Most of the electronic devices in your home contain **transformers**.
Transformers have several different uses. Their **most common use is
to change the size of an ac voltage**. For example, the sinusoidal
voltage waveform that's present at an electrical outlet in your home
has an effective value of about 120 V rms. That is far larger
than the voltage required by most electronic devices. Therefore, these
devices contain transformers to reduce the voltage from 120 V to
the voltage that they require (maybe 5 or 10 V rms). As we'll
see, a transformer can change not only voltage, but also current and
the apparent size of a resistor.

Like inductors, transformers rely on the close relationship between electricity and magnetism. In fact, if you looked at a transformer's internal construction, you'd see that it consists basically of two inductors wound on a single core.

In this unit we'll learn some simple but useful equations related to transformers.

We'll also look at some math that is very useful for analyzing
AC circuits. This math involves what are called **complex numbers.** You
may have learned about complex numbers in a high school or college math
course. If so, much of this unit will be a
review for you. But I'm not going to assume that you already know this
material, so even if you've never seen complex numbers in a math class,
you'll be able to pick up what we need pretty easily. A lot of students
are confused or scared by the names "complex number" and "imaginary number." Don't
let those names bother you. There's nothing particularly complicated
(complex) about this material, and you don't need to have a great imagination
to understand imaginary numbers.

Another thing that confuses students is the idea of taking the square root of a negative number. If this were a math class, we'd have to spend some time thinking about how that's possible. But for what we want to do in electronics, we don't need to get bogged down on this point. The main thing that we need is to learn some simple rules for how to work with these complex numbers. You can learn those rules without getting into deep philosophical discussions about what a complex number really is.

- This unit will build on material that you studied in Unit 4. So let's begin by taking this self-test to review what you learned in that unit.

- A
**transformer**is an electrical device that can change the magnitude of an AC voltage applied to it. - It contains two coils (windings) wound on the same core.
- Here's the schematic symbol for a transformer, which just looks like
two inductors next to each other:

Notice that it has four terminals, two connected to the coil on the left and two connected to the coil on the right. The coil on the left is called the**primary winding**, and the coil on the right is called the**secondary winding**. - The symbol shown above is for an air-core transformer. Often you'll see the same symbol but with a pair of vertical lines drawn between the two coils; this is the symbol for an iron-core transformer.

- Transformers come in a very wide range of sizes.
- If you followed the electric lines from your house back to the power-generating plant, you would see a number of huge transformers along the way. At the power plant, transformers boost the voltage up to 138 kV rms. Electricity is distributed at this very high voltage to substations, where other transformers bring the voltage down to 12.5 kV rms. Closer to your home, smaller transformers bring the voltage down even further, to the 120 V rms or 240 V rms on the lines that come into your house.
- On a much smaller scale, transformers are contained in most of the
electronic devices in your home, such as televisions, computers, and
stereos. The photograph below shows a computer power supply, which
contains a transformer (the large component to the left), a large capacitor,
a resistor, and some diodes.

- Another application of transformers is the AC adapter that you use
to plug a battery-operated device, such as a portable DVD player or
a laptop computer, into a wall outlet:

- A transformer's primary winding is usually connected to an AC voltage source, which is the transformer's input.
- The secondary winding is connected to a load, or output.
- For example, the schematic diagram below shows a very simple case
in which the load is simply a resistor, R
_{L}.

- If you stop and think about the circuit shown just above, you might
be surprised to learn that any current will flow through the load resistor.
After all, the voltage source is connected to the transformer's primary
winding, and the transformer's secondary winding is connected to the
load resistor, but
**there are no wires that connect the voltage source to the secondary winding or the load resistor**. - So if the voltage source is turned on, you might expect current to flow through the primary winding, but you wouldn't expect any current to flow through the secondary winding or the load resistor, would you?
- In fact, though, because of the magic of electromagnetism, current
**will**flow through the resistor when the source voltage is turned on. Here's how it works: When current flows through the transformer's primary winding, a changing magnetic field is created which induces a voltage across the transformer's secondary winding. This means that the secondary winding will behave like a voltage source, causing current to flow through the load resistor. - As we'll soon see, depending on the design of the transformer, the voltage across the secondary winding may be equal to the source voltage, or it may be greater than the source voltage, or it may be less than the source voltage.

