In EET 1150 you learned how to analyze circuits containing **DC** (direct-current)
sources. In this course you’ll learn how to analyze circuits containing **AC** (alternating-current)
sources. AC electricity is more complicated than DC, so this unit
will introduce some concepts and terms that we’ll use for the rest
of the course.

But we will pick up some new ideas, such as angular frequency and a way to write mathematical expressions for sinusoidal waves. For instance, here’s a mathematical expression that describes a particular sinusoidal voltage:

*v* = 4.62 V
sin(968*t* + 50°).

At first sight, this may look complicated, but with a bit of practice you’ll be able to interpret and use expressions of this kind.

In this unit we’ll get a start on analyzing AC circuits. Fortunately, **resistors** in
AC circuits behave pretty much the same way they behave in DC circuits.
Therefore, you can use circuit rules that you already know–such as
Ohm’s Law, Kirchhoff’s Voltage Law, and Kirchoff’s Current Law–in pretty
much the same way that you used them in DC circuits. (But as we’ll see
in later units, **inductors and capacitors** behave one way in DC
circuits, and they behave entirely differently in AC circuits.)

- This unit will build on material that you studied in Unit 1. So let’s begin by taking this self-test to review what you learned in that unit.

**Direct current**(**DC**) is current that flows in one direction only.- A
**DC voltage source**is a voltage source that produces direct current.- Examples: Batteries and dc power supplies (such as the power supply built into the trainer that you use in lab) are DC voltage sources.

**Alternating current**(**AC**) is current whose direction periodically reverses.- An
**AC voltage source**is a voltage source that produces alternating current.- Examples: Electrical outlets in the walls of your home provide alternating current. The trainer that you use in lab also contains an AC voltage source called a function generator.

- In most
**DC**circuits, current and voltage remain constant as time passes. But in**AC**circuits the voltage and current change as time passes. - In Unit 2 of this course we’ll see how to write down mathematical expressions that describe how the values change in time. But a simpler and more common way is to draw diagrams showing how the voltage or current changes in time.
- Such a graph of a current or voltage versus time is called
a
**waveform**. Below are several examples of voltage waveforms. Notice that each of these diagrams plots voltage (measured in volts or millivolts) on the vertical axis, and time (measured in microseconds or milliseconds) on the horizontal axis. - First, here is an example
of a
**triangle waveform**. In a triangle waveform, voltage increases gradually along a straight line to its maximum value, and then gradually decreases along a straight line to its lowest value, and then starts increasing again.

- Next, here is an example of a
**square waveform**. In a square waveform, voltage remains constant at a high value for a while, then suddenly drops to a low value, where it stays for a while before suddenly jumping back to its high value.

- Both of the examples just shown are
**AC waveforms**, because the voltage actually changes polarity. (In other words, the voltage is positive sometimes and negative sometimes.) - On the other hand, you could have a waveform whose polarity does
not change, even though its value does change as time passes. This
would be a
**pulsating DC waveform**. Here is an example:

- The waveform studied most frequently in electrical circuit theory
is the
**sine wave**. Here’s an example:

- It’s called a sine wave because this is the same shape that you get if you make a plot of the mathematical function y = sin(x), with x plotted on the horizontal axis and y on the vertical axis.

- Strictly speaking, a sine wave must pass through the origin (the point where the x-axis crosses the y-axis).
- The more general term
**sinusoid**is used to describe any waveform that has the same shape as a sine wave but that may be shifted to the right or to the left along the x-axis. The waveform shown just above is an example of a sinusoid. So is the following waveform, which has the same shape but is shifted horizontally so that it does not pass through the origin:

- Here’s a remarkable feature of sinusoidal waveforms that makes them particularly easy to work with. Suppose you connect a function generator to any circuit containing resistors, inductors, and capacitors. If the function generator is set to produce a sinusoidal waveform, then every voltage drop and every current in the circuit will also be a sinusoid.
- The same thing is
**not**true for waveforms of other shapes. For instance, if the function generator is set to produce a triangle waveform, and if the circuit contains any inductors or capacitors, then the voltage drops across the components will be complicated waveforms, not simple triangle waveforms. - Fortunately, it turns out that sinusoids are not only the easiest waveforms to work with, they’re also the most useful. Therefore we’ll concentrate primarily on analyzing AC circuits that have sinusoidal voltage sources, rather than triangle or square-wave or other voltage sources.

