|Review of Complex Numbers|
|AC Steady State Analysis|
|Definition of a Phasor|
|AC Analysis using Phasors|
|This AC Circuit is in Steady State.
This means that all of the AC voltages and currents
are constant (not flat like DC but unchanging).
Note there are many detailed steps, but if you can
follow this problem then you will be able to grasp
what a phasor is in the next section.
Note that in this example the ej2t cancelled out of the math.
Therefore to determine B we did not need the ej2t.
The radian frequency (the '2' in this case) is only needed when differentiating the KVL equation. (see step 3)
In summary, the radian frequency determines the coefficients of the equation
but after step 3 the radian frequency is simply in the way.
|The phasor for V is:
V = 30/-45
V = -30/135
|The phasor for I is:
I = -5/30
I = 5/-150
|Here we will convert from sine to cosine:
As seen in section one, we subtract 90o to convert from sine to cosine.
I = -10 cos(3t -30o - 90o) = -10 cos(3t - 120o)
Also recall from math that cos(wt) = -cos(wt +/- 180o)
So I = 10 cos(3t - 120o + 180o) = 10 cos(3t + 60o)
The phasors for I are:
I = -10/-120
I = 10/60
|Again we convert from sine to cosine:
V = 30 cos(10t - 180o - 90o) = 30 cos(10t - 270o)
V = 30 cos(10t - 270o + 360o) = 30 cos(10t + 90o)
The phasors for V are:
V = 30/-270
V = 30/90
Note that there are many ways of writing one phasor. As long as each phasor value
represents the same vector when graphed, then all are equivalent.
One can see that Ohm's Law, V=IR, holds true for phasors.|
The above Voltage-Current relationships for the Inductor, the Capacitor,
and the resistor will significantly simplify AC analysis.