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.