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12
Fundamentals of Alternating Current
In this chapter, we lead you through a study of the mathematics and physics of
alternating current (AC) circuits
...

Write the general equation for a sinusoidal signal based on its amplitude,
frequency, and phase shift
...

Manipulate the general equation of a sinusoidal signal to determine its amplitude,
frequency, phase shift at any time
...

Define root-mean-squared amplitude, angular velocity, and phase angle
...

Convert between polar and rectangular form
...

Discuss the phase relationship of voltage and current in resistive, inductive, and
capacitive loads
...

Define components of power and realize power factor in AC circuits
...


FOCUS ON MATHEMATICS
This chapter relates the application of mathematics to AC circuits, covering complex
numbers, vectors, and phasors
...


REFERENCES
•
•
•
•

Stephan J
...

Stephan J
...

Bosels, Electrical Systems Design, Prentice Hall
...
Harter and Wallace D
...


1

2

Chapter 12

12
...
It is used for household and industrial applications such as television
sets, computers, microwave ovens, electric stoves, to the large motors used in the
industry
...
The major advantage of AC is the fact
that it can be transformed, however, direct current (DC) cannot
...

Transmission of high voltage (in terms of kV) is that less current is required to
produce the same amount of power
...

In this chapter, we will introduce a sinusoidal signal and its basic
mathematical equation
...
The phasor analysis techniques will be used to
analyze electric circuits under sinusoidal steady-state operating conditions
...


12
...
The
most common AC waveform is a sine (or sinusoidal) waveform
...

In discussing AC signal, it is necessary to express the current and voltage in
terms of maximum or peak values, peak-to-peak values, effective values, average
values, or instantaneous values
...
Figure 12-1 is a plot
of a sinusoidal wave
...
1)

Where Vp is the peak voltage, ω = 2πf is the angular speed expressed in
radians per second (rad/s), f is the frequency expressed in Hertz (Hz), t is the time
expressed in second (s), and θ is phase of the sinusoid expressed in degrees
...
They then fall, just as they rose, back to 0o at 180o
...
The function then
returns symmetrically to 0o at 360o
...


12
...
1 Radian and Degree
A degree is a unit of measurement in degree (its designation is ° or deg), a turn
of a ray by the 1/360 part of the one complete revolution
...

A radian is defined as the central angle, for which lengths of its arc and radius
are equal (AB = A0)
...
These terms are shown in Figure 13-8
...
1), a length of a circumference C and its radius r can
be expressed as:
2π =

C
R

(12
...
Hence, we receive a value of one radian:
360
≈ 57
...
3)

2π
≈ 0
...
4)

1 rad =

and,
1 deg =

4

Chapter 12

The following comparative table of degree and radian provides measure for
some angles we often deal with:

B

0

A

Figure 12-2 Radian and arc length
...
2
...

The peak value is measured from zero to the maximum value obtained in
either the positive or negative direction
...
This value is twice the maximum
or peak value of the sine wave and is sometimes used for measurement of ac
voltages
...


12
...
3 Instantaneous Value
The instantaneous value of an AC signal is the value of voltage or current at
one particular instant
...
It may also be the same
as the peak value, if the selected instant is the time in the cycle at which the
voltage or current stops increasing and starts decreasing
...


Fundamentals of Alternating Current

5

12
...
4 Average Value
The average value of an AC current or voltage is the average of all the
instantaneous values during one alternation
...
The
average value is the amount of voltage that would be indicated by a DC voltmeter
if it were connected across the load resistor
...
It is possible to determine the average value by adding together a series of
instantaneous values of the alternation (between 0° and 180°), and then dividing
the sum by the number of instantaneous values used
...
636
times the peak value
...
636Vmax

(12
...
Similarly, the formula for average current is
I av = 0
...
6)

Where Iav is the average current for one alteration, and Imax is the maximum or
peak current
...
2
...

