W9_parallel_resonance.eps

310 Week 9: Alternating Current Circuits

Note that the solution to the homogeneous ODEdecays in timeexponentially. It is atransient

contribution to the overall solution and after many lifetimesτL=R/Lit will generally be

negligible.

The remaining particular part 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

chargeQto 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) (673)

then solving the problem is easy^87. We begin by noting the voltage drops across all three

circuit elements in terms ofI(t):

VR = I 0 Rsin(ωt−φ) (674)

VL = I 0 χLsin(ωt−φ+π/2) (675)

VC = I 0 χCsin(ωt−φ−π/2) (676)

or

I 0 Rsin(ωt−φ) + I 0 χLsin(ωt−φ+π/2)

+ I 0 χCsin(ωt−φ−π/2) =V 0 sin(ωt) (677)

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

ωt

Vosin( )ωt

Vo

IχL

I χC

o

o

φ I Ro

Figure 118: A phasor diagram for theLRCcircuit.

the phasors on the diagram that are proportional toI 0 must add up to produceV 0 sin(ωt),

and thismust 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:

(^87) 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.