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I did a circuit to test it using the application everycircuit (https://play.google.com/store/apps/details?id=com.everycircuit.free). But when I did the following circuit I've got the following error message: Voltage source Is shortedenter image description here

Could someone explain it to me?

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    \$\begingroup\$ the red dots might signify the wire is running behind the source. Did you perhaps forget to cut up the vertical line before adding the voltage source? \$\endgroup\$
    – Marcus Müller
    May 13, 2018 at 14:16
  • \$\begingroup\$ It's probably complaining about the fact that a capacitor with no series resistance is a short-circuit to high frequency AC signals. And at turn-on, there's a broad spectrum of high-frequency components due to the step function. \$\endgroup\$
    – Hearth
    May 13, 2018 at 14:19
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    \$\begingroup\$ @Felthry At 50 Hz a 10uF capacitor has an impedance of 318 ohms. \$\endgroup\$
    – DerStrom8
    May 13, 2018 at 14:22
  • \$\begingroup\$ What do you mean with your question? I cut the wire up the voltage source and it works. But when I connect it on that way it doesn't work. And I have changed the components with a resistance, for example, and it still not working \$\endgroup\$
    – F. Riggio
    May 13, 2018 at 14:22
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    \$\begingroup\$ @F.Riggio I asked because the big junction dots only appear in three-way crossings of wires, and the source is visibly not connected to two wires where there are the red junction dots, so I was assuming that you simply had more wires, just hidden. \$\endgroup\$
    – Marcus Müller
    May 13, 2018 at 14:25

2 Answers 2

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You have an inductor in parallel with the source, which, in DC, is represented as a short. Combined with a voltage source, whose internal resistance is also zero, results in an infinite current. Unless you provide some resistance between them , it will complain. The capacitor has nothing to do with it, you can see this by adding the series resistance after the capacitor.

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  • \$\begingroup\$ But the voltage source depicted is clearly an AC source. \$\endgroup\$
    – Ariser
    May 13, 2018 at 14:53
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    \$\begingroup\$ The inductor doesn't immediately reach infinite current, though. Current through an inductor can't change instantaneously. \$\endgroup\$
    – Hearth
    May 13, 2018 at 15:01
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    \$\begingroup\$ @Felthry, what analysis were you trying to do? DC, AC, or transient? \$\endgroup\$
    – The Photon
    May 13, 2018 at 15:09
  • \$\begingroup\$ @ThePhoton Whatever this program does, which I believe is transient from what I've seen of it elsewhere. \$\endgroup\$
    – Hearth
    May 13, 2018 at 15:19
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    \$\begingroup\$ Try removing the capacitor and leave only the inductor, see if it works. Then the reverse. \$\endgroup\$
    – a concerned citizen
    May 13, 2018 at 15:26
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This same problem with calculating the initial operating point can be demonstrated with more professional simulation packages such as LTspice, however it generally doesn't arise because the inductor model defaults to a series resistance of 1m\$\Omega\$, which is reasonable enough in most cases.

enter image description here

If you take a simple AC source and parallel it with a perfect inductor (as I did above by explicitly setting the resistance to 0 instead of leaving it blank) it will choke as follows:

enter image description here

So add 1m\$\Omega\$ (or whatever seems reasonable) in series with your inductor and it will simulate just fine. Since this program is aimed at beginners it's a shame they didn't do that as a default.

enter image description here

I have run into this when simulating circuits involving superconducting inductors where the resistance actually is zero, not a situation that most EEs encounter.

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