It is possible to compute the effective value of a sine wave of current to a good
degree of accuracy by taking equally spaced instantaneous values of current
along the curve and extracting the square root of the average of the sum of the
squared values
...
Therefore,
I eff = Average of the sum of the squares of I ins

(12
...
707 times the
maximum value of current (Imax)
...
707 × Imax
...
414 × Ieff
...
414 comes from
...
Assume that the DC in Figure is maintained at 1 A and the resistor

6

Chapter 12

temperature at 100°C
...
At this point it is found that a maximum AC
value of 1
...

Therefore, in the AC circuit the maximum current required is 1
...

When a sinusoidal voltage is applied to a resistance, the resulting current is
also a sinusoidal
...
Ohm’s law, Kirchhoff’s law, and the various
rules that apply to voltage, current, and power in a DC circuit also apply to the
AC circuit
...
8)

Importantly, all AC voltage and current values are given as effective values
...
2
...

Increasing the number of revolutions to two per second will produce two
complete cycles of ac per second (2 Hz)
...
Event frequency is always measured and
expressed in hertz
...
9)

Where ω is the angular velocity in radians per second (rad/s)
...
The frequency is an important term to
understand since most AC electrical equipment requires a specific frequency for
proper operation
...
0 seconds
80 cycles in 200 milliseconds
1000 revolutions in 0
...
0 = 10 cycles per second = 10 Hz
80/0
...
5 = 2000 cycles per second = 4000 Hz (4 kHz)
600/60 = 10 cycles per second = 10 Hz

Example 12-2

Express each of the following as angular velocity in radians per second
a)
b)
c)
d)

80 rad in 10 s
2
...
0 rad/s
ω = (40×106)/10 = 4
...
5 rad/s
ω = 2π× (10×103) = 62
...
0×106) = 6
...
2
...
It
is measured in seconds
...

The function repeats itself every 2π radians, and its period is therefore 2π radians
...
10)

Example 12-4

Express each of the following periods in seconds
a)
b)
c)
d)

500 Hz
90 kHz
900 MHz
5 Hz

Solution: Use Equation (12
...
11 µs
T = 1/(900×106) = 1
...
2 s

12
...
8 Phase
When two sinusoidal waves, such as those represented by Figure 12-3, are
precisely in step with one another, they are said to be in phase
...

To further describe the phase relationship between two sinusoidal waves, the
terms lead and lag are used
...
According to Figure 12-3, the sinusoid
VP sin (ωt + θ) occur θ rad, θ degrees seconds, earlier
...
Also, we may say that VP sin ωt lags VP sin (ωt + θ)
by θ
...
When the latter condition exists, the
two waves are said to be in phase
...

To determine the phase difference between two sine waves, locate the points
on the time axis where the two waves cross the time axis traveling in the same
direction
...
The wave that crosses the axis at the later time (to the right on the
time axis) is said to lag the other wave
...
2
...
For example, sin ω t = cos (ωt – 90o)
...
To realize this, let us consider
v1 = VP1 cos (10t + 20 o )
= VP1 sin (10t + 90 o + 20 o )

(12
...
12)

by 150o
...
13)

10

Chapter 12

v
VP
VP sin ωt

VP sin (ωt + θ)

ωt

θ

-VP

Figure 12-3 The sine wave VP sin (ωt + θ) leads VP sin ωt
...
3 PHASORS
We have learnt from the previous section how to define and express in a single
equation the magnitude, frequency, and phase shift of a sinusoidal signal
...
However, the linear circuit does change the
amplitude of the signal (amplification or attenuation) and shift its phase (causing
the output signal to lead or lag the input)
...

Accordingly, signal can be expressed as a linear combination of complex
sinusoids
...

The phasor is similar to vector that has been studied in mathematics
...
As the vector rotates it generates an angle
...

