W9_parallel_resonance.eps

308 Week 9: Alternating Current Circuits

V(t)

I(t)

C

Figure 115: AC voltage acrossC

• AC voltage across a capacitanceC:
  We use Kirchhoff’s voltage rule and the definition of capacitance to get:

V 0 sin(ωt)−

Q

C= 0 (658)

We can solve forQ(t):

Q(t) =CV 0 sin(ωt) (659)

Finally, we note that:

IC(t) =

dQ(t)

dt

= (ωC)V 0 cos(ωt)

= (ωC)V 0 sin(ωt+π/2) =I 0 sin(ωt+π/2) (660)

where

I 0 = (ωC)V 0 =

V 0

χC

(661)

We see that the current isπ/ 2 ahead in phaseof the voltage drop across the capacitor. We

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

capacitor isπ/ 2 behindthe current through it. We call the quantityχC=ωC^1 (which clearly

has the units of Ohms) thecapacitative reactance, the “resistance” of a capacitor to alternating

voltages.

• AC voltage across an inductanceL:

V(t)

I(t)

L

Figure 116: AC voltage acrossL

We use Kirchhoff’s voltage rule and the definition of capacitance to get:

V 0 sin(ωt)−LdI

dt

= 0 (662)

We can solve fordI(t):

dI=

V 0

L

sin(ωt)dt (663)

(^86) Not really.