1
Unit 1
REVIEW OF PLANE TRIGONOMETRY
LEARNING OUTCOMES
At the end of this unit,
you are expected to:
1.
2.
3.
4.
5.
6.
7.
analyze a right triangle.
add vectors in parallelogram and polygon method.
use trigonometric identities.
analyze oblique triangles.
draw graphs of trigonometric functions.
define a complex number.
illustrate the graphical representation of a complex
number.
8. convert complex numbers to different forms.
9. perform mathematical operations of complex numbers.
10. apply complex numbers.
2
Important Terms
complex number
conjugate
cosine law
exponential form
oblique triangle
polygon method
quadrant
rectangular form
right triangle
sine law
trigonometric
form
vector addition
parallelogram
polar form
1.1 The Right Triangle
𝐵
𝑐
𝐴
𝑎
𝑏
𝐶
Figure 1.1
sin A =
csc A =
a
c
c
a
cos A =
b
c
c
sec A = b
tan A =
cot A =
a
b
b
a
1.1.1 Solution of a Right Triangle
Referring to Figure 1.1, the relationships of the sides a, b, and c can be expressed as
c 2 = a2 + b 2
(Pythagorean Theorem)
3
To find c, a, and b
c = √a2 + b 2
𝑎 = √c 2 − b 2
𝑏 = √c 2 − a2
Example 1.1. Solve for b and the angles A and B.
𝐵
𝑐 = 13
𝐴
Given:
𝑎 = 12
𝑏
a = 12 units
c = 13 units
Find: side b and angle B
Known:
b = √c 2 − a2
sin A =
a
c
;
A = sin−1
a
c
A + B = 90 ; B = 90 – A
Solution:
b = √132 − 122
b = 5 units
12
A = sin−1 13
A = 67.38
B = 90 – 67.38
B = 22.62
Answer: b = 5 units , B = 22.62°
𝐶
4
Example 1.2. Find the sides a and b.
𝐵
𝑐 = 13
𝑎
40
𝐴
𝑏
𝐶
Solution:
sin A =
Solving for a
a
c
a = c sin A
= 13 sin 40
= 8.36 units
cos A =
To find b
b
c
b = c cos A
= 13 cos 40
= 9.96 units
1.2 Vector Addition
1.2.1 Parallelogram Method
y
F1
x
FT
F2
5
1.2.2
Polygon (Head to Tail method) Method
y
FT
F1
F2
Note: F1 and F2 are the given vectors, and FT is the equivalent (total or resultant) vector.
Example 1.3. Determine graphically the resultant of the following forces using:
(a) parallelogram method.
(b) polygon method.
F1
F2
F3
Solution:
6
(a) By parallelogram method, get first the resultant of F1 and F2 and call it FA. Then
use again parallelogram method to determine the resultant vector of FA and F3 to
get the total or equivalent force.
FA
F1
FT
F2
F3
(b) Using polygon method
F2
F1
F3
FT
.
7
1.3 Formulas Involving Addition or Subtraction of Functions
sin2  + cos2  = 1
sec2  = 1 + tan2 
csc2
 = 1 +
cot2


1.4 Sine and Cosine of the Sum or Difference of Two Angles
sin (  ) = sin  cos   cos  sin 
cos ( +) = cos  cos  - cos  sin 
cos ( - ) = cos  cos  + cos  sin 
sin (-A) = -sin A
cos (-A) = cos A
sin (x + 90) = cos x
cos (x + 90) = - sin x
sin (x – 90) = - cos x
sin (180 - x) = sin x
cos (180 - x) = - cos x
1.5 Double Angle Formulas
sin 2 = 2 sin  cos 
cos 2 = cos2  – sin2  = 2cos2  – 1 = 1 – 2 sin2 
tan 2 =
cot 2 =
2 tan 
1−tan2 
cot 2  − 1
2 cot 
cos2  =
1 + cos2
2
sin2  =
1 − cos2
2
8
1.6 Oblique Triangles
1.6.1 Sine Law
B
c
a
A
b
C
a
b
c
=
=
sin A
sin B
sin C
1.6.2 Cosine Law
c
A
a
b
a2 = b2 + c 2 − 2bc cos A
Example 1.4. Find the x and y components of F1.
F1 = 18 N
F2= 20 N
FT = 30 N
9
Solution:
F1 = 18 N
FT = 30 N
This is also
equal to F1



F2= 20 N
 can be found by applying cosine law:
FT2 = F12 + F22 - 2 F1 F2 cos 
302 = 182 + 202 – 2(18)(20) cos 
 = 104.15
Then  = 180 -  = 180 - 104.15 = 75.85
Considering F1
F1x = F1 cos 
= 18 cos 75.85
F1 = 18 N

F1y
= 4.4 N
F1y = F1 cos 

F1x
= 18 sin 75.85
= 17.45 N
10
1.7
Graphs of Trigonometric Functions
150
100
50
0
0
1
2
3
4
5
6
7
-50
-100
-150
(a) y = A sin x
A = 100 units, x is expressed in radians
-2
(b)
20
15
10
5
0
-5 0
-10
-15
-20
2
y = A cos x
4
6
8
or y = A sin(x + /2)
A = 15 units, x is expressed in radians
.
11
Assessment No. 1
PLANE TRIGONOMETRY
Name:
Score: _________ Rating: ______
Aviso, Myrell Jud A.
1. Sketch the graph of this equation: y = 60 sin x, where x is expressed in radians.
60
40
20
0
0
1
2
3
4
5
6
7
-20
-40
-60
𝝅
2. Sketch the graph of this equation: y = 150 sin(𝒙 + 𝟐 ) , where x is expressed in
radians.
150
100
50
0
-2
-50
-100
-150
-1
0
1
2
3
4
5
6
7
12
3. Find the x and y components of the force exerted by the father in pulling the cart
with his child riding. Given F = 100 N and  = 30.
13
4. Two horses are pulling a barge along a canal as shown in the figure below. The cable
connected to the first horse makes an angle of 30° with respect to the direction of
the canal (x-axis), while the cable attached to the second horse makes an angle of
45°. The first horse exerts a force of 50 N and the second horse exerts a force 70 N.
Find the total force exerted by the two horses and the direction of the barge with
respect to the x axis.
.
14
1.8
The Quadrants
-270
90
(/2)
II
180
()
3
(− 2 )
II
I
III
0 or 360
(0 or 2)
IV
-180
(-)
3
( )
2
Counterclockwise rotation
I
0
.
III
IV
0 or 360
(0 or 2)
-90
(-/2)
270
0
I
Clockwise rotation
II
90
ð
2
III
IV
180
270

3

2
360
2
15
Problem Set No. 1
REVIEW OF PLANE TRIGONOMETRY
1.
A ladder leans against a wall with its foot 10 ft from the wall. If the ladder makes
an angle of 60 with the ground, how long is the ladder.
A. 20 ft
B. 15 ft
C. 10 ft
D. 5 ft
2.
In no. 1, how far above the ground is the top of the ladder.
A. 12.73 ft
B. 13.72 ft
C. 17.32 ft
D. 19.73 ft
3.
A tower 160 ft casts a shadow 92.4 ft long on the ground. Find the angle of
elevation of the sun. (The angle of elevation is the angle with its rays make with
level ground.)
A. 35°
B. 60°
C. 45°
D. 75°
4.
From the top of a tower at C the angle of depression of point A is 56. If the
distance between point A and point B (the foot of the tower) is 180 meters, how
high is the tower?
A. 168.94 m
B. 178.5 m
C. 230.89 m
D. 266.86 m
5.
A boy pushes a force of 15 lb toward north. Another boy pushes the same box
toward east with a force of 20 lb. Find the resultant force.
A. 25 lb
B. 30 lb
C. 35 lb
D. 40 lb
6.
One of the diagonals of a parallelogram makes angles of 2235’ and 4848’,
respectively, with two adjacent sides. If the diagonal is 18.54 ft long, find the
lengths of the sides of the parallelogram.
A. 7,25 ft, 13.73 ft
B. 7.51 ft, 14.72 ft
C. 8.34 ft, 15.83 ft
D. 9.35 ft, 16.72 ft
16
7.
In the figure shown below, find the magnitude of F1 if F2 = 100 N and makes an
angle of 30 with the respect to the x axis and the resultant of the two forces is 167
N and makes an angle of 50 with respect to the x axis.
F1
F2
A.
B.
C.
D.
.
80.64 ft
82.64 ft
89.49 ft
90.29 ft
17
1.9 Complex Numbers
y (imaginary axis)
+ √−1 = + jA
-A
A
x (real axis)
−√−1 = - j
A complex plane.
A complex number is one of the form
x  jy
that is, it is the sum of a real number and an imaginary one.
A complex number is the conjugate of another if their imaginary parts are of
opposite signs. For example, -5 + j6 is the conjugate of –5 – j6.
1.9.1 Graphical Representation of Complex Numbers
P (x, y)
r
y

x
Let z⃑ be the vector connecting origin and point P whose coordinates are x and y. Then, in
complex form
z⃑ = x + jy
18
To find r and  in terms of x and y:
r = √x 2 + y 2
Note that r is the magnitude of z⃑
y
 = tan-1 x
To find x and y in terms of r and :
x = r cos 
y = r sin 
1.9.2 Forms of a Complex Number
Rectangular Form:
z⃑ = x + jy
Trigonometric Form:
z⃑ = r cos  + j r sin 
= r (cos  + j r sin  )
Polar Form:
z⃑ = r cjs  = r 
Exponential Form:
z⃑ = rej where  is in radians
1.9.3 Addition and Subtraction of Complex Number
1.9.3.1 Algebraic Method
(a) Addition. To add two or more complex numbers, add the real parts and
then the pure imaginary parts.
Example 1.5.
Add z⃑1 = 5 + j3 and z⃑2 = −2 − j5
z⃑1 + z⃑2 = (5 + j3) + (−2 − j5)
Solution:
= 3 − j2
(b) Subtraction. To subtract one complex number from another, subtract the
real parts and then subtract the pure imaginary parts.
Example 1.6.
Solution:
Subtract z⃑1 = 5 + j3 by z⃑2 = −2 − j5
z⃑1 − z⃑2 = (5 + j3) − (−2 − j5)
= 7 + j8
19
1.9.3.2 Graphical Method
(a) Addition. Use parallelogram method to find the sum of two or more
complex numbers.
Example 1.7.
Add z⃑1 = 5 + j3 and z⃑2 = −2 − j5
Solution:
First plot z⃑1 and z⃑2 .
1. Use parallelogram method to determine the sum.
2. Count the x and y components of the sum.
8
7
6
5
4
3
2
1
0
-9 -8 -7 -6 -5 -4 -3 -2 -1
1 2 3
-2
-3
-4
-5
-6
-7
-8
-9
4
5
6
7
8
9
As you can see the sum is 3 - j2
(b) Subtraction.
Example 1.8. Subtract z⃑1 = 5 + j3
by z⃑2 = -2 – j5
Solution:
1. First plot z⃑1 and z⃑2 .
2. Take the negative of z⃑2 . This is the opposite of z⃑2 .
3. Find the difference using parallelogram method. This is similar to
addition.
20
4. Counting the x and y components, the result is 7 + j8.
8
7
6
5
4
3
2
1
0
-9 -8 -7 -6 -5 -4 -3 -2 -1
1 2 3
-2
-3
-4
-5
-6
-7
-8
-9
4
5
6
7
8
9
1.9.3 Transformation of a Complex Number from One Form to Another
Example 1.9. Express each of the following in rectangular, polar and exponential forms.
z2
8
7
6
5
4
3
2
1
2
r2
0
-9 -8 -7 -6 -5 -4 -3 -2 -1
r3
3
z3
z1
r1
1
1 2 34 4
-2
-3
r4
-4
-5
-6
-7
-8
-9
5
6
7
4
z4
8
9
21
Solution:
Note: Always use the smaller angle between the positive real axis (the reference) and
the vector.
For z1: r1 = √42 + 52
1 = tan-1
5
4
= 6.4
For z2: r2 = √62 + 32 = 6.71
3
1 = 180 - tan-1 6
= 51.34
= 153.43
For z3: r3 = √42 + 52 = 6.4
For z4: r4 = √62 + 42 = 7.2
5
3 = 180 - tan-1 4 = 128.66
4
4 = tan-1 6 = 33.69
Answers
Rectangular Form
Polar Form
Exponential Form
z⃑1
4 + j5
6.4 51.34
6.4 ej0.8961
z⃑2
-6 + j3
6.71 153.43
6.71 ej2.6779
z⃑3
-4 - j5
6.4 -128.66
6.4 e–j2.2455
z⃑4
6 – j4
7.2 -33.69
7.2 e-j0.588
1.9.4 Multiplication and Division of Complex Numbers
(a) Multiplication of Complex Numbers in Rectangular Form
Example 1.10.
Multiply z⃑1 = 5 + j2 by z⃑2 = 3 – j4
Solution:
z⃑1 z⃑2 = (5 + j2)(3 – j4)
= (5)(2) + (j2)(3) + (5)(-j4) + (j2)(-j4)
= 10 + j6 – j20 – j2 8
= 10 + j6 – j20 – (-1)(8)
where j2 = -1
22
= 18 – j14
(b) Multiplication of Complex Numbers in Polar Form
z⃑1 = r11
Let
z⃑2 = r22
z⃑1z⃑2 = r11  r22 = r1  r21 + 2
Example 1.11.
Multiply z⃑1 = 2045 by z⃑2 = 2-30
Solution:
z⃑1 z⃑2 = (2045 )( 2-30)
= 20245+(-30)
= 4015
(c) Division of Complex Numbers in Rectangular Form
Example 1.12.
Divide z⃑1 = 5 + j2
by z⃑2 = 3 – j4
Solution:
⃑⃑1
z
⃑⃑2
z
=
5 + j2
3 – j4
Rationalize, i.e., multiply the numerator and denominator by the conjugate
of the denominator. The conjugate of z⃑2 = 3 – j4 is ⃑⃑⃑⃑
z̅2 = 3 + j4
⃑⃑1
z
⃑⃑2
z
⃑⃑1
z
⃑⃑2
z
=
=
5 + j2 3 + j4
3 – j4

7 + j26
32 + 42
3 + j4
=
7 + j26
25
= 0.28 + j1.04
(d) Division of Complex Numbers in Polar Form
Let z⃑1 = r11
⃑⃑1
z
⃑⃑2
z
=
r1 1
r2 2
Example 1.13. Divide
Solution:
⃑⃑1
z
⃑⃑2
z
=
=
z⃑2 = r22
r1
r2
 1 + 2
z⃑1 = 2045 by z⃑2 = 2-30
2045
2−30
=
20
2
 45 − (−30) = 1075
23
Assessment No. 2
COMPLEX NUMBERS
Name:
Score: _________ Rating: ______
Aviso, Myrell Jud A.
1. Express each of the following in rectangular, polar and exponential forms.
8
7
6
5
4
3
2
1
z2
z1
0
-9 -8 -7 -6 -5 -4 -3 -2 -1
z3
1 2 34 4
-2
-3
-4
-5
-6
-7
-8
-9
5
z4
6
7
8
9
24
2. Solve this problem using complex numbers: In the figure shown below, find the
magnitude of F1, its angle with respect to the + x axis, its x and y components , if F2 =
100 N and makes an angle of 30 with the respect to the x axis and the resultant of
the two forces is 167 N and makes an angle of 50 with respect to the x axis.
F1
F2
3. Add the following using algebraic and graphical methods.
(a) 8 + j4, 4 – j6
4. Convert the following to polar and exponential forms:
(a) (8 + j4)
(b) (-7 – j7)
25
5. Convert the following to rectangular and exponential forms.
(a) 660
(b) 10120
6. Convert the following to rectangular and polar forms.
(a) 9e-j0.9823
(b) 12ej/3
.
26
Unit 2
INTRODUCTION TO ELECTRICITY
LEARNING OUTCOMES
At the end of this unit,
you are expected to:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
define electricity.
identify the methods of producing electricity.
identify the effects of electricity.
name some scientists who contributed to the development of
electricity and electronics.
discuss the scientist’s contributions to electricity and
electronics.
quote some applications of electricity and electronics.
identify various electrical components.
use metric prefixes in simplifying large and small numbers.
perform mathematical operations involving powers of ten and
metric prefixes.
discuss the difference between direct current and alternating
current.
27
Important Terms
Electricity
static electricity
dynamic
electricity
Resistor
Resistance
Inductor
Inductance
Transformer
semiconductor
active element
passive element
electrical quantities
metric prefixes
direct current
alternating current
2.7 Definition of Electricity
Electricity is a physical phenomenon arising from the existence
and interaction of electric charge. It is a form of energy generated
by friction, heat, light, magnetism, chemical reaction, etc.
2.1.1 Two Types of Electricity:
a. Static electricity – electricity at rest. It cannot flow from one place to another.
b. Dynamic electricity – also known as current electricity. Electricity in motion. It
can be transmitted from one place to the other.
2.1.2 Methods of Producing Electricity
There are six methods for producing electricity:
1. Magnetism
2. Chemical action
3. Pressure
4. Heat
5. Friction
6. Light
2.1.3 Electrical Effects
28
With the exception of friction, electricity can be used to cause the same effects that
cause it.
1. Magnetism
2. Chemical action
3. Pressure
4. Heat
5. Light
2.7 History of Electricity and Electronics
Prehistoric people experienced the properties of magnetite – permanently
magnetized pieces of ore, often called lodestones. These magnetic stones were strong
enough to lift pieces of iron.
The philosopher Thales of Miletus (640 – 546 B.C.) is thought to have been the first person
who observed the electrical properties of amber. He noted that when amber was rubbed,
it acquired the ability to pick up light objects such as straw and dry grass. He also
experimented with the lodestones and knew of its power to attract iron. By the thirteenth
century, floating magnets were used for compasses.
One of the first important discoveries about static electricity is attributed to
William Gilbert (1540-1603). Gilbert was an English Physician who, in a book published
in 1600, described how amber differs from magnetic loadstones in its attraction of certain
materials. He found that when amber was rubbed with a cloth, it attracted only
lightweight materials, whereas loadstones attracted only iron. Gilbert also discovered
that other substances, such as sulfur, glass, and resin, behave as amber does. He used the
Latin word electron for amber and originated the word electrical for the other substances
that acted similar to amber. The word electricity was used for the first time by Sir Thomas
Browne (1605-82), an English physician.
Following Gilbert’s lead, Robert Boyle published his many experimental results in
1675. Boyle was one of the early experiments with electricity in a vacuum.
Otto von Guericke (1602 – 1686) built an electrical generator and reported it in his
Experimental Nova of 1672. This device was a sulfur globe on a shaft that could be turned
on its bearing . When the shaft was turned with a dry hand held on the surface, an
electrical charge gathered on the globe’s surface. Guericke also noted small sparks when
the globe was discharges. In his studies of attraction and gravitation, Guericke devised
the first electrical generator. When a hand was held on a sulfur ball revolving in its frame,
the ball attracted paper, feathers, chaff, and other light objects.
29
Another Englishman, Stephen Gray (1696-1736), discovered that some substances
conduct electricity and some do not. Following Gray’s Lead, a Frenchman named Charles
du Fay experimented with the conduction of electricity. These experiments led him, to
believe that there were two kinds of electricity. He called one type vitreous electricity and
the other type resinous electricity. He found that objects having vitreous electricity
repelled each other and those having resinous electrify attracted each other. It is known
today that two types of electrical charge exist, positive and negative. Negative charge
results from an excess of electrons in a material and positive charge results from a
deficiency of electrons.
A major advance in electrical science was made in Leyden, Holland, in 1746, when
Peter van Musschenbroek introduces a jar that served as a storage apparatus for
electricity. The jar was coated inside and out with a tinfoil, and a metallic rod was
attached to the inner foil lining and passed through the lid. Leyden jar were gathered in
groups (called batteries) and arranged with multiple connections, thereby further
improving the discharge energy.
Benjamin Franklin (1706-90) conducted studies in electricity in the mid-1700s. He
theorized that electricity consisted of a single fluid, and he was the first to use the terms
positive and negative. In his famous kite experiment, Franklin showed that lightning is
electricity.
Charles Augustin de Coulomb (1736-1806), a French physicist, in 1785 proposed
the laws that govern the attraction and repulsion between electrically charged bodies.
Today, the unit of electrical charge is called the coulomb.
Luigi Galvani (1737-98) experimented with current electricity in 1786. Galvani
was a professor of anatomy at the University of Bologna in Italy. Electrical current was
once known as galvanism in his honor.
In 1800, Alessandro Volta (1745-1827), an Italian professor of physics, discovered
that the chemical action between moisture and two different metals produced electricity.
Volta constructed the first battery, using copper and zinc plates separated by paper that
had been moistened with a salt solution. This battery, called the voltaic pile, was the first
source of steady electric current. Today, the unit of electrical potential energy is called
the volt in honor of Professor Volta.
A Danish scientist, Hans Christian Oersted (1777-1851), is credited with the
discovery of electromagnetism, in 1820. He found that electrical current flowing through
a wire caused the needle of a compass to move. This finding showed that a ,magnetic field
exists around a current-carrying conductors and that the field is produced by the current.
The modern unit of electrical current is the ampere (also called amp) in honor of
the French physicist André Ampère (1775-1836). In 1820, Ampère measured the
magnetic effect of an electrical current. He found that two wires carrying current can
30
attract and repel each other, just as magnets can. By 1822, Ampère had developed the
fundamental laws that are basic to the study of electricity.
One of the most well known and widely used laws in electrical circuits today is
Ohm’s law. It was formulated by Georg Simon Ohm (1787-1854), a German teacher, in
1826. Ohm’s law gives us the relationship among the three important electrical quantities
of resistance, voltage, and current.
Although it was Oersted who discovered electromagnetism, it was Michael Faraday
(1791-1867) who carried the study further. Faraday was an English physicist who
believed that electricity could produce magnetic effects, then magnetism could produce
electricity. In 1831 he found that a moving magnet cause an electric current in a coil of
wire placed within the field of the magnet. This effect, known today as electromagnetic
induction, is the basic principle of electric generators ands transformers.
Joseph Henry (1797-1878), an American physicist, independently discovered the
same principle in 1831, and it is in his honor that the unit of inductance is called the henry.
The unit of capacitance, the farad, is named in honor of Michael Faraday.
A paper published by James Prescott Joule in 1841 claimed the discovery of the
relationship between a current and the heat or energy produced, which today we call
Joule’s law. The unit of energy is called the joule in his honor.
In the 1860s, James Clerk Maxwell (1831-79), a Scottish Physicist, produced a set
of mathematical equations that expressed the laws governing electricity and magnetism.
These formulas are known as Maxwell’s equations.
Maxwell also predicted that
electromagnetic waves (radio waves) that travel at the speed of light in space could be
produced.
It was left to Heinrich Rudolph Hertz (1857-94), a German physicist, to actually
produce these waves that Maxwell predicted. Hertz performed this work in the late
1880s. Today, the unit of frequency is called the hertz.
The Beginning of Electronics
The early experiments in electronics involved electric currents in glass tubes. One of the
first to conduct such experiments was a German named Heinrich Geissler (1814-79).
Geissler found that when he removed most of the air from a glass tube, the tube glowed
when an electrical potential was placed across it.
Around 1878, Sir William Crookes (1832-1919), a British scientist, experimented
with tubes similar to those of Geissler. In his experiments, Crookes found that the current
in the tubes seemed to consist of particles.
Thomas Edison (1847 – 1931), experimenting with the carbon-filament light bulb
that he had invented, made another important finding. He inserted a small metal plate in
31
the bulb. When the plate was positively charged, a current flowed from the filament to
the plate. This device was the first thermionic diode. Edison patented it but never used
it.
The electron was discovered in the 1890s. The French physicist Jean Baptiste
Perrin (1870 – 1942) demonstrated that the current in a vacuum tube consists of
negatively charged particles. Some of the properties of these particles were measured by
Sir Joseph Thomson (1856 – 1940), a British physicist, in experiments he performed
between 1895 and 1897. These negatively charged particles later became known as
electrons. The charge on the electron was accurately measured by an American physicist,
Robert A. Millikan (1868 – 1953), in 1909. As a result of these discoveries, electrons could
be controlled, and the electronic age was ushered in.
Putting the Electron to Work
A vacuum tube that allowed electrical current in only one direction was constructed in
1904 by John A. Fleming, a British scientist. The tube was used to detect electromagnetic
waves. Called the Fleming valve, it was the forerunner of the more recent vacuum diode
tubes.
Major progress in electronics, however, awaited the development of a device that
could boost, or amplify, a weak electromagnetic wave or radio signal. This device was the
audion, patented in 1907 by Lee de Forest, an American. It was a triode vacuum tube
capable of amplifying small electrical signals.
Two other Americans, Harold Arnold and Irving Langmuir, made great
improvements in the triode tube between 1912 and 1914. About the same time, de Forest
and Edwin Armstrong, an electrical engineer, used the triode tube in an oscillator circuit.
In 1914, the triode was incorporated in the telephone system and made the
transcontinental telephone network possible.
The tetrode tube was invented in 1916 by Walter Schottky, a German. The tetrode,
along with the pentode (invented in 1926 by Tellegen, a Dutch engineer), provided great
improvements over the triode. The first television picture tube, called the kinescope, was
developed in the 1920s by Vladimir Zworykin, an American researcher.
During World War II, several types of microwave tubes were developed that
made possible modern microwave radar and other communications systems. In 1939,
the magnetron was invented in Britain by Henry Boot and John Randall. In the same
year, the klystron microwave tube was developed by two Americans, Russell Varian and
his brother Sigurd Varian. The traveling-wave tube was invented in 1943 by Rudolf
Komphner, an Austrian-American.
The Computer
32
The computer probably has had more impact on modern technology than any other single
type of electronic system. The first electronic digital computer was completed in 1946 at
the University of Pennsylvania. It was called the Electronic Numerical Integrator and
Computer (ENIAC). One of the most significant developments in computers was the
stored program concept, developed in the 1940s by John von Neumann, an American
mathematician.
Solid State Electronics
The crystal detectors used in the early radios were the forerunners of modern solid state
devices. However, the era of solid state electronics began with the invention of the
transistor in 1947 at Bell Labs. The inventors were Walter Brattain, John Bardeen, and
William Shockley.
In the early 1960s, the integrated circuit was developed. It incorporated many
transistors and other components on a single small chip of semiconductor material.
Integrated circuit technology continues to be developed and improved, allowing more
complex circuits to be built on smaller chips.
The introduction of the microprocessor in the early 1970s created another
electronics revolution: the entire processing portion of a computer placed on a single,
small, silicon chip. Continued development brought about complete computers on a
single chip by the late 1970s.
Major Events in Electrical Sciences and Engineering
1672
1675
1746
1750
1767
1786
1800
1801
1820
1821
1825
1828
Ottoo von I published I Nova.
Robert Boyle was published Production of Electricity.
The Leyden jar was demonstrated in Holland.
Benjamin Franklin invented the lightning conductor.
Joseph Priestley published the Present State of Electricity.
Luigi Galvani observed electrical convulsion in the legs of dead frogs.
Alessandro Volta announced the voltaic pile.
Henry Moyes was the first to observe an electric arc between carbon
rods.
Hans Oersted discovered the deflection of a magnetic needle by
current on a wire.
Michael Faraday produced magnetic rotation of a conductor and
magnet- the first electric motor.
Andre-Marie Ampere defined electrodynamics.
Joseph Henry produced silk-covered wire and more powerful
electromagnets.
33
1831
1836
1836
1841
1843
1850
1858
1861
1862
1873
1874
1877
1877
1881
1881
1883
1884
1885
1886
1897
1898
1904
Michael Faraday discovered electromagnetic induction and carried
out experiments with an iron ring and core. He also experimented
with a magnet and rotating disk.
Samuel Morse devised a simple relay.
Electric light from batteries was shown at Paris Opera.
James Joule stated the relation between current and energy produced.
Morse transmitted telegraph signals from England to France.
First channel telegraph signals from Baltimore to Washington, D.C.
Atlantic telegraph cable was completed and the first message sent.
Western Union established telegraph service from New York to San
Francisco.
James Clerk Maxwell determined the ohm.
Maxwell published Treatise on Electricity and Magnetism.
Alexander Graham Bell invented the telephone.
Thomas Edison invented the telephone.
Edison Electric Light Company was formed.
First hydropower station was brought into use at Niagara, New York.
Edison constructed the first electric power station at Pearl Street,
New York.
Overhead trolley electric railways were started at Portrush and
Richmond, Virginia.
Philadelphia electrical exhibition was held.
The American Telephone and Telegraph Company was organized.
H. Hallerith introduced his tabulating machine.
J.J. Thomson discovered the electron.
Guglielmo Marconi transmitted radio signals from South Foreland to
Wimereux, England.
John Ambrose Fleming invented the thermionic diode.
2.7 Applications of Electricity and Electronics
Computers
Communications
34
Medicine
Automation
Consumer Products
2.4 Circuit Components and Measuring Instruments
Resistors
These can be the carbon-composition type or wound with special resistance wire. Their
function is to limit the amount of current or divide the voltage in a circuit.
35
Capacitors
A capacitor is constructed of two conductor plates separated by an insulator (called a
dielectric). Its basic function is to concentrate the electric field of voltage across the
dielectric. As a result, the capacitor can accumulate and store electric charge from the
voltage source. Furthermore, the dielectric can discharge the stored energy when the
charging source is replaced by a conducting path.
When ac voltage is applied, the capacitor charges and discharges as the voltage varies.
The practical application of this effect is the use of capacitors to pass an ac signal but to
block a steady dc voltage.
Capacitors
Inductors
Inductors
An inductor is just a coil of wire. Its basic function is to concentrate the magnetic field of
electric current in the coil. Most important, an induced voltage is generated when the
current with its associated magnetic field changes in value or direction. Inductors are
often called chokes.
In the practical application of a choke, the inductor can pass a steady current better than
alternating current. The reason is that a steady current cannot produce induced voltage.
Note that the effect of a choke, passing a steady current, is the opposite of that of a
coupling capacitor, which blocks dc voltage.
Transformers
36
A transformer consists of two or more coil windings in the same magnetic field. Induced
voltage is produced when the current changing in any winding. The purpose of a
transformer is to increase or decrease the amount of ac voltage coupled between the
windings. However, the transformer operates only with alternating current.
Transformers
Semiconductor Devices
Semiconductor Devices
These include diode rectifiers and transistor amplifiers, either as separate, discrete
components or as part of an IC chip. A diode has two electrodes; the transistor has three.
In addition, the silicon controlled rectifier (SCR) and triac are used for power-control
circuits.
Active and Passive Elements
Active elements - are capable of delivering power to some external device.
Examples: dependent and independent voltage and current sources
Passive elements – are capable of receiving power. They are able to store to store finite
amounts of energy and then return that energy later to various external devices.
Examples are resistors, inductors, and capacitors.
Electronic Instruments
Typical instruments include the power supply, for providing voltage and current; the
voltmeter, for measuring voltage; the ammeter, for measuring current; the ohmmeter, for
measuring resistance; the wattmeter, for measuring power; and the oscilloscope for
observing and measuring AC voltages.
37
2.5 Electrical Quantities and Units with SI symbols.
Quantity
Capacitance
Charge
Conductance
Current
Energy
Frequency
Impedance
Inductance
Power
Reactance
Resistance
Voltage
Symbol
Unit
Symbol
C
Q
G
I
W
f
Z
L
P
X
R
V
farad
coulomb
siemen
ampere
joule
hertz
ohm
henry
watt
ohm
ohm
volt
F
C
S
A
J
Hz