Representing sinusoidal signals by phasors is useful since circuit analysis laws
such as KVL and KCL and familiar algebraic circuit analysis tools, such as series
and parallel equivalence, voltage and current division are applicable in the phasor
domain, which have been studied in DC circuits can be applied
...
The only
difference is that circuit responses are phasors (complex numbers) rather than DC
signals (real numbers)
...
There are two basic
forms of complex number notation: polar and rectangular
...
(b) A vector with its end fixed at the
origin and rotating in a counterclockwise (CCW) direction representing the
varying conditions of the AC signal
...
3
...
Standard orientation for vector angles in AC circuit
calculations defines 0o as being to the right (horizontal), making 90o straight up,
180o to the left, and 270o straight down
...
For example, a vector angled ∠ 270o
(straight down) can also be said to have an angle of -90o
...


12

Chapter 12

(b)

(a)

Figure 12-6 (a) A vector (5
...
4 ∠ -34o
...
Figure 12-6a shows a
vector with positive angle (5
...
4 ∠
-34o) with negative angle
...
14)

Where the uppercase V, indicates that the quantity is a phasor, having both
magnitude and phase
...
The phase angle is in
degrees
...

Example 12-5

Write the phasor form for the following signal and draw the phasor diagram
...


(

v = 300 V p sin 377t- 45o

Solution:
Vp
Vrms =
2
300
=
= 212
...
16 ∠ - 45o

The phasor diagram is shown in Figure 12-7
...


12
...
2 Rectangular Form
The horizontal and vertical components denote a complex number
...
These two dimensional figures (horizontal and
vertical) are symbolized by two numerical figures
...
Figure 12-8 shows
that a point on a complex plane located by a phasor could be described in
rectangular form
...


14

Chapter 12

Focus on Mathematics
Complex Algebra
A complex number is the sum of a real number and an imaginary
number [A = Real (A) + j Imaginary (A)]
...
What are imaginary
numbers? The answer to this question is related to another question
...
Sometimes, the
letter i is used to define the imaginary number
...

Example 12-6

Express

− 16 as an imaginary number
...
3
...
The hypotenuse is labeled as C
...
A represents the real value and B represents the imaginary value of the
rectangular form
...
15)

Fundamentals of Alternating Current

15

C

B

A
Figure 12-9 Relation between polar and rectangular forms
...
From trigonometry, the cosine of an included angle
relates the length of the adjacent side and the length of the hypotenuse
...
16)

To convert from rectangular form to polar form requires a different set of
trigonometric relationships
...
17)

Taking the inverse tangent of each side leaves θ as
B

 A

θ = tan -1 

(12
...
19)

16

Chapter 12

Example 12-7

Convert each of the following polar phasors into their rectangular form
...
6 Vrms
V = 50 Vrms - j 86
...

a)

V = 2 Vrms ∠45o

b)

V = 240 Vrms ∠ - 160 o

Solution:

a)
b)

V = 1
...
414 Vrms
V = - 225
...
084 Vrms

12
...
4 Euler’s Identity
Euler’s identity forms the basis of phasor notation
...
It states, the identity defines the complex
exponential ejθ as a point in the complex plane
...
20)

Figure 12-10 shows how the complex exponential may be visualized as a point
(or vector, if referenced to the origin) in the complex plane
...


(12
...
22)

cosθ + sin θ = cos 2θ + sin 2θ = 1

Remember that writing Euler’s identity corresponds to equating the polar form
of a complex number to its rectangular form
(12
...

To see how complex numbers are used to represent sinusoidal signals, we may
rewrite the expression for a generalized sinusoid using Euler’s equation:

(

A cos (wt + θ ) = Re A e j (wt + θ )

)

(12
...
24) is simplified as

(

)

(

A cos (wt + θ ) = Re A e j (wt + θ ) = Re A e jθ e

jwt

)

(12
...
Use a scale of 1 cm = 100 Vrms
...

