This circuit is also a band pass filter. High frequencies are shorted by the capacitor and low frequencies by the inductor.
It can be proven that:
where Zr is the impedance at resonance freqency and Q is the quality factor of the inductor. This equation is an approximation, but the Q of most inductors is so large that it can almost always be used.
Some complex math proves that the impedance reaches its maximum value at:
Again, this is an approximation that assumes the series resistance of the inductor is so small that it doesn't affect the frequency.
In our case f is 159kHz again. If rL is 20 ohms, Q is 1k/20=50. Zr will be 50∙1k=50k. The current and the input voltage are in phase, so:
The image below shows the frequency response.
The bandwidth is about 160.7kHz-157.5kHz=3.2kHz.
If R»Zr and Q is large enough, then: B=f/Q. In our case B=159kHz/50=3.18kHz.
And this is finally the best way to determine the Q of an inductor:
Adjust the source frequency to the frequency you want to know the Q of;
Adjust C to maximum output voltage;
Turn the source frequency down until the output voltage decreased by 3dB (=factor 1.41);
Increase the source frequency until the output voltage is again 3dB less than the maximum voltage;
Subtract both frequencies. This is the bandwidth;
Calculate Q: Q=f/B
With this method, the internal resistance of the source doesn't matter, and we don't run the risk of a high current flow.