H
W


V
Value
one billion
one million
one thousand
one-thousandth
one-millionth
one-billionth
one-trillionth
Metric Prefix
giga
mega
kilo
milli
micro
nano
pico
Metric Symbol
G
M
k
m

n
p
2.6 Metric Prefixes
Power of Ten
109
106
103
10-3
10-6
10-9
10-12
38
Problem Set No. 2
METRIC PREFIXES
I. Express each of the following as quantity having a metric prefix:
1) 31 𝑥 10−3 𝐴
A. 0.31 mA
B. 3.1 mA
C. 31 mA
D. 310 mA
2) 5.5 𝑥 103 𝑉
A. 5.5 kV
B. 55 kV
C. 550 kV
D. 5.5 MV
3) 200 𝑥 10−12 𝐹
A. 200 pF
B. 200 nF
C. 200 µF
D. 2000 pF
4) 0.000003 𝐹
A. 3 µF
B. 30 µF
C. 3 nF
D. 30 nF
5) 3,300,000 
A. 3.3 kΩ
B. 33 kΩ
C. 3.3 MΩ
D. 33 MΩ
6) 350 𝑥 10−9 𝐴
A. 35 nF
B. 350 nF
C. 3.5 pF
D. 35 pF
II. Express each quantity as a power of ten:
7) 5 𝐴
A. 5 x 10-3 A
B. 5 x 10-6 A
C. 5 x 10-9 A
D. 50 x 10-3 A
39
8) 43 𝑚𝑉
A. 43 x 10-3 V
B. 43 x 10-6 V
C. 43 x 10-9 V
D. 43 x 10-12 V
9) 275 𝑘
A. 275 x 106 V
B. 275 x 103 V
C. 275 x 10-3 V
D. 275 x 10-6 V
10) 10 𝑀𝑊
A. 10 x 1012 W
B. 10 x 109 W
C. 10 x 106 W
D. 10 x 103 W
III. Add the following quantities:
11) 6 𝑚𝐴 + 3 𝐴 = _________ mA
A.
6.03
B.
60.03
C.
6.003
D.
6.3
12) 550 𝑚𝑉 + 3.2 𝑉 = _________ V
A.
375
B.
37.5
C.
3.75
D.
0.375
13) 12 𝑘 + 6800  = ________ k
A.
1880
B.
188
C.
18.8
D.
1.88
14) 15 𝑀𝑊 + 7500 𝑘𝑊 = __________ MW
A.
0.0225
B.
0.225
C.
2.25
D.
22.5
.
40
2.7 Comparison of AC and DC
+ DC Voltage
AC Voltage
AC
Direct Current. The DC electricity, flows in one direction. The flow is said to be from
negative to positive. The normal source of a DC electricity, is the dry cell or storage
battery.
Alternating Current. The AC electricity constantly reverses its direction of flow. It is
generated by machine called generator. This type of current is universally accepted
because of its limited number of applications with the following advantages.
1.
2.
3.
4.
5.
It is easily produced.
It is cheaper to maintain.
It could be transformed into higher voltage.
It could be distributed to far distance with low voltage drop.
It is more efficient compared with the direct current.
Comparison of DC Voltage and AC Voltage
DC Voltage
AC Voltage
Fixed Polarity
Reverses polarity
Can be steady or vary in magnitude
Varies between reversals in polarity
Steady value cannot be stepped up or
Can be stepped up or down for electric
down by a transformer
power distribution
Easier to measure
Easier to amplify
Heating effect is the same for direct or alternating current
The War of Currents
Thomas Edison, a prolific inventor in the 19th and 20th centuries, holds the record
for the most U.S. patents by one person. When it came to technology, he was
usually a winner. But there was one important area where Edison lost - and lost
badly. It was called the 'War of Currents,' and it pitted Edison and his support for
direct current (DC) electricity against engineers like George Westinghouse and
Nikola Tesla, who supported alternating current (AC).
41
In the 1880s, incandescent lighting was the main goal, and DC was just as good as
AC. But a storm was brewing, and it centered on which type of power, AC or DC,
would be best for power generation, electric motors, and power transmission.
The war came to a head in 1893, when the contract to provide electricity to the
Chicago World's Fair was awarded to Westinghouse, whose proposal, using AC,
came in over 30% cheaper than Edison's. In that same year, the Niagara Falls
Power Company decided to go with AC power generation for the city of Buffalo and
signed with Westinghouse and Tesla as well. These two major victories were part
of the changes taking place rapidly in the 1890s that set our country on a path
toward AC power.
.
42
Objective Test No. 1
INTRODUCTION TO ELECTRICITY
1.
Which of the following is not an electrical quantity?
A. time
B. power
C. current
D. voltage
2.
The unit of current is
A. volt
B. watt
C. joule
D. ampere
3.
The unit of voltage is
A. ohm
B. volt
C. watt
D. farad
4.
The unit of resistance is
A. ohm
B. hertz
C. henry
D. ampere
5.
15,000 W is the same as
A. 15 W
B. 15 MW
C. 15 kW
D. 15 mW
6.
The quantity 4.7 x 103 is the same as
A. 0.0047
B. 470
C. 4700
D. 47,000
7.
The quantity 56 x 10-3 is the same as
A. 0.056
B. 0.560
C. 560
D. 56,000
8.
The number 3,300,000 can be expressed as
43
A.
B.
C.
D.
9.
3.3 x 10-6
3.3 x 106
3.3 x 109
3.3 x 1012
Ten milliamperes can be expressed as
A. 10 A
B. 10 Ma
C. 10 Ka
D. 10 MA
10. Five thousand volts can be expressed as
A. 5 kV
B. 50 MV
C. 500 kV
D. 5000 kV
11. Twenty million ohms can be expressed as
A. 20 
B. 20 m
C. 20 M
D. 20 MW
12. Hertz is the unit of
A. time
B. power
C. frequency
D. inductance
13. An oscilloscope is usually used to measure
A. rms voltage
B. average voltage
C. maximum voltage
D. effective voltage
14. A step-down transformer,
A. lowers both the voltage and current.
B. increases both the voltage and current.
C. lowers the voltage and increases the current.
D. lowers the current and increases the voltage.
15. The prefix pico means
A. 10-12 of a unit
B. 10-9 of a unit
44
Objective Test No. 2
INTRODUCTION TO ELECTRICITY
1.
Discovered the difference between conductors and insulators in 1729.
A. Luigi Galvani
B. Stephen Gray
C. Alessandro Volta
D. Gottfried Wilhelm Leibniz
2.
Discovered Galvanic action in 1780.
A. Luigi Galvani
B. Stephen Gray
C. Alessandro Volta
D. Gottfried Wilhelm Leibniz
3.
Invented the electric dry cell in 1800.
A. Luigi Galvani
B. Stephen Gray
C. Alessandro Volta
D. Gottfried Wilhelm Leibniz
4.
Discovered electromagnetism.
A. J W Ritter
B. Luigi Galvani
C. William Herschel
D. Hans Christian Oersted
5.
Formulated Ohm’s Law in 1826.
A. T J Seebeck
B. George S. Ohm
C. Hans Christian Oersted
D. Jean-Baptiste-Joseph Fourier
6.
Discovered electromagnetic induction in 1831.
A. T J Seebeck
B. George S. Ohm
C. Michael Faraday
D. Jean-Baptiste-Joseph Fourier
7.
Invented the transformer in 1831.
A. T J Seebeck
B. George S. Ohm
C. Michael Faraday
D. Jean-Baptiste-Joseph Fourier
45
8.
Invented the electric motor in 1837.
A. Samuel Morse
B. Charles Babbage
C. Michael Faraday
D. Thomas Davenport
9.
Invented the differential resistance measurer in 1843.
A. John Herschel
B. William Grove
C. Charles Wheatstone
D. Edmond Becquerel
10.
Formulated KCL and KVL in 1845.
A. John Herschel
B. George S. Ohm
C. Gustav Kirchhoff
D. Charles Wheatstone
.
46
Assessment No. 3
INTRODUCTION TO ELECTRICITY
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
Answer the questions in your own words:
1.
What is electricity?
2. Write at least five scientists who contributed to the development of electricity? Include their
contributions to electricity.
3. What are the applications of electricity and electronics?
4. Express the following using metric prefixes:
a. 0.00000625 F
b. 23000000000 Ω
c. 0.0000000000234 A
d. 126000 V
5. Add the following without using a calculator. Use metric prefixes to simplify.
a. 0.00000625 F + 0.0000835 F
b. 23000000000 Ω + 45 GΩ
c. 0.0000000000234 A + 0.000000002446 A
d. 126000 V + 923800 V
6. Write the SI units of mass, force, weight, work, power and energy.
7. What is the difference between AC and DC?
47
Assessment No. 4
Name:
Score: _________ Rating: ______
Aviso, Myrell Jud A.
How Much Have You Learned?
Direction: Search for names of scientist who contributed to the development of
electricity; electrical components and measuring instruments found in this puzzle.
Encircle the name or word vertically, horizontally, backward, upward or downward.
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L
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T
C
T
A
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E
C
M
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T
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L
D
R
B
X
Q
W
W
D
Z
A
U
B
K
E
F
V
Y
V
O
Y
W
T
K
L
F
Y
L
O
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T
V
C
W
L
E
F
T
R
A
N
S
F
O
R
M
E
R
G
E
E
X
R
E
T
E
M
M
A
C
V
E
Z
R
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A
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A
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G
E
X
B
48
Unit 3
INTRODUCTION TO ALTERNATING CURRENT AND VOLTAGE
LEARNING OUTCOMES
At the end of this unit,
you are expected to:
1. identify sine waves and measure their characteristics.
2. explain how frequency and period are related.
3. measure the following voltage or current values of a
sine wave: instantaneous, peak, peak-to-peak, rms, and
average.
4. define a form factor.
5. explain how sine waves are generated.
6. measure points on a sine wave in terms of angular units.
7. determine the phase angle lead and phase lag.
8. express sine waves with a mathematical formula.
9. discuss phasors and how they can be used to represent
sine waves.
10. apply Ohm’s law and Kirchhoff’s laws to ac circuits as
well as to dc circuits.
49
Important Terms
alternating
current
power
transmission
electricity
sine wave
waveform
cycle
polarity
alternation
period
frequency
instantaneous value
peak value
peak-to-peak
value
rms value
average value
Faraday's law
electromagnetic induction
Phase
Phasor
Ohm's law
Kirchhoff's law
3.1 The Alternating Current
Alternating current circuits improves the
versatility and usefulness of electrical
power system. Alternating current plays a
vital role in today’s energy generation
􀂩
Once a big controversy ensued between the proponents of the of
the DC electricity led by Thomas Edison and the advocates of the
AC electricity led by George Westinghouse. According to Thomas
Edison,
“The AC electricity is dangerous, because it involves high voltage transmission
lines.”
The AC advocates on the other hand, countered that:
50
“The AC alternation is just like a handsaw which cuts on the upstroke and the
down stroke. The high voltage in the transmission line could be reduced to the
desired voltage as it passes the distribution line.”
3.2 The Power Transmission
(1)
Electricity leaves the power plant. (2) Its voltage is increased at a step-up
transformer. (3) The electricity travels along a transmission line to the area where
power is needed. (4) There, in the substation, voltage is decreased with the help of stepdown transformer. (5) Again the transmission lines carry the electricity. (6) Electricity
reaches the final consumption points.
51
More than 90 per cent of the electrical energy used for commercial purposes is generated
as alternating current. This is not due primarily to any superiority of alternating over
direct current so far as applicability to industrial and domestic uses is concerned. In fact,
there are many instances where direct current is absolutely necessary for industrial
purposes, such as municipal traction, electrolytic processes, and certain types of arc
lamps; also, direct current motors are superior for elevators, printing presses, and many
variable-speed drives. However, for these various purposes the energy is generated and
transmitted almost always as alternating current and then converted to direct current.
Some of the reasons for generating electrical energy as alternating current are
the following:
 Alternating current can be generated at comparatively high voltages, and these
voltages can be raised and lowered readily by means of static transformers.
 For constant-speed work, the alternating-current induction motor is more efficient
than the direct-current motor and is less in first cost and in maintenance, owing in
part to the fact that the induction motor has no commutator.
A circuit operating at increased voltage, has a lower power loss, power voltage drop, and
economically constructed for using smaller copper wires. On transmission and
distribution line, power loss is the most important problem to resolve. This is the main
reason why Alternating Current (AC) gained more favor and acceptance during the
middle part of the 19th century. In the USA, an ordinary house current is described as 120
volts 60 Hertz.
3.3 The Sine Wave
Waveforms are a graphical representation of how voltage or current varies with time. The
most common type of a waveform is the sine wave. Other types of waveforms are the
sawtooth wave, triangular wave, and the square wave.
The sine wave is one very common type of alternating (ac) and alternating voltage. It is
also referred to as a sinusoidal wave or, simply, sinusoid. The electrical service provided
by the power companies is in the form of sinusoidal voltage and current.
Sine waves are produced by two types of sources: rotating electrical machines (ac
generators) and electronic oscillator circuits, which are in instruments known as
electronic signal generators.
52
v
o
l
t
a
g
e
o
r
15
c
u
r
r
e
n
t
10
5
0
0
1
2
3
4
5
6
-5
7
time (sec)
-10
-15
The Sine Wave
Voltage (V) or
Current (I)
Peak (maximum) value
Positive-going
zero crossing
Negative-going zerocrossing
Parts of a Sine Wave
3.3.1 Cycle
A cycle is a complete set of positive and negative values.
3.3.2 The Polarity of a Sine Wave
Vs
Positive
+
alternation
+
Vs
_
Vs
_
0
(a) Positive voltage: current direction as shown.
t
53
VS
_
Negative
Vs
alternation
+
(b) Negative voltage: current reverses direction
3.3.3 The Period of a Sine Wave
The period (T) of a sine wave is the time required to complete one cycle.
3.3.4 The Frequency of a Sine Wave
Frequency is the number of cycles that a sine wave completes in 1 sec.
The Unit of Frequency
Frequency (f) is measured in units of hertz, abbreviated Hz. One hertz is equivalent
to one cycle per second; 60 Hz is 60 cycles per second; and so on.
3.3.5 Relationship of Frequency and Period
The relationship between frequency and period is very important. The formulas for this
relationship are as follows:
f=
1
T
f=
1
T
;
T=
1
f
Example 3.1. The period of a certain sine wave is 10 ms. What is the frequency?
Solution:
f =
1
𝑇
=
1
10 𝑚𝑠
= 100 Hz
Example 3.2. The frequency of a sine wave is 60 Hz. What is the period?
Solution:
T =
1
𝑓
=
1
160 𝐻𝑧
= 16.67 ms
54
Assessment No. 5
SINE WAVES
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1.
Describe one cycle of a sine wave.
2.
At what point does a sine wave change polarity?
3.
How many maximum points does a sine wave have during one cycle?
4.
How is the period of a sine wave measured?
5.
Define frequency, and state its unit.
.
6.
Determine f when T = 5 s.
7.
Determine T when f = 120 Hz.
8.
A sine wave goes through 5 cycles in 10 s. What is its period?
9.
A sine wave has a frequency of 50 kHz. How many cycles does it complete in 10 ms?
55
3.4 THE 60-Hz AC POWER LINE
Practically all homes in the Unites States are supplied alternating voltage between 115
and 125 V rms, at a frequency of 60 Hz. This is a sine-wave voltage produced by a rotary
generator. The electricity is distributed by high voltage power lines from the generating
station and reduced to the lower voltages used in the home. Here the incoming voltage is
wired to all the wall outlets and electrical equipment in parallel. The 120-V source of
commercial electricity is the 60-Hz power line or the mains, indicating it is the main line
for all the parallel branches.
Advantages. The incoming electric service to residences is normally given as 120 V rms.
With an rms value of 120 V, the ac power is equivalent to 120-V dc power in heating effect.
If the value were higher, there would be more danger of a fatal electric shock. Lower
voltages would be less efficient in supplying power.
Higher voltage can supply electric power with less I2R; sine the same power is
produced with less I. Note that the I2R power loss increases as the square of the current.
For applications where large amounts of power are used such as central air-conditioners
and clothes dryers, a line voltage of 240 V is often used.
The advantage of ac over dc power is greater efficiency in distribution from the
generating station. Alternating voltages can easily be stepped up by means of a
transformer, with very little loss, but a transformer cannot operate on direct current. The
reason is that a transformer needs the varying magnetic field produced by an ac voltage.
Using a transformer, the alternating voltage at the generating station can be stepped
up to values as high as 500 kV for high-voltage distribution lines. These high-voltage lines
supply large amounts of power with much less current and less I2R loss, compared with a
120-V line. At the home, the lower voltage required is supplied by a step-down
transformer. The step-up and step-down characteristics of a transformer refer to the
ration of voltages across the input and output connections.
The frequency of 60 Hz is convenient for commercial ac power. Much lower
frequencies would require much bigger transformers because larger windings would be
56
necessary. Also, too low a frequency for alternating current in a lamp would cause the
light to flicker. For the opposite case, too high a frequency results in excessive iron-core
heating in the transformers because of eddy currents and hysteresis losses. Based on
these factors, 60 Hz is the frequency of the ac power line in the United States. It should be
noted that the frequency of the ac power mains in England and most European counters
is 50 Hz.
3.5 Voltage and Current Values of a Sine Wave
3.5.1 Instantaneous Value
Figure 3.1 illustrates that any point in time on a sine wave, the voltage (or current)
has an instantaneous value. This instantaneous value is different at different point in
the curve. Instantaneous values are positive during the positive alternation and
negative during the negative alternation. Instantaneous values of voltage or current are
symbolized by lower case v and i, respectively.
10
8
v2
V
v
o
l
t
s
v1
6
4
2
0
0
1
t1
-2
2
3
t2
4
5
6
7
time (ms)
-4
-6
-8
-10
Figure 3.1. Example of instantaneous value of a sine voltage.
3.5.2 Peak Value
The peak value of a sine wave is the value of voltage (or current) at the positive or
the negative maximum (peaks) with respect to zero. Since the peaks are equal in
magnitude, a sine wave is characterized by a single peak value, as is illustrated in Figure
57
3.2. For a given sine wave, the peak value is constant and is represented by VP or Vm for
voltage and IP or Im for current.
15
v
o
l
t
a
g
e
o
r
c
u
r
r
e
n
t
10
Vp
5
Vp-p
0
0
1
2
3
4
5
6
7
time (sec)
-5
-10
-15
Figure 3.2. Example of a peak value of a sine wave.
3.5.3 Peak-to-Peak Value
The peak-to-peak value of a sine, as illustrated in Figure 3.2, is the voltage (or current)
form the positive peak to negative peak. Of course it is always twice the peak value as
expressed in the following equations:
VPP = 2VP
IPP = 2IP
3.5.4 rms Value
The term rms stands for root mean square. It refers to the mathematical procedure which
this value is derived. The rms value is also referred to as the effective value. Most ac
voltmeters display rms voltage. The 220 V at your wall outlet is an rms value.
The rms value of a sine wave is actually a measure of the heating effect of a sine
wave. For an example, when a resistor is connected across an ac (sine wave) source, as
shown in Figure 3.3(a), a certain amount of heat is generated by the resistor. Part (b)
shows the same resistor connected across a dc voltage source.
58
XWM1
V
I
R
30Ω
E
(a)
12 Vrms
60 Hz
0°
XWM1
V
I
R
30Ω
V
12 V
(b)
Figure 3.3. When the same amount of heat is being produced in both cases, the sine wave
has an rms value equal to the DC voltage.
The value of the AC voltage can be adjusted so that resistor gives off the same
amount of heat as it does when connected to the DC source.
The rms value of a sine wave is equal to the dc voltage that produces the
same amount of heat in a resistance as the sinusoidal voltage.
The peak value of a sine wave can be converted to the corresponding rms value using
the following relationships for either voltage or current:
𝑉𝑟𝑚𝑠 =
𝐼𝑟𝑚𝑠 =
𝑉𝑝
√2
𝐼𝑝
√2
Using these formulas, we can also determine the peak value knowing the rms
value, as follows:
VP = √2 Vrms
Ip =
√2 Irms
59
3.5.5 Average Value
The average value of a sine wave when taken over one complete cycle is always zero,
because the positive values (above the zero crossing) offset the negative values (below
the zero crossing).
To be useful for comparisons, the average value of a sine wave is defined over a
half-cycle rather than over a full cycle. The average value is the total area under the halfcycle divided by the distance of the curve along the horizontal axis. The result is expressed
in terms of the peak value as follows for both voltage and current sine waves:
2
Vavg = 𝑉𝑃

2
Iavg = 𝐼𝑃

.
60
Assessment No. 6
VALUES OF SINE WAVE VOLTAGE AND CURRENT
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1.
Derive the formula for the rms value of current.
2.
Derive the formula for the average value of a sine wave in half-cycle.
61
3.
A sine wave has a peak value of 12 V. Determine the following values:
(a) rms
(b) peak-to-peak
(c) half-cycle average
4.
A sinusoidal current has an rms value of 5 mA. Determine the following value:
(a) peak
(b) half-cycle average
(c) peak-to-peak
5.
A direct current of 12.5 A flows in a 25-ohm noninductive resistance. Determine
maximum value of an alternating current that will produce heat at the same rate in
this resistance.
6.
Number 6 AWG underground cable, which supplies a series incandescent lamp
system with alternating current is guaranteed to operate safely with 5,000 volts
(rms) alternating. If the system were changed to direct current, at what voltage
would it be safe to operate the system?
.
62
3.6 Sine Wave Voltage Sources
A. An AC Generator
An AC generator is a rotating electrical machine that uses the principle of
electromagnetic induction. It converts mechanical energy into electrical energy.
1
2
4
3
5
An elementary generator.
3.6.1 Parts of an Elementary Generator
1.
2.
3.
4.
5.
Magnetic Poles – provide the magnetic field
Loop of wire – cuts the magnetic field so that emf will be induced.
Slip rings – connected to the loop of wire.
Brushes – slide with the slip rings to collect the current from the loop of wire.
Terminals - connected to the load
3.6.2 Generation of Alternating EMFs.
63
The generation of emf started with the discovery of electromagnetism by Hans Christian
Oersted in 1820. He found that when current flows through a coil a magnetic field is
established around it as shown in Figure 3.4.
Figure 3.4. Magnetic lines of force around a current-carrying conductor.
After the discovery (by Oersted) that electric current produces a magnetic field,
scientists began to search for the converse phenomenon from about 1821 onwards. The
problem they put to themselves was how to ‘convert’ magnetism into electricity. It is
recorded that Michael Faraday was in the habit of walking about with magnets in his
pockets so as to constantly remind him of the problem. After nine years of continuous
research and experimentation, he succeeded in producing electricity by ‘converting
magnetism’. In 1831, he formulated basic laws underlying the phenomenon of
electromagnetic induction (known after his name), upon which is based the operation of
most of the commercial apparatus like motors, generators and transformers etc.
3.6.3 Faraday’s Laws of Electromagnetic Induction
Faraday summed up the above facts into two laws known as Faraday’s Laws of
Electromagnetic Induction.
First Law. It states that whenever the magnetic flux linked with a circuit changes, an
e.m.f. is always induced in it.
or
whenever a conductor cuts magnetic flux, an e.m.f. is induced in that conductor.
64
Second Law. It states that the magnitude of the induced e.m.f. is equal to the rate of
change of flux-linkages.
Explanation. Suppose a coil has N turns and flux through it changes from an initial value
of Φ1 webers to the final value of Φ2 webers in time t seconds. Then, remembering that
by flux-linkages mean the product of number of turns and the flux linked with the coil,
we have
Initial flux linkages = NΦ1, and final flux linkages = NΦ2
induced e.m.f. e =
𝑁𝛷2 − 𝑁𝛷1
Wb/s or volt e =
𝑡2 − 𝑡1
𝑁(𝛷2 − 𝛷1 )
𝑡2 − 𝑡1
volt
Putting the above expression in its differential form, we get
e=
𝑑(𝑁𝛷)
=
𝑑𝑡
𝑁𝑑𝛷
𝑑𝑡
volt
Usually, a minus sign is given to the right-hand side expression to signify the fact that the
induced e.m.f. sets up current in such a direction that magnetic effect produced by it
opposes the very cause producing it . (Lenz’s Law)
e= −
𝑁𝑑𝛷
𝑑𝑡
volt
3.6.4 Three Ways of the Generation of Voltage.
A voltage can be developed in a coil of wire in one of three ways; these are (1) by
changing the flux through the coil as shown in Figure 3.5, (2) by moving the coil through
a magnetic field so that flux cutting results as in Figure 3.6, and (3) by altering the
direction of the flux with respect to the coil as shown in Figure 3.7. In the first of these
the voltage is said to be an induced emf and, in accordance with Faraday’s law, its
magnitude at any instant of time is given by the equation
In CGS
e = N
𝑑
𝑑𝑡
𝑥 10−8
𝑣𝑜𝑙𝑡𝑠
where N = number of turns in the coil
 = flux in maxwells
t = time in seconds
𝑑
𝑑𝑡
= rate at which flux, in maxwells, changes through the coil
65
In SI
𝑒=𝑁
𝑑
𝑑𝑡
volts
where  = flux in weber
Figure 3.5. Induction of voltage due to the change of flux which results from the changing
current.
Figure 3.6. Electromagnetic Induction in a coil with coil moving.
66
Figure 3.7. Electromagnetic induction in a coil with magnetic field moving.
Note particularly that, by this method (the fist method) of developing an emf, there is no
physical motion of coil or magnet; the current though the exciting coil that is responsible
for the magnetism is altered to change the flux through the coil in which the voltage is
induced. By the second or third method there is actual physical motion of coil or magnet,
and in altered positions of coil or magnet flux through the coil changes. A voltage
developed in either of these ways is said to be a generated emf and is given by the equation
e = Blv x 10-8 volts
where: B = flux density, lines per square inch
l = length of the wire, in., that is moved relative to the flux
v = velocity of the wire, in. per second, with respect to the flux
In SI:
e = Blv volts
where:
B = flux density in Tesla or Wb/m2
l = length of the wire that is moved relative to the flux, in meters
v = velocity of the wire with respect to the flux, in meters per second
In CGS:
e = Blv x 10-8 volts
where: B = flux density in Gauss
l = length of the wire that is moved relative to the flux, in centimeters
67
v = velocity of the wire with respect to the flux, in centimeters per
second
Note that formulas above are applied to a length of conductor as shown in Figure 3.11.
If there N number of conductors they are just multiplied by N. Also take note that B, l,
and v are mutually perpendicular so that
e = N Blv volts
Example 3.3. A coil has 500 turns in which a current rises a current from 0 to 0.25 A in
0.1 s. As the current rises linearly, it produces a flux of 2.5 Wb. What is the induced
emf in the coil?
Solution:
𝑒 =
𝑁(𝛷2 − 𝛷1 )
𝑡2 − 𝑡1
=
500(2.5 𝑊𝑏 − 0)
0.1 − 0
= 0.0125 𝑉
Example 3.4. A piece of conductor 10 cm long moves across a magnetic field of 10,000
gauss at a velocity of 120 cm/sec. What is the voltage across the conductor?
Solution:
e = Blv x 10-8 = (10,000)(10)(120) x 10-8 = 0.12 V
3.7 Generation of a Sine Wave Voltage
There are actually many turns of wire in the coil of a generator but in Figure 3.9shows
only a single turn. Considering one side of the single-turn coil, as this cuts magnetic field
a voltage is induced across it. Figure 3.8 shows the direction of flux, induced emf, and
the movement of conductor. These directions are determined by following the righthand rule.
Direction of
induced emf
Motion of
the
conductor
Direction of
flux
Figure 3.8. Generation of emf in a conductor.
68
Now, the cross-section of the conductor is being considered as shown in Figure 3.9. As
the conductor rotates around the magnetic field flux cutting results and an emf is
induced across it. As this continually rotated a sine wave voltage is produced.
N
S
Figure 3.9. Sine wave voltage produced by the rotating conductor.
3.7.1 Reasons for using alternating current (or voltage) of sinusoidal form:
An alternating current (or voltage) sinusoidal form is normally used because of
the following reasons:
1.
2.
3.
4.
Mathematically, it is quite simple.
Its integrals and differentials both are sinusoidal.
It lends itself to vector representation.
A complex waveform can be analyzed into a series of sine waves of various
frequencies, and component can be dealt with separately.
5. This waveform is desirable for power generation, transmission and
utilization.
3.7.2 Equations of the Alternating Voltages and Currents
69
Nv