300 Vrms

150 Vrms

Figure 12-11 Rectangular phasor plot of Example 12-9
...
Add the real parts of the phasors
...
Add the imaginary parts of the phasors
...
Form the sum as a phasor written in rectangular form
...
Change the sign of both the real and the imaginary part of the
phasor to be subtracted
...
Add the phasors following the steps in the previous box
...
Accordingly the answer is

-(10- j4) = -10 + j4, Now add

15 + j8
− 10 + j 4
5 + j12

The answer is 5 + j12

20

Chapter 12

Focus on Mathematics
Multiplying Phasors
Rectangular Form

To multiply phasor quantities in rectangular form, multiply the
numbers as if they were two binomials
1
...

2
...

3
...

4
...

5
...

Example 12-12

Multiply 3 + j2 and 4 – j5
Solution: Follow steps 1 to 5

Distribute (3 + j2) over (4- j5)
...
This yields
12 – j15 + j8 + 10
Combine like terms to obtain the answer
22 – j7

Fundamentals of Alternating Current

Focus on Mathematics
Multiplying Phasors
Polar Form

To multiply phasor quantities in polar form
1
...

2
...

3
...

Example 12-13

Multiply 4∠15o and 6∠25o
Solution: Follow steps 1 to 3

Multiply by magnitudes:
4 × 6 = 24
Add the angles:
15o + 25o = 40o
The answer is 24∠40o

21

22

Chapter 12

Focus on Mathematics
Dividing Phasors
Rectangular Form

To divide phasor quantities in rectangular form
1
...

4
...

5
...

Example 12-14

Divide (15 + j10) by (2 +j1)
Solution: Follow steps 1 to 5

15 + j10
2 + j1
Multiply numerator and denominator by 2 - j1, the complex
conjucate of 2 + j1
...
Divide the magnitudes
...
Subtract the angle of the denominator from the angle of the
numerator
...
Form the quotient as a phasor written in polar form
...
That is 40o – 20o = 20o
The answer is 5∠20o

24

Chapter 12

Focus on Mathematics
Power of a Phasor
To raise a phasor to a power, express the phasor in polar form, first
and then:
1
...

2
...

3
...


Example 12-17

Solve (3 + j4)2
Solution: Follow steps 1 to 3

First, express (3 + j4) in polar form:
3 + j4 = 5∠53
...
13 by 2: 53
...
26o
The answer is: 25∠106
...
4 ELECTRICAL SYSTEMS OF PHASORS
Phasors may be added, subtracted, multiplied, and divided
...
They provide significant meaning to the
systems under study
...

Vtotal = V1 + V2

(12
...
4
...

Example 12-18

Find the total voltage across the terminals of the circuit shown in Figure 12-12
...

Solution: Total length = 3 + 4 = 7 V (angle is 0o)

26

Chapter 12

3 V, 0o

+

4 V, 0o

7 V, 0o

-

Figure 12-12 Two sources connected in series
...

The circuit contains two sources connected in series: 3 V with 0o and 4 V with
180o
...


Fundamentals of Alternating Current

27

12
...
2 Complex Vector Addition
If vectors with uncommon angles are added, their magnitudes (lengths) add up
quite differently than that of scalar magnitudes:

5 V, 53
...


If two AC voltages, 90o out of phase, are added together by being connected in
series, their voltage magnitudes do not directly add or subtract as with scalar
voltages in DC
...
For example, in
Figure 12-14, a 3 V source at 0o added to a 4 V source at 90o results in 5 V at a
phase angle of 53
...

There is no suitable DC analogy for what we're seeing here with two AC
voltages slightly out of phase
...
With AC, two voltages can be aiding or
opposing one another to any degree between fully-aiding and fully-opposing,
inclusive
...

Example 12-20

Add the following phasors in rectangular form and then express the total in polar
...
099 ∠78
...

(250 A rms ∠ - 45 o ) - (100 A rms ∠75 o )

Solution:

250 A rms ∠ - 45 o = 176
...
77 A rms
100 A rms ∠75 o = 25
...
60 A rms

(176
...
77 A rms ) - (25
...
60 A rms )

150
...
17 A rms = 171 A rms ∠ - 28 o

12
...
This is not
true in AC circuit
...
Each of these loads produces a different circuit
condition
...