- A
**periodic waveform**is a waveform whose values are repeated at regular intervals. - All of the waveforms shown above are periodic waveforms.

- Important parameters associated with periodic waveforms include:
- Period
- Frequency
- Instantaneous Value
- Peak Value
- Peak-to-Peak Value
- RMS Value (also called effective value)
- Average Value

- Each of these terms is explained below.

- The plot of a periodic waveform shows a regularly repeating pattern
of values, each of which is called a
**cycle**. **Example**: In the picture of the sine wave shown below, we see a little less than two full cycles. The first cycle extends from 0 ms to 50 ms, and the second (incomplete) cycle extends from 50 ms to the edge of the chart, where it is cut off:

- The time required for the values to rise and fall through one complete
cycle is called the
**period**of the waveform. - The symbol for period is
.*T* - Period is measured in units of seconds, abbreviated
**s**. **Example**: The sine wave shown above has a period of 50 ms.

- The
**frequency**of an ac waveform is the number of cycles that occur in one second. - The symbol for frequency is
.*f* - Frequency is measured in units of cycles per second, or Hertz, abbreviated
**Hz**.

- Period and frequency are the reciprocal of each other:

*f*= 1 ÷*T*

and

*T*= 1 ÷*f* **Example**: The sine wave shown earlier has a period of 50 ms. Therefore, its frequency is 20 Hz.

- The
**instantaneous value**of an ac waveform is its value at a specific instant of time. - We’ll see below how you can use the mathematical expression for a waveform to find the waveform’s instantaneous values at specific times.
- You can read off approximate instantaneous values from the graph of a waveform.
- Example: At 20 ms, the sine wave shown earlier has an instantaneous value of about 300 mV.

- The maximum value reached by an ac waveform is called its
**peak value.**- If the waveform is a voltage waveform, then its peak value
is also called its
**peak voltage**, abbreviated*V*._{p} - If the waveform is a current waveform, then its peak value
is also called its
**peak current**, abbreviated*I*._{p}

- If the waveform is a voltage waveform, then its peak value
is also called its
- The peak value of a waveform is sometimes also called its
**amplitude**, but the term “peak value” is more descriptive. **Example**:The sine wave shown earlier has a peak value of 500 mV_{p}. Notice that I write a "p" after the unit to show that I’m talking about a peak value.

- The
**peak-to-peak value**is the difference between a waveform’s positive peak value and its negative peak value.- If the waveform is a voltage waveform, then its peak-to-peak
value is also called its
**peak-to-peak voltage**, abbreviated*V*._{pp} - If the waveform is a current waveform, then its peak-to-peak
value is also called its
**peak-to-peak current**, abbreviated*I*._{pp}

- If the waveform is a voltage waveform, then its peak-to-peak
value is also called its
- If the waveform is symmetrical about the time axis, then the peak-to-peak value equals twice the peak value.
**Example**: The sine wave shown earlier has a peak-to-peak value of 1 V_{pp}. That’s the difference between the maximum positive value (which is 500 mV) and the maximum negative value (which is -500 mV). Notice that I wrote a "pp" after the unit to show that I’m talking about a peak-to-peak value.

- As mentioned in the two examples above, we write
**p**as the subscript of a quantity or unit to show that we’re talking about a peak value, and we write**pp**as the subscript of a quantity or unit when we’re talking about a peak-to-peak value. **Example**: For the sine wave above, we could write

or*V*_{p}= 500 mV_{p}*V*_{pp}= 1 V_{pp}.- Similarly, if we were dealing with a current waveform whose peak
value is 20 mA and whose peak-to-peak value is 40 mA, we could write

or*I*_{p}= 20 mA_{p}*I*_{pp}= 40 mA_{pp}. - Some textbooks use
**pk**(instead of**p**) as the abbreviation for peak values, and**p-p**(instead of**pp**) as the abbreviation for peak-to-peak values.