- A transformer's
**turns ratio**is defined as the ratio of the number of turns in its secondary winding to the number of turns in its primary winding. Using symbols, if*N*is the number of turns in the primary winding and_{pri}*N*is the number of turns in the secondary winding, then the_{sec}**turns ratio***n*is defined as:*n*=*N*÷_{sec}*N*_{pri}- Example: if a transformer has 250 turns in its primary winding and 750 turns in its secondary winding, then its turns ratio is equal to 3.

- The turns ratio is often written in the form
*N*:_{pri}*N*._{sec}- Example: if a transformer has 250 turns in its primary winding and 750 turns in its secondary winding, then its turns ratio could be expressed as 250:750. Or you might also see it expressed as 1:3, which indicates that there are three turns in the secondary winding for every one turn in the primary winding.

- If
*V*is the input voltage across the primary winding, and_{pri}*V*is the output voltage across the secondary winding, then we have the important equation:_{sec}*V*÷_{sec}*V*=_{pri}*N*÷_{sec}*N*_{pri} - So the ratio between the voltages is equal to the turns ratio.

- According to the equation you just learned, If the secondary winding has more turns than the primary winding, then the secondary voltage will be greater than the primary voltage.
- In that case, we have a
**step-up transformer**. - The ability to do this is wonderful, if you think about it. It means that if you have an AC voltage source whose maximum output voltage is 20 V rms, but you have an application where you need a higher voltage--say 40 V rms--then you can use a transformer to transform the source's output voltage to that higher voltage.

- If the secondary winding has fewer turns than the primary winding, then the secondary voltage will be less than the primary voltage.
- In that case, we have a
**step-down transformer**. - Step-down transformers are very common in everyday devices. As mentioned
at the beginning of this lesson, wall outlets in the United States
deliver 120 V rms, but most electronic devices in your home
require voltages much less than this. So these devices usually contain
transformers to step the voltage down from 120 V rms to a
lower value.
- Most electronic devices in your home also require DC voltage rather than AC voltage, so they also contain a circuit called a rectifier that converts AC to DC. You'll study rectifier circuits in another course, EET 2201.

- Letting
*I*stand for the current in the primary winding and_{pri}*I*for the current in the secondary winding, here's another important equation:_{sec}*I*÷_{pri}*I*=_{sec}*N*÷_{sec}*N*_{pri} - Notice that in the left hand side of this equation, the
**primary**current is in the numerator. This is different from the earlier equation for voltage transformation, which had the**secondary**voltage in the numerator.- This means that a step-up transformer (with a turns ratio greater than 1) steps the voltage up, but also steps the current down.
- Conversely, a step-down transformer (with a turns ratio less than 1) steps the voltage down, but also steps the current up.

- Recall that AC power is equal to rms voltage times rms current.
- In a transformer, we distinguish the primary power and the secondary
power, which are given by

and*P*=_{pri}*V*_{pri}_{(rms)}×*I*_{pri}_{(rms)}*P*=_{sec}*V*_{sec}_{(rms)}×*I*_{sec}_{(rms)} - An ideal transformer is 100% efficient, which means that
*P*=_{sec}*P*_{pri} - A real transformer will always have an efficiency less than 100%, which means the secondary power is less than the primary power. But many real transformers have efficiencies close to 100%, so we often treat them as if they're ideal.
- Herre's a nice summary of the formulas you've studied so far:

- In general, a circuit's
**input impedance**,*Z*_{in}, is defined as the ratio of its input voltage to its input current:*Z*_{in}=*V*_{in}÷*I*_{in} - This is a general concept that applies to any circuit that's connected to a voltage source. In the circuits that you've learned how to analyze up to now, the input impedance is simply equal to the circuit's total impedance. But as we'll see in a minute, things are a little trickier if there's a transformer in the circuit.
- Example: The computer monitor on my desk draws 1.5 A from the
120-V wall outlet. Therefore, this monitor has an input impedance of
80 Ω. (I got that by taking 120 V divided by 1.5 A.)
- We can say that the wall outlet "sees" an impedance of 80 Ω when I plug in my monitor and turn it on.