S
Figure 3.10. Conductor that rotates at an angle of .
In a single-coil generator the e.m.f. generated in one side of the coil which contains N
conductors, is given by,
e = N Bl v
Using SI units, B = flux density in Wb/m2 or Tesla, N = number of turns, l length of the
wire that is moved relative to the flux, in meters, and v = velocity of the wire with
respect to the flux, in meters per second
Total e.m.f. generated in both sides of the coil is
e = 2BNl v sin θ volt
From Figure 3.10, getting the v sin  -component because that is the one which is
perpendicular to the field, the equation becomes,
e = 2BNl v sin θ volt
Now, e has maximum value of Em (say) when θ = 90. Hence,
Em = 2BNl v
Therefore
e = Em sin θ
If b = width of the coil in meters ; f = frequency of rotation of coil in Hz, then v = π bf
70
Em = 2 B N l ( π b f) = 2 π f N B A volts
Example 3.5. A square coil of 10 cm side and 100 turns is
rotated at a uniform speed of 1000 revolutions per minute,
about an axis at right angles to a uniform magnetic field of
0.5 Wb/m2. Calculate the instantaneous value of the
induced electromotive force, when the plane of the coil is
(i) at right angles to the field (b) in the plane of the field.
Solution:
Let the magnetic field lie in the vertical plane and the coil
in the horizontal plane. Also, let the angle θ be measured from X-axis.
Maximum value of the induced e.m.f., Em = 2 π f N Bm A volt.
Instantaneous value of the induced e.m.f. e = Em sin θ
Now f = 100/60 = (50/3) rps, N = 100, Bm = 0.5 Wb/m2, A = 10−2 m2
(a) In this case, θ = 0
e = 0
(b) Here θ = 90,  e = Em sin 90 = Em
Substituting the given values, we get
e = 2π x (50/3) x 100 x 0.5 x 10−2 = 52.3 V
Answer
3.7.3 Factors That Affect Frequency in ac Generator
(a) rate of rotation of the loop
(b) magnetic poles
60 Hz
120 Hz
N
1 revolution = 2 cycles
S
S
f = number of pole pairs x revolution per second
N
71
Example 3.6 A four-pole generator has a rotation speed of 100 rev/sec. Determine the
frequency of the output voltage.
Solution:
f = number of pole pairs x revolution per second
4
𝑓 = 2 (100) = 200 cycles per second or 200 Hertz (Hz)
3.7.4 Factors That Affect Amplitude in ac Generator


Number of turns (N)
Rate of change with respect to the magnetic field
B. Signal Generator- a source of sine wave and other waveforms which uses
electronic oscillator circuits.
3.7.5 Derivation of the RMS Value of Voltage or Current
Mid-ordinate Method
In Fig. 3.11 are shown the positive half cycles for both symmetrical sinusoidal and nonsinusoidal alternating currents. Divide time base ‘t’ into n equal intervals of time each of
duration t/n seconds. Let the average values of instantaneous currents during these
intervals be respectively i1, i2, i3 .... in (i.e. mid-ordinates in Fig. 3-16). Suppose that this
alternating current is passed through a circuit of resistance R ohms. Then,
Current
i3
i4
Current
i3
i2
i4
i2
i1
in
i1
in
t/n
t
t
Figure 3.11.
Heat produced in 1st interval = 0.24 x 10-3 𝑖12 Rt/n kcal (where 1/J = 1/4200 = 0.24 x
10-3
72
Heat produced in 2nd interval = 0.24 x 10-3 𝑖22 Rt/n kcal
:
:
:
:
:
:
:
:
Heat produced in nth interval = 0.24 x 10-3 𝑖𝑛2 Rt/n kcal
2
𝑖 2 + 𝑖22 + …+𝑖𝑛
Total heat produced in t seconds is = 0.24 x 10-3 Rt ( 1
𝑛
)
Now, suppose that a direct current of value I produces the same heat through the same
resistance during the same time t. Heat produced by it is = 0.24 x 10−3 I2Rt kcal. By
definition, the two amounts of heat produced should be equal.
2
𝑖 2 + 𝑖22 + …+𝑖𝑛
 0.24 x 10-3 I2 Rt = 0.24 x 10-3 Rt ( 1
2
𝑖 2 + 𝑖22 + …+𝑖𝑛
 I2 = ( 1
𝑛
𝑛
)
2)
(𝑖12 + 𝑖22 + …+𝑖𝑛
) and I = √
𝑛
= square root of the mean of the squares of the
instantaneous currents
Similarly, the r.m.s. value of alternating voltage is given by the expression
V= √
2)
(𝑣12 + 𝑣22 + …+𝑣𝑛
𝑛
Analytical Method
The standard form of a sinusoidal alternating current is i = Im sin ωt = Im sin θ.
The mean of the squares of the instantaneous values of current over one complete cycle
is (even the value over half a cycle will do).
2 𝑖 2 𝑑
= ∫0
(2 −0)
2 𝑖 2 𝑑
The square root of this value is =√(∫0
2
)
put i = Im sin θ and integrate
Hence, the r.m.s. value of the alternating current is
I=
𝐼𝑚
√2
= 0.707 Im
Hence, we find that for a symmetrical sinusoidal current
73
r.m.s. value of current = 0.707 . max. value of current
The r.m.s. value of an alternating current is of considerable importance in practice,
because the ammeters and voltmeters record the r.m.s. value of alternating current and
voltage respectively. In electrical engineering work, unless indicated otherwise, the
values of the given current and voltage are always the r.m.s. values.
It should be noted that the average heating effect produced during one cycle is
= I2R = ( Im/√2)2 R = ½ Im2 R
Average Value
The average value Ia of an alternating current is expressed by that steady current which
transfers across any circuit the same charge as is transferred by that alternating current
during the same time
.
In the case of a symmetrical alternating current (i.e. one whose two half-cycles are
exactly similar,
whether sinusoidal or non-sinusoidal), the average value over a complete cycle is zero.
Hence, in their case, the average value is obtained by adding or integrating the
instantaneous values of current over one half-cycle only. But in the case of an
unsymmetrical alternating current (like half-wave rectified current) the average value
must always be taken over the whole cycle.
(i)
Mid-ordinate Method
With reference to Fig. 11.16, Iav =
𝑖1 + 𝑖2 + …+ 𝑖𝑛
𝑛
This method may be used both for sinusoidal and non-sinusoidal waves, although it is
specially convenient for the latter.
(ii)
Analytical Method
The standard equation of an alternating current is, i = Im sin θ
2 𝑖𝑑
Iav = ∫0
(2 −0)
putting the value of i and integrate
Iav =
Form Factor
2

Im = 0.637 Im
74
𝑟𝑚𝑠 𝑣𝑎𝑙𝑢𝑒
0.707𝐼
It is defined as the ratio, Kf = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 = 0.637 𝐼𝑚 = 1.1. (for sinusoidal alternating
𝑚
currents only)
𝑟𝑚𝑠 𝑣𝑎𝑙𝑢𝑒
0.707𝐸
In the case of sinusoidal alternating voltage also, Kf = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 = 0.637 𝐸𝑚 = 1.11
𝑚
As is clear, the knowledge of form factor will enable the r.m.s. value to be found from the
arithmetic mean value and vice-versa.
Crest or Peak or Amplitude Factor
It is defined as the ratio Ka =
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒
𝑟𝑚𝑠 𝑣𝑎𝑙𝑢𝑒
=𝐼
𝐼𝑚
𝑚 /√2
= √2 = 1.414 (for sinusoidal AC
only)
For sinusoidal alternating voltage also, Ka = 𝐸
𝐸𝑚
𝑚 /√2
= 1.414
Knowledge of this factor is of importance in dielectric insulation testing, because the
dielectric stress to which the insulation is subjected, is proportional to the maximum or
peak value of the applied voltage. The knowledge is also necessary when measuring iron
losses, because the iron loss depends on the value of maximum flux.
3.8 Angular Relationship of a Sine Wave
3.8.1 The Quadrants
90
(/2)
II
180
(/2)
I
III
IV
270
3
(2 )
0 or 360
(0 or 2)
75
0
0
90
ð
2
180
270

3

2
360
2
3.8.1 Radian-Degree Conversion
rad =
degrees =
𝜋 𝑟𝑎𝑑
180
180
𝜋 𝑟𝑎𝑑
x degrees
x radians
3.8.2 Sine Wave Angles
3.9 Phase of a Sine Wave
The phase of a sine wave is an angular measurement that specifies the position of the
sine wave relative to a reference.
(a) In Phase
Two or more waveforms are said to be in phase if they reach their zero and
maximum values at the same time.
76
30
I2m 20
I1m
current
10
0
0
1
2
3
4
5
6
7
i1
i2
-10
time (sec)
-20
-30
(b) Out-of-Phase – one wave either leads or lags the other.
o Lagging Phase Difference
o Leading Phase Difference
A
B
In these two waveforms, sine wave A leads sine wave B by 90 or sine wave B lags sine
wave A by 90
Note: To get the phase angle always consider the positive-going zero crossing nearest to
the reference or to other wave.
Example 3.7 What is the phase angle between the two sine waves in the figures below?
Which one is lagging? Which one is leading?
Solution:
B
The phase angle  between the
A
two sine waves is 45 ( = 45
0 45
- 0). Sine wave A leads B by
45 or sine wave B lags sine
wave A by 45.
77
B
Solution:
A
The phase angle  between the
0 30 90
two sine waves is 60( = 90 30). Sine wave A lags B by 60
or sine wave B leads sine wave
3.10 The Sine Wave Equation
A by 60.
A
y

Figure 3-17. Sine wave with phase angle of 0.
The equation of a sine wave is
y = A sin (  )
where: y = ordinate; the value of the sine wave at an angle .
A = amplitude ( the peak or maximum value)
 = the phase angle which is the angle between the reference (the origin) and
the given sine wave.
In the sine wave of Figure 3.17, the phase angle  is 0 since its positive-going zerocrossing is at 0 (or at the origin), so that the equation becomes,
y = A sin 
Example 3.8 Assuming that the sine wave below is a voltage, find its value at 30 and
60.
78
V
90 V
The equation is v = Vm sin 
Solution:
At  = 30,
v = 90 sin 30 = 45 V
At  = 60,
v = 90 sin 60 = 77.94 V
3.11 Expressions for Phase-Shifted Sine Waves

Phase-shifted to the right
y = A sin ( - )

Phase-shifted to the left
y = A sin ( + )
Example 3.9 Determine the instantaneous value at the 90 reference point on the
horizontal axis of each sine wave current.
79
15
E1m
E2m
c
u
r
r
e
n
t
A 10
B
C 5
c
-1
30°
0
0
1
0°
2
3
4
5
6
7
b
60°
a
-5
-10
-15
 (degrees)
Solution:
The equation of sine wave current a is, a = A sin ( - 60); where A = 10 A
so that a = 10 sin ( - 60) A;
When  = 90
a = 10 sin ( - 60) = 10 sin (90 - 60) = 5 A
The equation of sine wave current b is b = B sin ( + 30) A; where B = 8 A
so that b = 8 sin ( + 30)
When  = 90
b = 8 sin ( + 30) = 8 sin (90 + 30) = 6.93 A
The equation of sine wave current c is c = C sin ( + 0); where C= 5 A
so that b = 5 sin  A
When  = 90
c = 5 sin  = 5 sin 90 = 5A
80
3.12 Expression of Voltage/Current in terms of Time
In Figure 3.13,
 = 2
𝑟𝑎𝑑𝑖𝑎𝑛𝑠
𝑐𝑦𝑐𝑙𝑒
x f (cycle/second) x t ( time in sec)
 = 2ft radians
 = 360 ft degrees
where 2f =  (angular velocity in radians/second)
or
 = 360 f (degrees per second)
and so from e = Em sin, this equation becomes
e = Em sint
This equation can also be written as
e(t) = Em sint
meaning e is a function of t.
and for current
i = Im sint
Vm or Im
v or i

t
 (degrees or radians)
t (seconds)
Example 3.10 A 60-Hz alternating current has an rms value of 42.42 A making its
maximum value of 60 A. Draw current wave to scale. Find the time corresponding to an
angle of 45, 90, 135 , 225 , and 270 .
Solution
81
60 A
45
90 135
225
270 
Use the formula  = 360 ft
t=

360 𝑓
when  = 45, t =
when  = 90, t =
when  = 135, t =
when  = 225, t =
when  = 270, t =
=

360 (60)

360 (60)

360 (60)

360 (60)

360 (60)

360 (60)
=
=
=
=
=
45
360 (60)
90
360 (60)
135
360 (60)
225
360 (60)
270
360 (60)
= 2.08 ms
= 3.85 ms
= 6.25 ms
= 10.42 ms
= 12.5 ms
Example 3.11 A sine-wave alternating voltage has a maximum value of 170 volts and a
frequency of 25 Hz. Determine (a) value of voltage 0.001, 0.004, 0.01 sec after crossing
zero axis in a positive direction; (b) angles corresponding to each value of time; (c) rms
and average (for a half-cycle) values.
Solution
82
170 V
T = 40 ms
(a) The equation of the voltage is, v = 170 sin 225t V
(1) when t = 0.001 sec; v = 170 sin [225(0.001)] = 26.59 V
(2) when t = 0.004 sec; v = 170 sin [225(0.004)] = 99.92 V
(3) when t = 0.01 sec, v = 170 sin [225(0.01)] = 170 V
(b) The formula is  = 360ft degrees = 360(25)t
(1) when t = 0.001 sec;  = 360 x 25
(c)
𝑐𝑦𝑐𝑙𝑒𝑠
𝑠𝑒𝑐
(2) when t = 0.004 sec;  = 360x 25
𝑐𝑦𝑐𝑙𝑒𝑠
(3) when t = 0.01 sec;  = 360x 25
𝑐𝑦𝑐𝑙𝑒𝑠
Vrms =
170
√2
𝑠𝑒𝑐
𝑠𝑒𝑐
x 0.001 sec = 9
x 0.004sec = 36
x 0.01 sec = 90
= 120.21 V
2
Vavg =  (170) = 108. 23 V
Example 11.4.
An alternating current varying sinusoidally with a frequency of 50 Hz has an rms value
of 20 A. Write down the equation for the instantaneous value and find this value 0.0125
second after passing through a positive maximum value. At what time, measured from a
positive maximum value, will the instantaneous current be 14.14 A ?
(Elect. Science-I Allahabad Univ. 1992)
Solution:
Im = 20 2 = 28.2 A,
ω = 2π × 50 = 100 π rad/s.
The equation of the sinusoidal current wave with reference to point O (Fig. 11.17) as
zero time point is i = 28.2 sin 100 πt ampere
83
Assessment No. 7
EXPRESSION AND PHASE RELATIONSHIP OF SINE WAVES
Name:
Score: _________ Rating: ______
Aviso, Myrell Jud A.
Solve the following problems. Draw the corresponding sine waves.
1.
Sine wave A has a positive-going zero crossing at 30. Sine wave B has a positivegoing zero crossing at 45. Determine the phase angle between the two signals.
Which signal leads?
150
100
50
0
-90
0
90
180
270
360
-50
-100
-150
2.
One sine wave has a positive peak at 75, and another has a positive peak at 100.
How much is each sine wave shifted in phase from the 0 reference? What is the
phase angle between them?
150
100
50
0
-90
-50
-100
-150
0
90
180
270
360
84
3.
Make a sketch of two sine waves as follows: Sine wave A is the reference, and sine
wave B lags A by 90. Both have equal magnitude.
150
100
50
0
-90
0
90
180
270
360
-50
-100
-150
4.
Convert the following angular values from degrees to radians:
(a) 30
(b) 135
5.
Convert the Following angular value from radians to degrees:
(a) 3/5
(b) 6/5
6.
A certain sine wave has a positive-going zero crossing at 0 and an rms value of 20
A. Calculate its instantaneous value at each of the following angles:
(a) 15
(b) 300
85
7.
Sine wave A lags sine wave B by 30. Both have peak values of 15 V. Sine wave A is
the reference with a positive-going zero crossing at 0. Determine the instantaneous
value of sine wave B at 45, and 300.
8.
A 50-Hz alternating current has a maximum instantaneous value of 42.42 A. It
crosses the zero axis in a positive direction when time is zero. Determine (a) time
when current first reaches a value of 30 A; (b) time when current, after having gone
through its maximum positive value, reaches a value of 36.7 A; (c) value of current
when the time is 1/120 sec; (d) value of time when current first reaches a negative
value of 21.21 A.
86
9.
A current is given by i = 22.62 sin 377t. Determine (a) maximum value; (b) rms
value; (c) frequency; (d) radians through which its vector has gone when t = 0.01
sec; (e) number of degrees in (d); (f) value of current at instant in (d).
10. A 25-Hz emf has an rms value of 250 volts, is zero and increasing positively when t
= 0. Determine (a) maximum value; (b) equation; (c) radians at t = 1/75 sec; (d)
degrees in (c); (e) emf at time t in (c).
.
87
3.13 Addition of Sine Waves
(a) In Phase Sine Waves
25
I2m
20
I1m
15
current
10
5
0
i1
0
1
2
3
4
5
6
7
i2
-5
time (sec)
-10
-15
-20
-25
The equation of i1 = Im1 sin t and that of i2 = Im2 sin t
The sum (total, resultant, or equivalent) of the two waveforms is
iT = (Im1 + Im2) sin t
Example 3.12 If the maximum values of i1 and i2 are 15 A and 20 A respectively, and the
frequency is 60 Hz, find the equation of the resultant of the two waves.
Solution
The equation of i1 = 15 sin 260t A and that of i2 = 20 sin 260t A
The resultant equation is
𝑖 𝑇 = i1 + i2
= 15 sin 260t A + 20 sin 260t A
𝑖 𝑇 = 35 sin 260t A
Answer
88
The figure below shows the waveform of the resultant current
40
30
20
current
10
i1
0
-10
0
1
2
3
4
5
6
7
i2
iT
time (sec)
-20
-30
-40
Example 3.13 Find the equation of the resultant of the two 60-Hz sine waves below.
E1m
E2m
100
80
60
voltage
40
20
0
-20
0
1
2
3
4
5
6
7
8
30°
-40
e1
e2
 (degrees)
-60
-80
-100
The equation of 𝑒1 = 90 sin (260t – 30) V and that of 𝑒2 = 70 sin (260t 30) V
or converting 30 to radians we have, /6 radians
So that 𝑒1 = 90 sin (260t – /6) V and 𝑒2 = 70 sin (260t - /6) V
The resultant equation is
89
𝑒𝑇 = 160 sin (260t - /6) V
Answer
The figure below shows the resultant voltage.
200
ETm
150
100
voltage
50
e1
0
0
1
2
3
4
5
6
7
30°
-50
8
 (degrees)
e2
eT
-100
-150
-200
(b) Waves Differing in Phase by 90 degrees
20
C
B
15
10
A
5
-2
-1
90°

0
a
0
1
0°
2
90°
3
180°
4
5
270°
6
7
360°
-5
-10
-15
-20
Let
A = be the maximum value of sine wave a
B = the maximum value of sine wave b
C = the maximum value of the sum or resultant of the two sine waves
 = the phase angle of the resultant.
b
c
90
The equation of sine wave a = A sin t and b = B sin (t + 90) = B cos t and
c = C sin(t + )
Now, find C and  in terms of A and B
a+b=c
A sin t + B cos t = C sin(t + )
Expanding sin(t + ) = sint cos  + cos t sin 
A sin t + B cos t = C (sint cos  + cos t sin )
A sin t + B cos t = C sint cos  + Ccos t sin 
Equating coefficients of sin t,
A = C cos 
(1)
Equating coefficients of cos t,
B = C sin 
(2)
To find C, square equations 1 and 2 then add,
A2 + B2 = C2 (cos2  + sin2 )
A2 + B2 = C2
C = √𝐴2 + 𝐵 2
To find , divide equation 2 by equation 1
𝐵
𝐴
=
𝐶 𝑠𝑖𝑛 
𝐶 𝑐𝑜𝑠 
= tan  ;
 = tan-1
𝐵
𝐴
Hence,
𝐵
A sin t + B cos t = c = √𝐴2 + 𝐵 2 sin (t + tan-1 )
𝐴
Example 3.14 A 60-Hz current i1 = 9 sint A is added to 60-Hz current i2 = 15cost,
where  = 260. Determine (a) the equation of the resultant current; (b) time at
which the two currents are equal.
Solution
91
20
ITm
I2m
15
I1m
10
current
5
i1
-2
-1

90°
0
0
1
2
0°
3
90°
4
5
180°
6
7
270°
i2
360°
-5
iT
 (degrees)
-10
-15
-20
𝑖 𝑇 = √𝐼𝑚1 2 + 𝐼𝑚2 2 sin (t + tan-1
𝐼𝑚2
𝐼𝑚1
)
Im1 = 9 A , Im2 = 15 A ,  = 260 rad/sec.
where
𝑖 𝑇 = √92 + 152 sin (260 t + tan-1
so that ,
15
9
)
𝑖 𝑇 = 17.49 sin (260t + 59.04) or 𝑖 𝑇 = 12.04 sin (260t + 1.03)
(c) Waves Differing in Phase Other than 90 degrees
.
150
100
voltage
50
e1
-2
-1
60°
0
0
0° 45°
-50
-100
-150
1
2
3
4
5
6
7
8
 (degrees)
e2
eT
92
Example 3.15 Two 25-Hz emfs differing in phase by 105 are given by
𝑒1 = 120 sin (t + 60) V and 𝑒1 = 60 sin (t - 45) V
Solution
𝑒𝑇
=
e1 + e2
= 120 sin (t + 60) + 60 sin (t - 45)
= 120 (sin tcos60 + cost sin 60) + 60 (sin t cos 45 - cos t sin 45)
= 60sint + 103.92 cost + 42.43sin t - 42.43cos t
= 102.43 sint + 61.49 cost
Following the formula
𝐵
c = A sin t + B cos t = √𝐴2 + 𝐵 2 sin (t + tan-1 𝐴)
where A = 102.43, B = 61.49
and let c = be the eT
Therefore,
61.49
eT = √(102.43)2 + (61.49)2 sin (t + tan-1 102.43)
eT
=
119.47 sin (t + 30.98) V
In this equation, ITm = 119.47 V and  = 30.98
.
.
.
.
.
.
.
.
.
93
Assessment No. 8
ADDITION OF SINE WAVES
Name:
Score: _________ Rating: ______
Aviso, Myrell Jud A.
1. An emf e1 = 100 sin 260t is in series with an emf e2 = 120 sin 260t. Sketch the
waveform of e1, e2 and the resultant eT.
150
100
50
0
-90
0
90
180
270
360
-50
-100
-150
2. Two currents i1 = 12 sin 260t and i2 = 9 cos 260t flow in a wire. Sketch the
waveforms of i1, i2 and the total current iT.
15
10
5
0
-90
-5
-10
-15
0
90
180
270
360
94
3. Two 50-Hz currents i1 = 2.5 sin (ωt - 15) and i2 = 3.5 sin (ωt - 75) flow in a
common wire. Sketch the waveforms of i1, i2 and the total current iT.
4
3
2
1
0
0
-1
-2
-3
-4
.
90
180
270
360
95
Problem Set No. 3
ADDITION OF SINE WAVES
1. An emf e1 = 100 sin 260t is in series with an emf e2 = 120 sin 260t. Determine
(a) their resultant e3; (b) angle between e1 and e3.
A. e3 = 220 sin 260t, 0°
B. e3 = 156.2 sin 260t, 0°
C. e3 = 20 sin 260t, 0°
D. e3 = 220 sin 260t, 0°
2. Two currents i1 = 12 sin 260t and i2 = 9 cos 260t flow in a wire. (a) Determine
the equation of resultant current i3 ; (b) Determine rms value of i3.
A. i3 = 21 sin (260t + 0.6435) , 14.85 A
B. i3 = 15 sin (260t + 0.6435) , 10.61 A
C. i3 = 21 sin (260t + 0.6818) , 14.85 A
D. i3 = 15 sin (260t + 0.6818) , 10.61 A
3. Two 50-Hz currents i1 = 2.5 sin (ωt - 15) and i2 = 3.5 sin (ωt - 75) flow in a
common wire. Determine (a) their resultant i3; (b) angle between i1 and i3 .
A. i3 = 6 sin (ωt - 90°) , 90°
B. i3 = 5.22 sin (ωt – 78.69°) , 63.69°
C. i3 = 6 sin (ωt – 50.5°) , 35.5°
D. i3 = 5.22 sin (ωt – 50.5°) , 35.5°
4. Two 50-Hz emf having rms values of 22.6 and 33.9 volts differ in phase by an angle of
60, the latter current lagging and its positive-going zero crossing is at 0.
Determine the resultant emf (in wave equation) if they are connected in series.
A. eT = 49.26 sin (250t + 0.4086)
B. eT = 69.66 sin (250t + 0.4086)
C. eT = 49.26 sin (260t + 0.4086)
D. eT = 69.66 sin (260t + 0.4086)
.
96
3.14 Phasors
Phasors can be used to represent time-varying quantities, such as sine waves, in terms of
their magnitude and angular position (phase angle)
Sine Waves
Phasor Diagram
i
60 A
𝐈⃗
i = 60 sin (t + 0)
A
𝐈⃗ = 𝟒𝟐. 𝟒𝟑 𝐀𝟎
120 V
e
60
60
=
e =60
120 sin (t - 60)
⃑𝑬
⃑⃗ = 84.85 V -60
120 V
e
=
⃑E⃗
60
60
e = 120 sin (t + 60)
⃑⃗ = 84.85 V 60
𝑬
97
Note: For the magnitude of the phasors always use the rms value of the sine wave
voltage and current.
3.15 Addition of Phasors
Example 3.16 Find the total current if the maximum values of i1 and i2 are 20 and 15
amp, respectively.
40
30
20
current
10
i1
0
-10
0
1
2
3
4
5
6
time (sec)
-20
-30
-40
Solution
For the maximum values of the currents:
I1 = 14.14 A
I2 = 10.61 A
I1 = 14.14 A
24.75 A
⃑⃑⃑⃗
𝐼𝑇 = ⃑⃑⃗
𝐼1 + ⃑⃑𝐼⃑2⃗
= 14.140 + 10.610
= 24.75 A0
I2 = 10.61 A
7
i2
iT
98
To convert this to a sine wave equation, the maximum value of the total current
ITm =2( 24.75) = 35 A. The equation of the total current is
iT = 35 sin(t + 0)
Example 3.17
200
ETm
150
E1m100
E2m
voltage
50
e1
0
0
1
2
3
4
5
6
30°
-50
-150
-200
E1 =
E2 =
90 𝑉
√2
70 𝑉
√2
8
 (degrees)
-100
Solution
7
= 63.64 A
= 49.5 A
30
E1 = 63.64 V
E2 = 49.5 V
ET = 113.14 V
⃑⃑⃑⃑⃗
𝐸𝑇 = ⃑⃑⃑⃑⃗
𝐸1 + ⃑⃑⃑⃑⃗
𝐸2
= 63.64 V-30 + 49.5 V-30
= 113.14 V-30
e2
eT
99
To convert this to a sine wave equation, the maximum value of the total current
ETm = √2 ( 113.14) = 160 A. The equation of the total current is
𝑒𝑇 = 160 sin(t - 30)
Example 3.18 A 60-Hz current i1 = 9 sint A is added to 60-Hz current i2 = 15cost,
where  = 260. Determine (a) the equation of the resultant current; (b) time at
which the two currents are equal.
Solution
20
ITm
I2m
15
10
I1m
current
5
i1
-2
-1
90°