- The
**function generator**, or**signal generator**, is an instrument designed to produce ac waveforms, such as square waves, triangle waves, and sine waves. Using it, you can set the peak value and the frequency of these waveforms. - Here is a photo of the built-in function generator on the trainers
that we use in our electronics labs. It provides the basic controls
that any function generator must have:
- an
**Amplitude**control to set the waveform’s peak value **Frequency**and**Range**controls to set the waveform’s frequency- a
**Function**control to set the shape of the waveform.

- an
- The photo below shows a professional-quality function generator made
by Tektronix. It provides all of the controls discussed above, as well
as more advanced features.

- The
**oscilloscope**is an instrument designed to display waveforms. Using it, you can measure period, frequency, peak values, peak-to-peak values, and other important quantities. - Shown here is a Tektronix 2213, one of the types of oscilloscopes
that we have in Sinclair’s electronics labs. At the left is the screen
on which waveforms are displayed. To the right are the knobs and switches
that you can adjust to control the waveform’s appearance.

- Below are three separate photos showing how a triangle wave, a square
wave, and a sine wave look on the oscilloscope screen.

- The oscilloscope displays a graph of voltage versus time, with
**voltage plotted on the vertical axis**and time plotted on the horizontal axis. - To measure a waveform’s peak-to-peak voltage, you count how many vertical divisions (squares) the waveform covers on the oscilloscope’s screen, and then you multiply this number times the setting of the oscilloscope VOLTS-PER-DIVISION knob.
- This lesson will show you how to do it:

- Remember, the oscilloscope displays a graph of voltage versus time,
with voltage plotted on the vertical axis and
**time plotted on the horizontal axis**. - To measure a waveform’s period, you count how many horizontal divisions (squares) the waveform covers on the oscilloscope’s screen, and then you multiply this number times the setting of the oscilloscope SECONDS-PER-DIVISION knob.
- Once you know the waveform’s period, you can use the formula
*f*= 1 ÷*T*to find its frequency. - This lesson will show you how to do it:

- The oscilloscope is a complicated piece of equipment. You’ll need plenty of practice to learn how to use it correctly.
- To brush up on your oscilloscope skills, take some time right now to play Oscilloscope Challenge. In particular, work through the game’s "Study" section, which is a tutorial containing several pages of notes to help you identify and use the oscilloscope’s controls. This will be a good preparation for lab work in which you’ll use a real oscilloscope to make measurements.

- We have a few different ways to specify the size of an ac current or voltage.
- We can give either
- the peak value, or
- the peak-to-peak value, or
- something called the effective value (also called rms value).

**Example**: As you can see, the sine wave shown below has a**peak voltage of 6 V**. Also, its_{p}**peak-to-peak voltage is 12 V**. And, as we’ll see below, it’s_{pp}**effective voltage****is 4.24 V**. So you can’t just say something like "The sine wave had a voltage of 6 V." You’ve got to be careful to say whether you’re talking about peak voltage, peak-to-peak voltage, or effective voltage._{rms}

- This may seem confusing, but you have to be able to deal with it. It’s similar to the situation that we have with temperatures or distances: when you give the distance between two cities, you can give it either in miles or in kilometers. It’s the same distance, but expressed with two different units.
- These distinctions apply only to ac, not to dc.

- The
**root-mean-square**(**rms**) value or**effective**value of an ac waveform is a measure of how effective the waveform is in producing*heat*in a resistance. **Example**: If you connect a 5 V_{rms}source across a resistor, it will produce the same amount of heat as you would get if you connected a 5 V dc source across that same resistor. On the other hand, if you connect a 5 V**peak**source or a 5 V**peak-to-peak**source across that resistor, it will**not**produce the same amount of heat as a 5 V dc source.- That’s why rms (or effective) values are useful: they give us a way to compare ac voltages to dc voltages.
- To show that a voltage or current is an rms value, we write
**rms**after the unit: for example,*V*_{rms}= 25 V rms. - A multimeter set to AC mode measures rms values.

