W9_parallel_resonance.eps

Week 9: Alternating Current Circuits 325

R

Z

φ

χL− χC

Figure 128: The impedance diagram for theLRCcircuit.

so that

I 0 =

V 0

Z (744)

and

φ= tan−^1

(

χL−χC

R

)

(745)

Power in a SeriesLRCCircuit

Power in this circuit is worth a section all its own, as understanding power delivery to the circuit
is essential to the understanding ofwhythis circuit is useful. The seriesLRCcircuit functions as a
band pass filterto an applied harmonic voltage. Basically, it only allows a large current to flow (and
deliver power to the circuit) when the frequency of the applied voltage is (nearly) thesameas the
resonant frequency for the circuit:

ω 0 =

√

1

LC

(746)

obtained above for the undrivenLCcircuit. For frequencies far from resonance, the current delivered
and power dissipated in the circuit rapidly goes to zero. Let us understand this.

First we consider the power delivered to or by each circuit element. The power deliveredbythe
voltagetothe circuit is just:

P(t) =V(t)I(t) =V 0 sin(ωt)I 0 sin(ωt−φ) (747)

If we use the trig identity:

sin(A−B) = sin(A) cos(B)−cos(A) sin(B) (748)

(which can be trivially proven with complex exponentials) andI 0 =V 0 /Zwe get:

P(t) =

V 02

Z

(

sin^2 (ωt) cos(φ)−sin(ωt) cos(ωt) sin(φ)

)

(749)

We don’t usually care about theinstantaneouspower delivered to the circuit (although there
are very definitely exceptions, such as when the peak power ismuch largerthan the average power,
which can stress e.g. transformers that might be providing the power). If we time average this to
obtain theaveragepower we get:

< P(t)>=Pav=

V 02

2 Z

cos(φ) =

V 02 R

2 Z^2

=

1

2

I^20 R (750)

where we used the fact that the time average of the square of anyharmonic function of time is 1/2,
the fact that cos(φ) =R/Zfrom the impedance triangle above, and the fact thatI 0 =V 0 /Z. Note
as well that the time average of sin(ωt) cos(ωt) iszero(why?) so that the second term does not
contribute.

Now consider each of theothercircuit elements separately:

PR(t) = VR(t)I(t) =I 02 Rsin^2 (ωt−φ) (751)

PL(t) = VL(t)I(t) =I 02 χLsin(ωt−φ+π/2) sin(ωt−φ) (752)

PC(t) = VC(t)I(t) =I 02 χCsin(ωt−φ−π/2) sin(ωt−φ) (753)