0
0
1
0°
2
3
90°
180°
4
5
270°
6
7
360°
-5
i2
iT
 (degrees)
-10
-15
-20
The resultant equation of the two waves is,
𝑖 𝑇 = 17.49 sin (260t + 59.04) A
Converting these sine waves as phasors,
I2 =
IT = 12.37 A
10.61 A
 = 59.04
I1 = 6.36 A
100
Adding the phasors,
⃑⃑⃑⃑⃗
𝑰𝑻 = ⃑⃑⃑⃑⃗
𝑰𝟏 + ⃑⃑⃑⃑⃗
𝑰𝟐
= 6.36 A0 + 10.61 A 90
= (6.36 + j0) + (j10.61)
= 6.36 + j 10.61
= 12.37 A 59.04
.
101
Assessment No. 9
PHASOR ANALYSIS
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. Two 50-Hz currents i1 = 2.5 sin (ωt - 15) and i2 = 3.5 sin (ωt - 75) flow in a
common wire. Draw the phasor diagram of ⃑⃑⃗
𝐼1 , ⃑⃑⃑⃗
𝐼2 , and the total current ⃑⃑⃑⃗
𝐼3
2. Two 50-Hz emf having rms values of 22.6 and 33.9 volts differ in phase by an angle of
60, the latter current lagging and its positive-going zero crossing is at 0. Draw the
corresponding phasor diagram.
.
102
Problem Set No. 4
PHASOR ANALYSIS
1. Find the rms value of the resultant.
E1 = 25 V
45
A.
B.
C.
D.
ET
ET
ET
ET
=
=
=
=
85.38 V
60.37 V
30.19 V
42.69 V
E2 = 15 V
60
E3 = 20 V
2. An emf e1 = 100 sin 260t is in series with an emf e2 = 120 sin 260t. Determine
(a) their resultant ⃑⃑⃑⃑⃗
𝐸3 in polar form (b) angle between ⃑⃑⃑⃑⃗
𝐸1 and ⃑⃑⃑⃑⃗
𝐸3 .
A. ⃑⃑⃑⃑⃗
𝐸3 = 220 V0°, 0°
B. ⃑⃑⃑⃑⃗
𝐸3 = 155.56 V0°, 0°
C. ⃑⃑⃑⃑⃗
𝐸3 = 155.56 V50.19°, 50.19°
D. ⃑⃑⃑⃑⃗
𝐸3 = 220 V 50.19°, 50.19°
3. Two currents i1 = 12 sin 260t and i2 = 9 cos 260t flow in a wire. (a) Determine
the resultant current ⃑⃑𝐼⃑3⃗ in polar form ; (b) Determine rms value of ⃑⃑𝐼⃑3⃗.
A. ⃑⃑𝐼⃑3⃗ = 14.85 A36.87° , 10.5 A
B. ⃑⃑𝐼⃑3⃗ = 21 A36.87° , 14.85 A
C. ⃑⃑𝐼⃑3⃗ = 10.61 A36.87° , 10.61 A
D. ⃑⃑𝐼⃑3⃗ = 10.61 A39.06° , 10.61 A
4. Two 50-Hz currents i1 = 2.5 sin (ωt - 15) and i2 = 3.5 sin (ωt - 75) flow in a
common wire. Determine (a) their resultant ⃑⃑𝐼⃑3⃗ (b) angle between ⃑⃑⃗
𝐼1 and ⃑⃑⃑⃗
𝐼3 .
⃑⃑
⃑
⃗
A. 𝐼3 = 4.24 A-90°, 90°
B. ⃑⃑𝐼⃑3⃗ = 5.22 A-78.69°, 63.69°
C. ⃑⃑𝐼⃑3⃗ = 4.24 A-50.5°, 35.5°
D. ⃑⃑𝐼⃑3⃗ = 3.69 A-50.5°, 35.5°
5. Two 50-Hz emf having rms values of 22.6 and 33.9 volts differ in phase by an angle of
60, the latter current lagging and its positive-going zero crossing is at 0.
Determine the resultant emf if they are connected in series.
A. ⃑⃑⃑⃑⃗
𝐸𝑇 = 49.26 V23.41°
B. ⃑⃑⃑⃑⃗
𝐸𝑇 = 49.26 V-23.41°
C. ⃑⃑⃑⃑⃑⃑⃑⃗
𝐸𝑇 = 69.66 V-23.41°
D. ⃑⃑⃑⃑⃗
𝐸𝑇 = 69.66 V-23.41°
103
3.16 Ohm’s Law and Kirchhoff’s Laws in AC Circuits
When time-varying ac voltages such as the sine wave are applied to resistive
circuits, the circuit laws that you studies earlier still apply. Ohm’s law applies to
resistive ac circuits in the same way that it applies to dc circuits. If a sine wave voltage is
applied across a resistor as shown in the figure below, a sine wave current flows. The
current is zero when the voltage is zero and is maximum when the voltage is maximum.
When the voltage changes polarity, the current reverses direction. As a result, the
voltage and current are said to be in phase with each other.
When using Ohm’s law in ac circuits, remember that both the voltage and the
current must be expressed consistently, that is, both as peak values, both as rms values,
both as average values, and so on.
v
I
i
VAC
R
Sine wave voltage produces a sine wave current.
In DC circuit,
110 V dc
𝑉 = 𝑅𝐼
R1 = 1 kohm
I=
V
R
𝑅=
𝑉
𝐼
104
In AC circuit,
R1 = 1 kohm
V ac
Vrms = RIrms
Irms =
or 𝑉 = 𝑅𝐼
𝑉𝑟𝑚𝑠
R
or I =
V
R
Note: A quantity with no rms subscript is understood as rms quantity. For
instance, in 110 V (rms) the rms can be deleted and it is still understood as an rms
quantity; in I(rms) = 110 V or I = 110 V (rms) can be written as I = 110 V.
The Power formulas P = I2R =
𝑉2
𝑅
= IV are applicable to both DC and AC circuits,
where I and V = are either dc values and rms values
P = power in watts
R = resistance
Example 3.19 Determine the rms voltage across each resistor and the rms current. The
source voltage is given as rms value.
R1 = 1 kohm
Vs = 110 V
R2 = 560 ohms
Solution
The total resistance of the circuit is
RT = R1 + R2 = 1 k + 560  = 1.56 k
Use Ohm’s law to find the rms current,
Irms =
𝑉𝑠(𝑟𝑚𝑠)
𝑅𝑇
=
110 𝑉
1.56 𝑘Ω
= 70.5 mA
The rms voltage drop across each resistor is
V1(rms) = Irms R1 = (70.5 mA)(1 k) = 70.5 V
V2(rms) = Irms R2 = (70.5 mA)(560 ) = 39.5 V
105
Problem Set No. 5
OHM’S LAWS AND KIRCHHOFF’S LAWS IN AC CIRCUITS
1. Three resistors are connected in series across a 120-V (rms) source. The voltage
drop across R1 is 60 V (rms) and across R2 is 40 V (rms). Find the peak voltage
across R3.
A. 14.14 V
B. 20 A
C. 28.28 V
D. 40 V
2. Three resistors that are connected in parallel have currents of 6 A, 5 A, and 8 A. Find
the total rms current. All values are given in rms.
A. 19 A
B. 20 A
C. 21 A
D. 22 A
3. A 117- V 60-Hz source is connected to a series circuit consisting of three resistors. If
the ohmic values of the latter are 20, 30, and 40 ohms, respectively, calculate the
current through the circuit and the voltage drop across each resistor.
A. 1.0 A. 21 V, 32 V, 49 V
B. 1.0 A. 22 V, 36 V, 52 V
C. 1.2 A. 25 V, 39 V, 50 V
D. 1.3 A, 26 V, 39 V, 52 V
4. Three incandescent lamps (resistors) are connected in parallel, and to a 115-volt 60Hz source. If the lamp ratings are 75, 100, and 150 watts, (a) calculate the rms value
of the resultant current, (b) write the equation for the resultant current.
A. 57.54 A, iT = 81.37 sin 377t
B. 57.54 A, iT = 81.37 sin 314.2t
C. 40.69 A, iT = 57.54 sin 314.2t
D. 40.69 A, iT = 57.54 sin 377t
5. A resistance load of 4 ohms is connected to a 220-V 60-Hz line which leads back to
the source through a pair of wires, the resistance of each of which is 0.08 ohm. What
is the voltage at the source?
A. 228.8 V
B. 224.4 V
C. 222.8 V
D. 220 V
.
106
Objective Test No. 3
INTRODUCTION TO ALTERNATING CURRENT AND VOLTAGE
1.
2.
3.
4.
5.
6.
7.
8.
The difference between alternating current (AC) and direct current (DC) is
A. AC changes value and DC does not.
B. DC changes direction and AC does not.
C. DC changes value and AC does not.
When using circuit laws and rules we must use
A. maximum value.
B. effective value
C. average value
D. peak-to-peak value
If the peak value of a sine wave is 20 V, the rms value is
A. 14.14 V
B. 6.37 V
C. 7.07 V
D. 0.707 V
The average value of a 10-V peak sine wave over one complete cycle is
A. 0 V
B. 6.37 V
C. 7.07 V
D. 5 V
The instantaneous value of a 15-A peak sine wave at a point 32º from its positivegoing zero crossing is
A. 7.95 A
B. 7.5 A
C. 2.13 A
D. 7.95 V
The value of ac that would have the same effect in power produced as a similar
value of DC is known as
A. peak value.
B. rms value
C. average value
D. peak-to-peak value
If e1 = A sin t and e2 = B sin (t - ) then
A. e1 lags e2 by 
B. e2 lags e1 by 
C. e2 leads e1 by 
The equation for 25 cycles current sine wave having rms value 30 amperes will be
A. 30 sin 25t
107
B. 30 sin 50t
C. 42.4 sin 25t
D. 42.4 sin 50t
9. A phasor is a
A. line representing the magnitude and direction of an alternating current.
B. line which represents the magnitude and phase of an alternating current.
C. color band for distinguishing between different phases of a 3-phase supply.
D. Instrument used for measuring phases of an unbalanced 3-phase load.
10. A 60-Hz frequency would cause an electric light to
A. turn on and off 120 times per second.
B. flicker noticeable.
C. turn and off 180 times per second.
D. turn and off 60 times per second.
11. The difference between the positive peak value and the negative peak value of a
sine wave is called the
A. maximum value
B. average value
C. effective value
D. peak-to-peak value
12. The relationship between frequency f, number of revolutions per second n and pair
of poles p is given by
A. f = n/p
B. f = n/2p
C. f = np
D. f = 2np
13. The root-mean-square (rms) value of ac is the same as
A. instantaneous value
B. effective value
C. average value
D. maximum value
14. The rms value of sine wave is equal to
A. 0.637 max. value
B. 0.707 max. value
C. 0.506 max. value
D. 1.414 max. value
15. Form factor is defined as
A. rms value/peak value
B. max. value/rms value
C. rms value/average value
D. effective value/rms value
16. The value of form factor for a pure sine wave is
108
A. 1.414
B. 0.707
C. 0.637
D. 1.11
17. The value of peak factor for a pure sine wave
A. 1.414
B. 0.707
C. 0.637
D. 1.11
18. Which of the following statement concerning the sinusoidal waveform, is most
correct?
A. it represents AC
B. it represents DC
C. it represents half-wave rectified AC
D. it represents sum of AC and DC
19. The average value of a sine wave is 2 times the maximum value.
A. true
B. false
20. The form factor of dc supply voltage is always
A. infinite
B. zero
C. 0.5
D. unity
21. The rms value of a sinusoidal AC current is equal to its value at angle of ______
degrees.
A. 90
B. 60
C. 45
D. 30
.
109
Assessment No. 10
Name:
Score: _________ Rating: ______
Aviso, Myrell Jud A.
How Much Have You Learned?
Directions: Solve the crossword puzzle. Use the given clues to arrive at the
right answer.
1
2
3
2
4
3
5
4
6
6
7
7
9
8
8
DOWN
1 number of cycle per second
2 0.637 x peak value
3 a set of positive and negative values
4 Gustav _____________
5 positive and negative
6 common waveform
7 relative time difference
8 alternating current
ACROSS
1 electromagnetic induction
2 maximum value
3 effective value
4 V = IR
5 a line representing an AC voltage or current
6 symbol T
7 flow of electrons
8 saw tooth, sine wave, triangular wave
9 value of voltage or current at any instant
110
QUESTIONS:
1.
2.
3.
4.
5.
6.
7.
8.
What is a sine wave ? Describe its characteristics.
Explain how frequency and period are related.
Explain the following: instantaneous, peak, peak-to-peak, rms, and average.
What is a form factor?
Discuss how sine waves are generated.
What are the phase angle leading and phase lagging?
What is the expression of a sine wave?
What is a phasor? How can they be used to represent sine waves?
.
111
PRACTICAL APPLICATION No. 1
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. You are an electrician working on an overhead crane. The crane uses a large
electromagnet to pick up large metal pipes. The magnet must have a minimum of
200 V DC to operate properly. The crane has an AC source of 240 V. You are given
four diodes that have a peak voltage rating of 400 V each. These diodes are to be
used to form a bridge rectifier to convert the AC voltage into DC voltage. Is the
voltage rating of the diodes sufficient? To the nearest volt, what will be the DC
output voltage of the bridge rectifier?
2. You are a journeyman electrician working in a large office building. The fluorescent
lighting system is operated at 277 V. You have been instructed to replace the
existing light ballasts with a new electronic type that is more efficient. The
manufacturer of the ballast states that the maximum peak operating voltage for the
ballast is 350 V. Will the new electronic ballast operate on the building’s lighting
system without harm?
112
3. You are working as an electrician installing fluorescent lights. You notice that the
lights were made in Europe and that the ballasts are rated for operation on a 50-Hz
system. Will these ballasts be harmed by overcurrent if they are connected to 60 Hz?
If there is problem with these lights, what will be the most likely cause of the
trouble?
.
113
Unit 4
INDUCTOR AND INDUCTANCE
LEARNING OUTCOMES
At the end of this unit,
you are expected to:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
describe the basic construction of an inductor.
define inductance.
determine the voltage induced in an inductor.
explain how an inductor stores energy.
determine the energy stored in an inductor.
relate various physical parameters to inductance value.
solve inductance using the various physical parameters.
explain why practical inductors have both resistance and
capacitance.
state Lenz’s law and Faraday’s law.
identify various types of inductors.
determine total series inductance.
determine total parallel inductance.
solve combination circuit consisting of inductors.
define time constant as related to an inductive circuit.
analyze electrical transients in an inductor in DC circuit.
describe some common inductor applications.
check out an inductor with an ohmmeter.
114
Important Terms
inductor
electromagnetic field
self-inductance
henry
induced voltage
energy storage
core
winding resistance
winding capacitance
Faraday’s law
Lenz’s law
permeability
time constant
4.1 The Inductor
When a length of wire is formed into a coil, as shown in Figure 4.1, it becomes a basic
inductor. Current through coil produces an electromagnetic field. The magnetic lines of
force form a strong magnetic field within and around a coil.. The net direction of the total
magnetic field creates a north and a south pole, as indicated.
Figure 4.1. A coil of wire forms an inductor. When current flows through it, a three-dimensional
electromagnetic field is created, surrounding the coil in all directions.
4.2 Self-Inductance
When there is current through an inductor, an electromagnetic field is established. When
the current changes, the electromagnetic field also changes. An increase in current
expands the field, and a decrease in current reduces it. Therefore, a changing current
produces a changing electromagnetic field around the inductor (coil). In turn, the
changing electromagnetic field produces a voltage across the coil in a direction to oppose
115
the change in current. This property is called self-inductance, but it is usually referred to
as simply inductance. Inductance is symbolized by L.
Inductance is a measure of a coil’s ability to establish an induced voltage as a result
of a change in its current and that induced voltage is in direction to oppose that change in
current.
4.2.1 The Unit of Inductance
The henry, symbolized by H, is the basic unit of inductance. By definition, the inductance
is one henry when current through the coil, changing at the rate of one ampere per second,
induces one volt across the coil. In many practical application, millihenries (mH) and
microhenries (H) are the most common units. A common schematic symbol for the
inductor is shown in Figure 4.2.
L
Figure 4.2. Symbol of inductor
4.3 The Induced Voltage in an Inductor
A changing current in an inductor causes a changing magnetic field though it. Since
according to Faraday’s law a changing magnetic field results to the induction of voltage
across the inductor.
The formula for the induced emf (or voltage) across the coil or inductor is,
𝑒𝐿 = 𝐿
where
𝑑𝑖
𝑑𝑡
eL = the induced emf across a coil or inductor in volts (V)
L = the inductance in henry (H)
di/dt = rate of change of current in amp/sec.
Example 4.1 A group of electromagnets that create a flux in a dc generator – the field
circuit – has an inductance of 15 henrys. If the 2.6 amp excitation is interrupted in 0.04
sec by the opening of the field switch, what average voltage is induced in the winding?
Solution
eL = 15 x
2.6
0.04
= 975 volts
116
4.4 Energy Storage
An inductor stores energy in the magnetic field created by the current. The energy stored
is expressed as follows
W = ½ LI2
As you can see, the energy stored is proportional to the inductance and the square of the
current. When I is in amperes and L is in henries, the energy is in joules.
4.5 Physical Characteristics
The following characteristics are important in establishing the inductance of a coil, the
core material, the number of turns of wire, the length, and the cross-sectional area.
4.5.1 Core Material
As discussed earlier, an inductor is basically a coil of wire. The material around which the
coil is formed is called the core. Coils are wound on either nonmagnetic or magnetic
materials. Examples of nonmagnetic materials are air, wood, copper, plastic, and glass.
The permeabilities of these material are the same as for a vacuum. Examples of magnetic
materials are iron, nickel, steel, cobalt, or alloys. These materials have permeabilities that
are hundreds or thousands of times greater than that of a vacuum and are classified as
ferromagnetic. A ferromagnetic core provides a better path for the magnetic lines of force
and thus permits a stronger magnetic field.
The permeability () of the core material determines how easily a magnetic field
can be established. The inductance is directly proportional to the permeability of the core
material.
4.5.2 Parameters
As indicated in the Figure 4.3, the number of turns of wire, the length, and the cross
sectional area of the core are factors in setting the value of inductance. The inductance is
inversely proportional to the length of the core and directly proportional to the crosssectional area. Also, the inductance is directly related to the number of turns squared.
117
Figure 4.3. Factors that determine the inductance of a coil.
This relationship is as follows:
L =
𝑵𝟐  𝑨
𝒍
where L is the inductance in henries, N is the number of turns,  is the permeability, A is
the cross-sectional area in meters squared, and l is the core length in meters.
Example 4.2
is 0.25 x 10-3.
Determine the inductance of the coil below. The permeability of the core
0.01 m
0.1 m2
N=4
Solution
L =
𝑵𝟐  𝑨
𝒍
=
42 (0.25 𝑥 10−3)(0.1)
0.01
= 40 mH
4.6 Winding Resistance
When a coil is made of a certain material, for example, insulated copper wire, that wire
has a certain resistance per unit of length. When many turns of wire are used to construct
a coil, the total resistance may be significant. This inherent resistance is called the dc
resistance of the winding resistance (Rw). Although this resistance is distributed along
the length of the wire, it effectively appears in series with the inductance of the coil, as
shown in Figure 4.4. In many applications, the winding resistance can be ignored and the
coil considered as an ideal inductor. In other cases, the resistance must be considered.
Rw
(a) The wire has resistance
(b) Equivalent circuit
Figure 4.4. Winding resistance of a coil.
L
118
4.7 Winding Capacitance
When two conductors are placed side by side, there is always some capacitance between
them. Thus, when many turns of wire are placed close together in a coil, a certain amount
of stray capacitance is a natural side effect. In many applications, this stray capacitance is
very small and has no significant effect. In other cases, particularly at high frequencies, ti
may become quite important.
The equivalent circuit for an inductor with both its winding resistance (Rw) and its
winding capacitance (Cw) is shown in Figure 4.5. The capacitance effectively acts in
parallel.
Cw
Rw
(a) Stray capacitance between each
loop appears as a total parallel capacitance
L
(b) Equivalent circuit
Figure 4.5. Winding capacitance of a coil.
4.8 Faraday’s Law
Faraday found that by moving a magnet through a coil of wire, a voltage was introduced
across the coil, and that when a complete path was provided, the induced voltage an
induced current.
The amount of induced voltage is directly proportional to the rate of change of the
magnetic field with respect to the coil.
This principle is illustrated in the Figure 4.6, where a bar magnet is moved through a coil
of wire. An induced voltage is indicated by the voltmeter connected across the coil. The
faster the magnet is moved, the greater is the induced voltage.
When a wire is formed into a certain number of loops or turns and is exposed to a
changing magnetic field, a voltage is induced across the coil. The induced voltage is
proportional to the number of turns of wire in the coil, N, and to the rate at which the
magnetic field changes.
119
Figure 4.6. Induced voltage is created by a changing magnetic field.
4.9 Lenz’s Law
Lenz’s law adds to Faraday’s law by defining the direction of induced voltage as follows:
When the current through a coil changes and an induced voltage is created as a
result of the changing magnetic field, the direction of the induced voltage is such that it
always opposes the change in current.
In Figure 4.7 (a), the current is constant and is limited by R1. There is no induced
voltage because the magnetic field is unchanging. In part (b), the switch suddenly is
closed, placing R2 in parallel with R1 and thus reducing the resistance. Naturally, the
current tries to increases and the magnetic field begins to expand, but the induced voltage
opposes this attempted increases in current for an instant.
In part (c), the induced voltage gradually decreases, allowing the current to
increase. In part (d), the current has reached a constant value as determined by the
parallel resistors, and the induced voltage is zero. In part (e), the switch has been
suddenly opened, and, for an instant, the induced voltage prevents any decreases in
current. In part (f), the induced voltage gradually decreases, allowing the current to
decreases back to a value determined by R1. Notice that the induced voltage has a polarity
that opposes any current change. The polarity of the induced voltage is opposite that of
the battery voltage for an increases in current and aids the battery voltage for a decreases
in current.
120
(a) Switch open: Constant current
and constant magnetic field;
no induced voltage.
(c) Right after switch closure: The rate
of expansion of the magnetic field
decreases, allowing the current to
increase as induced voltage
decreases.
(b) At instant of switch closure:
Expanding magnetic field
induces voltage, which prevents
increase in total current.
(d) Switch remains closed: Current
and magnetic field reach
constant value.
121
(e) At instant of switch opening:
Magnetic field begins to collapse,
creating an induced voltage, which
prevents decrease in current.
(f) After switch opening: Rate of
collapse of magnetic field decrease,
allowing current to decrease back to
original value.
Figure 4.7. Demonstration of Lenz’s law: When the current tries to change suddenly, the
electromagnetic field changes and induces a voltage in a direction that opposes that change in
current.
4.10 Classifications of Inductor
Inductors are made in a variety of shapes and sizes. Basically, they fall into two general
categories: fixed and variable.
(a) Fixed
(b) Variable
Both fixed and variable inductors can be classified according to the type of core material.
Three common types are the air core, the iron core, and the ferrite core. Each has a unique
symbol, as shown.
(a) Air core
(b) Iron core
(c) Ferrite core
Adjustable (variable) inductors usually have a screw-type adjustment that moves
a sliding core in and out, thus changing the changing the inductance.
122
4.11 Types of Inductor:
(a) fixed molded inductors
(b) variable coils
(c) toroid inductor
4.12 Inductors in Series and Parallel
When inductors are connected in series, the total inductance, LT, is the sum of the
individual inductances. The formula for LT is expressed in the following equation for the
general case of n inductors in series:
LT = L 1 + L 2 + L 3 + . . . + L n
Notice that the formula for inductance in series is similar to the formula for resistance in
series.
L1
L2
L3
Ln
Inductors in Series
When inductors are connected in parallel, the total inductance is less than the smallest
inductance. The formula for total inductance in parallel is similar to that for total
parallel resistance.
1
𝐿𝑇
=
1
𝐿1
+
L1
15H
1
𝐿2
+
1
𝐿3
+ …+
1
𝐿𝑛
L2
20H
L3
30H
Ln
60H
Inductors in parallel
The general formula states that the reciprocal of the total inductance is equal to the sum
of the reciprocals of the individual inductances. LT can be found by taking the reciprocals
of both sides of the equation.
123
LT =
or
𝟏
𝟏
𝟏
𝟏
𝟏
+ + + …+
𝑳𝟏 𝑳𝟐 𝑳𝟑
𝑳𝒏
LT = (L1-1 + L2-1 + L3-1 + . . . + Ln-1)-1
4.13 Inductors in DC Circuits
When there is constant direct current in an inductor, there is no induced voltage. There
is, however, a voltage drop due to the winding resistance of the coil. The inductance itself
appears as a short to DC. Energy is stored in the magnetic field according to the formula
W = ½ LI2. The only energy loss occurs in the winding resistance (P = I2Rw).
This condition is illustrated in Figure 4.8.
Figure 4.8. Energy storage and loss in an inductor. The only dc voltage drop across the
coil is due to the winding resistance.
4.14 Time Constant
Because the inductor’s basic action is to oppose a change in its current, it follows that
current cannot change instantaneously in an inductor. A certain time is required for the
current to make a change from one value to another. The rate at which the current
changes is determined by the time constant. The time constant for a series RL circuit is
 =
𝐿
𝑅
where  is in seconds when L is in henries and R is in ohms.
Example 4.3 A series RL circuit has a resistance of 1 k and an inductance of 1 mH.
What is the time constant?
124
 =
𝐿
𝑅
1 𝑚𝐻
=
1 𝑘Ω
= 1 s
4.15 Energizing Current in an Inductor
In a series RL circuit, the current will increase to 63% of its value in one time constant
interval after the switch is closed. The buildup of current is analogous to the buildup of
capacitor voltage during the charging in an RC circuit; they both follow an exponential
curve and reach the approximate percentages of final value as indicated in the Figure 4.9
Final current
98%
99%
Considered 100%
95%
86%
63%
0
1
2
3
4
5
Figure 4.9. Energizing current in an inductor
The change in current over five time constant intervals is illustrated in Figure 4.10. When
the current reaches its final value at approximately 5, it ceases to change. At this time,
the inductor acts as a short (except for winding resistance) to the constant current. The
final value of the current is Vs/Rw = 10 V/10  = 1 A.
125
Figure 4.10. Illustration of the exponential buildup of current in an inductor. The current
increases another 63% during each time constant interval. A winding resistance of 10 is assumed.
A voltage (VL) is induced in the coil that tends to oppose the increase in current.
Example 4.4 Calculate the time constant for the circuit shown below. Then determine
the current and the time at each time constant interval, measured from the instant the
switch is closed.
R
100 ohms
L
50 mH
20 V
Solution
Ifinal =
𝑉
𝑅
20 𝑉
= 100 Ω = 0.2 A
126
 =
𝐿
𝑅
=
50 𝑚𝐻
100 Ω
= 0.5 ms
At 1 = 0.5 ms:
i = 0.63(0.2 A) = 0. 126 A
At 2 = 1.0 ms:
i = 0.86(0.2 A) = 0. 172 A
At 3 = 1.5 ms:
i = 0.95(0.2 A) = 0. 190 A
At 4 = 2.0 ms:
i = 0.98(0.2 A) = 0. 196 A
At 5 = 2.5 ms:
i = 0.99(0.2 A) = 0. 198 A  0.2 A
4.16 Electrical Transient Analysis
E
R
ER
L
EL
3 1Space
Key
=
2J1
1.0kΩ
12
V
4
1.0mH
Let ER be the voltage across the winding resistance.
EL be the voltage induced due to the inductance of the inductor.
ER = RI
dI
EL = L
dt
By KVL
ER + EL = E
dI
RI + L dt = E
Rearranging the equation.
dI
L dt + RI = E
Divide the L
127
dI
dt
𝑅
𝐸
+ 𝐿I = 𝐿
With the initial values I = 0 when t = 0,
The particular solution is
I=
E
R
R
(1 − e− L t )
To find ER and EL in a given time
ER = RI
R
dI
E
− t
d[ (1− e L )]
EL = L dt = L
R
dt
R
= Ee− L t
4.17 Inductor Applications
Power Supply Filter
rf Choke
Tuned Circuits
4.18 Testing Inductors
The most common failure in an inductor is an open coil. To check for an open, remove the
coil from the circuit. If there is an open, an ohmmeter check will indicate infinite
resistance, as shown in Figure 4.11. If the coil is good, the ohmmeter will show the
winding resistance. The value of winding resistance depends on the wire size and length
of the coil. It can be anywhere from one ohm to several hundred ohms.
Occasionally, when an inductor is overheated with excessive current, the wire insulation
will melt, and some coil will short together. This produces a reduction in the inductance
by reducing the effective number of runs and a corresponding reduction in winding
resistance.
Figure 4.11. Checking a coil by measuring the resistance.
128
QUESTIONS:
1. Describe the basic construction of an inductor.
2. What is inductance?
3. Explain how an inductor stores energy.
4. What are the various physical parameters that effect inductance?
5. Explain why practical inductors have both resistance and capacitance.
6. What are Lenz’s law and Faraday’s law?
7. What are the various types of inductor?
8. How do you determine the total inductance of series inductors?
9. How do you determine the total inductance of parallel inductors?
10. Define time constant as related to an inductive circuit.
11. What are some common inductor applications?
12. How do you check an inductor?
.
129
Assessment No. 11
INDUCTOR AND INDUCTANCE
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. Derive the formula for the energy stored in an inductor.
2. Derive the formula for the inductance.
3. Derive the formula for the total inductance of series inductors.
130
4. Derive the formula for the total inductance of parallel inductors.
5. Convert the following to millihenries:
a. 1 H
b. 250 H
c. 10 H d. 0.0005 H
131
Problem Set No. 6
INDUCTOR AND INDUCTANCE
1. How many turns are required to produce 30 mH with a coil wound on a cylindrical
coil having a cross-sectional area of 10 x 10-5 m2 and a length of 0.05 m? The core
has a permeability of 1.2 x 10-6?
A. 3536 turns
B. 3679 turns
C. 3987 turns
D. 4502 turns
2. A 12-V battery is connected across a coil with a winding resistance of 12 . How
much current is there in the coil?
A. 1 A
B. A
C. A
D. A
3. How much energy is stored by a 100-mH inductor with a current of 1 A?
A. 0.02 J
B. 0.04 J
C. 0.05 J
D. 0.09 J
4. The current through a 100-mH coil is changing at a rate of 200 mA/s. How much
voltage is induced across the coil?
A. 0.02 V
B. 0.04 V
C. 0.08 V
5. Suppose that you require a total inductance of 50 mH. You have available a 10-mH
coil and a 22-mH coil. How much additional inductance do you need?
A. 16 mH
B. 18 mH
C. 20 mH
D. 36 mH
6. Determine the total parallel inductance for the following coils in parallel: 75 H, 50
H, 25 H, and 15 H.
A. 7.14 µH
B. 9.28 µH
C. 10.67 µH
7. You have a 12-mH inductor, and it is your smallest value. You need an inductance of
8 mH. What value can you use in parallel with the 12-mH to obtain 8 mH?
A. 12 mH
B. 18 mH
C. 20 mH
D. 24 mH
.
132
Objective Test No. 4
INDUCTOR AND INDUCTANCE
1. When the current though an inductor increases, the amount of energy stored in the
electromagnetic field
A. Decreases
B. remains constant
C. increases
D. doubles
2. When the current though an inductor doubles, the stored energy
A. Doubles
B. Quadruples
C. is halved
D. does not change
3. The winding resistance of a coil can be decreased by
A. increasing the number of turns
B. using a larger wire
C. changing the core material
4. The inductance of an iron-core coil increases if
A. the number of turns is increased
B. the iron core is removed
C. the length of the core is increased
D. larger wire is used
5. An inductor, a resistor, and a switch are connected in series to a 12-V battery. At the
instant the switch is closed, the inductor voltage is
A. 0 V
B. 12 V
C. 6 V
D. 4 V
6. An ohmmeter is connected across an inductor and the pointer indicates an infinite
value. The inductor is
A. Good
B. Open
C. shorted
D. resistive
7. The property that opposes any change in current
A. mutual inductance
B. friction
C. self-inductance
D. losses
8. An open coil has
A. infinite resistance and inductance
B. zero resistance and inductance
C. zero resistance and infinite inductance
D. infinite resistance and zero inductance
133
9. If the number of turns in an inductor is increased, its inductance will
A. Vary
B. Decrease
C. increase
D. remain the same
10. At DC steady state, an inductor acts like ___________.
A. an open circuit
B. a short circuit
C. a capacitor
D. an insulator
11. Unit of inductance
A. Farad
B. Ohm
C. Henry
D. siemen
.
134
Unit 5
CAPACITOR AND CAPACITANCE
LEARNING OUTCOMES
At the end of this unit,
you are expected to:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
describe the basic construction of a capacitor.
define capacitance and tell how it is measured.
explain how a capacitor stores energy.
determine the energy stored in a capacitor.
state Coulomb’s law and discuss how it relates to an electric field and the
storage of energy.
illustrate the charging and discharging of a capacitor.
relate various physical parameters to capacitance value.
determine total series capacitance.
determine total parallel capacitance.
solve a combination circuit consisting of capacitor.
define time constant as related to a capacitive circuit.
relate the charging and discharging of a capacitor to the time constant.
Analyze electrical transients in series resistance and capacitance in DC
circuit.
explain why a capacitor blocks dc.
explain why a capacitor produces no energy loss.
explain the significance of reactive power in a capacitive circuit.
describe some common capacitor applications.
check out a capacitor with an ohmmeter.
135
Important Terms
dielectric
voltage rating
charging
dielectric strength
discharging
temperature coefficient
capacitance
leakage
Farad
dielectric constant
Coulomb’s law
time constant
energy storage
permittivity
5.1 The Capacitor: Basic Construction
dielectric
terminal
Parallel metal plates
Basic Construction of Capacitor
In its simplest form, a capacitor is an electrical device constructed of two parallel
conductive plates separated by an insulating material called the dielectric.
5.2
Charge Storage of Capacitor
In the neutral state, both plates of a capacitor have an equal number of free electrons, as
indicated in Figure 5.1. When the capacitor is connected to a voltage source through a
resistor, electrons (negative charge) are removed from plate A, and an equal number are
deposited on plate B. As plate A loses electrons , plate B gains electrons, plate A becomes
136
positive with respect to plate B. During the charging process, electrons flow only through
the connecting leads and the source. No electrons flow through the dielectric of the
capacitor because it is an insulator. The movement of electrons ceases when the voltage
across the capacitor equals the source voltage. If the capacitor is disconnected from the
source retains the stored charge for a long period of time (the length depends upon the
type of capacitor) and still has the voltage across it. Actually charged capacitor can be
considered as a temporary battery.
Figure 5.1. Illustration of a capacitor storing charge.
5.3 Capacitance
The amount of charge per unit of voltage that a capacitor can store is its capacitance,
designated C. That is, capacitance is a measure of capacitor’s ability to store charges. The
more charge per unit of voltage that a capacitor can store, the greater its capacitance, as
expressed by the following formula:
C =
𝑄
𝑉
where C is the capacitance, Q is charge, and V is voltage
137
The Unit of Capacitance
The farad (F) is the basic unit of capacitance. By definition,
One farad is the amount of capacitance when one coulomb of charge is stored with
one volt across the plates.
Most capacitors that you will use in electronics work have capacitance values in
microfarads (F) and picrofarads (pF).
5.4 How a Capacitor Stores Energy
A capacitor stores energy in the form of an electric field that is established by the opposite
charges on the two plates. The electric field is represented by lines of force between the
positive and negative charges and concentrated within the dielectric.
Lines of force
Q1
Q2
d
Electric field exists between the plates of a charged capacitor.
Q1
F
Q2
d
A force exists between a charged bodies.
d
5.5 Coulomb’s law
A force exists between two charged bodies that is directly proportional to the
product of the twp charges and inversely proportional to the square of the distance
between the bodies.
138
This relationship is expressed as
F =
𝑘𝑄1 𝑄2
𝑑2
where F is the force in newtons, Q1 and Q2 are the charges in coulombs, d is the distance
between the charges in meters, and k is a proportionality constant equal to 9 x 109.
5.6 The Energy Stored in a Capacitor
The formula for the energy stored by a capacitor is as follows:
W = ½ CV2
where the energy, W, is in joules when C is in farads and V is in volts.
5.7 Voltage Rating
Every capacitor has a limit on the amount of voltage that it can withstand across its plates.
The voltage rating specifies the maximum dc voltage that can be applied without risk of
damage to the device. If this maximum voltage, commonly called the breakdown voltage
or working voltage, is exceeded, permanent damage to the capacitor can result.
Both the capacitance and the voltage rating must be taken into consideration
before a capacitor is used in a circuit application. The choice of capacitance value is based
on particular circuit requirements (and or factors that are studied later). The voltage
rating should always be well above the maximum voltage expected in a particular
application.
5.8 Dielectric Strength
The breakdown voltage of a capacitor is determined by a dielectric strength of the
dielectric material used. The dielectric strength is expressed in volts/mil (1 mil = 0.001
in.) Table 5.1 below typical values for several materials. Exact values vary depending on
the specific composition of the material.
139
Table 5.1. Some Common Dielectric Materials and their Dielectric Strengths.
Material
Dielectric Strength
(volts/mil)
Air
80
Oil
375
Ceramic
1000
Paper
1200
Teflon
1500
Mica
1500
Glass
2000
5.9 Temperature Coefficient
The temperature coefficient indicates the amount and direction of a change in capacitance
value with temperature. A positive temperature coefficient means that the capacitance
increases with an increase in temperature or decreases in temperature. A negative
coefficient means that the capacitance decreases with an increases in temperature or
increases with a decreases in temperature.
Temperature coefficients typically are specified in parts per million per degree
Celsius (ppm/C). For example, a negative temperature coefficient of 150 ppm/C for a
1-F capacitor means that for every degree rise in temperature, the capacitance decreases
by 150 pF (there are one million picofarads in one microfarad).
5.10 Leakage
No insulating material is perfect. The dielectric of any of any capacitor will conduct some
very small amount of current. Thus, the charge on a capacitor will eventually leak off.
Some types of capacitors have higher leakages than others. An equivalent circuit for a
nonideal capacitor is shown in Figure 5.2. The parallel resistor represents the extremely
high resistance of the dielectric material through which leakage current flows.
140
R leak
C
Figure 5.2. Equivalent Circuit for a Nonideal Capacitor
5.11 Physical Characteristics of a Capacitor
The following parameters are important in establishing the capacitance and the voltage
rating of a capacitor: plate area, plate separation, and dielectric constant.
5.11.1 Plate Area
Capacitance is directly proportional to the physical size of the plates as determined by
the plate area.
5.11.2 Plate Separation
Capacitance is inversely proportional to the distance between the plates.
5.11.3 Dielectric Constant
As you know, the insulating material between the plates of a capacitor is called the
dielectric. Every dielectric material has the ability to concentrate the lines of force of the
electric field existing between the oppositely charged plates of a capacitor and thus
increase the capacity for energy storage. The measure of a material’s ability to establish
an electric field is called the dielectric constant or relative permittivity , symbolized by
r (the Greek letter epsilon).
Capacitance is directly proportional to the dielectric constant. The dielectric
constant (relative permittivity) is dimensionless, because it is a relative measure and is
a ration of the absolutely permittivity, , of a material to the absolute permittivity, o, of a
vacuum, as expressed by the formula:

