W9_parallel_resonance.eps

324 Week 9: Alternating Current Circuits

The remaining particular solutionQp(t) is therefore called thesteady statepart of the solution,
and it persists indefinitely, as long as the driving voltage remains turned on. We expect that the
time dependence of the steady state solution beharmonic(like the applied voltage) and to have the
same frequencyas the applied voltage. However, there is no particular reason to expect the charge
Qto bein phasewith the applied voltage.

We will find it slightly more convenient to work at first with the currentIthan the chargeQ–
we can always findQ(t) (orVC) by integration andVLby differentiation – although when we go to
a complex formulation it won’t matter. If we make theguess:

I(t) =I 0 sin(ωt−φ) (738)

then solving the problem is easy^93. We begin by noting the voltage drops across all three circuit
elements in terms ofI(t) (where we use the rules we derived abovebackwardsas we are given the
current and seek the voltage):

VR = I 0 Rsin(ωt−φ) (739)

VL = I 0 χLsin(ωt−φ+π/2) (740)

VC = I 0 χCsin(ωt−φ−π/2) (741)

or (substituting into Kirchoff’s loop rule for the voltage):

I 0 Rsin(ωt−φ) + I 0 χLsin(ωt−φ+π/2)

+ I 0 χCsin(ωt−φ−π/2) =V 0 sin(ωt) (742)

Our goal, then, is to find values ofI 0 andφfor which this equation istrue. This is quite simple.
Suppose I use aphasor diagramto add the trig functions graphically: They-components of the

ωt

Vosin( )ωt

Vo

IχL

I χC

o

o

φ I Ro

Figure 127: A phasor diagram for the seriesLRCcircuit.

phasors on the diagram that are proportional toI 0 must add up to produceV 0 sin(ωt), and this
must be trueif we add up the phasors as shown, taking advantage of our knowledge of the phase of
the voltage drop across the various elements relative to the current through those elements.

If we letV 0 =I 0 ZwhereZis called theimpedanceof the circuit, we cancanceltheI 0 and get
the following triangle for the impedance: From this triangle we can easily see that:

Z=

√

R^2 + (χL−χC)^2 (743)

(^93) This isn’t really a guess. If we were to solve the differentialequation ”properly” using fourier transforms and
using a complex exponential sourceV 0 eiωtwe would discover that the complex solution for the current has a complex
amplitude and phase determined from an algebraic equation.We are simply making the guess here because many
students don’t know enough math yet to handle this approach,although this may change in some future edition of
this book.