330 Week 9: Alternating Current Circuits
Far from resonance on either side, eitherχCorχLwill be verysmall– in particular much less
thanR. The current produced by the voltage will thus find either the capacitor (for high frequencies)
or the inductor (for low frequencies) to be a much easier path to ground, provided only that the
load resistanceRis bigger.
If the voltage were anidealvoltage withr= 0, capable of deliveringanyamount of current,
this wouldn’t matter. As the impedance of the parallelLCcombination drops, it would simply
provide more current and maintain its voltage, while continuing to deliver as much current to the
resistor as before. However, many voltage sources – in particulararadio antenna– have a signficant
impedance/resistance of their own, and if they are provided with aneasy path to ground thisshorts
outthe antenna by pulling enough current from it so that its pole voltagedrops to zero (or at any
rate a very small number), reducing the current through the resistor to zero at the same time.
This suffices to show that there should be amaximumpower delivered to the resistance when one
is at resonance and the current has no alternative pathway to ground through theLCcombination,
but it does not suffice to show what the characteristics of the power curve are. To solve this problem
exactly, one has to write Kirchoff’s laws for the entire circuit, reduce them to an algebraic form, and
then solve that form. This is rather painful to do working with trig functions, somewhat easier with
complex exponentials, and beyond the scope of this course.
However, we can at least comment on certain aspects of the solution and show a curve or two
(for the benefit of any would-be crystal radio builders). First, although it is far from obvious, the
power delivered to the load resistor (headphones) will be maximum if its resistance more or less
matchesthe resistance of the antenna. This is called “impedance matching” (impedance because in
general one has to account for more than just resistance). Onecan in fact prove a result known as
theMaximum Power Theorem^94 orJacobi’s Lawthat states that in general when a power source
has a complex internal impedanceZSand the load has a complex impedanceZL, maximum power
is transferred when
ZL=Z∗S (773)or the impedance of the load has the same amplitude but the oppositephase of the source. This
theorem works for purely resistive loads – in fact in its simplest application it simply describes the
energy distribution between two resistorsRSandRLin series! Hence one needs to design a radio
(when possible) to match the impedance of the antenna one hopes to use with it; if one doesn’t one
either burns too much of the received energy in the antenna itself (when the impedence of the load
is too small) or one eliminates one’s ability to discriminate the signal.
We can do a somewhat sloppy job of estimating the power delivered tothe load resistor with the
following argument. SupposeZis the impedance of the parallel circuit above andris the resistance
of the source. Then we expect the total impedance of the circuit to beZ′=r+Z(where if we don’t
use complex numbers we will have to separate out and add separately the resistive component ofZ
tor). Thetotalcurrent drawn from the source is thus approximatelyI 0 =V 0 /Z′. We can then find
the “corrected” source voltage across the resistanceRas a (phase shifted)VR=V 0 −I 0 r, and the
power delivered to it is thus approximately:
PR=
VR^2
R (774)
We plot this very approximate function, computed in just this way, for a range of values ofω
around the resonant frequencyω 0 = 1/LC= 1.0 as before in figure 133. The voltage and resistance
have been mutually adjusted to make the picture pleasing, withr= 10Ω. Note that we do indeed
see peak power delivery to the load whenR≈10Ω as expected at least for the three values forR
shown. Note how theQvalue of the circuit visibly changes withRfor the fixedLas well.
(^94) Wikipedia: http://www.wikipedia.org/wiki/Maximum Power Theorem.