r = 
𝑜
The value of o is 8.85 x 10-12 (farads per meter).
141
Table 5.2. Some Common Dielectric Materials and their Dielectric Strengths.
Material
Typical r values
Air (vacuum)
1.0
Teflon
2.0
Paper (paraffined)
2.5
Oil
4.0
Mica
5.0
Glass
7.5
Ceramic
1200
5.12 Formula for Capacitance in Terms of physical Parameters
An exact formula for calculating the capacitance in terms of the three quantities
mentioned is as follows:
C =
𝐴 𝑟 (8.85 𝑥 10−12 𝐹/𝑚)
𝑑
where A is in square meters (m2), d is in meters (m), C is in farads (F) and r is the
relative permittivity.
Example 5.1 Determine the capacitance of a parallel plate capacitor having a plate area
of 0.01 m2 and a plate separation of 0.02 m. The dielectric is mica, which has a dielectric
constant of 5.0.
Solution
C =
=
𝐴 𝑟 (8.85 𝑥 10−12 𝐹/𝑚)
𝑑
(0.01 𝑚2 )(5.0)(8.85 𝑥 10−12 𝐹/𝑚)
= 22.13 pF
0.02 𝑚
142
5.13 Types of Capacitors








Mica Capacitors
Ceramic Capacitors
Paper/Plastic capacitors
Electrolytic Capacitors
Variable Capacitors
Air Capacitors
Trimmers and Padders
Varactors
5.14 Series Capacitors
C2
C1
C3
Vs
While charging, I = Q/t is the same at all points so that all capacitors store the same
amount of charge (QT = Q1 = Q2 = Q3)
By Kirchhoff’s voltage law,
Vs = V1 + V2 + V3
Using the fact that V = Q/C, we can substitute into the formula for Kirchhoff’s law and
get the following relationship (where Q = QT = Q1 = Q2 = Q3):
𝑄
𝐶𝑇
=
𝑄
𝐶1
+
𝑄
𝐶2
+
𝑄
𝐶3
Canceling out Q we have,
1
𝐶𝑇
=
1
𝐶1
+
1
𝐶2
+
1
𝐶3
Taking the reciprocal of both sides gives the formula for the total capacitance:
143
CT =
1
1
1
1
+ +
𝐶1 𝐶2 𝐶3
Voltage Division in Series Capacitors
Vx =
𝐶𝑇
𝐶𝑥
x Vs
where Vx is the voltage across Cx which is any capacitor, such as C1, C2, and so on.
5.15 Parallel Capacitors
Vs
C1
C2
C3
The charged stored by the capacitors together equals the total charge that was delivered
from the source:
QT = Q1 + Q2 + Q3
Using the fact that Q = CV, we can substitute into the preceding formula and get the
following relationship:
CTVs = C1V1 + C2V2 + C3V3
Since Vs = V1 = V2 = V3, they can be canceled, leaving
CT = C1 + C2 + C3, etc.
5.16 Capacitors in DC Circuits
In this section, the response during charging and discharging of a simple capacitive circuit
with a dc source is examined. Figure 5.3 shows a capacitor connected in series with a
resistor and a switch to a dc voltage source. Initially, the switch is open and the capacitor
is uncharged with zero volts across its plates. At the instant the switch is closed, the
144
current jumps to its maximum value and the capacitor begins to charge. The current is
maximum initially because the capacitor has zero volts across it and therefore, appears as
a short; thus, the current is limited only by the resistance. As time passed and the
capacitor charges, the current decreases and the voltage VC across the capacitor
increases. The resistor voltage is proportional to the current during this charging period.
Figure 5.3 Charging and discharging of a capacitor.
After a certain period of time, the capacitor reaches full charge. At this point, the current
is zero and the capacitor voltage is equal to the dc source voltage, as shown in Figure 53(b). If the switch were opened now, the capacitor would retain its full charge (neglecting
any leakage).
In Figure 5.3(c) , the voltage source has been removed. When the switch is closed, the
capacitor begins to discharge. Initially, the current jumps to a maximum but in a direction
opposite to its direction during charging. As time passes, the current and capacitor
145
voltage decrease. The resistor voltage is always proportional to the current. When the
capacitor has fully discharged, the current and the capacitor voltage are zero.
Remember the following about capacitors in DC circuits:
1.
2.
3.
4.
Voltage across a capacitor cannot change instantaneously.
Current in a capacitive circuit can change instantaneously.
A fully charged capacitor appears as an open to nonchanging current.
An uncharged capacitor appears as a short to an instantaneous change in current.
5.17 The RC Time Constant
As you have seen, when a capacitor charges or discharges through a resistance, a certain
time is required for the capacitor to charge fully or discharge fully. The voltage across a
capacitor cannot change instantaneously, because a finite time is required to move charge
from one point to another. The arte at which the capacitor charges or discharges is
determined by the time constant of the circuit.
The time constant of a series RC circuit is a time interval that equals the product of the
resistance and the capacitance.
The time constant is symbolized by  , and the formula is as follows

