Can I measure resistance of an inductor on a bridge?
Your bridge measurement tends to give unloaded Q. Perhaps you have determined the "R" from your bridge measurement, and used that value in your RCL resonant circuit. Your bridge may measure inductor parameters at a fixed frequency (often 1 kHz) whereas your circuit resonance is at a far different frequency. Be aware that inductor characteristics are often not frequency independent. So that is one caveat. You are probably correct to assume that the capacitors are lossless.
What you've neglected is the coupling from your signal source, and load, which also contributes to "R". So "loaded Q" is always less than raw inductor Q.





simulate this circuit &ndash; Schematic created using CircuitLab

The circuit at left has an inductor whose self-resistance yields a Q of 15 at resonance. Yet the resonant voltage across R2 yields a \$ \frac{\omega}{\Delta\omega}\$ loaded Q much less, because total R is now 171 ohms.
The circuit on right has much lower Q to the point where it really cannot be called a resonant circuit. R4 loads the resonator (L2,R5,C2) so much that loaded Q&lt;&lt;1.