Week 9: Alternating Current Circuits 327
Next, let’s factor the power into a slightly more convenient form:Pav =Vrms^2 R
R^2 + (ωL−ωC^1 )^2=Vrms^2 R
R^2 +Lω^22 (ω^2 −LC^1 )^2= V
rms^2 Rω^2
R^2 ω^2 +L^2 (ω^2 −ω^20 )^2(762)
This function is plotted, forL=C=V 0 = 1.0 and several values ofR(and henceQ) in figure 129.
Whenω→0,Pav→0 likeω^2. Whenω→ ∞,Pav→0 like 1/ω^2. In between, it clearly peaks at
ω=ω 0 , resonance, with a peak power as given above. We are thus almost ready to draw ageneric
shapefor theresonance curve, the power delivered to the circuit as a function of frequency.
(^00) 0.5 1 1.5 2
5
10
15
20
omega
Power
Q = 3
Q = 10
Q = 20
Resonance of Series LRC Circuit
Figure 129: A typical series of resonance curves forQ= 3, 10 ,20, plotted on a scale such thatω 0 = 1
:L=C= 1.0, andR= 0. 3333 , 0. 1 , 0 .05.
To do so, however, we need to learn one last concept that is extremely useful in understanding the
behavior of electrical band pass circuits: theQ-factor(quality factor) of the circuit. TheQ-factor
of a circuit is defined to be:
Q=
ω 0
∆ω
(763)
where ∆ωis thefull widthof the resonance curve athalf-maximum. TheQ-factor is a measure of the
sharpnessof the resonance. A circuit with a lowQ-factor delivers significant power to the circuit for
frequencies far from resonance (although asymptotically the power still vanishes at zero and infinity
as given above). A circuit with a highQ-factor has asharply peaked resonance curvethat goes to
zeroquicklywhenωis far from resonance – power is delivered to a circuit only for frequenciesvery
close to the resonance frequency.
In the homework you will be asked to derive the relation:Q=
ω 0 L
R=
L
√
LCR
=
1
R
√
L
C
=
ω 0
∆ω(764)
In figure 129, you can see how decreasingR(withV 0 ,L, andCall fixed at one) causes the resonance
tosharpen up– become much narrower at half-max – at the same time it increases the maximum
power delivered at peak dramatically.