= RC
Recall that I = Q/t. The current is the amount of charge moved in a given time. When the
resistance is increased, the charging current is reduced, thus increasing the charging time
of the capacitor. When the capacitance is increased, the amount of charge increases; thus,
for the same current, more time is required to charge the capacitor.
Example 5.2
A series RC circuit has a resistance of 1 M and a capacitance of 5 F.
What is the time constant?
Solution
 = RC = (1 M)(5 F) = 5s
146
5.18 The Charging Curve
ic
Initial value
100%
37%
14%
5%
2%
1%
0
1
2
3
4
5
Voltage and Current in a Capacitor during charging
Voltage across the
capacitor
A.
Electrical Transient Analysis
Current through the capacitor
R
E
ER
time
3 1Space
Key
=
2J1
1.0kΩ
12
V
4
1.0mH
time
L
Let ER be the voltage across the resistor
EC be the voltage across the capacitor
EL
147
dQ
ER = RI = R dt
Q
EC =
C
By KVL
ER + EC = E
dQ
R dt +
Q
C
= E
Divide the C
dQ
dt
+
Q
RC
𝐸
=𝑅
With the initial values Q= 0 when t = 0,
The particular solution is
t
Q = EC (1 − e− RC )
To find ER and EC at a given time
t
−
d[EC (1− e RC )]
ER = R
Q
EL =
C
dt
QUESTIONS:
1. Describe the basic construction of a capacitor.
2. What is capacitance? What are the factors that affect capacitance?
3. Explain how a capacitor stores energy.
4. Explain Coulomb’s law and discuss how it relates to an electric field and the storage
of energy.
5. How does capacitor charges and discharges?
6. How do you determine total series capacitance?
7. How do you determine total parallel capacitance?
8. What is time constant as related to a capacitive circuit.
9. Relate the charging and discharging of a capacitor to the time constant.
10. Explain why a capacitor blocks dc.
11. Explain why a capacitor produces no energy loss.
12. Explain the significance of reactive power in a capacitive circuit.
13. What are some common capacitor applications?
14. How do you check out a capacitor with an ohmmeter?
.
148
Assessment No. 12
CAPACITOR AND CAPACITANCE
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. Derive the formula for the energy stored in a capacitor.
2. Derive the formula for the capacitance.
3. Derive the formula for the total capacitance of series capacitors.
149
4. Derive the formula for the total capacitance of parallel capacitors.
5. Convert the following to microfarad:
a. 0.000000657 F
b. 0.0000982 F
.
150
Problem Set No. 7
CAPACITOR AND CAPACITANCE
1. Two capacitors connected in parallel across a 250-V mains have charges of 3,000 µC
and 5,000 C, respectively. Find the total capacitance of the combination.
A. 32 µF
B. 45 µF
C. 56 µF
D. 76 µF
2. A 0.4-F capacitor has a charge of 20 C. How much is the voltage across it?
A. 20 V
B. 30 V
C. 45 V
D. 50 V
3. The equivalent capacitance of two capacitors in series is 2.4 F. If one of the
capacitors has a capacitance of 4 F, what is the capacitance of the other?
A. 2 µF
B. 4 µF
C. 6 µF
D. 8µF
4. Three capacitors having capacitance of 4 F , 6 F and 8 F respectively are
connected in series. Find the equivalent capacitance of the combination.
A. 1.08 µF
B. 2.84 µF
C. 1.85 µF
D. 4.84 µF
5. The energy stored in a 0.125 F capacitor is 50 J, solve for the charge accumulated.
A. 3.54 mC
B. 5.34 mC
C. 6.23 mC
D. 8.45 mC
6. A certain capacitor is charged at 48 volts after its stored energy is 5.76 x 10-2 joules.
What is the capacitance of the capacitor?
A. 25 µF
B. 50 µF
C. 75 µF
D. 89 µF
7. Calculate the capacitance between two plates each of which is 100 cm2 and 2 mm
apart in air.
A. 44.27 pF
B. 48.90 pF
151
C. 56.84 pF
D. 76.43 pF
8. A capacitor whose plates is 20 cm x 3.0 cm and is separated by a 1.0-mm air gap is
connected across a 12-V battery. Determine the charge accumulated on each plate
after a long time.
A. 43.89 nC
B. 637.5 nC
C. 0.6375 nC
D. 89.56 pC
9. Three capacitors A, B, and C are charged as follows: A: 10F, 100 volts; B: 15 F,
150 volts; C: 25 F, 200 volts. They are then connected in parallel with terminals of
like polarity together. What is the voltage across the combination?
A. 165 V
B. 175 V
C. 185 V
D. 195 V
10. A given capacitor has a capacitance of 100 F. Calculate its elastance.
A. 1,000 D
B. 10,000 D
C. 100,000 D
D. 1,000,000 D
11. Three capacitors of 5 F , 10 F and 15 F respectively are connected in series
across a 100-V supply. Solve for the voltage across the 15-F capacitor.
A. 18.2 V
B. 19.6 V
C. 20.7 V
D. 25.4 V
.
152
Objective Test No. 5
CAPACITOR AND CAPACITANCE
1. Which of the following statements does not describe a capacitor?
A.
B.
C.
D.
2.
3.
4.
5.
6.
The plates are conductive.
The dielectric is an insulator between the plates.
Constant dc flows through a fully charged capacitor.
A practical capacitor stores charge indefinitely when disconnected from the
source.
The capacity of a condenser is proportional to ___________.
A. area of its plates
B. volume of its plates
C. the specific resistance of the plate material
D. the temperature coefficient of the plate material
The capacity of the capacitor is inversely proportional to ________.
A. the temperature of the dielectric
B. the material of the dielectric
C. the thickness of the dielectric
D. the permeability of the material and inductance
The capacitors in series have the same _______.
A. voltage
B. capacity
C. charge
D. energy loss
The capacitors are named according to the _______ used.
A. material of the plate
B. dielectric used
C. enclosures used
D. voltage
When one of the following statements is true?
A. There is current through the dielectric of a charging capacitor.
B. When a capacitor is connected to a DC voltage source, it will charge to the value
of the source.
C. An ideal capacitor can be discharged by disconnecting it from the voltage source.
7. A capacitance of 0.01 F is larger than
A. 0.00001 F
B. 100,000 pF
C. 1000 pF
D. all of the above
8. When the voltage across a capacitor is increased, the stored charge
153
A. increases
B. decreases
C. remains constant
D. fluctuates
9. The voltage rating of a capacitor is increased by
A. decreasing plate area
B. increasing plate separation
C. increasing the plate area
D. a and b
10. The capacitance value is increased by
A. decreasing plate area
B. increasing plate separation
C. decreasing plate separation
11. An uncharged capacitor and a resistor are connected in series with a switch and a
12-V battery. At the instant the switch is closed, the voltage across the capacitor is
A. 12 V
B. 6 V
C. 24 V
D. 0 V
12. In Question 12, the voltage across the capacitor when it is fully charged is
A. 12 V
B. 6 V
C. 24 V
D. -6 V
13. An ohmmeter is connected across a discharged capacitor and the needle stabilizes
at approximately 50 k. The capacitor is
A. good
B. charged
C. too large
D. leaky
14. A good capacitor has a __________ resistance.
A. negligible
B. very high
C. negative
D. none of these
15. A capacitor opposes any change in __________.
A. current
B. voltage
C. Resistance
D. flux
154
16. The capacitance of a capacitor is directly proportional to
A. area of its plate
B. thickness of dielectric
17. Capacitors are used to
A. filter AC currents and pass DC currents
B. filter AC and DC currents
C. filter DC currents and pass AC currents
D. pass AC and DC currents
18. A capacitor consists of two _________
A. insulators separated by a conductor.
B. conductors separated by an insulator
C. conductors
D. insulators
19. Capacitors designed to be used in places where a high dielectric breakdown voltage
is important .
A. paper capacitors
B. ceramic capacitors
C. electrolytic capacitors
D. mica capacitors
20. Reciprocal of capacitance.
A. Inductance
B. Elastance
C. Reluctance
D. Daraf
21. The capacitance of a capacitor is NOT affected by _____.
A. type of dielectric material
B. distance between plates
C. area of the plates
D. type of material used in the plates
22. The capacitor stores the electricity in the shape of _____.
A. dynamic charge
B. static charge
C. current electricity
D. molecules
23. Capacitors are used in electric circuits to ____________.
A. store energy
B. introduce a voltage drop
C. produce a low opposition path to high frequencies
D. all of these
155
Unit 6
ALTERNATING CURRENT CIRCUITS
LEARNING OUTCOMES
At the end of the lesson, you are expected to:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Apply Ohm’s law and Kirchhoff’s laws to ac circuits as well as to dc circuits.
explain the behavior of a resistor and other purely resistive load in ac circuit.
explain the behavior of an inductor in ac circuit.
discuss the behavior of a capacitor in ac circuit.
analyze the response of an RL, RC, and RLC circuit with a sine wave input.
define impedance.
determine the impedance of both series and parallel RL, RC, and RLC circuit.
convert a parallel circuit to an equivalent series circuit.
determine the effects of frequency on an RL, RC, and RLC circuits in terms of
changes in impedance and phase angle.
define conductance, capacitance, capacitive and inductive susceptance, and
admittance.
Apply Ohm’s law to RC circuits in order to find current and voltage value.
determine the true power, reactive power, and apparent power.
define power factor and explain its significance.
explain how RL, RC, and RLC circuits are used as phase-shifting circuits (lead
and lag network).
156
6.1 AC Circuit with Resistance Only
This is a circuit in which the applied voltage is ac connected to any resistive load.
What is a resistive load?
Any component which convert electrical energy to heat and light energy such as lamp, flat
iron, electric stove, etc. These components are made up of metals which posses certain
amount of resistance. The most common type of resistive load is resistor in which the
common function is to resist and limit the amount of current. Heat is also produced by
the resistor when current flows through it. Motor also have resistance because their
windings
are
made
up
of
copper
wire.
.
R
Vs
Let
VR be the voltage across the resistor (or any resistive load) in rms value
Vs be the source voltage in rms value.
Note:
 The load here represents any purely resistive load, that is, a load which has
resistance only.
 rms value is used because it is the effective value in the circuit.
 VR = Vs because there is only component in the circuit.
If the sine wave equation of the of Vs is vs = Vsm sint the equation of VR is definitely vR
= VRm sint.
To find the current through the resistor using Ohms’ law,
𝑖 =
Let
𝑣𝑅
𝑅
𝐼𝑅𝑚 =
=
𝑉𝑅𝑚 𝑠𝑖𝑛 𝑡
𝑅
𝑉 𝑅𝑚
𝑅
so that the equation of the current i = IRm sint.
where
i = is the instantaneous value of the current
157
VS and Vr = the instantaneous value of voltage source and the voltage across the
purely load, respectively.
IRm , VSm, VRm = the maximum value of the current through the load, maximum
value of the voltage source, and the maximum value of the voltage across the resistor.
As you can see, the phase angle of current and voltage are both equal to zero.
6.1.1 The Phase Relationship of Current and Voltage
When the equation of the voltage across the resistor is vR = VRm sint the current is i =
IRm sint, therefore,
In a purely resistive load the current is in phase with the voltage.
This phase relationship is illustrated below where the phase angles of the voltage and
current are both zero meaning one of the positive-going zero crossings is at 0.
80
60
40
current through the
resistor
20
0
0
1
2
0°
3
90°
4
180°
-20
5
270°
6
360°
7
voltage across the
resistor
-40
-60
-80
IR
VR
If in case the phase angle of voltage is 30, then the phase angle of current is also 30
158
Note: The phase angle that is being referred here in the angle of positive-going zero
crossing of a sine wave nearest to the origin.
80
VRm
60
40
IRm
20
iR
0
-1
0
1
2
3
4
5
6
7
vR
30°
-20
-40
-60
-80
VR
IR
30
If the equation of the voltage across the resistor is v = VSm sin (t + 30) then the
current is
i = IRm sin (t + 30).
From IRm =
𝑉 𝑅𝑚
𝑅
since, IRm = √2 IR rms and VRm = √2 VR rms
√2 IR rms =
√2 𝑉𝑅 𝑟𝑚𝑠
𝑅
Canceling √2 , we have,
IR rms =
𝑉𝑅 𝑟𝑚𝑠
𝑅
or simply IR =
𝑉𝑅
𝑅
159
Also
VR = IR R
R=
𝑉𝑅
𝐼
Note: These formulas are the same as in DC circuit and these are applicable regardless of
the frequency of the source because frequency has no effect on the resistance.
Example 6.1 In the circuit below, find the sine wave equation of the current if the
source is 120 V (rms).
Vs
120 V
60 Hz
R
100 ohms
Solution
The maximum value of the source is √2 (120) = 169.71 V. Assuming that its
positive-going zero-crossing is at 0, therefore its equation is vS = 169.70 sin 260t and
the voltage across the resistor is vR= 169.70 sin 260t.
Using the formula
IR =
𝑉𝑅
𝑅
=
169.70 𝑠𝑖𝑛 260𝑡 𝑉
100 
,
so that the equation of the current
is
iR = 1.697 sin 260t
A
Another solution is simply using the formula IR =
answer
𝑉𝑅
𝑅
where VR = Vs = 120 V
IR =
120 𝑉
100 Ω
= 1.2 V the rms value of which is √2 (1.2) = 1.6971 A,
The equation is also
iR = 1.697 sin 260t A
answer
160
Example 6.2
In the circuit below, find the value of the resistance if the vR = 127.28
sin (t - 60) V and iR = 14.14 sin(t - 60) A.
R
Vs
Solution
R =
Example 6.3
𝑣𝑅
𝑖𝑅
=
127.28 𝑠𝑖𝑛 (𝑡 − 60) 𝑉
14.14 𝑠𝑖𝑛(𝑡 − 60) 𝐴
= 9
Find the rms value of the current in the circuit shown below
R
10 ohms
Vs
100 V
Solution
𝑉𝑅
IR =
𝑅
=
100 𝑉
10 Ω
= 10 
6.1.2 The Power in a Circuit with Resistance Only
E
R
161
80
70
60
50
iR
40
new axis
vR
30
Pave
P
20
10
0
0
1
2
0°
3
90°
4
180°
5
270°
6
7
360°
-10
-20
Let p = be the instantaneous power absorbed by the resistor, where p = vRiR
In the sine waves below iR = IRm sin t and
vR = VRm sin t, so that
p = (IRm sin t )(VRm sin t)
= IRm VRm sin2 t
p = IRm VRm (
where sin2t =
1 − 𝑐𝑜𝑠2𝑡
2
1 − 𝑐𝑜𝑠2𝑡
2
)
𝐼𝑅 𝑚 𝑉𝑅𝑚
The average value of the power wave is
2
P =
𝐼𝑅 𝑚 𝑉𝑅𝑚
2
where P = average value of the power. This average value is also known as true or
active power.
6.1.3 What is meant by true power?
This is the rate at which energy is being dissipated by the resistor or any resistive load.
6.1.4 The Formula of True Power
From P =
𝐼𝑅 𝑚 𝑉𝑅𝑚
2
, IRm = √2 IRrms or √2 IR and VRm = √2 VRrms or √2 VR
162
Note: Eliminating rms does not change the meaning of the voltage and current; they are
still rms values.
Then the power formulas are
P = IRVR = I2R =
where
𝑉𝑅 2
𝑅
P = power in watts
IR = rms value in case of ac circuit and dc value in case of dc circuit.
VR = rms value in ac circuit and dc value in dc circuit
R = resistance in ohms
Example 6.4 The emf and current waves of a circuit having resistance only are
e = 170 sin 260t and i = 14.14 sin 260t. Determine (a) equation of power wave; (b)
frequency of power wave; (c) maximum value of power wave; (d) average power; (e)
power when t = 1/480 sec, 1/240 sec, 1/90 sec. (f) Draw voltage, current, and power
waves.
Solution
(a) p = ei
= (170 sin 260t)(14.14 sin 260t)
= 2403.8 sin2 260t
= 2403.8 (
=
2403.8
2
1 − 𝑐𝑜𝑠2260𝑡
2
)
(1 - cos2260t)
= 1201.9 (1 - cos2260t) watts
= 1201.9 (1 - cos2120t) watts
(d)
(e)
(f)
(g)
f = 120 Hz
Pmax = 2403.8 watts
Pave = 1201.9 watts
From the power wave equation p = 1201.9 (1 - cos2120t) watts
when t = 1/480 sec ; p = 1201.9 [1 - cos2120(1/480)] = 1201.9 watts
when t = 1/240 sec p = 1201.9 [ 1 - cos2120(1/240)] = 2403.8 watts
when t = 1/90 sec ; p = 1201.9 [ 1 - cos2120(1/90)] = 1802.85 watts
163
3000
2500
2000
iR
1500
vR
1000
p
500
0
0
1
2
3
4
5
6
7
-500
Example 6.5 A lamp load consist of thirty 100-watt lamps each taking rated power
from a 120-V supply. Determine (a) power when supply is DC; (b) power when supply
is 60-Hz sinusoidal AC; (c) equation of AC voltage wave if voltage is zero and increasing
when t = 0; (d) equation of current wave; (e) equation of power wave.
Solution
(a) P = 30 lamps x 100 watts/lamp = 3,000 watts
(b) 3,000 watts also because the frequency has no effect on the resistance.
(c) e = vR = 2(120) sin 260t = 169.71 sin 260t V
𝑃
3,000 𝑊
(d) I = =
= 25 A
𝐸
120 𝑉
Im =√2 (25) = 35.36 A
The equation of the current wave is i = 35.36 sin 260t A
(e) p = ei
= (120 sin 260t)(35.36 sin 260t)
= 4243.2 sin2 260t
= 4243.2 (
=
4243.2
2
1 − 𝑐𝑜𝑠2260𝑡
2
)
(1 - cos2260t)
= 2121.6 (1 - cos2260t) watts
= 2121.6 (1 - cos2120t) watts
164
6.2 AC Circuit with Inductance Only
Practically, all inductors have resistance in them because they are made of metals and
metals have resistance. In this lesson inductors are considered as ideal, meaning the
resistance is zero or negligible.
6.2.1 What are some examples of inductive loads?
Inductors, motors, and everything that is made of coils of conductors .
Vs
L
(a)
25
VLm
20
15
10
ILm
current through the
inductor
5
0
0
1
0°
2
90°
3
180°
4
5
270°
6
7
360°
-5
8
9
voltage induced by the
inductor
-10
-15
-20
-25
(b)
Figure 6.1. (a) An inductor connected across an ac source; (b) The waveforms of voltage
and current in the inductor.
Note: L represents the inductance of a load which is considered as purely inductive.
165
6.2.2 Phase relationship of Current and Voltage in a Purely Inductive Load
In the waveforms shown in Figure 6-1(b) the current lags the voltage by 90 or the voltage
leads the current by 90. The equation of voltage across the capacitor is vL = VLmsin t V
and the current is iL = ILm sin (t - 90) A. Just subtract 90 from the phase angle of the
voltage (that is 0 - 90 = -90).
In case the phase angle of the voltage is other than 0, say +30, the phase angle of the
current would be 30 - 90 = -60. The equation of the voltage would be vL = VLm sin (t
+ 30) V and that of the current is iL = ILm sin (t - 60) A.
If the equation of current is iL = ILm sin (t + 30) A, the equation of the voltage is vC =
VCmsin (t + 120) V. Add 90 from 30 since the voltage is leading the current by 90.
Example 6.6 The waveform below is the voltage across an ideal inductor. Draw the
waveform for its current if its maximum value is 5 A. Write also their sine wave
equations.
150
VLm = 120 V
100
50
0
-2
-1
0
60°
-50
-100
-150
Solution
1
2
3
4
5
6
7
8
166
150
VLm = 120 V
100
50
ILm = 5A
0
-2
-1
0
60°
1
2
3
4
5
6
7
8
30°
-50
-100
-150
The equation of the voltage is vL = 120 sin (t + 60) V and that of the current is iL = 5
sin (t - 30) A.
6.2.3 Why the Current Lags the Inductor Voltage by 90
As you know, a sine wave voltage has a maximum rate of change at its zero crossings and
a zero rate of change at the peaks. From Faraday’s Law you know that the amount of
voltage induced across a coil is directly proportional to the rate at which the current is
changing. Therefore, the coil voltage is maximum at the zero crossings of the current
where the rate of change of the current is the greatest. Also, the amount of voltage is zero
at the peaks of the current where its rate of change is zero. This relationship is shown in
Figure 6-1(b) . As you can see, the current peaks occur a quarter cycle after the voltage
peaks. Thus, the current lags the voltage by 90.
6.2.4 The Inductive Reactance, XL
In Figure 6.2, an inductor is connected to a sine wave source. Note that when the source
voltage is held at a constant amplitude value and its frequency is increased, the amplitude
of the current decreases. Also, when the frequency of the source is decreased the current
amplitude increases. The reason is as follows: when the frequency of the applied voltage
increases, its rate of change also increases, as you already. Now if the frequency of the
applied voltage is increased, the frequency of the current also increases. According to
Faraday’s law and Lenz’s law, this increase in frequency induces more voltage across the
inductor in a direction to oppose the current and cause it to decrease in amplitude.
Similarly, a decrease in frequency will cause an increase in current.
167
A decrease in the amount of current for a fixed amount of voltage indicates that
opposition to the current has increased. Thus, the inductor offers opposition to current
which varies directly with frequency.
Figure 6.2. The current in an inductive circuit varies inversely with the frequency of the applied
voltage.
The opposition to sinusoidal current in an inductor is called inductive reactance.
The symbol for inductive reactance is XL, and its unit is the ohm ()
Formula:
XL = 2fL
where XL = the inductive reactance in ohms ()
f = frequency in hertz (Hz)
L = the inductance in henry (H)
XL is proportional to the frequency and inductance
Example 6.7 A sinusoidal voltage is applied to the circuit in Figure. The frequency is 1
kHz. Determine the inductive reactance.
Vs
Solution
L = 5 mH
XL = 2fL
= 2(1 kHz)(5 mH) = 31.4 
168
6.2.5 Ohm’s law in Inductive Circuits
The reactance of an inductor is analogous to the resistance of a resistor. In fact, XL, just
like R, is expressed in ohms. Since inductive reactance is a form of opposition to current,
Ohm’s law applies to inductive circuits as well as to resistive circuits and capacitive
circuits, and it is stated as follows:
VL = ILXL
where VL = the voltage across the inductor in volts (V).
IL = the current through the inductor (A)
XL = the inductive reactance in ohms ()
Example 6.8 Determine the rms current.
5V
10 kHz
L = 100 mH
Solution
First calculate XL:
XL = 2fL = 2(10 kHz)(100 mH) = 6283 
Using Ohm’s law, we obtain
IL =
𝑉𝐿
𝑋𝐿
=
5𝑉
6283 
= 795.8 A
6.2.6 Power in an Inductor
Vs
L
169
60
40
20
vL
0
0
1
0°
2
90°
3
4
180°
5
270°
6
360°
7
8
iL
p
-20
-40
-60
As discussed earlier, an inductor stores energy in its magnetic field when there is
current through it. An ideal inductor (assuming no winding resistance) does not
dissipate energy; it only stores it. When an ac voltage is applied to an inductor, energy is
stored by the inductor during a portion of the cycle; then the stored energy is returned
to the source during another portion of thee cycle. There is no net energy loss. Figure
above shows the power curve that results from one cycle of inductor current and
voltage.
6.2.7 Instantaneous power (p)
The product of instantaneous voltage, vL, and instantaneous current, IL, gives
instantaneous power, p. At points where vL and iL is zero. When both vL and iL are positive,
p is also positive. When either vL or iL is positive and the other negative, p is negative.
When both vL and iL are negative, p is positive. As you can see, the power follows a
sinusoidal-type curve. Power values of power indicates that energy is stored by the
inductor. Negative values of power indicate that energy is returned from the inductor to
the source. Note that the power fluctuates at a frequency twice that of the voltage or
current, as energy is alternately stored and returned to the source
6.2.8 True Power (Ptrue or simply P)
Ideally, all of the energy stored by the inductor during the positive portion of the power
cycle is returned to the source during the negative portion. No net energy is consumed in
the inductance, so the power is zero. Actually, because of winding resistance in a practical
inductor, some power is always dissipated. But since in this lesson, the inductor is
considered ideal so its resistance is negligible so its power dissipation or true power is
zero.
170
6.2.9 Inductive Reactive Power (QL)
The rate at which an inductor stores or returns energy is called its reactive power, QL.
The inductive reactive power (so called because this is the reactive power of the
inductor) is nonzero quantity, because at any instant in time, the inductor actually
taking energy from the source or returning energy to it. Reactive power does not
represent an energy loss. The following formula apply:
QL = VL IL
QL =
𝑉𝐿 2
𝑋𝐿
QL = IL2 XL
where QL = the inductive reactive power in volt-ampere reactive (VAR or var)
VL = the rms voltage across the inductor in volts (V)
IL = the rms current through the inductor in amperes (A)
XL = the inductive reactive in ohm ()
Example 6.9 A 10-V rms signal with a frequency of 1 kHz is applied to a 10-mH coil
with a negligible resistance. Determine the reactive power (QL) and the true power (P).
Solution
XL = 2fL = 2(1 kHz)(10 mH) = 62.83 
I =
𝑉𝐿
𝑋𝐿
=
10 𝑉
62.83 Ω
= 159.16 mA
We can find the inductive reactive using the three formulas:
QL = VL IL = (10 V)(159.16 mA) = 1.59 VAR
QL =
𝑉𝐿 2
𝑋𝐿
=
(10 𝑉)2
62.83 Ω
= 1.59 VAR
QL = IL2 XL = (159.16 mA)2(62.83) = 1.59 VAR
For the true power, P = 0 because there is no resistance in the inductor.
6.3 AC Circuit with Capacitance Only
171
Vs
C
60
Vcm
40
Icm
20
iC
0
-4
-2
0
90°
0°
2
90°
4
180°
6
270°
8
10
360°
vC
-20
-40
-60
Note: The capacitor in the circuit above is considered ideal, meaning the resistance
is negligible.
What are some of the components that are considered capacitive?
Basically, all capacitors and synchronous motors.
6.3.1 The Phase Relationship of Current and Voltage in a Purely Capacitive Load
The current leads the voltage by 90, or the voltage lags the current by 90.
In the waveforms above the equation of voltage across the capacitor is vC = VCmsin t V
and the current is iC = ICm sin (t + 90) A. Just add 90 from the phase angle of the voltage
(that is 0 + 90 = 90).
In case the phase angle of the voltage is other than 0, say +30, the phase angle of the
current would be 30 + 90 = 120. The equation of the voltage would be vC = VCm sin (t
+ 30) V and that of the current is iC = ICm sin (t + 120) A.
172
If the equation of current is iC = ICm sin (t + 30) A, the equation of the voltage is vC =
VCmsin (t - 60) V. Subtract 90 from 30 since the voltage is lagging the current by 90.
Example 6.10 The waveform below is the voltage across an ideal capacitor, draw the
waveform for its current if its maximum value is 5 A. Write also their sine wave
equations.
120 V
60
60
Solution
60
The equation of the voltage is vC = 120 sin (t + 60) V and that of the current is iC
= 5 sin (t + 150) A
6.3.2 Why the current leads the voltage by 90?
A sine wave voltage is shown in the figure below. Notice that the rate at which the voltage
is changing varies along the sine wave curve, as indicated by the “steepness” of the curve.
At the zero crossings, the curve is changing at a faster rate than anywhere else along the
curve. At the peaks,
173
Zero rate of change
Maximum negative
rate of change
(steepest)
Maximum negative
rate of change
(steepest)
Zero rate of change
the curve has a zero rate of change because it has just reached its maximum and is at the
point of changing direction.
The amount of charge stored by a capacitor determines the voltage across it.
Therefore, the rate at which the charge is moved (Q/t = current) from one plate to the
other determines the rate at which the voltage changes. When the current is changing
at its maximum rate ( at the zero crossings), the voltage is at its maximum value
(peak). When the current is changing at its maximum rate (zero at the peaks), the
voltage is at its minimum value (zero). This relationship is illustrated in the figure. As
you can see, the current peaks occur a quarter of a cycle before the voltage peaks. Thus
the current leads the voltage by 90.
6.3.3 Capacitive Reactance, XC
Capacitive reactance is the opposition to sinusoidal current, expressed in ohms and
symbolized by XC.
In Figure, a capacitor is connected to a sine wave voltage source. Note that when the
source voltage is held at a constant amplitude value and its frequency is increased, the
amplitude of the current increases. Also, when the frequency of the source is
decreased, the current amplitude decreases. The reason is as follows: When the
frequency of the voltage increases, its rate of change also increases. This relationship
in the figure, where the frequency is doubled.
The current in a capacitive circuit varies directly with the frequency of the applied
voltage.
174
B has the greater
rate of change
B
A
(steeper slope,
more cycles)
Rate of change increases with the frequency.
Now, if the rate at which the voltage is changing increase, the amount of charge moving
though the circuit in a given period of time must also increase. More charge in a given
period of time means more current. For example, a tenfold increase in frequency means
that the capacitor is charging and discharging 10 times a much in a given time interval.
Therefore, since the rate of charge movement has increased 10 times, the recurrent
must increases by 10 because I = Q/t.
An increase in the amount of current for a fixed amount of voltage indicates that
opposition to the current has decreased. Therefore, the capacitor offers opposition to
current, which varies inversely with frequency. The opposition to sinusoidal current
called capacitive reactance. The symbol for capacitive reactance is XC, and its unit is the
ohm ().
You have just seen how frequency affects the opposition to current (capacitive
reactance) in a capacitor. Now let’s see how the capacitance affects the reactance. Figure
shows that when a sine wave voltage with a fixed amplitude and frequency is applied to a
1-F capacitor, a certain amount of current flows. When the capacitance value is
increased to 2 F, the current increases. Thus, when the capacitance increases, the
opposition to current (capacitive reactance) decreases. Therefore, not only is the
capacitive reactance inversely proportional to frequency, but it is also inversely
proportional to capacitance:
XC is proportional to
1
𝑓𝐶
175
It can be proven that the constant of proportionality is 1/2. Therefore, the formula for
XC is
XC =
1
2𝑓𝐶
where XC = the capacitive reactance in ohms ()
f = frequency in hertz (Hz)
C = capacitance in farad (F)
The 2 term comes from the fact that a sine wave can be described in terms of rotational
motion, and one revolution contains 2 radians.
Example 6.11 A sinusoidal voltage is applied to a capacitor. The frequency of the sine
wave is 1 kHz. Determine the capacitive reactance.
Vs
0.005 µF
Solution
XC =
1
2𝑓𝐶
=
1
2(1 𝑘𝐻𝑧)(0.005 𝐹)
= 31.83 k
6.3.4 Ohm’s Law in Capacitive Circuits
The reactance of a capacitor is analogous to the resistance of a resistor. In fact, both are
expressed in ohms. Since both are forms of opposition to current, Ohm’s law apply to
capacitive circuits as well as to resistive circuits and is stated as follows for capacitive
reactance:
VC = IC XC
where VC = the rms voltage across the capacitor in volts (V)
IC = the rms current through an inductor in amperes (A)
176
XC = the capacitive reactance in ohms ()
Example 6.12 Determine the rms current through the circuit shown below.
Vs = 5 V
10 kHz
0.005 µF
Solution
1
XC =
2𝑓𝐶
=
1
2(10 𝑘𝐻𝑧)(0.005 𝐹)
= 3.183 k
Applying Ohm’s law:
IC =
𝑉𝐶(𝑟𝑚𝑠)
𝑋𝐶
=
5𝑉
3.183 𝑘
= 1.57 mA
6.3.5 Power in a Capacitor
As discussed earlier, a charged capacitor stores energy in the electric field within the
dielectric. An ideal capacitor does not dissipate energy; it only stores it. When an ac
voltage is applied to a capacitor, energy is stored by a capacitor during a portion of the
voltage cycle. There is no net energy loss. Figure below shows the power curve that
results from one cycle of capacitor voltage and current.
Vs
C
177
200
Pm
150
100
VCm
Icm
50
iC
0
-2
-1
90°
0
1
0°
2
3
90°
180°
4
5
270°
6
360°
7
8
vC
p
-50
-100
-150
-200
6.3.6 Instantaneous Power (p)
The product of instantaneous voltage, VC, and instantaneous current, IC, gives
instantaneous power, p. At point where vC or IC is zero, p is also zero. When both vC or IC
is positive and the other negative, p is negative. When both vC and IC are negative, p is
positive. As you can see, the power follows a sinusoidal-type curve. Positive values of
power indicate that energy is stored by the capacitor. Negative values of power indicate
that energy is returned from the capacitor to the source. Note that the power fluctuates
at a frequency twice that of the voltage or current, as energy is alternately stored and
returned to the source.
6.3.7 True Power (P)
Ideally, all of the energy stored by a capacitor during the positive portion of the power
cycle is returned to the source during the negative portion. No net energy is consumed
in the capacitor, so the true power is zero. Actually, because of leakage and foil
resistance in a practical capacitor, a small percentage of the total power is dissipated.
But in this lesson, the capacitor is considered as ideal (meaning the capacitor has infinite
resistance) so the true power is zero.
6.3.8 Capacitive Reactive Power (QC)
The rate at which a capacitor stores or returns energy is called its capacitive reactive
power, QC. The capacitive reactive power is a nonzero quantity, because at any instant
in time, the capacitor is actually taking energy from the source or returning energy to it.
Reactive power does not represent an energy loss. The following formulas apply:
QC = VC IC
QC =
𝑉𝐶 2
𝑋𝐶
178
QC = IC2XC
where QC = capacitive reactive power in volt-ampere reactive (VAR or var)
IC = current through the capacitor
VC = rms voltage across the capacitor
XC = capacitive reactance
Example 6.13 Determine the true power and the reactive power in the circuit shown.
110 V
60 Hz
110 µF
Solution
P = 0
XC =
QC =
.
1
2𝑓𝐶
𝑉𝐶 2
𝑋𝐶
=
=
1
2(60)(110𝐹)
(110 𝑉)2
24.114 
= 24.114 
= 501.78 VARS
179
PRACTICAL APPLICATION No. 2:
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. You are an electrician working in an industrial plant. You discover the problem with
a certain machine is a defective capacitor. The capacitor is connected to a 240-V AC
circuit. The information on the capacitor reveals that it has a capacitance value of 10
µF and a voltage rating of 240 V AC. The only 10-µF capacitor in the storeroom is
marked with a voltage rating 350 VDC. Can this capacitor be used to replace the
defective capacitor? Explain your answer.
2. You find that a 25-µF capacitor connected to a 480 V AC is defective. The storeroom
has no capacitors with a 480 V AC rating. However, you find two capacitors rated at
50 µF and 370 V AC. Can these capacitors be connected in such a manner that they
can replace the defective capacitor? If yes, explain how they are connected and why
the capacitors will not be damaged by the lower voltage rating. If no, explain why
they cannot be used without damage to the capacitor.
180
3. You are working as an electrician in an industrial plant. You are given an AC oil-filled
capacitor on a 480-V, 60-Hz AC line. The capacitor has the following marking: (15 µF
600 V DC). Will this capacitor be damaged if it is installed? Explain your answer.
.
181
Assessment No. 13
AC CIRCUIT WITH RESISTANCE, INDUCTANCE, AND CAPACITANCE ONLY
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. Prove that the average value of power in a purely resistive load is P = VRIR.
2. By applying integration, prove that the average power or true power in a purely
inductive load or ideal inductor is equal to zero.
182
3. Derive XL = 2fL.
4. By applying integration prove that the average power or true power of an ideal
capacitor is equal to zero.
5. Prove that XC =
1
2𝑓𝐶
183
.
6. The current in a pure resistor circuit is given by the equation i = 4 cos 5,000t. If the
rms value of the impressed emf is 110 volts, what are equations for the voltage and
power?
7. The current in a pure capacitor circuit is given by the equation i = 4 cos 5,000t. If
the rms value of the impressed emf is 110 volts, what are equations for the voltage
and power?
8. The current in a pure inductor circuit is given by the equation i = 4 sin 5,000t. If the
rms value of the impressed emf is 110 volts, what are equations for the voltage and
power?
184
.
Problem Set No. 8
AC CIRCUIT WITH RESISTANCE, INDUCTANCE, AND CAPACITANCE ONLY
1. An electric flatiron whose heating element is practically a pure resistance takes 480
watts when connected across 115-V DC mains. Determine power that it takes from
120-V 60-Hz mains.
A. 522.69 W
B. 480 W
C. 440. 83 W
D. 420 W
2. Find the resistance of the heating element in no. 1.
A. 27.55 ohms
B. 30 ohms
C. 35.84 ohms
D. 40.25 ohms
3. In no. 1 determine the equation of AC voltage, current, and power waves (draw
waves), zero time being when voltage is going through zero and increasing
positively.
A. e = 169.71 sin 377t, 6.16 sin 377t, 522.7 (1 – cos 377t)
B. e = 169.71 sin 377t, 6.16 sin 314.2t, 522.7 (1 – cos 754t)
C. e = 162.63 sin 377t, 5.9 sin 377t, 522.7 (1 – cos 754t)
D. e = 169.71 sin 377t, 6.16 sin 377t, 522.7 (1 – cos 754t)
4. Find (a) maximum value of power; (b) instantaneous power when t is 1/480 sec in
no. 2
A. 1045.4 W, 1045.4 W
B. 522.7 W, 522.7 W
C. 1045.4 W, 0 W
D. 1045.4 W, 522.7 W
5. A large room is illuminated by twenty 150-watt lamps and thirty 100-watt lamps. If
the circuit voltage is 116, calculate the total current.
A. 50.92 A
B. 51.72 A
C. 52.86 A
D. 54.61 A
6. How much power is represented by a circuit in which the voltage and current
equations are e = 160 sin314t and i = 42.5 sin 314t?
A. 3,400 W
B. 6,000 W
C. 6,800 W
D. 13,345 W
7. Calculate XL for f = 5 kHz and L = 100 mH.
185
A. 500 Ω
B. 1000 Ω
C. 3141.59 Ω
D. 6283.18 Ω
8. At what frequency is the reactance of a 50-H inductor equal to 800 ?
A. 1.54 kHz
B. 2.54 kHz
C. 40 kHz
D. 125.66 kHz
9. A 50-mH inductor is connected to a 12-V rms source. What is the true power? What is
the reactive power at a frequency of 1 kHz?
A. 0 W, 458.36 mVAR
B. 0.4584 W, 458.4 VAR
C. 0 W, 458,36 VAR
D. 600 mW, 0 VAR
10. Calculate the rms current in the figure below.
A. 0.6256 A
B. 636.6 A
L
C. 63.66 mA
120
120 V
E 2 Vpk
500mH
D. 0.6366 A
601Hz
60 Hz
0°
11. The emf and current waves of a circuit having inductance only are e = 141.4 sin 225t
and i = -17 cos 225t. Determine (a) equation of power wave; (b) frequency of power
wave;
A. p = -1201.9 sin425t W, 50 Hz
B. p = 1201.9 sin425t W, 60 Hz
C. p = 600.95 sin425t W, 60 Hz
D. p = 600.95 sin425t W, 50 Hz
12. In no. 11 determine (a) maximum value of power wave; (b) average power;
A. 1201.9 W, 0
B. 2403.8 W, 0
C. 1699.74 W, 0
D. 0 W, 1201.9 W
13. In no. 11, find (a) power when t = 1/200 sec. (b) Draw voltage, current, and power
waves.
A. 1101.9 W
B. 2403.8 W
C. 1699.74 W
D. 1201.9 W
14. A pure inductance takes 4 amp from 120-V(rms) 60-Hz mains. Determine (a)
equation of voltage and current waves, zero time being when current is going through
zero and increasing positively;
186
A.
B.
C.
D.
169.71 sin (260t + /2), 4 sin 260t
169.71 sin (250t + /2), 5.66 sin 250t
169.71 sin (260t + /2), 5.66 sin 260t
120 sin (260t + /2), 5.66 sin 260t
15. In no. 14 determine (a) equation of power wave; (b) maximum instantaneous power;
(d) average power; (g) Plot all three waves.
A. 480 sin 377t, 480 W, 480 W
B. 480 sin 754t, 960 W, 0 W
C. 480 sin 377t, 960 W, 480 W
D. 960 sin 754t, 960 W, 0 W
16. A reactance coil whose resistance is negligible takes 1.060 amp from 120-V 60-Hz
mains. Determine (a) inductance; (b) current when connected to 220-V 60-Hz mains.
A. 300 mH, 1.95 A
B. 400 mH, 2 A
C. 500 mH, 2.85 A
D. 600 mH, 1.96 A
17. Calculate XC for f = 5 kHz and C = 50 pF.
A. 500.79 kΩ
B. 636.62 kΩ
C. 1570.20 kΩ
D. 2500 kΩ
18. At what frequency is the reactance of a 0.1-F capacitor equal to 2 k?
A. 726 Hz
B. 896 Hz
C. 796 Hz
D. 996 Hz
19. Calculate the rms current for the circuit shown below.
A.
B.
C.
D.
0.6283 A
0.6583 A
62.83 mA
658.3 mA
1V
1 MHz
0.1 µF
20. A 1-F capacitor is connected to an ac voltage source of 12 V, 500 Hz. What is the
true power?
A. 0
B. 0.4524 VAR
C. 12 VAR
187
D. 6000 VAR
21. In problem 20, determine reactive power at a frequency of 500 Hz.
A. 0
B. 0.4524 VAR
C. 12 VAR
D. 6000 VAR
22. A capacitance of 4 F is connected across a 40-V (rms) 1,000 Hz power supply.
Determine (a) rms current; (b) maximum instantaneous current.
A. 1.01 A, 2.02 A
B. 1.01 A, 1.01A
C. 1.43 A, 1.43 A
D. 1.01 A, 1.43 A
23. In no. 22 find the equations of current and emf waves, zero time being when current
is crossing zero axis in a positive direction;
A. 1.43 sin 2500t, 56.57 sin (21000t - /2)
B. 1.01 sin 21000t, 113.14 sin (21000t - /2)
C. 1.43 sin 21000t, 56.57 sin (21000t - /2
D. 1.43 sin 22000t, 113.57 sin (21000t - /2)
24. The emf and current waves in a circuit having capacitance only are e = 311 sin 250t
and i = 5.65 cos 250t. Determine (a) equation of power wave; (b) maximum
value of power wave; (c) average power; (d) power when t = 1/400 sec. (e) Draw
the waves.
A. 878.58 sin 425t, 1757.15 W, 0, 878.58 W
B. 878.58 sin 425t, 1757.15 W, 0, 878.58 W
C. 878.58 sin 450t, 878.58 W, 0, 1757.15 W
D. 878.58 sin 450t, 1757.15 W, 0, 878.58 W
25. In a power circuit it is desired to obtain a 90 leading current of 60 amp by the use of
capacitors, the voltage supply being 600 volts, 60 Hz. (Capacitors are used on power
systems to correct power factor.) Determine (a) required capacitance; (b) current
that capacitance would take at 440 volts, 120 Hz.
A. 262.67 µF, 88 A
B. 265.26 µF, 88 A
C. 278.95 µF, 62.23 A
D. 296.35 µF, 124.45 A
.
6.4 Power Factor
188
The Power factor is the ratio of true power and apparent power. The power factor
ranges from 0 to 1.
pf =
𝑃
𝑆
where pf – is the power factor which can expressed as a decimal or percentage.
P – true power in watts (W)
S – apparent power in volt-amperes (VA). The apparent power is just the
product of the total voltage and the total current in an ac circuit.
From the power triangle below which will be learned later, it can be seen that P/S is cos
. Therefore
pf =
𝑃
𝑆
= cos 
P – true power
S
Q
Q – reactive power
S – apparent power

