W9_parallel_resonance.eps

Week 9: Alternating Current Circuits 321

To go on, we need to introduce a classical harmonic oscillating voltagelike that produced by
an AC generator. Our first step is to determine what the relationship is between voltage (provided
by the generator) acrosseachcircuit element, one at a time, and the currentthroughthat circuit
element as a function of time. We begin with the resistor, as the easiest to understand and as a
model for the other two.

A Harmonic AC Voltage Across a ResistanceR.

V(t) R

I(t)

Figure 123: AC voltage acrossR

Consider the circuit diagram in figure 123, portraying an alternatingvoltageV(t) =V 0 sin(ωt)
placed across a resistanceR. Applying Kirchhoff’s voltage/loop rule and Ohm’s Law to the circuit
loop, we get the following equation of motion for the circuit:

V 0 sin(ωt)−IR= 0 (719)

or (solving for the desired current):

IR(t) =

V 0

R

sin(ωt) (720)

and we see that the current isin phasewith the voltage drop across a resistor.

A Harmonic AC Voltage Across a CapacitanceC

V(t)

I(t)

C

Figure 124: AC voltage acrossCR

Proceeding the exact same way, we use Kirchhoff’s voltage rule and the definition of capacitance
to get an equation of motion:

V 0 sin(ωt)−

Q

C

= 0 (721)

We wish to findIC(t), the current through the capacitor. To get it, first we solve forQ(t):

Q(t) =CV 0 sin(ωt) (722)

and then we differentiate (and use the trigonometric identity cos(θ) = sin(θ+π/2) to express the
result in terms of the original harmonic function and a phase):

IC(t) =

dQ(t)

dt

= (ωC)V 0 cos(ωt)

= (ωC)V 0 sin(ωt+π/2) (723)