Electronic – the output impedence of this RC Circuit

Your electret microphone probably uses a JFET transistor to provide hi-impedance to the condenser element whilst acting as a constant current circuit to the output. The constant current might be 0.3mA for instance and if you measure the current at two different voltages you'll get approximately the same current but, if you calculated the supposed resistance at these two different voltages you will get different values.

When powered by your circuit via the 10k resistor what you might find is that the (guessed-at) 0.5mA drawn by the microphone drops 3V across the 10k ohm and leaves about 5V across the microphone - this would appear to be like a 16.7kohm resistor where the microphone is. Obviously you got 17.5kohm and I've tailored the numbers to try and match this. Try measuring at the top of the pot and the 10k to see what voltage it actually is.

Now, when you measure with a meter you won't be supplying 5V - it'll be more like 1V (another guess) and the JFET may be less than perfectly biased and it will start to conduct more than 0.2mA - remember, when a JFET loses its bias it conducts heavily compared to the biased condition. So this might mean 1V and 1mA (not 0.2mA) which equals 1kohm.

The transfer function is

$$\frac{V_{out}}{V_{in}} = \frac{j\omega \frac{L}{R}}{(1 - \omega^2LC) + j\omega \frac{L}{R} }$$

The Q is given by

$$Q = R\sqrt{\frac{C}{L}}$$

So the Q is proportional to the series resistance \$R\$ and, for your values, is quite high

$$Q = 1k \sqrt{\frac{244.8p}{41.3n}} \approx 77$$

With this high \$Q\$, the transfer function magnitude rapidly changes near the resonance frequency so simulation round off may greatly affect the results.

I tried \$R=100 \Omega\$ which reduced the \$Q\$ by a factor of 10 and the simulation output is very nearly the input after a number of cycles.

Update:

I adjusted the "step ceiling" parameter for the transient simulation down to \$10^{-12}\$ and I get a good simulation with \$R=1k\Omega\$.