However, for AC circuits it is called impedance
...
It is a phasor quantity
...
The current and voltage may not have the
same amplitude, but they are in phase
...

Voltage and current are in phase with each other in a pure resistive circuit as
shown in Figure 12-15 (a)
...
When like signs are multiplied, the
product is positive and when unlike signs are multiplied the product is negative
...

The impedance in AC circuits is defined through Ohm’s law
Z=

V
I

(12
...


Fundamentals of Alternating Current

29

Voltage
Current

VP

IP

(a)

(b)

Figure 12-15 (a) Voltage in phase with current
...


When sinusoidal current flows through the impedance, we have
v(t ) = i (t ) × R

(12
...
26), we obtain

(R × I ∠0 )
(I ∠0 )
o

Z=

rms

o

rms

30

Chapter 12

To divide phasors, divide the magnitudes and subtract the angles
Z = R∠0 o

(12
...
Therefore, its
units are ohms
...
The 0o phase shift indicates that
the voltage across the resistor is in phase with the current through it
...

Z R = R + j0

Example 12-22

Calculate the phasor current through a 100-Ω resistor assuming a voltage of (100
Vrms∠0o) applied across it
...
6 INDUCTIVE LOADS
12
...
1 Inductance
The inductance of an inductor (L) is measured in henries (H)
...
This is expressed mathematically as
L=

N2µ A
l

(12
...
Some amount of inductance is
present in all AC circuits because of the continually changing magnetic field
...
Loads such as motors, transformers, and chocks all
contain coils of wire
...
If the amount of current decreases, the magnetic field
will collapse
...
In a DC circuit containing pure inductance the current takes
time to rise to maximum even though the full-applied voltage is immediately at
maximum
...
6
...
This
causes the magnetic field to continually increase, decrease, and reverse polarity
...
This induced voltage is 180o out of phase with
the applied voltage
...
This current-limiting property of the
inductor is called reactance (X)
...
It is measured in ohms just as the resistance is

(a)

(b)
Figure 12-16 (a) Magnetic field increases around the coil as current flows
through the coil
...


32

Chapter 12

Induced
voltage

Applied
voltage

Figure 12-17 Applied voltage and induced voltage across a coil
...
30)

The voltage and current relationship for an inductor involves a derivative
...
31)

where
i = I P sin (ω t ) = I rms ∠0 o

The derivative of a sinusoidal current is calculated as

(

di
= ω I P sin ω t + 90 o
dt

)

Substitute this into Equation (12
...
(b) Phasor diagram shows the lead by 90o
...
Figure 12-18 (a)
shows the voltage leading the current when AC current passes through an
inductor The current is at 0o and the voltage drop across the inductor is at + 90o
(leads) as shown in Figure 12-18 (b)
...
26) to find the impedance
ZL =

(X

I rms ∠90 o
I rms ∠0 o

L

(

)

)

Divide the magnitude and subtract the angle

(

Z L = X L ∠90 o

)

(12
...
32) indicates that the opposition an inductor presents to a
sinusoidal current is proportional to the size of the inductor (L) and the value of
the frequency
...

In rectangular form, the impedance of the inductor contains a real and
imaginary component
...
The length
of the phasor XL lies entirely along the imaginary (+y) axis
...


12
...
3 Power in Inductive Load
In a pure resistive circuit, the true power is equal to the product of the voltage
and current
...
In
order to produce true power, voltage and current must both be either positive or
negative
...

