at two different voltage levels (I'm referring to circuits here, not real circuits connected by copper conductors).

KVL forbids this. Just as KVL forbids voltage sources in parallel, etc. In fact, KVL can be restated as the law that says voltage changes must take place across devices, not conduction paths.

Even if they are at the same voltage level, basic linear circuit theory no longer applies because you won't be able to come up with a unique solution to the current drawn from each of the capacitor and the voltage source.

However, in real circuits, all conductors have some resistance, so you're not going to have an issue if you connect a 9V battery to a suitable capacitor using basic copper wire. If you did the same using a DC power supply you won't charge up the capacitor as the power supply detects the lack of resistance and will not operate.

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