- You need to be able to convert peak values to rms values, and vice versa. There are some standard conversion factors that let you do this.
- For a sine wave, to convert
**from peak values to rms values**, use these equations:For voltages,

*V*_{rms}≈ 0.707 ×*V*_{p}For currents,

To convert in the other direction (*I*_{rms}≈ 0.707 ×*I*_{p}**from rms values to peak values**), use these equations:For voltages,

*V*_{p}≈ 1.414 ×*V*_{rms}For currents,

*I*_{p}≈ 1.414 ×*I*_{rms} - Note: the "squiggly" equals sign means
**approximately**equal. These approximate conversion factors are close enough for our purposes. - These equations are
**valid only for sinusoidal waveforms**. For other wave shapes, there are other numbers (which we won’t need in this course) that you would use to convert between peak values and rms values. **Example**: Earlier I said that a sine wave with a peak value of 6 V_{p}has an effective value of 4.24 V_{rms}. I got the number 4.24 by using the first equation above.- The following animated lesson gives you more practice with peak values, peak-to-peak values, and rms values. It also introduces average values, which we’ll discuss below.

- When you use a multimeter to measure ac voltage or current, it gives
you
**effective (or rms) values**, not peak values or peak-to-peak values. - So if you measure the same voltage with both the multimeter and the oscilloscope , you’ve got to realize that you’re getting an effective voltage from the meter and you’re getting a peak (or peak-to-peak) voltage from the oscilloscope. To compare the two values, you’d need to convert one of them using the equations given earlier.

- Some inexpensive multimeters measure peak values of AC waves, and then use the equations given above to compute rms values. Since these equations hold only for sine waves, these meters give incorrect rms values for non-sinusoidal waves.
- A
**true rms meter**gives correct rms value for any AC wave. The multimeters in Sinclair’s EET labs are true rms meters.

- The
**average value**of a waveform is the average of its values over a time period. - Any waveform that is symmetrical about the time axis has zero average
value over a complete cycle.
- For example, consider the sine wave
shown below. Over one cycle, this waveform’s positive values exactly
cancel out its negative values, so its average value
**over a complete cycle**is zero. - Sometimes, though, it’s useful to refer to the waveform’s average value
**over a half cycle**. So, just looking at the positive "hump" of the sine wave below, what is its average value? Using calculus, it can be shown that a sine wave’s average value over a half ccyle is equal to 0.636 times its peak value.

- For example, consider the sine wave
shown below. Over one cycle, this waveform’s positive values exactly
cancel out its negative values, so its average value
- The average value of a waveform is also called its
**DC value**. - When a waveform is measured with a multimeter set to DC mode, the meter indicates the waveform’s average value over a complete cycle, which would be zero for the sine wave pictured above.

- The pictures of sinusoidal waveforms shown above had voltage on the vertical axis and time on the horizontal axis. That’s a useful picture when you’re interested in knowing how the voltage changes as time goes by.
- Recalling that one complete cycle of a sine wave is generated when
a generator’s rotor rotates through an angle of 360° (a full rotation),
here’s another way of plotting a sine wave:

- Here we’re plotting voltage on the vertical axis, just as before, but now we’re plotting degrees of the rotor’s rotation on the horizontal axis. Notice that one complete cycle of the sine wave corresponds to 360°.
- The quantity on the horizontal axis in this sort of picture is called
the
**phase**of the sine wave. For instance, we can see that this particular sine wave has a voltage of 0 V when its phase is 0°, and a voltage of 300 V when its phase is 90°, and so on.

- Often we measure angles in degrees, but the
**radian**(rad) is another unit for measuring angles. Since both units are widely used, it’s important to be comfortable with both units and to be able to convert from degrees to radians (or vice versa). - A full circle is equal to 360°, and it’s also equal to 2p radians.
And since p is approximately equal to 3.14,
this means that 360° is approximately equal to 6.28 radians. As
an equation:
360° = 2p rad ≈ 6.28 rad

- Dividing both sides of that equation by 2, we can also see that
180° = p rad ≈ 3.14 rad

- Dividing both sides again by 2, we can also see
that
90° = p/2 rad ≈ 1.57 rad

- Using these equalities, we can say that the sine wave pictured above has a voltage of 300 V when its phase is p/2 rad, and a voltage of 0 V when its phase is p rad, and so on.

- To convert any angle from radians to degrees, multiply by 180 ÷ p.
- In the other direction, to convert from degrees to radians, multiply by p ÷ 180.

