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
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
Unit 4 Review
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:
Primary & Secondary
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, RL.
Could This Be Magic?
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 Npri is
the number of turns in the primary winding and Nsec is
the number of turns in the secondary winding, then the turns ration is
n = Nsec ÷ Npri
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 Npri:Nsec.
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 Vpri is the input voltage across the primary
winding, and Vsec is the output voltage across
the secondary winding, then we have the important equation:
Vsec ÷ Vpri = Nsec ÷ Npri
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
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,
Letting Ipri stand for the current in the primary
winding and Isec for the current in the secondary
winding, here’s another important equation:
Ipri ÷ Isec = Nsec ÷ Npri
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
Conversely, a step-down transformer (with a turns ratio
less than 1) steps the voltage down, but also steps the current
Power in an Ideal Transformer
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
Ppri = Vpri(rms) × Ipri(rms)
Psec = Vsec(rms) × Isec(rms)
An ideal transformer is 100% efficient, which means that
Psec = Ppri
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, Zin,
is defined as the ratio of its input voltage to its input current:
Zin = Vin ÷ Iin
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 Zin is often referred to as
the input resistance, abbreviated Rin. In the circuits
that you learned how to analyze in EET 1150, the input resistance was
simply equal to the circuit’s total resistance.
Impedance Transformation and Reflected Load
Consider the simple circuit shown below, in which a voltage source
is connected across a transformer’s primary winding and a load resistor RL is
connected across the transformer’s secondary winding.
The left-hand side of this equation is equal to the circuit’s input
resistance Rin. Also, the quantity Npri ÷ Nsec in
parentheses is equal to the reciprocal of the transformer’s turns ratio n.
So we can rewrite this equation as
Rin = (1 ÷ n)2RL
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
It’s common to use the term reflected load in
this context. In the case above, we’d say that the real load
is RL, but the reflected load (the load
that the voltage source "sees") is (1 ÷ n)2RL.
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 RT,
the voltage source will "see" a resistance equal to
Rin = (1 ÷ n)2RT
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 Nsec1 (the number of turns in the first
secondary winding), Nsec2 (turns in the second
secondary winding), and so on.
This lets us get two or more different output voltages from a single
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
Vsec ÷ Vpri = Nsec ÷ Npri
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:
Vsec1÷ Vpri = Nsec1 ÷ Npri
for the first winding and
Vsec2÷ Vpri = Nsec2 ÷ Npri
where Vsec1 is the voltage across the first secondary
winding and Vsec2 is the voltage across the second
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
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.
Transformers are for AC Only
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.
Real Numbers Versus Imaginary and Complex Numbers
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
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.
Imaginary Unit, j
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 j5,
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 5j, and 3.47j, and (√2)j.
But we’ll follow the convention that many textbooks use, which
is to write the j first.
A complex number can be written as the sum of a real part and
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 + j35, 100 is the real part and 35 is the imaginary part.
The Complex Plane
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+j40
and -20+j10. Can you tell which one is which?
Rectangular & Polar Forms
When we write a complex number in the form A + jB,
as we did above, we’re writing it in rectangular
Any complex number can also be written in polar form:
where C is the complex number’s magnitude and θ is
For example, consider the complex number 30 + j40. 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.
Converting from Rectangular Form to Polar Form
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 = √ (A2 + B2)
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
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 + j40 in rectangular
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
This point applies to the procedure given just above, since we must
use the tan-1 button to find θ, the complex number’s
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
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.
Converting from Polar Form to Rectangular Form
To convert a complex number from its polar form C∠θ to
its rectangular form A + jB, use the
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
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.
Adding Complex Numbers
To add two complex numbers in rectangular form, add their
real parts to each other and then add their imaginary parts to each
Example: When you add 4+j3 plus 8+j2, you
To add complex numbers that are in polar form, first convert them
to rectangular form, and then follow the rule given above.
Subtracting Complex Numbers
To subtract two complex numbers in rectangular form, subtract
their real parts and then subtract their imaginary parts.
Example: When you subtract 4+j3 from 11+j13,
you get 7+j10.
To subtract complex numbers that are in polar form, first convert
them to rectangular form, and then follow the rule given above.
Multiplying Complex Numbers
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.
Dividing Complex Numbers
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.
Complex Numbers on Your Calculator
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+j1.73.
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+j7, 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
To add two complex numbers, you use the same + key that you
use for regular numbers. For instance, to add 3+j7 to
5-j9, 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).
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.
Representing Phasors with Complex Numbers
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
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 v1 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 v2 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
Here’s another example: Suppose you have a voltage waveform,
v = 8.59 V sin(314t − 40°)
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°.
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,
v = 8.59∠−40° V
Easy, isn’t it?
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.
Many Uses for Phasors
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.
Using Phasors to Add Sinusoids
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(200t)
5 V sin(200t + 30°)
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
17 V sin(200t + 30°)
But you’d be wrong.
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(200t)
plus 5 V sin(200t + 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
Vp sin(200t + φ)
The tricky part is figuring out the values of Vp (the
peak voltage) and φ (the phase angle).
Before we look at the procedure for figuring out Vp and φ,
let’s take a quick graphical look at what we’re doing here.
Here’s a plot of 12 V sin(200t), which we’ll
And here’s a plot of 5 V sin(200t + 30°),
which we’ll call v2:
Here’s a plot showing both v1 and v2:
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 v1 and v2,
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.)
Procedure for Adding Sinusoids
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
v1 = 12 V sin(200t)
v2 = 5 V sin(200t + 30°).
To find v1(t) + v2(t),
we’ll follow the five steps just listed.
Convert from sinusoidal form to polar form. This will give
v1 = 12∠0° V
and v2 = 5∠30° V.
Next, convert from polar form to rectangular form. This will
v1 = 12 + j0 V
and v2 = 4.33 + j2.5 V.
Next, add these rectangular forms to get the sum. This will
v1 + v2 = 16.33 + j2.5 V.
Then convert the sum back from rectangular form to polar form.
This gives us
v1 + v2 = 16.5∠8.70° V.
Finally, convert back from polar form to sinusoidal form.
This gives us our answer, which is
v1 + v2 = 16.5 V sin(200t + 8.70°)
Here’s a plot showing the two original sinusoids and their sum:
Unit 5 Review
This e-Lesson has covered several important topics, including:
rectangular and polar forms of complex numbers
adding, subtracting, multiplying, and dividing complex numbers
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.