- In circuits that contain only resistors (and no capacitors or inductors),
the input impedance
*Z*_{in}is often referred to as the input resistance, abbreviated*R*_{in}. In the circuits that you learned how to analyze in EET 1150, the input resistance was simply equal to the circuit's total resistance.

- Consider the simple circuit shown below, in which a voltage source
is connected across a transformer's primary winding and a load resistor
*R*is connected across the transformer's secondary winding._{L}

- Using Ohm's law and the equations given above for voltage
transformation and current transformation,
it's easy to show that
*V*÷_{pri}*I*= (_{pri}*N*÷_{pri}*N*)_{sec}^{2}*R*_{L} - The left-hand side of this equation is equal to the circuit's input
resistance
*R*. Also, the quantity_{in}*N*÷_{pri}*N*in parentheses is equal to the reciprocal of the transformer's turns ratio_{sec}*n*. So we can rewrite this equation as*R*= (1 ÷_{in}*n*)^{2}*R*_{L} - So when you connect a transformer with a load resistor to a voltage
source, the impedance that the voltage source "sees" depends
not only on the size of the load resistor, but also on the transformer's
turns ratio.
- It's common to use the term
**reflected load**in this context. In the case above, we'd say that the real load is*R*, but the reflected load (the load that the voltage source "sees") is (1 ÷_{L}*n*)^{2}*R*._{L}

- It's common to use the term
- The same relationship applies to cases where, instead of having a
single load resistor connected to the transformer's secondary winding,
we have a more complicated circuit connected to the transformer's secondary
winding. In such a case, if that circuit's total resistance is
*R*, the voltage source will "see" a resistance equal to_{T}*R*= (1 ÷_{in}*n*)^{2}*R*_{T} - Example: Let's return to the previous example of a computer monitor
with an 80-Ω input impedance. If I connect this device directly
to a voltage source, then the voltage source will "see" an
impedance of 80 Ω. But if instead I insert a transformer
between the voltage source and the monitor, the voltage source will "see" an
impedance that depends not only on the monitor's input impedance but
also on the transformer's turns ratio.
- If the turns ratio is less than 1, then the voltage source will "see" an impedance greater than 80 Ω.
- If the turns ratio is greater than 1, then the voltage source will "see" an impedance less than 80 Ω.

- As we'll see next, this can be useful because it lets you use a transformer to match a load's impedance to the voltage source that you're connecting it to.

- Recall the
**maximum-power-transfer theorem**from EET 1150: maximum power is delivered to a load when the load resistance is equal to the Thevenin resistance of the source to which it is connected. - Because transformers can transform impedances, they are widely used
to achieve
**impedance matching**between a source and load. For example, suppose you want to connect a load (such as a speaker) to a voltage source (such as an amplifier), but the load's resistance is not very close to the voltage source's Thevenin resistance. In such a case, you might insert a transformer between the source and the load, so that the source "feels" as if it is connected to a load resistance that's equal to its own internal resistance. By "faking out" the source in this way, we can reach a better match between the source and the load, which will result in more power being delivered to the load. - Here a nice animated explanation:
- And here's a nice example of a practical application:

- Some iron-core transformers have more than one secondary winding.
Usually these secondary windings have different numbers of turns, which
we'll call
*N*(the number of turns in the first secondary winding),_{sec1}*N*(turns in the second secondary winding), and so on._{sec2} - This lets us get two or more different output voltages from a single
transformer.

- In such cases, the secondary windings operate independently of each other, and the equations given earlier still work, as long as you're careful not to mix up your secondary windings.
- For example, you're familiar with the equation for voltage transformation,
which says that

Suppose now that we have a transformer with two secondary windings. The equation for voltage applies separately to each of these windings, so we have:*V*÷_{sec}*V*=_{pri}*N*÷_{sec}*N*_{pri}

for the first winding and*V*÷_{sec1}*V*=_{pri}*N*÷_{sec1}*N*_{pri}

where*V*÷_{sec2}*V*=_{pri}*N*÷_{sec2}*N*_{pri}*V*is the voltage across the first secondary winding and_{sec1}*V*is the voltage across the second secondary winding._{sec2} - We've been considering the case of a transformer with more than one secondary winding. Similar comments apply to transformers with more than one primary winding: the basic equations you've learned still apply in these cases, as long as you're careful to treat the separate windings separately.

