# Why this circuit has a DC current?

Consider the circuit below:

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:

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!

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.