- Often we’ll find ourselves dealing with two or more sinusoids that
have the same frequency, one of which is shifted to the right or the
left of the other one. For example:

- In the case pictured here, notice that the pink sinusoid is rising through zero (the horizontal axis) at the same time that the blue sinusoid is reaching its maximum value. Therefore the pink sinusoid has a phase of 0° at the same time that the blue sinusoid has a phase of 90°.
- We express this by saying that there’s a 90°
**phase shift**(also called a 90°**phase angle**) between the two waveforms.

- In cases like the one shown above, we’re interested not only in how far apart the two waveforms are shifted, but which one "comes before" the other one.
- The waveform shifted farther to the left is said to
**lead**the other waveform.- So in the picture shown above, the blue waveform leads the pink waveform by 90°.

- As another way of saying the same thing, the waveform shifted farther
to the right is said to
**lag**the other waveform.- So in the picture shown above, the pink waveform lags the blue waveform by 90°.

- By displaying two waveforms simultaneously and then measuring the time interval between corresponding points on the two waveforms, we can determine the phase shift between them.
- In particular, if
*T*is the period of the waveforms, and*t*is the time interval between corresponding points on the two waveforms, then the phase shift φ is given by the equation:φ = (

*t*÷*T*) × 360° - The following animated lesson also shows how to measure
phase angles. Their explanation is a little different from
mine, but you should be able to see it’s the same idea.
They first count the number of "spaces" (by which they mean
hashmarks) in a complete cycle, then divide that by 360° to find
the number of degrees per "space," and then multiply that
times the number of "spaces"
between the two waveforms. The end result will be the same that you’d
get using the formula φ = (
*t*÷*T*) × 360°.

- From EET 1150 you know that uppercase letters are used to represent
most DC quantities.
- For example,
*V*represents DC voltage, and*I*represents DC current.

- For example,
- On the other hand, most AC quantities are represented by lowercase
letters.
- For example,
*v*represents AC voltage, and*i*represents AC current.

- For example,
- Within AC, uppercase letters
are also used to represent constants that don’t change with time.
- For example,
*V*represents peak voltage, and_{p}*I*represents peak current. Peak voltage and peak current don’t continually change as time passes, so that’s why they’re represented by uppercase letters._{p}

- For example,

- The mathematical expression for a voltage sine wave with peak value
*V*is_{p}*v*=*V*sin(θ)_{p} - Here the Greek letter θ (theta) stands for the phase of the sine wave. So in this expression we’re multiplying the peak value times the sine of the wave’s phase.
- For example, in the sine wave shown below, the peak value is 300
V, so the expression for this sine wave is
*v*= 300 V sin(θ) - In this graph we’re plotting θ on the horizontal axis and
*v*on the vertical axis. - With this expression, you can use your calculator to compute
instantaneous values of a sine wave at a particular phase. For example,
the wave shown above has an instantaneous value of 282 V at
a phase of 110°, since
300 V × sin(110°) = 282 V

- Similarly, the expression for a current sine wave with
peak value
*I*is_{p}*i*=*I*sin(θ)_{p}

- Above you learned that a phase-shifted sinusoid is one that is shifted to the right or to the left along the horizontal axis, so that the wave does not pass through the origin.
- Mathematically, this is shown by having a fixed angle added
to θ in the expression for the waveform. In other words, instead
of having an expression of the form

we’ll have an expression that looks like this:*v*=*V*sin(θ)_{p}

where φ is a constant angle called (not surprisingly) the phase shift.*v*=*V*sin(θ + φ)_{p} - Note: φ is the Greek letter phi, pronounced "fie."
- For
example, here is the graph of
*v*= 300 V sin(θ + 90°):

- Notice that this graph has the same shape as the sine wave shown just above, but this one is shifted 90° (one-quarter of a cycle) to the left.
- A
**positive phase shift causes the waveform to shift left**along the x-axis, and**a negative phase shift causes it to shift right**. - As another example, here is the graph of
*v*= 300 V sin(θ − 90°):

- Notice that this graph has the same shape as the sinusoids shown above, but this one is shifted 90° (one-quarter of a cycle) to the right of the origin.
- In each of these diagrams, we’re looking at only one cycle of the waveform. You should imagine the wave extending indefinitely to the left and to the right.