Example 12-23

The inductor shown in Figure has an inductance of 1 H and is connected to a 120
V 60 Hz line
...
1416 × 60 × 1
X L = 377 Ω

XL may be substituted for R in Ohm’s law
I=

120
V
=
= 0
...
7 CAPACTIVE LOADS
An inductor opposes a change in current
...
It
opposes a change in voltage
...
They are formed whenever two customers run side-by-side
...

iC = C

dv
dt

(12
...
34)

Equation (12
...
The voltage is at 0o
but the resulting current through the capacitance is at +90o (leads)
...
35)

The capacitive reactance is defined as
XC =

1
ωC

(12
...
36) may be rewritten as
IC =

(V

∠90 o
XC

rms

)

Using Equation (12
...
37)

Divide the magnitudes and subtract the angles
Z C = X C ∠ - 90 o
=

-j
1
=
ωC
jω C

(12
...
Therefore, the impedance of a capacitor is a
j
frequency-dependent complex quantity, with the impedance of the capacitor
varying as an inverse function of frequency
...

Equation (12
...
The current is shifted 90o ahead of the voltage
...


where

Z C = (0 - j X L )

(12
...
The length of
the phasor XC lies entirely along the imaginary (-y) axis
...

Example 12-24

Compute the reactance of a 10 µF capacitor at a frequency of (a) 0 Hz, (b) 10
kHz
...

Solution: Use Equation (12
...
59 Ω
2 π 10 × 103 10 × 10 −6

(

)(

)

(c) When f = 1 MHz
XC =

1
= 0
...
0 × 10 6 10 × 10 −6

(

)(

)

38

Chapter 12

12
...
Figure 12-21 depicts the
impedances of R, L, and C in the complex plane
...


All the rules and laws learned in the study of DC circuits apply to AC circuits
including Ohm's law, Kirchhoff's laws, and network analysis methods
...

It is necessary to emphasize that although the impedance of circuit elements is
either purely real (for resistors) or purely imaginary (for inductors and
capacitors), the general definition of impedance for an arbitrary circuit should
allow for the possibility of having both a real and imaginary part, since practical
circuits are made up of more or less complex interconnections of various circuit
elements
...


22 V, -64o

12 V, 35o

15 V, 0o

Figure 12-22 Circuit for Example 12-25
...
This is shown in Figure 1223
...


In order to determine what the resultant vector's magnitude and angle are
without resorting to graphic images, we can convert each one of these polar-form
complex numbers into rectangular form and add
...
These figures are

40

Chapter 12

added together because the polarity marks for the three voltage sources are
oriented in an additive manner:
22∠ − 64 = 9
...
8
12∠35 = 9
...
9
15∠0 = 15 + j 0

Etotal = (9
...
8) + (9
...
9 ) + (15 + j 0 )
Etotal = 34
...
90V

In polar form, this equates to 36
...
50o
...
80 V, lagging the 15 volt (0o phase reference) by 20
...
A voltmeter
connected across these points in a real circuit would only indicate the polar
magnitude of the voltage (36
...
An oscilloscope could be used
to display two voltage waveforms and thus provide a phase shift measurement,
but not a voltmeter
...

This is extremely important in relating calculated figures of voltage and
current to real circuits
...
Rectangular figures must be
converted to polar figures (specifically polar magnitude) before they can be
related to actual circuit measurements
...

22 V, -64o

12 V, 35o

Figure 12-24 Circuit for Example 12-26
...
Note how the 12 V supply’s phase angle is still
referred to as 35o, even though the leads have been reversed
...
Even
though the angle is still written as 35o, the vector will be drawn 180o opposite of
what it was before: The resultant (sum) vector should begin at the upper-left point
(origin of the 22 volt vector) and terminate at the right arrow tip of the 15-V
vector: The connection reversal on the 12-V supply can be represented in two
different ways in polar form: by an addition of 180o to its vector angle (making it
12 V ∠ 215o), or a reversal of sign on the magnitude (making it -12 V ∠ 35o)
...