- Here's another way to get more than one output voltage from a transformer. In some transformers, a lead is connected to the secondary winding at a point between the endpoints of the winding. This lead is connected to an external terminal, so secondary voltages can be obtained between that terminal and the end terminals of the winding.
- Such a connection is called a
**tap**.

- To find the voltage between the tap and one of the ends of the winding, use simple algebra. For instance, if the tap is located one-third of the way between the winding's top end and the winding's bottom end, then the voltage between the tap and the top end will be one-third of the total secondary voltage. And the voltage between the tap and the bottom end will be two-thirds of the total secondary voltage.
- When the connection is made at the center point of the secondary
winding, it is called a
**center tap**. In such a case, the voltage between the tap and either end is equal to half of the voltage between the two ends.

- Everything you've learned about transformers is true only when the
input voltage is an AC voltage:
Transformers don't work

with DC voltages. - Why is this true? Because transformers rely on the magic of electromagnetic
induction, which operates only when you have a
**changing**magnetic field. In DC circuits we have constant electric and magnetic fields, so electromagnetic induction is not a factor.

- This concludes our discussion of transformers. We've restricted our attention to ideal transformers, which means we've ignored the fact that real transformers lose some energy because of winding resistance, heating of the transformer's core, and various other factors. In many cases, our idealized picture of transformers works just fine.
- Let's turn now to a quick look at complex numbers, which we'll use repeatedly in the weeks ahead.

- In mathematical operations that we've performed up to now in this
course, we've used
**real numbers**. Examples of real numbers include 5, and 3.47, and √2. Real numbers are perfectly adequate for most everyday mathematics and for analyzing DC circuits. But for analyzing AC circuits, we need more powerful mathematics. In particular, we need the system of numbers that mathematicians call**complex numbers**. - The complex numbers include the real numbers as a subset, but they also include numbers called imaginary numbers, as well as numbers that have a real part and an imaginary part.

- The system of complex numbers is based on the so-called
**imaginary unit**, the square root of −1. - Many math textbooks use the letter
*i*used to denote this imaginary unit. In circuit theory, we use*j*instead of*i*:

*j*= √−1 - Whenever you multiply a real number times this imaginary unit, you
get an
**imaginary number**. Examples of imaginary numbers include*j*5, and*j*3.47, and*j*√2.- As you know, whenever we multiply two numbers, the order in which we
multiply them does not matter. Therefore, we could just as well write the
imaginary numbers above as 5
*j*, and 3.47*j*, and (√2)*j*. But we'll follow the convention that many textbooks use, which is to write the*j*first.

- As you know, whenever we multiply two numbers, the order in which we
multiply them does not matter. Therefore, we could just as well write the
imaginary numbers above as 5

- A complex number can be written as the sum of a
**real part**and an**imaginary part**:*A*+*jB**A*and*B*are both real numbers, but*A*is the complex number's real part and*B*is its imaginary part.- For example, in the complex number 100 +
*j*35, 100 is the real part and 35 is the imaginary part.

- For example, in the complex number 100 +

- The
**complex plane**is a rectangular coordinate system with real numbers plotted along the horizontal (real) axis and imaginary numbers along the vertical (imaginary) axis. - So every point in the complex plane corresponds to a complex number, and vice versa.
- For instance, the diagram below shows the complex plane with two
complex numbers plotted on the plane. These two numbers are 30+
*j*40 and -20+*j*10. Can you tell which one is which?

