W9_parallel_resonance.eps

322 Week 9: Alternating Current Circuits

We observe in this result a quantity that behaves like the “resistance” of the capacitor in an AC
circuit, regulating the magnitude of the current as the frequencychanges much the way a resistor
would as it increases or decreases. We will give this quantity its own name – thecapacitive reactance
of the capacitor at the angular frequencyω– and define it to be:

χC=^1

ωC

(724)

Note well that the units ofχCareohms.

Using the capacitive reactance, the peak current in the circuit takes on a more familiar form:

I 0 = (ωC)V 0 =

V 0

χC

(725)

so that
IC(t) ==I 0 sin(ωt+π/2) (726)

We see that the current isπ/ 2 ahead in phaseof the voltage drop across the capacitor. We will
actually usually use this in series AC circuits with capacitors the otherway around and note that
the voltage drop across the capacitor isπ/ 2 behindthe current through it.

A Harmonic AC Voltage Across an InductanceL

V(t)

I(t)

L

Figure 125: AC voltage acrossL

We repeat this process one more time for an inductanceL. The methodology is basically the
same: We use Kirchhoff’s voltage rule and the definition of capacitance to get:

V 0 sin(ωt)−L

dI

dt

= 0 (727)

This time we solve fordI(t):

dI=V^0

L

sin(ωt)dt (728)

and integrate both sides to get:

IL(t) =

∫

V 0

Lsin(ωt)dt

=

∫

V 0

ωL

sin(ωt)ωdt

=

V 0

ωL

cos(ωt) (729)

=

V 0

ωL

sin(ωt−π/2) (730)

We define theinductive reactance:

χL=ωL (731)

(in ohms once again) so that:
IL(t) =I 0 sin(ωt−π/2) (732)