ac theory question

I'm really struggling with this question. You're given an LCR circuit which has resonance at a frequency of 15kHz which gives a current of 0.2A. And then, at a 15.1 kHz frequency, you get a current of 0.03A.

Now the question I'm struggling with says: Explain briefly in terms of Q factor why the current at 15.1 kHz is far lower than at resonance

What does the Q factor have to do with this? Doesn't it just describe the sharpness of the peak? There isn't a mark scheme for this question so I'm really struggling with how you'd answer it. Any help would be much appreciated thanks!

Reply 1

Ferret!

The Q factor is proportional to the inverse of bandwidth.

Q factor is also proportional to the inductance of the coil (L) in this circuit over the internal resistance of the coil (r_L), that is Q α L / r_L.

Then, the dynamic resistance (R_D) of the inductor-capacitor circuit is proportional to L / r_L.

From Ohm's law, this dynamic resistance is proportional to the inverse of current flowing through.

So, as the circuit seems to have a very small bandwidth, the Q factor will be very high -> L / r_L will be high -> R_D will be high -> current flowing through the parallel circuit will be low.

This is what we want at the resonant frequency, so most current goes through to the load instead.

The opposite will happen though for frequencies outside the bandwidth, that is the current in the LC circuit will be high, so the load will have far lower current than at resonance.