- When we write a complex number in the form
*A*+*jB*, as we did above, we're writing it in**rectangular form**. - Any complex number can also be written in
**polar form**:

*C*∠θ

where*C*is the complex number's**magnitude**and θ is its**angle**. - For example, consider the complex number 30 +
*j*40. I just wrote it in its rectangular form. The same number can be written in polar form as 50 ∠53.1°. (We'll see in a minute how I got these numbers.) - Displayed graphically, here's the situation we're dealing with:

- In this diagram, the point
*p*represents the complex number that we're interested in. We can specify*p*either by saying how many units over and up it is from the origin (that's rectangular form:*A*+*jB*) or by saying how far it is from the origin, and in which direction (that's polar form:*C*∠θ). - This is similar to two different ways of giving someone directions
in everyday life. Rectangular form is similar to telling someone,

"Go 30 feet east, and then go 40 feet north."

Polar form is similar to saying,

"Go 50 feet at an angle of 53.1° north of east."

Either way, the person will get to the same point if he follows your directions.

- When you're working with complex numbers, sometimes it's more convenient to have them in rectangular form, and sometimes it's more convenient to have them in polar form. So you need to be able to convert complex numbers back and forth between rectangular and polar forms.
- To convert a complex number from its rectangular form
*A*+*jB*to its polar form*C*∠θ, use the equations

*C*= √ (*A*^{2}+*B*^{2})

and

θ = tan^{-1}(*B*÷*A*)

- Does this look complicated? It shouldn't! It's just basic trigonometry
on a right-angled triangle. Looking again at the diagram from above,
you'll see that we have a right-angled triangle in which
*A*is the adjacent side,*B*is the opposite side,*C*is the hypotenuse, and θ is the angle:

- So now, with the help of your calculator, you should be able to see
why I said above that 50 ∠53.1° is
the polar form of the same number that is 30 +
*j*40 in rectangular form.

- Recall from our trigonometry review (in Unit 4) that when
you use a calculator's tan
^{-1}to calculate an inverse tangent, the calculator sometimes gives answers in the wrong quadrant. In such cases you must adjust the calculator's answer by adding or subtracting 180°. - This point applies to the procedure given just above, since we must
use the tan
^{-1}button to find θ, the complex number's angle part. - In particular, when you're converting a complex number in
**Quadrant II**from rectangular to polar form, the tan^{-1}button will give you an angle in Quadrant IV, which you must adjust by**adding 180°**.- In case you've forgotten how the four quadrants are numbered,
here's a reminder:

- In case you've forgotten how the four quadrants are numbered,
here's a reminder:
- Also, when you're converting a complex number in
**Quadrant III**from rectangular to polar form, the tan^{-1}button will give you an angle in Quadrant I, which you must adjust by**subtracting 180°**. - To be on the safe side, always draw a quick sketch so that you can see which quadrant the answer should be in; then, if necessary, correct the calculator's answer.

- To convert a complex number from its polar form
*C*∠θ to its rectangular form*A*+*jB*, use the equations

and*A*=*C*cos(θ)*B*=*C*sin(θ)

- Again, this is just basic trigonometry on our right-angled triangle. So don't think of these as new equations that you must memorize--just remember your trigonometry, and you'll be okay.

- You'll need to be able to add, subtract, multiply, and divide complex numbers.
- Next we'll learn the procedures for doing these operations. But first, I'll mention that for some of these operations you want the numbers to be in rectangular form, and for the other operations you want the numbers to be in polar form. In particular:
- To
**add**or**subtract**complex numbers, you should first make sure the numbers are in**rectangular form**. - To
**multiply**or**divide**complex numbers, you should first make sure the numbers are in**polar form**.

- To add two complex numbers in rectangular form,
**add their real parts to each other and then add their imaginary parts to each other**.- Example: When you add 4+
*j*3 plus 8+*j*2, you get 12+*j*5.

- Example: When you add 4+
- To add complex numbers that are in polar form, first convert them to rectangular form, and then follow the rule given above.

- To subtract two complex numbers in rectangular form,
**subtract their real parts and then subtract their imaginary parts**.- Example: When you subtract 4+
*j*3 from 11+*j*13, you get 7+*j*10.

- Example: When you subtract 4+
- To subtract complex numbers that are in polar form, first convert them to rectangular form, and then follow the rule given above.