- A
**phasor**is a vector that represents an AC electrical quantity, such as a voltage waveform or a current waveform. - The phasor’s length represents the voltage’s or current’s peak value.
- The phasor’s angle represents the voltage’s or current’s phase.
- For example, in the following diagram, think of the phasor as representing
a particular voltage waveform that has, let’s say, a peak voltage
of 4 V p. If you wanted to draw on this diagram another
phasor representing a voltage waveform with a peak voltage of 8 V p,
then you would draw a new phasor that’s twice as long as the original
one.

- Phasors have two main uses in studying AC circuits. In the first use, we imagine a phasor rotating counterclockwise around the origin at a speed that depends on the waveform’s frequency. (Higher-frequency waveforms rotate more quickly than lower-frequency waveforms.)
- As the phasor rotates, the waveform’s instantaneous value at any time is equal to the phasor’s y-component.
- For example, looking at our sample diagram again, imagine the phasor
to be rotating counterclockwise around the origin, and think of this
diagram as a "snapshot" of the phasor at one instant in
time. As we know from above, the phasor’s length represents the waveform’s
peak voltage,
*V*. The phasor’s y-component, which is equal to_{p}*V*sin(θ), represents the waveform’s instantaneous voltage._{p}

- The second, and more important, use of phasors is to represent the relationship between two or more waveforms with the same frequency.
- For example, consider the following diagram, which shows two phasors
labeled
*v*_{1}and*v*_{2}.- Phasor
*v*_{1}is drawn at an angle of 0°, and it has a length of 10 units. - Phasor
*v*_{2}is drawn at an angle of 45°, and it’s half as long as*v*_{1}. (Measure it with a ruler if you don’t believe me.)

- Phasor
- Such a diagram might represent two voltage waveforms in a circuit.
From the diagram we can see that
*v*_{1}‘s peak voltage is twice as great as*v*_{2}‘s peak voltage. (Assuming that the units are V, then*v*_{1}has a peak voltage of 10 V p, which means that*v*_{2}must have a peak voltage of 5 V p.) We can also see that*v*_{2}leads*v*_{1 }by a phase shift of 45°. - In terms of the equations for sinusoidal waveforms that you studied
above, this diagram would then be a pictorial representation
of the equations
*v*_{1}= 10 V sin(θ)*v*_{2}= 5 V sin(θ + 45°) - The equations and the phasor diagram convey the same information, but the diagram can be easier to understand and interpret, especially in cases where you’re dealing with half a dozen waveforms instead of just two.
- The same information can also be conveyed using a sinusoidal diagram
such as the following:

- Carefully compare the phasor diagram, the equations, and the sinusoidal diagram given above, until you’re convinced that they all describe the same pair of waveforms.
- So we can use sinusoidal diagrams, mathematical equations, or phasor diagrams to describe the relationships among waveforms in a circuit. Of these three representations, phasor diagrams are the probably the easiest to use and to understand, so they are widely used in AC circuit analysis.

- As mentioned above, when you imagine a phasor rotating about the
origin, the waveform’s frequency determines the speed of the phasor’s
rotation. In particular, if the waveform’s frequency is
*f*, then the phasor will rotate with an angular speed of 2p*f*. - This quantity 2p
*f*, which appears in many equations, is called the waveform’s**angular frequency**. - Its symbol is ω, and its unit is radians per second (rad/s):
ω

**=**2p*f* - Of course, since a waveform’s frequency is equal to the reciprocal
of its period (
*f*= 1 ÷*T*), we can also writeω

**=**2p÷*T* - So if you know a waveform’s frequency or period, you can easily compute its angular frequency (or vice versa).
- Note that ω is the Greek letter omega; it’s not a w.

