W9_parallel_resonance.eps

Week 9: Alternating Current Circuits 309

We integrate both sides to get:

IL(t) =

∫

V 0

L

sin(ωt)dt

=

∫

V 0

ωL

sin(ωt)ωdt

=

V 0

ωL

cos(ωt) (664)

= V^0

ωL

sin(ωt−π/2) (665)

= I 0 sin(ωt−π/2) (666)

(667)

where

I 0 =

V 0

ωL

=

V 0

χL

(668)

We see that the current isπ/ 2 behind in phaseof the voltage drop across the inductor. We

will actually usually use this the other way around and note that the voltage drop across the

inductor isπ/ 2 aheadof the current through it. We call the quantityχL=ωL(which clearly

has the units of Ohms) theinductive reactance, the “resistance” of an inductor to alternating

voltages.

• The Series LRC Circuit:We apply Kirchhoff’s voltage/loop rule to this circuit and get:

I(t)

0

C

R

L

V sin( t)ω

Figure 117: A seriesLRC(tank) circuit.

V 0 sin(ωt)−LdI

dt

−RI−Q

C

= 0 (669)

or

VL+VR+VC=V 0 sin(ωt) (670)

or

d^2 Q

dt^2 +

R

L

dQ

dt+

1

LCQ=

V 0

Lsin(ωt) (671)

There are a number of way to solve this second order, linear,inhomogeneousordinary differ-

ential equation. We will first show a simple one that relies on a “guess”, then we will show

how if we use complex exponentials we really don’t have to guess.

Our goal will be to solve for all voltage drops, the current in the circuit, the power delivered

to each circuit element and the entire circuit as a whole – pretty mucheverything.

The first thing to note that if we find at least one “particular” solutionQp(t) to the inhomoge-

neous ODE, we can construct a new solution by addinganysolution to thehomogeneousODE

(the undrivenLRCcircuit solved above) and still get a solution. That is, a general solution

can be written:

Q(t) =Qp(t) +Qh(t) (672)