P
 - the phase angle between the total
current and the total voltage or
6.4.1 Types of Power Factor
 Unity pf (pf = 1) - the voltage and current are in phase. A purely resistive load has
a unity power factor since the voltage and current are in phase. Any circuit whose
total inductive reactive power is equal to the total capacitive reactive power will
have a unity power factor.
 Lagging pf (pf is between 0 to 1) – the current lags the voltage by an acute angle
.
 Leading pf (pf factor is between 0 to 1) – the current leads the voltage by an acute
angle .
 Zero pf – if the voltage and current are out of phase by exactly 90. Any circuit
which has a zero true power has zero pf. The purely inductive or capacitive load
would have a zero pf since the true power is zero and the voltage and current are
out of phase by exactly 90.
189
6.5 Series Resistance and Inductance
R
Vs
L
30
VLm
20
VRm
10
ITm
iT
0
-2
-1
90°
0
0°
1
2
90°
3
180°
4
5
270°
6
360°
7
8
vL
vR
-10
-20
-30
Characteristics:
1. There is a common current in the circuit, that is, IT = IR = IL.
2. The total voltage is the algebraic sum of the voltage across the resistor and across
the inductor.
3. The apparent power is the algebraic sum of the true power (power taken by the
resistive load) and the inductive reactive power in the inductor.
Omit the current wave, and drawing the total (resultant) voltage. In this case, the total
or resultant voltage is the source or applied voltage.
190
40
VTm
VLm
30
20
VRm
10
vR
0
-2
-1
0
90°
1
0°
2
3
90°
4
180°
5
6
270°
7
360°
8
vL
vT
-10
-20
-30
-40
VL
VS

IT
VR
(ref)
The Phasor Diagram of a Series RL Circuit.
Note: The angle of the reference is not always zero. It can be any angle other than 0.
The figure below shows an example. This applies to the succeeding circuits.
VL
VS


VR
IT
(ref)
191
⃑⃑𝑇 = VR + jVL
𝑉
volts
VR = VT cos
⃑⃑𝑇 = VT 
𝑉
total voltage in rectangular form
VL = VT sin
volts
total voltage in polar form
where VT - is the magnitude of the total voltage
VR – the voltage across the resistance of the load
VL - the voltage across the inductance of the load.
 - the phase angle, the angle between the total voltage and the total current.
The magnitude of the total voltage determined by using Pythagorean Theorem.
VT = √𝑉𝑅 2 + 𝑉𝐿 2
and
𝑉𝐿
 = tan-1
𝑉𝑅
From the rectangular form of the total voltage, divide each parameter by I since it is
common in the circuit.
⃑⃑𝑇
𝑉
𝐼⃑𝑇
⃑⃑𝑇
𝑉
𝐼⃑𝑇
But,
=
= 𝑍⃑ ,
𝑉𝑅
𝐼𝑅
+ 𝑗
𝑉𝑅
𝐼𝑅
𝑉𝐿
𝐼𝐿
= R ,
𝑉𝐿
𝐼𝐿
= XL
so that,
𝑍⃑ = R + jXL
the rectangular form of the impedance
where 𝑍 – impedance in ohms ()
𝑅 - resistance in ohms ()
𝑋 – inductive reactance in ohms ()
6.5.1 What is impedance?
The impedance is the joint effect of combining resistive and reactance (inductive
or capacitive) in an AC circuit. Impedance is also defined as the total opposition to
192
alternating current. The figure below shows the impedance triangle of a series RL
circuit.
XL
XL
Z

R
R
The impedance triangle of a series RL circuit.
𝑍⃑ = 𝑍
polar form of the impedance
where Z – the magnitude of the impedance of the circuit.
 - the phase angle.
By Pythagorean Theorem, the magnitude of the impedance is,
𝑍 = √𝑅 2 + 𝑋𝐿 2
And
 = tan-1
𝑋𝐿
which is the same as in the phasor diagram
𝑅
now for the Power Triangle,
VL
VS
S
QL


IT
VR
P
(ref)
The Power Triangle in Series RL Circuit.
193
Multiply the voltages by the current,
𝑉𝐿 𝑥 𝐼𝐿
= 𝑄𝐿
𝑉𝑅 𝑥 𝐼𝑅 = 𝑃
𝑉𝑇 𝑥 𝐼𝑇 = 𝑆
𝑆⃑ = 𝑃 + 𝑗𝑄𝐿
apparent power
𝑆⃑ = 𝑆
where
the rectangular form of the
the polar form of the apparent power
𝑆 = the magnitude of the apparent power
 = power factor angle which is similar to the phasor diagram and impedance
triangle.
The magnitude of the apparent power can be found by Pythagorean Theorem.
𝑆 = √𝑃2 + 𝑄𝐿 2
and
 = tan-1
𝑄𝐿
which is the same as in the phasor diagram
𝑃
Also,
𝑆 =
𝑃
𝑐𝑜𝑠 
=
𝑃
𝑝𝑓
𝑃 = 𝑆 𝑐𝑜𝑠  = 𝐼𝑇 𝑉𝑇 𝑐𝑜𝑠 
Example 6.14 In the circuit below, determine the magnitudes of the following:
a.
b.
c.
d.
e.
the impedance
total current
voltage across the resistor and inductor
power, reactive power, and apparent power
angle between the total current and the total voltage
194
f.
power factor
R = 75
Vs= 24 V
60 Hz
L = 290 mH
VL = 19.79 V
VS = 24 V
 = 55.54
VR = 13.58 V
IT (ref)
Solution
(a) XL = 2fL = 2(60 Hz)(290 mH) = 109.33 
Z = √(75)2 + (109.33)2 = 132.58 
(b) IT =
(c)
𝑉𝑇
𝑍
=
24 𝑉
132.58 
= 181.02 mA = IR = IL
VR = IRR = (181.02 mA)(75 ) = 13.58 V
VL = ILXL = (181.02 mA)(109.33 ) = 19.79 V
(d) P = IR2R = (181.02 mA)2 (75) = 2.46 W
or P = IRVR = (181.02 mA) (13.58) = 2.46 W
QL = IL2XL = (181.02 mA)2 (109.33) = 3.58 VAR
or
QL = ILVL = (181.02 mA)(19.79) = 3.58 VAR
S = √𝑃2 + 𝑄𝐿 2 = √(2.46)2 + (3.58)2 = 4.34 VA
(e)
𝑉
 = tan-1 𝑉 𝐿 = tan-1
𝑅
 = tan-1
𝑋𝐿
𝑅
= tan-1
19.79
13.58
109.33
75
= 55.54
= 55.54
195
(f)
𝑃
pf = 𝑆 =
2.46
= 0.567 or 56.7% lagging (because the current is lagging)
4.34
or cos  = cos 55.55 = 0.567 lagging
Example 6.15 A 6-ohm resistor and an 8-ohm inductive reactance when connected in
series across a 60-Hz supply take 12 amp. Determine (a) impedance of circuit; (b)
voltage across resistor; (c) voltage across reactance; (d) circuit voltage; (e) power; (f)
angle between current and voltage; (g) power factor; (h) inductance. (i) Draw phasor
diagram.
Solution
12 A
R =6
Vs=?
60 Hz
XL = 8
(a)
(b)
(c)
Z = √62 + 82 = 10 
VR = IRR = (12 A)(6 ) = 72 V
VL = ILXL = (12 A)(8 ) = 96 V
(d)
VT = √𝑉𝑅 2 + 𝑉𝐿 2 = √(72)2 + (96)2 = 120 V
(e)
(f)
P = IRVR = (12 A)(72) = 864 W
 = tan-1
 = tan-1
(g)
𝑉𝐿
𝑉𝑅
𝑋𝐿
𝑅
96
= tan-1 72 = 53.13
8
= tan-1 6 = 53.13
pf = cos 53.13 = 0.6 or 60% lagging
196
(h)
Phasor diagram:
VL =
96 V
VS = 120 V
 = 53.13
VR = 72 V
IT (ref)
Example 6.16 The equation of the emf on an inductive circuit is e = 400 sin 377t and the
current is i = 40 sin (377t - 60). What is the inductance?
Solution:
𝑉𝑇 = 282.82 0
𝐼𝑇 = 28.28  − 60
𝑍⃗ =
⃑⃗𝑇
𝑉
𝐼⃗𝑇
=
282.82 0
28.28 −60
= 10 Ω60
In rectangular form,
𝑍⃗ = 5 + 𝑗8.66 Ω
so that R = 5 Ω and 𝑋𝐿 = 8.66 Ω
and
𝐿 =
𝑋𝐿

=
8.66
377
= 22.97 𝑚𝐻
Example 6.17 An induction motor which is connected to 120-V 50-Hz source draws a
current of 5 A. If the power factor of the motor is 95% lagging, find (a) the apparent
power, (b) the true power that it takes.
Solution:
(a) 𝑆 = 𝐼𝑇 𝑉𝑇 = (5)(120) = 600 𝑉𝐴
(b) 𝑃 = 𝐼𝑇 𝑉𝑇 𝑐𝑜𝑠 = 𝐼𝑇 𝑉𝑇 𝑝𝑓 = (5)(120)(0.95) = 570 𝑊
197
Assessment No. 14
SERIES RESISTANCE AND INDUCTANCE
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. In the circuit shown below, R = 20  and L = 200 mH are connected in series
across a 100-V 100-Hz source. Determine the magnitude of the following: (a)
impedance, (b) total current, (c) the voltage across the resistor and inductor,
(d) true power and reactive power, (e) apparent power, power factor, and
power factor angle.
198
2. A small AC motor used in a washing machine is, in effect, and RL circuit. If the
machine takes 311 watts and 4.5 amp from a 115-V source when operating
normally, calculate its power factor.
3. An impedance coil has a resistance of 20 ohms and an inductive reactance of 40
ohms. For what values of resistance will be connected in series with the coil to
make the overall power factor of the circuit be 0.6?
199
4. A series circuit takes 371.2 watts at a power factor of 0.8 from a 116-volt 60-Hz
source. What are the values of R and L?
5. An RL circuit takes a current of 7 amp that lags behind the 231-volt source by
35. Calculate the power factor, power, impedance, resistance, and inductive
reactance of the circuit.
200
PRACTICAL APPLICATION No. 3
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. You have the task of ordering a replacement inductor for one that has become
defective. The information on the nameplate has been painted over and cannot
be read. The machine that contains the inductor operates on 480 V at a
frequency of 60 Hz. Another machine has an identical inductor in it, but its
nameplate has been painted also. A clamp-on mater indicates a current of 18 A,
and a voltmeter indicates a voltage drop across the inductor of 324 V in the
machine that is still in operation. After turning off the power and locking out the
panel, you disconnected the inductor in the operating machine and measure a
wire resistance of 1.2 Ω with an ohmmeter. Using identical inductor in the
operating machine as an example, what inductance value should you order and
what would be the minimum VAR rating of the inductor? Should you be
concerned with the amount of wire resistance in the inductor when ordering?
Explain your answers.
201
2. An AC electric motor is connected to a 240-V, 60-Hz source. A clamp-on ammeter
with a peak hold function reveals that the motor has an inrush current of 34 A
when the motor is first started . Your job is to reduce the inrush current to a
value of 20 A by connecting a resistor in series with the motor. The resistor will
be shunted out of the circuit after the motor is started. Using an ohmmeter, you
find that the motor has a wire resistance of 3 Ω. How much resistance should be
connected in series with the motor to reduce the starting current to 20 A?
3. You are a journeyman electrician working in an industrial plant. Your task is to
connect an inductor to a 480-V, 60-Hz line. To determine the proper conductor
and fuse for this installation, you need to know the amount of current the
inductor will draw from the line. The nameplate on the inductor indicates that it
has inductance of 0.1 H. An ohmmeter reveals that it has a wire resistance of 10
Ω. How much current should the inductor draw when connected to the line?
202
4. You are working on a residential heat pump. The heat pump is connected to a
240-V, 60-Hz power line. The compressor has a current draw of 34 amperes
when operating. The compressor has a power factor of 70%. The back-up strip
heat is rated at 10 kW. You need to know the amount of total current draw that
will occur if the strip heat comes on while the compressor is operating.
.
203
Problem Set No. 9
SERIES RESISTANCE AND INDUCTANCE
1. A 0.0159-henry inductance coil and a 4-ohm resistor are connected in series across
240-V 60-Hz mains. Determine (a) reactance; (b) impedance; (c) current.
A. 5.99 Ω, 7.2 Ω, 33.33 A
B. 5.99 Ω, 7.2 Ω, 60 A
C. 5.99 Ω, 4 Ω, 33.33 A
D. 5.99 Ω, 5.99 Ω, 60 A
2. In no. 2 find (a) true power; (b) phase angle; (c) power factor.
A. 3142.07 W, 56.27°, 0.455 lagging
B. 6284.14 W, 36.27°, 0.555 lagging
C. 4443.56 W, 56.27°, 0.555 lagging
D. 8887.12 W, 56.27°, 0.655 lagging
3. In no. 2 determine (a) voltage across resistor; (b) voltage across inductance coil.
(c) Draw phasor diagram.
A. 33.32 V, 199.65 V
B. 33.32 V, 1200 V
C. 133.32 V, 199.65 V
D. 233.32 V, 199.65 V
4. When a 12-ohm resistor and an unknown inductance coil of negligible resistance are
connected in series across a 120-V 50-Hz supply, the current is 8 amp. Determine
(a) reactance; (b) inductance;
A. 7 Ω, 8.65 mH
B. 8 Ω, 18.65 mH
C. 9 Ω, 28.65 mH
D. 10 Ω, 38.65 mH
5. In no. 4 determine (a) phase angle; (b) power; (c) power factor
A. 36.87°, 788 W, 0.74
B. 56.87°, 768 W, 0.8
C. 36.87°, 768 W, 0.8
D. 45°, 768 W, 0.86
6. In no. 4 find the voltage across resistor and across inductance coil.
A. 56 V, 70 V
B. 66 V, 71 V
C. 86 V, 77 V
D. 96 V, 72 V
7. The corrected readings of a voltmeter, an ammeter, and wattmeter when connected
to measure the voltage, current, and power of a circuit known to consist only of
resistance and inductance coil in series are as follows: volts, 118; amperes, 3.27;
power, 320 watts. The frequency is 60 Hz. Determine (a) power factor; (b) circuit
phase angle;
A. 0.8293, 33.97°
B. 0.8093, 32.97°
C. 0.83293, 34.92°
D. 0.8693, 33.21°
8. In no. 7 find the following: (a) resistance; (b) reactance; (c) inductance;
204
A. 19.93 Ω, 20.16 Ω, 53.48 mH
B. 29.93 Ω, 20.86 Ω, 52.48 mH
C. 29.93 Ω, 20.16 Ω, 53.48 mH
D. 49.93 Ω, 25.16 Ω, 55.48 mH
9. In the no. 7 find the voltage across resistance and inductance coil. Draw phasor
diagram.
A. 57.87 V, 25.92 V
B. 77.87 V, 65.92 V
C. 87.87 V, 85.92 V
D. 97.87 V, 65.92 V
10. The primary of a telephone induction coil has an effective resistance of 60 ohms and
an inductance of 0.154 H. Determine current that it takes when 50 volts at 1,000 Hz
is impressed across it
A. 0.05157 A
B. 0.15157 A
C. 0.25157 A
D. 0.35157 A
11. The current in a circuit known to consist only of resistance and inductance in series
is 8.31 amp when the circuit is connected across 120-V 25-Hz mains; when
connected across 120-V 60-Hz mains the current is 5.30 amp. Determine the
resistance and inductance.
A. 30.88 mH, 15 Ω
B. 50.88 mH, 12 Ω
C. 60.88 mH, 18 Ω
D. 80.88 mH, 20 Ω
12. The current in a series inductive circuit is 7.5 amp at 25 Hz. The circuit takes 425,
watts, and the power factor is 0.47. Determine circuit voltage.
A. 220.57 V
B. 320.57 V
C. 120.57 V
D. 620.57 V
13. In no. 12 find (a) inductance; (b) resistance. (c) Draw phasor diagram.
A. 80.33 mH, 7.55 Ω
B. 90.33 mH, 7.55 Ω
C. 96.33 mH, 8.55 Ω
D. 85.33 mH, 7.85 Ω
.
205
6.6 Series Resistance and Capacitance
R
Vs
C
50
40
30
20
iT
10
vR
0
0
1
0°
2
90°
3
4
180°
5
270°
6
7
8
9
10
360°
vC
-10
-20
-30
-40
-50
60
VTm
VCm
40
VRm
20
vR
0
0
0° 
-20
-40
-60
1
2
90°
3
180°
4
5
270°
6
360°
7
8
9
10
vC
vT
206
VR
IT (ref)

VT
VC
Phasor Diagram of Series RC Circuit
⃑⃑𝑇 = VR – jVC volts
𝑉
total voltage in rectangular form
VR = VT cos
⃑⃑𝑇 = VT 
𝑉
VC = VC sin
volts
total voltage in polar form
where VT = is the magnitude of the total voltage
VR = the voltage across the resistance of the load or resistor
VC = the voltage across the capacitance of the load or capacitor
 = the phase angle, the angle between the total voltage and the total current.
The magnitude of the total voltage is determined by using Pythagorean Theorem.
VT = √𝑉𝑅 2 + 𝑉𝐶 2
𝑉
 = tan-1 𝑉𝐶
and
𝑅
From the rectangular form of the total voltage, divide each parameter by I since it is
common in the circuit.
⃑⃑𝑇
𝑉
𝐼⃑𝑇
=
But,
𝑉𝑅
𝑉𝐶
+ 𝑗
𝐼𝑅
𝐼𝐶
⃑⃑𝑇
𝑉
𝐼⃑𝑇
= 𝑍⃑
,
𝑉𝑅
𝐼𝑅
= R ,
𝑉𝐶
𝐼𝐶
= XL
So that,
𝑍⃑ = R - jXC
the rectangular form of the impedance
207
where Z = impedance in ohms ()
R = resistance in ohms ()
XL = capacitive reactance in ohms ()
R
R

XC
Z
XC
The impedance triangle of a series RC circuit.
𝑍⃑ = Z
polar form of the impedance
where Z = the magnitude of the impedance of the circuit.
 = the phase angle.
By Pythagorean Theorem, the magnitude of the impedance is,
Z = √𝑅 2 + 𝑋𝐶 2
and
 = tan-1
𝑋𝐶
𝑅
which is the same as in the phasor diagram
Now for the Power Triangle,
VR

IT (ref)
P

QC
VC
S
VT
Power Triangle in Series RL Circuit.
208
Multiply the voltages by the current,
VC x IC = QC
VR x IR = P
VT x IT = S
𝑆⃑ = P + jQC
the rectangular form of the apparent power
𝑆⃑ = S
the polar form of the apparent power
where
S = the magnitude of the apparent power

= power factor angle which is similar to the phasor diagram and impedance
triangle.
The magnitude of the apparent power can be found by Pythagorean Theorem.
S = √𝑃2 + 𝑄𝐶 2
and
 = tan-1
𝑄𝐶
𝑃
which is the same as in the phasor diagram
also,
S =
𝑃
𝑐𝑜𝑠 
=
𝑃
𝑝𝑓
Example 6.16 In the circuit below, determine the magnitudes of the following:
a.
b.
c.
d.
e.
f.
the impedance
total current
voltage across the resistor and capacitor
power, reactive power, and apparent power
angle between the total current and the total voltage
power factor
209
VR = 13.85 V
 = 54.74
75 ohms
24 V
60 Hz
VC =
25 µF
19.6 V
VS =
24 V
Solution
(a) XL =
1
2𝑓𝐿
1
=
2(60 𝐻𝑧)(25 𝐹)
= 106.1 
Z = √(75)2 + (106.1)2 = 129.93 
(b) IT =
(c)
𝑉𝑇
𝑍
24 𝑉
=
129.93 
= 184.71 mA = IR = IC
VR = IRR = (184.71 mA)(75 ) = 13.85 V
VC = ICXC = (184.71 mA)(106.1 ) = 19.6 V
(d) P = IR2R = (184.71 mA)2 (75) = 2.56 W
or
P = IRVR = (184.71 mA) (13.85) = 2.56 W
QL = IL2XL = (184.71 mA)2 (106.1) = 3.62 VAR
or
QL = ILVL = (184.71 mA)(19.6) = 3.62 VAR
S = √𝑃2 + 𝑄𝐿 2 = √(2.56)2 + (3.62)2 = 4.43 VA
 = tan-1
𝑉𝐶
𝑉𝑅
 = tan-1
= tan-1
𝑋𝐿
𝑅
19.60
13.85
= tan-1
106.1
75
= 54.75
= 54.74
IT(ref)
210
(e)
pf =
𝑃
𝑆
=
2.56
4.53
= 0.578 or 57.8% leading (because the current is leading)
cos  = cos 57.8 = 57.8%
or
Example 6.17 A 6-ohm resistor and an 8-ohm inductive reactance when connected in
series across a 60-Hz supply take 12 amp. Determine (a) impedance of circuit; (b)
voltage across resistor; (c) voltage across reactance; (d) circuit voltage; (e) power; (f)
angle between current and voltage; (g) power factor; (h) inductance. (i) Draw phasor
diagram.
Solution
12 A
6 ohms
Vs=?
60 Hz
8 ohms
(a)
Z = √62 + 82 = 10 
(b)
VR = IRR = (12 A)(6 ) = 72 V
(c)
VC = ICXC = (12 A)(8 ) = 96 V
(d)
VT = √𝑉𝑅 2 + 𝑉𝐶 2 = √(72)2 + (96)2 = 120 V
(e)
P = IRVR = (12 A)(72) = 864 W
(f)
 = tan-1
=
tan-1
𝑉𝐶
𝑉𝑅
𝑋𝐶
𝑅
= tan-1
= tan-1
96
72
8
6
= 53.13
= 53.13
211
(g)
pf = cos 53.13 = 0.6 or 60% leading
(h) Phasor diagram:
VR = 72 V
IT (ref)
 = 53.13
VS = 120 V
VL = 96 V
Example 6.16 The equation of the emf on a capacitive circuit is e = 400 sin 377t and the
current is i = 40 sin (377t + 60). What is the capacitance?
Solution:
⃑⃑𝑇 = 282.82 0
𝑉
⃑𝐼⃑𝑇 = 28.28 60
𝑍⃗ =
⃑⃗𝑇
𝑉
𝐼⃗𝑇
=
282.82 0
28.28 60
= 10 Ω − 60
In rectangular form,
𝑍⃗ = 5 − 𝑗8.66 Ω
so that R = 5 Ω and 𝑋𝐶 = 8.66 Ω
and
.
𝐶 =
1
𝑋𝐶
=
1
377(8.66)
= 306.3 µ𝐹
212
Assessment No. 15
SERIES RESISTANCE AND CAPACITANCE
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. In the circuit shown below, R = 50  and C = 30 µF are connected in series
across a 100-V 100-Hz source. Determine the magnitude of the following: (a)
impedance, (b) total current, (c) the voltage across the resistor and capacitor,
(d) true power and reactive power, (e) apparent power, power factor, and
power factor angle.
213
2. What should be the capacitance of a capacitor, in series with a 250-ohm resistor,
that will limit the current to 1.2 amp when the circuit is connected to a 600-V 60Hz source.
214
3. A series circuit consists of a 66.2-F and a variable resistor. For what two values
of resistance will the power taken by the circuit be 172.8 watts, if the impressed
60-Hz emf is 120 V?
215
.
216
Problem Set No. 10
SERIES RESISTANCE AND CAPACITANCE
1. A 50-ohm resistance and an 80-F capacitor are connected in series across 115-V 60Hz mains. Determine (a) current; (b) true power; (c) power factor;
A. 0.92 A, 186.32 W, 0.8354
B. 1.92 A, 184.32 W, 0.8333
C. 1.52 A, 114.32 W, 0.8453
D. 1.92 A, 189.32 W, 0.9367
2. In no. 1 find (a) voltage across resistor; (b) voltage across capacitor. (c) Draw
phasor diagram.
a. 96 V, 63.67 V
b. 99 V, 53.67 V
c. 120 V, 73.67 V
d. 220 V, 93.67 V
3. A current of 2 amp at 60 Hz flows in a circuit with the resistor and a capacitor in
series. The voltage across the resistor is 60 volts, and that across the capacitor is
90.8 volts. Determine circuit voltage.
A. 100 V
B. 108.83 V
C. 148.22 V
D. 148.83 V
4. In no. 3 find (a) true power; (b) power factor.
A. 100 W, 0.5513 leading
B. 110 W, 0.4513 leading
C. 120 W, 0.6513 leading
D. 120 W, 0.5513 leading
5. Find the capacitance in problem no 3. Draw vector diagram.
A. 38.43 µF
B. 48.43 µF
C. 58.43 µF
D. 78.43 µF
6. A circuit with a resistor and a capacitor in series takes 200 watts at a power factor of
0.40 from 200-Volt 50-Hz mains. Determine current.
A. 2 A
B. 2.5 A
C. 3 A
D. 3.5 A
7. Find the power-factor angle in problem no. 6.
A. 46.42°
B. 56.42°
C. 66.42°
D. 86.42°
8. In no. 6 determine (a) resistance; (b) impedance; (c) capacitance.
A. 30 Ω, 75 Ω, 43.41 µF
B. 31 Ω, 80 Ω, 41.41 µF
C. 32 Ω, 80 Ω, 43.41 µF
217
D. 35 Ω, 90 Ω, 45.41 µF
9. A circuit with a resistor and a capacitor in series takes 3.0 amp, 216 watts, at 0.6
power factor from a 60-Hz supply. Determine resistance;
a. 18 Ω
b. 20 Ω
c. 22 Ω
d. 24 Ω
10. Find the circuit voltage in no. 9.
A. 120 V
B. 130 V
C. 140 V
D. 150 V
11. Determine the capacitive reactance and capacitance in no. 9.
A. 30 Ω, 82.89 µF
B. 32 Ω, 82.89 µF
C. 32 Ω, 84.89 µF
D. 36 Ω, 82.89 µF
12. Determine the power-factor in no. 9.
A. 0.6
B. 0.7
C. 0.8
D. 0.9
13. A 33-ohm resistor is in series with a 35.3-F capacitor across a constant potential
source of 100 volts. Determine frequency that will give current of 2.0 amp;
A. 60 Hz
B. 100 Hz
C. 120 Hz
D. 150 Hz
14. Find the following in problem no. 13: (a) circuit true power; (b) power factor.
A. 112 W, 0.636
B. 132 W, 0.66
C. 133 W, 0.656
D. 167 W, 0.636
15. A circuit with a 50-F capacitor and an adjustable resistor in series is connected
across 120-V 60-Hz mains. To what value of ohms must the resistor should be
adjusted for the circuit to take 80 watts? (Two values of resistance will satisfy this
condition.)
A. 112.7 Ω, 12.3 Ω
B. 122.7 Ω, 15.3 Ω
C. 152.7 Ω, 16.3 Ω
D. 162.7 Ω, 17.3 Ω
.
218
6.7 Series Resistance, Inductance, and Capacitance
R
1.0kΩ
E
12 Vrms
60 Hz
0°
2 41
3
L
1.0mH
C
1.0µF
50
VLm
40
VCm
VRm
30
20
iT
ITm
10
vR
0
-2
-1
0
90°
1
0°
2
3
90°
4
180°
5
70°
6
360°
7
8
vL
-10
vC
-20
-30
-40
-50
VL
VR
VC
I (ref)
219
To find the resultant or total voltage which is also the voltage source, VS.
50
40
30
20
vR
10
vL
0
-2
-1
90°
0
1
0°
2
90°
3
4
180°
5
270°
6
7
360°
8
vC
-10
vT
-20
-30
-40
-50
VL
VX = VL - VC


VR
I (ref)
The phasor diagram if VL > VC
VX = the net reactive voltage.
Formulas
⃑⃑𝑇 = VR + jVL - jVC
𝑉
the rectangular form of the total voltage
220
⃑⃑𝑇 = VR + j(VL - VC)
𝑉
⃑⃑𝑇 = VR + j VX
𝑉
where VX = VL - VC
⃑⃑𝑇 = VT
𝑉
the polar form of the total voltage
where
VT = √𝑉𝑅 2 + 𝑉𝑥 2
 = tan-1
𝑉𝑋
𝑉𝑅
VX = net reactive voltage
From VT = VR + jVL - jVC we can divide this by the total current since it is common in
the circuit , that is, IT = IR = IL = IC
⃑⃑𝑇
𝑉
𝐼⃑𝑇
=
𝑉𝑅
𝐼𝑅
+j
𝑉𝐿
𝐼𝐿
− 𝑗
𝑉𝐶
𝐼𝐶
where
⃑⃑𝑇
𝑉
𝐼⃑𝑇
= 𝑍⃑ ,
𝑉𝑅
𝐼𝑅
= R,
𝑉𝐿
𝐼𝐿
= XL ,
𝑉𝐶
𝐼𝐶
= XC
So that
𝑍⃑ = R + jXL - jXC
the rectangular form of the impedance
𝑍⃑ = R + j(XL - XC)
𝑍⃑ = R + jXnet
Xnet = XL - XC
XL
𝑍⃑ = Z
where Z = √𝑅 2 + 𝑋𝑛𝑒𝑡 2
Xnet = XL - XC
 = tan-1
𝑋𝑛𝑒𝑡