- To multiply two complex numbers in polar form,
**multiply their magnitudes and add their angles**.- Example: When you multiply 3∠40° times 5∠12°, you get 15∠52°.

- To multiply complex numbers that are in rectangular form, first convert them to polar form, and then follow the rule given above.

- To divide two complex numbers in polar form,
**divide their magnitudes and subtract their angles**.- Example: When you divide 20∠70° by 4∠9°, you get 5∠61°.

- To divide complex numbers that are in rectangular form, first convert them to polar form, and then follow the rule given above.

- Most scientific calculators have some built-in functions that can save you a lot of work when you're dealing with complex numbers.
- Some of these calculators, such as the Texas Instruments TI−30
and the Casio fx−250, have keys to convert from polar form to
rectangular form, and vice versa.
- To convert a complex number from polar form to rectangular
form on the Casio fx−250, you use the
**P→R**and**X↔Y**keys. For instance, to convert the number 2∠60° to rectangular form, first make sure that your calculator is in degrees mode, and then type**2 P→R 60 =**. This will give you the real part of your answer, which is 1. Now press**X↔Y**. This will give you the imaginary part of your answer, which is 1.73 (after rounding to three significant digits). So your complete answer is 1+*j*1.73.

- To convert a complex number from polar form to rectangular
form on the Casio fx−250, you use the
- Some other calculators, such as the Texas Instruments TI−86,
not only let you convert numbers from one form to another, but also
let you add, subtract, divide and multiply complex numbers directly.
- To enter a complex number in rectangular form on the TI−86,
you use the parentheses and comma keys. For instance, to enter
the number 3+
*j*7, you would type**(3,7)**. - To enter a complex number in polar form on the TI−86,
you use the parentheses and angle keys. For instance, to enter
the number 2∠40°, you would
first make sure that the calculator is in degrees mode, and then
type
**(2∠40)**. - To add two complex numbers, you use the same + key that you
use for regular numbers. For instance, to add 3+
*j*7 to 5-*j*9, you would type**(3,7)+(5,-9)**. - The TI-86 even lets you add polar-form numbers directly to
each other! For instance, to add 2∠40° to
8∠25°, you would first make
sure that the calculator is in degrees mode, and then type
**(2∠40)+(8∠25)**.

- To enter a complex number in rectangular form on the TI−86,
you use the parentheses and comma keys. For instance, to enter
the number 3+
- Consult your calculator's manual to learn how to use these features. They might take a little while to learn, but they'll save you a lot of time in the long run. By the way, the user's manuals for many calculators are available online. Click the following links to find the manual for your brand of calculator:

This concludes our mathematical review of complex numbers. Now let's take a quick look at how complex numbers are used in electronics.

- Recall the following points from Unit 2:
- 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 quantity's peak value.
- The phasor's angle represents the quantity's phase angle.
- From Unit 2 onward we have occasionally drawn phasor diagrams such
as the following one, which shows two phasors representing voltage
waveforms.

- Each phasor has a magnitude and an angle. Every complex number also has a magnitude and an angle, so we can conveniently use complex numbers to represent phasors.
- Why is it useful to represent phasors by complex numbers? Because, as you saw earlier, we have simple rules for adding, subtracting, multiplying and dividing complex numbers, which means that we can use these same rules to add, subtract, multiply, and divide phasors.
- In the diagram shown above, phasor
*v*_{1}has a length of 10 V and an angle of 0°.- Expressing this phasor as a complex number, we would write it as 10∠0° V.

- In the same diagram, phasor
*v*_{2}has a length of 5 V and an angle of 45°.- Expressing this second phasor as a complex number, we would write it as 5∠45° V.

- Remember, in this notation, the complex number's magnitude
*C*represents the quantity's peak value. - And the complex number's angle θ represents the quantity's phase angle.
- Here's another example: Suppose you have a voltage waveform,

Remember, this means that we've got a peak value of 8.59 V, an angular frequency of 314 rad/s, and a phase angle of −40*v*= 8.59 V sin(314*t*− 40°)**°**. As I originally wrote it, it's in a form that we'll call**sinusoidal form**. To write it in**polar form**, we just write the peak value, followed by the angle symbol, followed by the phase angle, like this:

Easy, isn't it?*v*= 8.59∠−40° V - One thing that might be confusing at first is that we separate the unit (V in the previous example) from the number that it goes along with (8.59 in the previous example). But with a little practice you'll get used to this notation.

