W9_parallel_resonance.eps

338 Week 9: Alternating Current Circuits

We wish to evaluate theQ-factorfor this resonant circuit, as this is an important design parameter
for band-pass filters such as those used in radios.

If you did part b) of the previous problem correctly, you should have found that:

Pav(ω) =Iav^2 R=

Vrms^2 Rω^2

L^2 (ω^2 −ω 02 )^2 +ω^2 R^2

is the average power delivered to the circuit by the voltage and is alsothe average power “burned”
by the resistor, since the inductor and capacitor do not dissipate energy and there is no net work
done per cycle upon them. In this expressionω 0 =

√

1 /LCas you should fully understand at this
point.

Show that for a sharply peaked resonance (one with largeQ):

∆ω≈R

L

so that

Q=

ω 0

∆ω≈

ω 0 L
R
where ∆ωis the full width at half maximum of the power curve you derive in the first part.

To do this, set the expression above equal to thecomputedhalf-maximum power, and solve for
the two quadratic roots forω, assuming that both of them arevery closeto (but not equal to)ω 0
(this is the sharply peaked part). You may find the following factorization useful:

ω^2 −ω^20 = (ω−ω 0 )(ω+ω 0 )