Etotal = E1 + E 2 + E 3

Etotal = (15 + j 0 ) + (− 9
...
9 ) + (9
...
80) = 14
...
65

In polar form, this equates to 30
...
9368o
...
9 POWER AND POWER FACTOR
An understanding of load characteristics in electrical power systems involves
the concept of power and power factor
...
These components are
apparent power, reactive power, and active or real power
...
9
...
The reactive power component of the load is used to supply

42

Chapter 12

energy that is stored in either a magnetic or electrical field
...
An example of a device that
supplies reactive power is the capacitor
...

The relationship between these electrical power quantities is best realized by
using the power triangle shown in Figure 12-26 (a)
...

When an AC power is applied to a reactive load, the voltage is 90o out of
phase with the current
...
Accordingly, the average power is zero, which means that reactive
loads do not dissipate power
...
The hypotenuse of the triangle
represents the apparent power component, which is calculated as

S = P2 + Q2

(12
...
9
...
It is expressed as
PF =

P
= cos (θ )
S

(12
...
It is expressed as
RF =

Q
= sin (θ )
S

(12
...
A power factor equal to 1
...
However, power factor
of 0
...
In general, the power
factor of a load will be between 0
...
0
...
The ratio of the circuit resistance
to the amplitude of the circuit impedance is called power factor
...
(b) Impedance triangle
...
43)

According to Equation (12
...
Therefore, the power factor is 1
...
In general, the power factor is related to the phase angle
through the impedance diagram (Figure 12-26)
...
9
...
However, a load in which the current leads the applied voltage is
said to have a leading power factor
...
Therefore, an
inductive load will have a lagging power factor
...

A leading power factor is one in which the current leads the applied voltage by
certain angle as shown in Figure 12-27 (b)
...


S

Q

θ
P

(a)
P

θ
S

Q

(b)
Figure 12-27 (a) Power triangle for lagging power factor
...

Example 12-27

Compute the power factor for each of the following cases
...
866
Power factor is 0
...


Fundamentals of Alternating Current

45

b) Convert 100 +j50 to polar form
100 +j50 = 111
...
56o
cos (θ) = cos (26
...
894
Power factor is 0
...


Example 12-28

A three-phase load consumes 100 kW, and 50 kVAR
...

Solution: Use Equation (12-40) to find the apparent power
S=

(100,000)2 + (50,000)2

= 111
...
8944
111,803

Use Equation (12-42) to find the reactive power
RF =

50,000
= 0
...
10 THREE-PHASE AC CIRCUITS
Nearly all-electric power generation and most of power transmission in the
world today are in the form of three-phase AC circuits
...
The three-phase
system is the usual type of supply for commercial and industrial premises of
medium size: schools, hotels, blocks of apartments, hospitals, etc
...

A three-phase power system consists of three-phase generators, transmission
lines, and loads
...


12
...
1 Wye-Connected System
A three-phase system consists of three AC sources, with voltages equal in
magnitude but differing in phase angle from the others by 120o, and connected at
a common point called neutral as shown in Figure 12-28
...
44)

Accordingly, the currents flowing in the three phases are
IA =

V∠0 o
= I∠ − θ
Z∠θ

IB =

V∠ - 120 o
= I∠ − 120 o − θ
Z∠θ

IC =

V∠ - 240 o
= I∠ − 240 o − θ
Z∠θ

(12
...
(b) Voltage
waveforms of each phase of the generator
...
This type of
connection is called wye or Y
...

Ia

IL

a
Iφ

IN

N

Ib

Resistive
load

b

IC

c

Figure 12-29 Y-connected generator with a resistive load
...

Since the load connected to this generator is assumed to be resistive, the
current in each phase of the generator will be at the same angle as the voltage
...
46)

I c = Iφ ∠ − 240 o

It is obvious that the current in any line is the same as the current in the
corresponding phase
...
47)

The relationship between line voltage and phase voltage is given by the
following equation
(12
...
10
...
The ∆ connection
is possible because the sum of the three voltages VA + VB + VC = 0
...


