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Consider the circuit below:

schematic

simulate this circuit – Schematic created using CircuitLab

If I try to solve this circuit in the frequency domain, I get:

$$X_C = \frac{1}{\omega C} = \frac{1}{(120 \pi) (2.652)(10^{-3})}\approx1$$ $$X_L = \omega L = (120 \pi) (2.652)(10^{-3})\approx1$$

So, the equivalent impedance between A and B is:

$$Z_{eq} = \frac{(j)(-j)}{j-j} \approx \infty$$

Taking it into consideration, the current passing through AM1 is:

$$|I| = 0A$$

This seems reasonable, but if I simulate the same circuit in CircuitLab and check the current passing through AM1 I get:

enter image description here

(Please, click on the image)

As you can see, the current passing through AM1 is a DC current of 10A.

Why the current is not 0A and how can a linear circuit excited by an AC source generate a DC current of 10A? Since I remember, all signals in a linear circuit excited by sinusoidal sources must also be sinusoidal signals (steady state).

What am I missing here?

Thank you!

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From what I see your analysis is correct but for some reason you have an offset in the inductor current (it goes from 0A to 20A, when it should oscillate between -10A and 10A, just like the cap). I would guess that is the issue, maybe some simulation setup. I don't know where it is coming from, but I was able to simulate the circuit in LTSpice and I get the results you are expecting. enter image description here

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  • \$\begingroup\$ That's the result I was expecting. Idk why my simulation didn't work in CircuitLab. The inductor current was also wrong, but I didn't mention it in the question. I it's a CircuitLab problem, since I didn't set any offset. Thank you \$\endgroup\$ – felipeek Apr 25 '16 at 22:29
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The ammeter seems to measure the rms current, not the instantaneous current. This is defined as the value of DC that would generate the same power (i.e. would heat a resistor the same) as your AC does. In some cases one has to solve the integral, but for a sine wave this value is simply the peak-to-peak voltage divided by \$\sqrt{2}\$ (see e.g. here); in your case the value of the voltage source is already given as rms value, and the impedance is 1Ohm, so all is very reasonable.

Additionally I have found circuit lab to be a bit cumbersome for this application, I'd suggest simply reading on rcl circuits.

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  • \$\begingroup\$ The ammeter, cap, and inductor internal resistences are all set to 0Ohm in my simulation. Plus, CircuitLab sources does not have configurable internal resistence (only batteries). So there's no 1Ohm impedance in my simulation. Of course all you said would make perfect sense if this impedance was there, but it isn't. The current flowing through the ammeter should be 0, regardless if the simulation is giving the peak value or the RMS value, since there's no resistence. (You can simulate the circuit and see that all impedances are set to 0) \$\endgroup\$ – felipeek Apr 25 '16 at 22:36
  • \$\begingroup\$ There is, you even calculated it! > The current flowing through the ammeter should be 0 If this were true the ammeter would be an open circuit; but even your graph shows that there is current flowing through there. \$\endgroup\$ – caconyrn Apr 25 '16 at 23:07

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