- Until now, we’ve been writing the mathematical expressions for voltage
waveforms as
*v*=*V*sin(θ)_{p} - Recall that here
*V*is a constant equal to the waveform’s peak voltage. On the other hand, θ is a variable that changes continually as time passes. It can also be shown that θ=ω_{p}*t*, where ω is the waveform’s angular frequency and*t*is time (measured in seconds). - Making this subsitution for θ, the expression for a voltage
sine wave with peak value
*V*and angular frequency ω becomes_{p}*v*=*V*sin(ω_{p}*t*) - Similarly, for a current sine wave with peak value
*I*and angular frequency ω,_{p}*i*=*I*sin(ω_{p}*t*)

- The
**instantaneous value**of an ac waveform is its value at a specific instant of time. You can use the mathematical expression for a waveform to find the waveform’s instantaneous values at specific times. - Example: The instantaneous value of 10 V sin(377
*t*) at time 3 s is equal to 266 mV, since

10 x sin(377 x 3) = 266 mV - Since ω is in rad/s, your calculator must be in Radians mode when you do this calculation.

- We’ve just seen that the mathematical expression for a voltage or
current sine wave contains two important pieces of information: the
wave’s peak value (
*V*or_{p}*I*) and its angular frequency (ω)._{p} - If we now add in the possibility that the wave may also have a phase
shift (φ), we find that the general expression for a sinusoidal
voltage or current is

or*v*=*V*sin (ω_{p}*t*+ φ)*i*=*I*sin (ω_{p}*t*+ φ) - Often the phase angle φ is given in
**degrees**, but the angular frequency ω is almost always given in**radians**per second. So, before you can use the calculator to do a computation involving these quantities, you must convert φ from degrees to radians. If you don’t do this, you’ll be mixing degrees with radians, and you’ll get the wrong answer. (That would be like trying to add 2 inches plus 10 centimeters without first converting one of those quantities so that they both have the same unit.) - Of course, since we know that ω
**=**2p*f*, we could also write these two equations as*v*=*V*sin (2p_{p}*f**t*+ φ)*i*=*I*sin (2p_{p}*f**t*+ φ)

- In later units we’ll find that analyzing an AC circuit can get pretty tricky if the circuit contains capacitors or inductors. But if the circuit contains only resistors, then you can analyze it using the same techniques you’ve learned for DC circuits, with a couple of slight changes mentioned below. Once you understand how to apply Ohm’s law, Kirchhoff’s Laws, and the power laws to DC circuits, you already know almost everything you need to analyze resistive AC circuits.

- Here’s a remarkable feature of sinusoidal waveforms that makes them particularly easy to work with. Suppose you connect a function generator to any circuit containing resistors, inductors, and capacitors. If the function generator is set to produce a sinusoidal waveform, then every voltage drop and every current in the circuit will also be a sinusoid.
- The same thing is
**not**true for waveforms of other shapes. For instance, if the function generator is set to produce a triangle waveform, and if the circuit contains any inductors or capacitors, then the voltage drops across the components will be complicated waveforms, not simple triangle waveforms. - Fortunately, it turns out that sinusoids are not only the easiest waveforms to work with, they’re also the most useful. Therefore we’ll concentrate primarily on analyzing AC circuits that have sinusoidal voltage sources, rather than triangle or square-wave or other voltage sources.

- The voltage across any resistor and the current through that resistor have the same phase angle. They reach their peak values at the same instant.
- We say that the resistor’s voltage and current are
**in phase**with each other. - For instance, in the diagram below, suppose the tall blue waveform
represents the voltage across a resistor, and the short purple waveform
represents the current through that resistor. These two waveforms
are in phase with each other, since they reach their peak values at
the same instant.

- This means that in a phasor diagram, the phasor for a resistor’s voltage must point in the same direction as the phasor for that resistor’s current.

- Ohm’s law can be applied to resistors in AC circuits:

*I*_{p}=*V*_{p}÷*R*

- In words, this says that a resistor’s peak current is equal to the resistor’s peak voltage drop divided by the resistor’s resistance.
- The same law also applies to peak-to-peak values and rms values:
and
*I*_{pp}=*V*_{pp}÷*R*

*I*_{rms}=*V*_{rms}÷*R*

- Of course, you can also rearrange any of these equations algebraically
if you wish to solve for voltage or resistance. For example, if you
wish to calculate peak voltage based on known values of resistance
and peak current, you would rearrange the first equation to the form
*V*_{p}=*I*_{p}×*R.*

- Recall from your studies of DC circuits that Kirchhoff’s Voltage
Law (KVL) says that
**the sum of the voltage drops around any closed loop in a circuit equals the sum of the voltage rises around that loop.** - Recall also that Kirchhoff’s Current Law (KCL) says that
**the sum of all currents entering a point is equal to the sum of all currents leaving that point**. - You can also apply KVL and KCL to AC circuits that contain just resistors, as long as you’re careful to use all peak values, or all rms values, or all peak-to-peak values.

