Hi,
The simulation will probably work ok, in that you will see the correct waveform,
however the interpretation must be correct also or at least the way we
interpret the results has to be right also.
What i believe is happening here is that the exponential part of the oscillatory
part of the wave (I) is not being given enough time to damp out enough to
allow only the steady state part of the response to show itself.
Remember that the circuit is not only a 'passive' circuit, but it is also an
oscillator in itself, which will have a different frequency (most likely) than
the line frequency.
Keep in mind that when we calculate the steady state response
with a formula like Iout=Vin / sqrt(R^2 + (XL-XC)^2), that
calculation gives us the answer to the problem long *after* any exponentials have
died out. This means if we want to simulate the steady state response alone
(instead of the actual true wave form) we have to *wait* for a period of time
that makes the exponential part look negligible compared to the steady state
part.
For the circuit presented in the diagram, the exponential factor is R/(2*L),
and in order for this exponential part to die down to less than 1 percent
we have to force the exponent to be equal to about 5. This means we have
to wait:
t=5/(R/(2*L))=5*2*L/R=10*L/R seconds.
In this case it comes out to:
t=2832 seconds.
This means we have to allow the simulation to run to show the wave for
at least 2832 seconds in order to be sure we are seeing the steady state
response alone. At the end of this time the wave should be a steady state
sine wave.
Note that this is based on the steady state part being large enough to show
up after this time too, in that in relation to the steady state part the
exponential becomes low enough. Since the steady state peak is about 2.5 amps
we may not have to wait as long as 2832 seconds, but perhaps 1133 seconds
because 2832/2.5=1133 approximately.
If you can, try simulating for at least 1000 seconds and see what you get.
You should probably look at the actual wave and not RMS(I), then calculate
the RMS value yourself.
The true steady state response is 2.5 amps peak, which is about 1.8 amps RMS
as noted correctly by the other posts.