Calculating Q for a Series LC Circuit: Where Did I Go Wrong?

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SUMMARY

The discussion centers on calculating the quality factor (Q) for a series LC circuit using the formula Q = XL / R, where XL is the inductive reactance and R is the resistance. The user initially calculated Q using Q = 1/R * (L/C) with a 1.923 Henry inductor, a 4.4 µF capacitor, and a resistance of 52.5 ohms, resulting in an incorrect value of 111.86. Upon further analysis, the correct Q value at resonance was determined to be approximately 12.6, indicating a misunderstanding in the application of the Q formula.

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Idea04
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For a series LC circuit the Q I believe is Q= 1/R^(L/C). I have a 1.923 Henry inductor and a 4.4uF cap. The resistance in the inductor is 52.5 ohms. The answer I got was 111.86.
I know the circuit can't have a Q of that value. Where did I go wrong?
 
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This looks like homework, so I'll just describe the process.

The frequency wasn't given, so I guess you mean at resonance.

The circuit resonates at ... Hz.

The reactance of the coil and capacitor at resonant frequency is ... ohms.

Q= XL / R

I get a Q of about 12.6.
 
I thought it might be homework as well, but I'm not sure. Idea, could you please post your math so we can check it? Thanks.
 

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