- Recall from EET 1150 that,
**in DC circuits**, you can use any one of the following equations to find the power dissipated in a resistor*P*=*I*^{2}×*R*

*P*=*V*^{2}÷*R*

In these equations,*P*=*V*^{ }×*I*

*V*is the DC voltage drop across the resistor, and*I*is the DC current through the resistor. **In AC circuits**, similar equations apply, but you must be sure to use the resistor’s rms voltage and rms current.:*P*=*I*_{rms}^{2}×*R**P*=*V*_{rms}^{2}÷*R*

In these equations,*P*=*V*_{rms}^{ }×*I*_{rms}*V*_{rms}is the rms voltage drop across the resistor, and*I*_{rms}is the rms current through the resistor.**If you use peak values or peak-to-peak values instead of rms values, you’ll get the wrong answer for the power.**

- Many circuits that you’ll encounter in the field contain not just AC voltages sources or just DC voltage sources, but a combination of AC and DC voltage sources.
- As a very simple example of this, the following diagram shows a
series circuit containing a DC voltage source, and AC voltage source,
and a single resistor.

- In this case, the voltage across the resistor is a 10 V pp
sine wave superimposed on 8 V DC. What this means is that
we’ll have a 10 V pp sine wave that is "moved up" 8 V
from where it would normally be, as in the following diagram.

- Notice the scale on the vertical axis. We have a 10 V pp sine wave, but instead of ranging from −5 V to +5 V, the sine wave’s voltage ranges from +3 V to +13 V.
- When you use a multimeter or oscilloscope to make measurements in such a circuit, you must pay close attention to whether your equipment is set to measure AC values or DC values (or both). Unit 3 will be devoted to this topic.

- The table below summarizes the electrical quantities that we’ve studied so far. The table shows the abbreviation for each quantity, along with the standard unit for measuring the quantity and the abbreviation for the unit.
- You studied the first eleven quantities (through inductance) in
EET 1150. The others have been added in this course.

**Quantity****Abbreviation****Unit****Abbreviation for the Unit**charge*Q*coulombCcurrent*I*ampereAvoltage (or emf)*V*(or*E*)voltVresistance*R*ohmΩconductance*G*siemensSenergy (or work)*W*jouleJpower*P*wattWefficiencyηcapacitance*C*faradFtime constantτsecondsinductance*L*henryHperiod*T*secondsfrequency*f*hertzHzpeak voltage*V*_{p}volts peakV ppeak current*I*_{p}amps peakA ppeak-to-peak voltage*V*_{pp}volts peak-to-peakV pppeak-to-peak current*I*_{pp}amps peak-to-peakA pprms voltage*V*_{rms}volts rmsV rmsrms current*I*_{rms}amps rmsA rmstime*t*secondsvarying voltage*v*voltVvarying current*i*ampereAangleθradian or degreerad or °phase angleφradian or degreerad or °angular frequencyωradians per secondrad/s - Notice that the abbreviations for quantities are usually written with italicized letters, while the abbreviations for units are usually written with plain, non-italicized letters.
- We’ll continue to learn about more new quantities throughout this course. Unit 4 will include an expanded version of the table.

- This e-Lesson has covered several important topics, including:
- waveforms
- waveform parameters, including:
- period
- frequency
- instantaneous value
- peak value
- peak-to-peak value
- rms value (also called effective value)
- average value

- the oscilloscope
- mathematical expressions for waveforms
- phasors
- mathematical expressions for waveforms
- resistors in AC circuits
- average power
- measuring phase shifts.

- To finish the e-Lesson, take this self-test to check your understanding of these topics.

Congratulations! You’ve completed the e-Lesson for this unit.