- Phasors, and the complex numbers that represent them, have many uses in AC circuit theory. You'll see some of these uses in the remaining units of this course. For now, we'll just look at how phasors let us easily answer a type of question that might initially look very difficult.

- Phasors give us an easy way to add sinusoidal waves of the same frequency. As you'll see in future weeks, this is something that you need to be able to do in analyzing AC circuits.
- For example, suppose someone asked you to find the sum of
12 V sin(200

plus*t*)5 V sin(200

Would you know how to do it? You might guess that you could just add the peak values and add the phase angles, giving you an answer of*t*+ 30°)17 V sin(200

But you'd be wrong.*t*+ 30°) - Before we look at the correct procedure for doing this, let's note
that
**whenever you add two sinusoids of the same frequency, the sum is another sinusoid of the same frequency.**It may not be obvious that this is true, but it can be proved mathematically.- This means that if we add 12 V sin(200
*t*) plus 5 V sin(200*t*+ 30°), our answer will be another sinusoid with an angular frequency of 200 rad/s. So we know that our answer is of the form*V*sin(200_{p}*t*+ φ) - The tricky part is figuring out the values of
*V*(the peak voltage) and φ (the phase angle)._{p}

- This means that if we add 12 V sin(200
- Before we look at the procedure for figuring out
*V*and φ, let's take a quick graphical look at what we're doing here._{p}- Here's a plot of 12 V sin(200
*t*), which we'll call*v*_{1}:

- And here's a plot of 5 V sin(200
*t*+ 30°), which we'll call*v*_{2}:

- Here's a plot showing both
*v*_{1}and*v*_{2}:

- The question that we're trying to answer is: if we add the
instantaneous values of those two sinusoids at every instant,
what will the sum look like? So far, we know that the sum will
be another sinusoid with the same frequency as
*v*_{1}and*v*_{2}, but we don't know yet what this sinusoid's peak value or phase angle will be. (In other words, we don't know how tall it will be or how far to the left or right it will be shifted.)

- Here's a plot of 12 V sin(200

- Here's the procedure that we'll follow to add two sinusoids:
- Convert from sinusoidal form to polar form.
- Next, convert from polar form to rectangular form.
- Next, add these rectangular forms to get the sum.
- Then convert the sum back from rectangular form to polar form.
- Finally, convert back from polar form to sinusoidal form.

- Example: Suppose we wish to add the two sinusoidal voltages mentioned
above,

and*v*_{1}= 12 V sin(200*t*)

To find*v*_{2}= 5 V sin(200*t*+ 30°).*v*_{1}(*t*) +*v*_{2}(*t*), we'll follow the five steps just listed.- Convert from sinusoidal form to polar form. This will give
us
*v*_{1}= 12∠0° V and*v*_{2}= 5∠30° V. - Next, convert from polar form to rectangular form. This will
give us
*v*_{1}= 12 +*j*0 V and*v*_{2}= 4.33 +*j*2.5 V. - Next, add these rectangular forms to get the sum. This will
give us
*v*_{1}+*v*_{2}= 16.33 +*j*2.5 V. - Then convert the sum back from rectangular form to polar form.
This gives us
*v*_{1}+*v*_{2}= 16.5∠8.70° V. - Finally, convert back from polar form to sinusoidal form.
This gives us our answer, which is
*v*_{1}+*v*_{2}= 16.5 V sin(200*t*+ 8.70°)

- Convert from sinusoidal form to polar form. This will give
us
- Here's a plot showing the two original sinusoids and their sum:

- This e-Lesson has covered several important topics, including:
- transformers
- turns ratio
- voltage transformation
- current transformation
- impedance transformation
- rectangular and polar forms of complex numbers
- adding, subtracting, multiplying, and dividing complex numbers
- phasors
- using phasors to add sinusoids.

- 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.