At steady state current sources act as

This chapter covers analysis of circuits when the source is sinusoidal.
The analysis techniques are exactly the same as those used when the source
was DC. What makes AC analysis more difficult is the mathematics, as will
be seen in the next section. Fortunately there are mathematical tools and
short cuts, such as using Phasors.

Review of Sinusoidal Sources

• An amplitude
• A frequency
• A Phase Angle

The following expession represents an AC source.

Here are some examples of AC sources:

Review of Complex Numbers

In order to perform AC analysis, one must understand how to
work with complex numbers.

Complex Addition and Multiplication

Show why is j*j = -1?

AC Steady State Analysis

We are now ready to analyze a simple circuit with an AC Source.
Do not panic because after we introduce phasors this problem will
be significantly easier.

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 e j2t cancelled out of the math.
Therefore to determine B we did not need the e j2t .

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.

As we will see in the next section, the introduction of the phasor
helps simplify the math by eliminating the radian frequency from the math.

Definition of a Phasor

A phasor represents the magnitude and phase ONLY of a voltage or current.

In this table we will convert each voltage or current into its phasor form:

I = -5 cos(5t + 30 o ) The phasor for I is:

I = -10 sin(3t - 30 o ) Here we will convert from sine to cosine:
As seen in section one, we subtract 90 o to convert from sine to cosine.
I = -10 cos(3t -30 o - 90 o ) = -10 cos(3t - 120 o )

Also recall from math that cos(wt) = -cos(wt +/- 180 o )
So I = 10 cos(3t - 120 o + 180 o ) = 10 cos(3t + 60 o )

The phasors for I are:
I = -10/-120
or
I = 10/60

V = 30 sin(10t - 180 o ) Again we convert from sine to cosine:
V = 30 cos(10t - 180 o - 90 o ) = 30 cos(10t - 270 o )
V = 30 cos(10t - 270 o + 360 o ) = 30 cos(10t + 90 o )

The phasors for V are:
V = 30/-270
or
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.

Phasors are a very compact form of writing amplitude and phase for a voltage or current.

AC Analysis using Phasors

Let's begin analysis using phasors by considering single element circuits.
We will consider a resistor circuit, a capacitor circuit, and an inductor circuit.

The above Voltage-Current relationships for the Inductor, the Capacitor,
and the resistor will significantly simplify AC analysis.

We can now apply Ohm's Law to Inductors, Capacitors, and Resistors using Phasors.

Study Problems

After clicking on the following link enter 9-1 for the problem and 1 for the step:
Study Problem 9-1

After clicking on the following link enter 9-2 for the problem and 1 for the step:
Study Problem 9-2

After clicking on the following link enter 9-3 for the problem and 1 for the step:
Study Problem 9-3

After clicking on the following link enter 9-5 for the problem and 1 for the step:
Study Problem 9-5