W9_parallel_resonance.eps

326 Week 9: Alternating Current Circuits

Again we don’t much care about the peak values, but the averages are important. The time averages
of sin^2 (ωt−φ) is 1/2. The time average of sin(ωt−φ±π/2) sin(ωt−φ) is zero (why?)! Thus the
average power delivered to the circuit is delivered to theresistor only:

< PR(t)>=^1

2

I^20 R=< P(t)>=Pav (754)

from above. We see that the energy flowing in and out of the capacitor and inductor may be very
large (if e.g. I^20 χLis large), but theyuseno energy and hence in every cycle the net energy they
absorb from the circuit equals the net energy the return to the circuit!

Now let us considerPav, understanding that it is delivered to theresistor(or circuit element such
as an amplifier that behaves like a resistance in the circuit) and not the inductance or capacitance.

Pav=

V 02 R

2 Z^2

=

Vrms^2 R

R^2 + (χL−χC)^2

=

Vrms^2 R

R^2 + (ωL−ωC^1 )^2

(755)

where we have introduced theroot-mean square voltage

Vrms=

1

√

2

V 0 (756)

as a form of the voltage that lets us drop the pesky factor of 1/2 that frequently arises from
averages in harmonic circuits (and leaves us with quantities that lookmore like their direct current
counterparts, easier to remember).

We will often want to express this quantity in terms ofimpedance(which determines the current)
and a convenient quantity (that we saw arose quite naturally above):

Pav=

V 02

Z

R

Z=

1

2 V^0 I^0 cosφ=VrmsIrmsPf (757)

where

Pf= cos(φ) =

R

Z

(758)

is called thepower factorof the circuit. WhenPf= 1,Z=Rand the load is said to beentirely
resistive. A lightbulb plugged into a wall is an example of a purely resistive load. When the power
factor is different from one, in general thepeakpower delivered to the circuit is much greater than
theaveragepower, which means that the power supply has to deliver much largerpeak voltages
than you expect from the power rating of the appliance being used.This can in turn blow fuses or
circuit breakers for a load that a circuit “should” be able to manage.

Let us consider the variation of the average power with frequencyfor

fixed circuit elements. The first thing to note is that power is obviously at a max whenχL=χC:

ωL =

1

ωC

ω^2 =

1

LC=ω

2

0 (759)

whereω 0 is theresonant frequencyof the circuit. At resonance the power delivered to the circuit
is:

Pav,max=V

rms^2 R

R^2

=V

rms^2

R

(760)

just as we would expect for a DC circuit. If we use peak instead of rms voltage, of course, we have
to put back the factor of 1/2:

Pav,max=

V 02

2 R

(761)