In the case of the ∆ connection, it is obvious that the line-to-line voltage
between any two lines will be the same as the voltage in the corresponding phase
...
49)

The relationship between line current and phase current can be found by
applying Kirchhoff’s current law at a nodes of the ∆
I L = 3Iφ

(12
...
A sinusoidal signal is mathematically represented in one of two ways: a timedomain forms ( v (t ) = A cos (ω t + θ ) and a frequency-domain (phasor) form
( V ( jω ) = A e jθ = A∠θ )
...
A phasor is a complex number, expressed in rectangular form and polar form
...
In rectangular form, the phasor has both real and imaginary parts
...
In polar form, the phasor consists of a magnitude equal to the peak amplitude
of the sinusoidal signal and a phase angle equal to the phase shift of the
sinusoidal signal referenced to a cosine signal
...
The three basic types of AC loads are resistive, inductive, and capacitive
...
In AC resistive load, the current and voltage are in phase with each other
...
True power can be produced only during periods of time that both the voltage
and current have the same polarity
...
The impedance of a resistor is simply its resistance
...
Induced voltage is proportional to the rate of change of current
...
Induced voltage is always opposite in polarity to the applied voltage
...
Pure inductive load contains no true power
...
The impedance of an inductor is jωL Ω
...
The impedance of an inductor is 1/jωC Ω
...
Once a circuit is represented in phasor-impedance form, all analysis
techniques practiced in resistive circuits (Chapter 11) apply once all elements
are replaced by their frequency-domain equivalents
...
Phasor analysis can only be performed on single-frequency circuit
...

16
...

17
...

18
...
Both types of
sources have three terminals, one for each phase; Y-connected sources have a
neutral connection as well
...
What is the difference between AC and DC electricity?
2
...

3
...

4
...
How many degrees are the current and voltage out of phase with each other
in a pure inductive circuit?
6
...
What two factors determine the capacitive reactance of a capacitor?
8
...
What is meant by a leading and lagging power factor?
10
...

a
...

c
...


40 cycles in 5
...

120 cycles in 100 ms
...
5 s
...
0 min
...


I = 1
...


12-3 Determine the inductance of the inductor in henries whose phasor diagram is
given in Figure 12-32
...
4 kHz
...


12-4 Determine the capacitance of the capacitor in farads whose phasor diagram is
given in Figure 12-33
...


Fundamentals of Alternating Current

53

I = 20 mA

E = 40 V

Figure 12-33 Circuit for Problem 16-4
...
Find the
value of E
...


12-6 Consider a series circuit consisting of a 2
...
An AC source is powering the circuit with a current of 50∠0o
mA
...

12-7 Repeat Problem 16-6
...

12-8 A 100∠0o-V (200 kHz) is applied across a parallel circuit consisting of 5
...
2-H inductor
...


54

Chapter 12

12-9 Repeat Problem 12-8
...
2-kΩ resistance with a 3-µF capacitor
...

12-12 Determine the power factor for each of the following circuit conditions and
state if it is leading or lagging
a
...

c
...

I=2A

θ = -50o

E = 20 V

Figure 12-34 Circuit for Problem 12-35
...

b
...

100 kW, 0
...


Fundamentals of Alternating Current

55

MULTIPLE CHOICE QUESTIONS
•

The peak value of a sine wave occurs
a
...

c
...


•

One of the following is not a right format to express the sinusoid V cos ωt
...

b
...

d
...

b
...

d
...

f
...

h
...

b
...

d
...

Once each cycle at the negative maximum value
...

Twice each cycle at the positive maximum value
...

b
...

d
...

b
...

d
...

b
...

d
...

Voltage leads current by 90o
...

Current and voltage are in phase
...

Voltage leads current by 90o
...

Current and voltage are in phase
...
2 kΩ

16 sin (5000 t + 10o)

0
...


a
...

c
...

e
...
7∠10o
16∠-80o
12
...
The answer is
Is the ratio of true power to apparent power

a
...

c
...


Reactive power
Power ratio
Power factor
Phase angle

150 mH



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