𝑅
Xnet = net reactance
XC
R
I (ref)
221
For the power triangular, multiply each quantity in the phasor diagram by the each
corresponding current.
VL
QL
P
VR
I (ref)
QC
VC
QL
S
Qnet = QL - QC
P
QC
𝑆⃑ = P + jQL - jQC
rectangular form of the apparent power
𝑆⃑ = P + j(QL - QC)
𝑆⃑ = P + jQnet
where Qnet = QL - QC = net reactive power
222
𝑆⃑ = S
polar form of the apparent power
where S = √𝑃2 + 𝑄𝑛𝑒𝑡 2
 = tan-1
𝑄𝑛𝑒𝑡
𝑃
Now if VL < VC, the formulas above, that is for VL > VC, also apply. But the VT and VX will
have a different direction as shown below. The Power factor here is leading because the
total current is leading the total voltage by an angle .
VL
VR
I (ref)

VX = VC – VL
V
T
VC
The Phasor Diagram if VC > VL
Example 6.18 In the circuit below, determine the magnitudes of the following:
(a) the impedance
(b) total current
(c) voltage across the resistor, inductor and capacitor
(d) power, reactive power (inductive and capacitive), and apparent power
(e) angle between the total current and the total voltage
(f) power factor
50 ohms
Vs= 110 V
60 Hz
L = 129 mH
120 µF
223
Solution
(a) impedance
XL = 48.63 
XL = 48.63 
Z = 56.6 
Xnet =
26.53 
 = 27.95
R = 50 
R = 50 
XC = 22.10 
XC = 22.10 
By complex quantities we can get the rectangular and polar form of the impedance,
Z = 50 + j48.63 – j22.10 = 50 + j26.53 
Converting this to polar form
Z = 56.6 27.95
If the required is only the magnitude, Z = 56.6 
Or use the formula for the magnitude of the impedance
Z = √𝑅 2 + 𝑋𝑛𝑒𝑡 2
where Xnet = XL - XC
Z = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 ) 2
Z = √502 + (48.63 − 22.10) 2
= 56.6 
(b) IT =
𝑉𝑇
𝑍
=
110 𝑉
56.6 
= 1.94 A
(c) VR = IRR = (1.94 A)(50 ) = 97 V
VL = ILXL = (1.94 A)(48.63 ) = 94.34 V
VC = ICXC = (1.94 A)(22.10 ) = 42.87
224
VL = 94.34 V
VT = 110 V
Vnet =
 = 27.95
51.47 V
R = 50 
I (ref)
VC = 42.87 V
(d)
P = IR2 R = (1.94 A)2 (50 ) = 188.18 W
QL = IL2 XL = (1.94 A)2 (48.63 ) = 183.02 VARs
QC = IC2 XC = (1.94 A)2 (22.10 ) = 83.18 VARs
S = √𝑃2 + 𝑄𝑛𝑒𝑡 2 = √𝑃2 + (𝑄𝐿 − 𝑄𝐶 ) 2
= √(188.18)2 + (183.02 − 83.18) 2 = 213.03 VA
(e) To find , we can use any of these formulas, the value of  = 27.95
𝑋 −𝑋
 = tan-1 𝐿 𝐶
𝑅
 = tan-1
𝑉𝐿 − 𝑉𝐶
 = tan-1
𝑄𝐿 − 𝑄𝐶
𝑉𝑅
𝑃
(f) pf = cos 27.95 = 0.883 or 88.3% lagging; since the current is lagging the total
voltage as seen
in the phasor diagram.
.
225
Assessment No. 16
SERIES RESISTANCE, INDUCTANCE , AND CAPACITANCE
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. A 25-  resistor, a 300-mH inductor, and a 50-µF capacitor are connected in
series across a 120-V 50-Hz source. Determine the magnitude of the following:
(a) impedance; (b) total current; (c) the voltage across the resistor, inductor,
and capacitor; (d) true power , inductive reactive power, and capacitive reactive
power; (e) apparent power, power factor, and power factor angle.
226
2. A series RLC circuit consist of a 25-ohm resistor, a 0.221-henry inductor, and a
66.3-F capacitor. If the series circuit is connected to a 125-V variable-frequency
source, what will be the frequency at which the current is 5 A.
3. An RLC circuit consist of a 55.4-ohm resistor, a 170-mH inductor, and an 82.8-F
capacitor. If the potential drop across the resistor is 72 volts, calculate the emf of
the 60-Hz source.
227
Problem Set No. 11
SERIES RESISTANCE, INDUCTANCE AND CAPACITANCE
1. A series circuit with 12 ohms resistance, 32 ohms inductive reactance, and 20 ohms
capacitive reactance is connected across 240-volt 60-Hz mains. Determine (a)
impedance; (b) current.
A. 13.64 Ω, 17.6 A
B. 24 Ω, 10 A
C. 16.97 Ω , 14.14 A
D. 64 Ω, 3.75 A
2. Find the voltage across each circuit element in no. 1.
A. 169.68 V, 452.48 V, 282.8 V
B. 211.2 V, 563.2 V, 352 V
C. 120 V, 320 V, 200 V
D. 45 V, 120 V, 75 V
3. Determine the power, power factor and a power-factor angle in no. 1. Draw phasor
diagram.
A. 2986. 37 W, 0.7070, 45°
B. 2400 W, 0.687, 35°
C. 2890 W, 0.8329, 55°
D. 2399.28 W, 0.7071, 45°
4. A voltage of 220 volts at 60 Hz is impressed on a circuit having a 50-ohm resistor, 25F capacitor, and a 0.2-henry inductor in series. Determine (a) impedance; (b)
current;
A. 38.67 Ω, 1.75 A
B. 48.67 Ω, 2.75 A
C. 58.67 Ω, 3.75 A
D. 68.67 Ω, 6.75 A
5. Find voltage across resistor, inductor, capacitor in no. 4
A. 200 V, 212.75 V, 297.88 V
B. 187.5 V, 282.75 V, 397.88 V
C. 198.8 V, 252.75 V, 297.88 V
D. 186.5 V, 292.75 V, 197.88 V
6. Determine the true power, power factor and power factor angle in no. 4. Draw phasor
diagram to scale.
A. 703.125 W, 0.8522, 31.55°
B. 603.125 W, 0.6522, 21.55°
C. 602.125 W, 0.7522, 61.55°
D. 763.125 W, 0.5572, 51.55°
7. A 15-ohm resistor, a 0.25-henry inductor, and a 100-F capacitor are connected in
series across 200-volt 25-Hz mains. Determine (a) current; (b) power;
A. 6 A, 635 W
228
B. 7 A, 735 W
C. 8 A, 835 W
D. 9 A, 435 W
8. In no. 7, find the voltage across resistor, inductor, capacitor;
A. 105 V, 274.89 V, 445.62 V
B. 100 V, 264.89 V, 465.42 V
C. 102 V, 174.54 V, 345.62 V
D. 125 V, 276.85 V, 446.62 V
9. In no. 7 determine the apparent power, power factor and power factor angle. Draw
phasor diagram to scale.
A. 1200 VA, 0.5640, 48.34°
B. 1300 VA, 0.4248, 58.14°
C. 1400 VA, 0.5248, 58.34°
D. 1500 VA, 0.8648, 38.34°
10. A 20-ohm resistor, a 0.200-henry inductor, and an unknown capacitor are connected
in series across 200-volt 25-Hz mains. Determine the capacitance of the capacitor so
that the circuit will take 67.86 W.
A. 15 µF
B. 20 µF
C. 25 µF
D. 30 µF
.
229
6.8
Parallel Resistance and Inductance
RL parallel circuits exist throughout the electrical field. Many circuits contain
both inductive and resistive loads connected to the same branch circuit. A ceiling
fan, for example, often has a light kit attached to the fan. The fan is an inductive
load, and the incandescent lamps are a resistive load. Branch circuits can also
contain incandescent lamps, fluorescent lights, and motors connected on the
same circuit. All these loads are connected in parallel.
Vs
R
L
40
VTm
30
20
ILm
IRm
10
vT
iR
0
0
1
0°
2
90°
3
180°
4
5
6
270°
7
8
360°
iL
-10
-20
-30
-40
IR
VT (ref)
IR
VT (ref)

IL
IT
Let ⃑⃑⃑⃑⃑
𝑉𝑇 = VT0; since this is a parallel circuit, VR = VL = VT = VT0;
230
𝐼𝑅 =
𝐼𝐿 =
𝑉𝑅
⃑⃑⃑⃑
𝐼𝑅 =
𝑅
𝑉𝐿
𝑉𝑇 0
⃑𝐼⃑⃑𝐿 =
𝑋𝐿
𝑅
𝑉𝑇 0
𝑋𝐿 90°
=
𝑉𝑇 0
𝑗X𝐿
Based on the phasor diagram above
⃑⃑⃑⃑
𝐼𝑇 = IR - jIL
the rectangular form of the total voltage
⃑⃑⃑⃑
𝐼𝑇 = IT
the polar form of the total voltage
where
IT = √𝐼𝑅 2 + 𝐼𝐿 2
 = tan-1
From
𝐼𝑅
𝐼⃑⃑⃑⃑
𝑇 = IR - jIL
⃑⃑⃑⃑⃑
𝐼𝑇
⃑⃑⃑⃑⃑⃑
𝑉𝑇
where
𝐼𝐿
𝐼𝑇
𝑉𝑇
𝐼𝑅
𝑉𝑅
𝐼𝐿
𝑉𝐿
=
𝐼𝑅
𝑉𝑅
−𝑗
Divide by VT or by the corresponding voltage,
𝐼𝐿
𝑉𝐿
= admittance (Y ) in siemens (S)
= conductance (G ) in siemens (S)
= inductive susceptance (BL) in siemens (S)
⃑⃑ = G - jBL
𝑌
⃑⃑ = Y
𝑌
polar form of admittance
Where Y = √𝐺 2 + 𝐵𝐿 2
 = tan-1
𝐵𝐿
𝐺
Power Factor: pf = cos lagging (the total current is lagging the total voltage in this
circuit by an angle ).
231
Admittance, Conductance, and Susceptance
Admittance – the reciprocal of admittance; the property of a circuit to allow the flow of
alternating current.
Y =
1
𝑍
Conductance – the reciprocal of resistance; the property of a circuit to allow the flow of
current.
G =
1
𝑅
Susceptance - the reciprocal of reactance.
BL – inductive susceptance; the ability of inductor to allow the flow of alternating
current
BC – inductive susceptance; the ability of a capacitor to allow the flow of
alternating current
BL =
.
1
𝑋𝐿
BC =
1
𝑋𝐶
232
PRACTICAL APPLICATION No. 4
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. Incandescent lighting of 500 W is connected in parallel with an inductive load. A
clamp-on ammeter reveals a total circuit current of 7 A. What is the inductance of
the load connected in parallel with the incandescent lights? Assume a voltage of 120
V and a frequency of 60 Hz.
.
233
6.9 Parallel Resistance and Capacitance
E
12 Vrms
60 Hz
0°
C
1.0µF
R
1.0kΩ
42
40
VTm
30
20
ICm
IRm
10
vT
0
-2
-1
0
90°
1
0°
2
90°
3
180°
4
5
270°
6
7
8
360°
vR
vC
-10
-20
-30
-40
IC
IC
IT

IR
VT (ref)
IR
VT (ref)
234
Let ⃑⃑⃑⃑⃑
𝑉𝑇 = VT0; since this is a parallel circuit, VR = VC = VT = VT0;
𝐼𝑅 =
𝐼𝐶 =
𝑉𝑅
⃑⃑⃑⃑
𝐼𝑅 =
𝑅
𝑉𝐶
𝑉𝑇 0
⃑⃑⃑⃑
𝐼𝐶 =
𝑋𝐶
𝑅
𝑉𝑇 0
𝑋𝐶 −90
=
𝑉𝑇 0
−𝑗𝑋𝐶
Based on the phasor diagram above
⃑⃑⃑⃑
𝐼𝑇 = IR + jIC
the rectangular form of the total voltage
⃑⃑⃑⃑
𝐼𝑇 = IT
the polar form of the total voltage
where
IT = √𝐼𝑅 2 + 𝐼𝐶 2
 = tan-1
From
𝐼𝐿
𝐼𝐶
IT = IR + jIC
⃑⃑⃑⃑⃑
𝐼𝑇
⃑⃑⃑⃑⃑⃑
𝑉𝑇
where
=
𝐼𝑇
𝑉𝑇
𝐼𝑅
𝑉𝑅
𝐼𝐶
𝑉𝐶
𝐼𝑅
𝑉𝑅
+𝑗
Divide by VT or by the corresponding voltage,
𝐼𝐶
𝑉𝐶
= admittance (Y ) in siemens (S)
= conductance (G ) in siemens (S)
= capacitive susceptance (BC) in siemens (S)
Y = G + jBC
rectangular form of admittance
Y = Y
polar form of admittance
where Y = √𝐺 2 + 𝐵𝐶 2
 = tan-1
𝐵𝐶
𝐺
Power Factor: pf = cos  leading (because the total current is leading the total voltage
by an angle )
235
6.10 Parallel Resistance, Inductance, and Capacitance
E
4
2
12 Vrms
60 Hz
0°
L
1.0mH
R
1.0kΩ
C
1.0µF
40
VTm
30
20
ILm
ICm
IRm
vT
10
vR
0
-2
-1
0
90°
1
0°
2
90°
3
180°
4
5
6
270°
7
8
360°
vC
vL
-10
-20
-30
-40
IC
IC
IT

IR
IL
VT (ref)
IR
IL
VT (ref)
236
Let ⃑⃑⃑⃑⃑
𝑉𝑇 = VT0; since this is a parallel circuit, VR = VL = VC = VT0;
𝐼𝑅 =
𝐼𝐿 =
𝐼𝐶 =
𝑉𝑅
⃑⃑⃑⃑
𝐼𝑅 =
𝑅
𝑉𝐿
⃑⃑⃑⃑
𝐼𝐿 =
𝑋𝐿
𝑉𝐶
⃑⃑⃑⃑
𝐼𝐶 =
𝑋𝐶
𝑉𝑇 0
𝑅
𝑉𝑇 0
𝑋𝐿 90
=
𝑉𝑇 0
𝑋𝐶 −90
=
𝑉𝑇 0
𝑗𝑋𝐶
𝑉𝑇 0
−𝑗𝑋𝐶
Based on the phasor diagram above
⃑⃑⃑⃑
𝐼𝑇 = 𝐼𝑅 + 𝑗𝐼𝐶 − 𝑗𝐼𝐿
⃑⃑⃑⃑
𝐼𝑇 = 𝐼𝑅
the rectangular form of the total voltage
+ 𝑗(𝐼𝐶 − 𝐼𝐿 )
⃑⃑⃑⃑
𝐼𝑇 = 𝐼𝑇 
the polar form of the total voltage
where
𝐼𝑇
= √𝐼𝑅 2 + (𝐼𝐶 − 𝐼𝐿 )2
𝐼𝑇
= √𝐼𝑅 2 + (𝐼𝑛𝑒𝑡 )2 where 𝐼𝑛𝑒𝑡 (𝑛𝑒𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡) = 𝐼𝐶 − 𝐼𝐿
 = 𝑡𝑎𝑛−1
From
𝐼𝑛𝑒𝑡
𝐼𝑅
⃑⃑⃑⃑
𝐼𝑇 = 𝐼𝑅 + 𝑗𝐼𝐶 − 𝑗𝐼𝐿 Divide by VT or by the corresponding voltage,
⃑⃑⃑⃑
𝐼𝑇
𝐼𝑅
𝐼𝐶
𝐼𝐿
=
+ 𝑗
− 𝑗
⃑⃑⃑⃑⃑
𝑉𝑅
𝑉𝐶
𝑉𝐿
𝑉𝑇
⃑⃑ = 𝐺 + 𝑗𝐵𝐶 − 𝑗𝐵𝐿
𝑌
⃑⃑⃑⃑⃑
𝐼𝑇
⃑⃑ admittance in siemens (S)
where ⃑⃑⃑⃑⃑⃑
= 𝑌
𝑉
𝑇
𝐼𝑅
𝑉𝑅
𝐼𝐶
𝑉𝐶
𝐼𝐿
𝑉𝐿
= G conductance in siemens (S)
= BC
capacitive susceptance in siemens (S)
= BL inductive susceptance in siemens (S)
237
⃑⃑ = 𝐺 + 𝑗𝐵𝐶 − 𝑗𝐵𝐿
𝑌
⃑⃑ = 𝐺 + 𝑗(𝐵𝐶 − 𝐵𝐿 )
𝑌
⃑⃑ = 𝐺 + 𝑗𝐵𝑛𝑒𝑡
𝑌
𝑤ℎ𝑒𝑟𝑒 𝐵𝑛𝑒𝑡 is the net susceptance
⃑⃑ = 𝑌
𝑌
where 𝒀 = √𝐺 2 + 𝐵𝑛𝑒𝑡 2
 = 𝑡𝑎𝑛−1
𝐵𝑛𝑒𝑡
𝐺
6.11 True Power, Reactive Power, and Apparent Power
𝑆⃑ = 𝑃 + 𝑗𝑄𝐿 − 𝑗𝑄𝐶
rectangular form of the apparent power
𝑆⃑ = 𝑃 + 𝑗(𝑄𝐿 − Q𝐶 )
where 𝑄𝑛𝑒𝑡 = 𝑄𝐿 − 𝑄𝐶 = net reactive power
𝑆⃑ = 𝑆
polar form of the apparent power
where 𝑆 = √𝑃2 + 𝑄𝑛𝑒𝑡 2
 = 𝑡𝑎𝑛-1
𝑄𝑛𝑒𝑡
𝑃
QL
P
QC
These directions of P, QL and QC are applicable to any type of circuit
238
Power factor: pf = cos leading (because the total current leads the total voltage by an
angle ).
Note: These formulas can be used even if BC < BL only that that in this case the power
factor is lagging.
Example 6.19 A 24-ohm resistor and a 0.0796-henry inductor are connected in parallel
across 115-volt 60-Hz mains. Determine (a) current in resistor; (b) current in
inductor; (c) total current; (d) power factor; (e) power-factor angle. (f) Draw phasor
diagram.
Solution
115 V
60 Hz
24 ohms
0.0796 H
VT = VR = VL = 115 V
XL = 2fL = 2(60Hz)(0.0796 H) = 30 
(a) 𝐼𝑅 =
(b) 𝐼𝐿 =
𝑉𝑅
𝑅
𝑉𝐿
𝑋𝐿
=
115 𝑉
24 
= 4.79 A
115 𝑉
=
30 
2
= 3.83 A
(c)
𝐼𝑇 = √𝐼𝑅 + 𝐼𝐿 2 = √(4.79)2 + (3.83)2 = 6.13 𝐴
or 𝐼⃑⃑⃑⃑
𝑇 = 𝐼𝑅 – 𝑗𝐼𝐿 = 4.79 – 𝑗3.83 𝐴. Converting to polar form IT = 6.13 A-38.65 
The magnitude of the total current is IT = 6.13 A
𝐼
3.83
(d)
 = tan-1 𝐼 𝐿 = tan-1 4.79
(e)
pf = cos = cos 38.65 = 0.78
(f)
Phasor diagram
𝑅
= 38.65
239
IR = 4.79 A
VT(ref)
 = 38.65
IL =
3.83 A
IT = 6.13 A
Example 6.20 An 80-ohm resistor and a 4.0-F capacitor are connected in parallel across
a 240 volt 400-Hz supply. Determine (a) current in resistor; (b) current in capacitor;
(c) total current; (d) power-factor angle; (e) power-factor. (f) Draw phasor diagram.
240 V
400 Hz
80 ohms
4 µF
VT = VR = VC = 115 V
𝑋𝐶 =
1
(c) 𝐼𝑅 =
(b)
𝐼𝑇
=
2𝑓𝐶
𝑉𝑅
𝑅
𝐼𝐶 =
=
𝑉𝐶
𝑋𝐶
1
2(400 𝐻𝑧)(4 𝐹)
240 𝑉
80 
=
= 99.47 
= 3A
240 𝑉
99.47
= 2.41 A
= √𝐼𝑅 2 + 𝐼𝐶 2 = √(3)2 + (2.41)2 = 3.85 𝐴
Or 𝐼⃑⃑⃑⃑
𝑇 = 𝐼𝑅 + 𝑗𝐼𝐶 = 3 + 𝑗2.41𝐴. Converting to polar form IT = 3.85 A38.78 
The magnitude of the total current is IT = 3.85 A
(g)
𝐼
 = tan-1 𝐼𝐶 = tan-1
𝑅
2.41
3
= 38.78
pf = cos = cos 38.78 = 0.78
240
(f) Phasor diagram
IC = 2.41 A
IT = 3.85 A
 = 38.78
IR = 3 A
VT(ref)
Example 6.21 A 25-ohm resistor, a 0.1-henry inductor, and a 160-F capacitor are
connected in parallel across 200-volt 25-Hz mains. Determine (a) current to resistor,
inductor, capacitor; (b) total current; (c) power-factor angle; (d) power factor. (e)
Draw vector diagram.
IR = 4.79 A
200 V
25 Hz
25 ohms
VT(ref)
0.1 H
160 µF
 = 38.65
Solution
VT = VR = VL = VC
IL =
XL = 2fL = 2(25 Hz)(0.1 H) = 15.71 
3.83 A
𝑋𝐶
(a)
𝑉𝑅
𝐼𝑅 =
𝑅
𝑉𝐿
𝐼𝐿 =
𝑋𝐿
𝐼𝐶 =
(b)
IT = 6.13 A
1
1
=
=
= 39.79 
2𝑓𝐶
2(25 𝐻𝑧)(160 𝐹)
𝑉𝐶
𝑋𝐶
=
200 𝑉
25 
=
=
= 8𝐴
200 𝑉
15.71 
200 𝑉
39.79 
= 12.73 𝐴
= 5.03 𝐴
𝐼𝑇 = √𝐼𝑅 2 + (𝐼𝐶 − 𝐼𝐿 )2 = √(8)2 + (5.03 − 12.73)2 = 11.10 𝐴
or
⃑⃑⃑⃑
𝐼𝑇 = 𝐼𝑅 + 𝑗𝐼𝐶 − 𝑗𝐼𝐿
Converting to polar form
⃑⃑⃑⃑
𝐼𝑇
= 8 + 𝑗5.03 − 𝑗12.73 =
= 11.10 − 43.91
8 – 𝑗7.7 𝐴
241
so ,
𝐼𝑇
= 11.10 𝐴
(c)  = 43.91
(d) pf = cos 43.91 = 0.72
IL = 5.03 A
IR = 8 A
VT (ref)
 = 43.79
Inet = 7.7 A
IT = 11.10 A
6.12 Conversion of Parallel or Combination Circuits to Series Circuit
Example 6.22
Transform the circuit below to series circuit:
Vs= 110 V
60 Hz
50 ohms
1
XL = 48.63 
BL =
XC = 22.10 
BC = 22.10
R = 50 
G =
48.63
1
1
50
L = 129 mH
120 µF
= 20.56 mS
= 45.25 mS
= 20 mS
⃑⃑ = G + jBC - jBL = 20 mS + j45.25mS - j20.56 = 20 mS + j24.69 mS
𝑌
Converting to polar form
242
⃑⃑ = 31.774 mS 50.99
𝑌
where 𝑍⃑ =
𝟏
⃑⃑
𝑌
=
1
31.774 𝑚𝑆 50.99
= 31.47  − 50.99
Converting Z to rectangular form
𝑍⃑ = 19.81 − 𝑗 24.45 
In this form R = 19.81  and XC = 24.45  (capacitive reactance because the operator
is –j)
To find the capacitance;
1
1
𝐶 =
=
= 108.49 𝐹
2𝑓𝑋𝐶
2(60)(24.45)
Therefore, the equivalent series of the parallel RLC above is a series components that
consists R = 19.81  and capacitance of C = 108.49 F
19.81 ohms
Vs= 110 V
60 Hz
108.49 µF
Example 6.23 Determine the series components of an impedance 60 60 connected
across a 60-Hz source.
Given 𝑍⃑ = 60 60
Solution
Convert to rectangular form Z = 30 + j51.96 
From this, R = 30  and XL = 51.96 (inductive reactance because the operator is +j)
For 𝐿 =
𝑋𝐿
2𝑓
=
51.96
260
= 137.83 𝑚𝐻
Therefore the series components are R = 30  and L = 137.83 Mh
.
243
Assessment No. 17
PARALLEL RESISTANCE, INDUCTANCE , AND CAPACITANCE
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. A 25-  resistor and a 300-mH inductor are connected in parallel across a 120-V
50-Hz source. Determine the magnitude of the following: (a) admittance; (b)
current through resistor and inductor; (c) the total current; (d) true power and
inductive reactive power; (e) apparent power, power factor, and power factor
angle.
244
2. A 25-  resistor and a 40-µF capacitor are connected in parallel across a 120-V 50Hz source. Determine the magnitude of the following: (a) admittance; (b) current
through resistor and capacitor; (c) the total current; (d) true power and capacitive
reactive power; (e) apparent power, power factor, and power factor.
245
3. A 25-  resistor, a 300-mH and a 40-µF capacitor are connected in parallel across a
120-V 50-Hz source. Determine the magnitude of the following: (a) admittance; (b)
current through resistor, inductor and capacitor; (c) the total current; (d) true
power, inductive and capacitive reactive power; (e) apparent power, power factor,
and power factor.
246
4. A resistor and an inductor are connected in parallel to a 120-V 60-Hz source. If the
total current ad power in the circuit are, respectively, 17 amp and 1,800 watts,
calculate the values of R and L.
5. A resistor and an inductor are connected in parallel to a 120-V 60-Hz source. The
total current ad power in the circuit are, respectively, 17 amp and 1,800 watts. If a
177-F capacitor is connected in parallel with the circuit, calculate the values of R
and L.
247
Problem Set No. 12
PARALLEL CIRCUITS
1.
2.
3.
4.
5.
6.
A 40-ohm resistor and an inductor are connected in parallel across 120-volt 50-Hz
mains, and the total current is 6.0 amp. Determine the inductance.
A. 53.46 mH
B. 63.46 mH
C. 73.46 mH
D. 83.46 mH
A circuit consists of a resistor and a 53.0-F capacitor in parallel across 120-volt
60-Hz mains. The total current is 4 amp. Determine ohms of resistor.
A. 22.5 Ω
B. 27.5 Ω
C. 37.5 Ω
D. 39.5 Ω
A 50-ohm resistor, an 80-ohm inductive reactor, and a 60-ohm capacitive reactor are
connected in parallel across 240-volt 60-Hz mains. Determine current to resistor,
inductor, capacitor and total current.
A. 2.8 A, 4 A, 4 A, 4.5 A
B. 4.9 A, 3.5 A, 5 A, 9.9 A
C. 3.8 A, 5 A, 2 A, 5.9 A
D. 4.8 A, 3 A, 4 A, 4.9 A
Find the power-factor angle and power factor of the circuit in no. 3. Draw phasor
diagram.
A. 11.77°, 0.979
B. 12.77°, 0.949
C. 13.77°, 0.879
D. 14.77°, 0.779
A 30-ohm resistor and a 0.0637-henry inductor are connected in parallel across a
120-V ac supply. Determine (a) frequency at which total current will be 7.22 amp;
(b) power factor of the circuit. (c) Draw phasor diagram.
A. 60 Hz, 0.354
B. 50 Hz, 0.554
C. 100 Hz, 0.654
D. 25 Hz, 0.754
A 25-ohm resistor and an unknown capacitor are connected in parallel across a 100volt 50-Hz supply, and the total current is 4.75 amp. When the resistor and capacitor
are connected across a 100-volt supply of unknown frequency, the total current is
6.41 amp. Determine frequency.
A. 26 Hz
B. 54 Hz
C. 60 Hz
D. 98 Hz
248
Unit 7
Combination Circuits and
Network Theorems
LEARNING OUTCOMES
At the end of the lesson,
you are expected to:
1. solve the required parameters of a series/parallel
combination circuit.
2. analyze a circuit using the network theorems such as
Maxwell’s Mesh Equations, Nodal Analysis, Thevenin’s
Theorem, Norton’s Theorem, Superposition Theorem, and
Millman’s Theorem
7.1 Combination Circuits
Example 7. 1. A non-inductive resistor and an impedance coil are connected in parallel
across 208-volt 60-Hz mains. The resistor takes a current of 2.25 amp; the impedance
coil takes a current of 1.5 amp; the total current is found to be 3.1 amp. Determine (a)
power-factor angle and power factor of circuit; (b) power to circuit; (c) power-factor
angle and power factor of impedance coil; (d) effective resistance of impedance coil; (e)
inductance of impedance coil.
249
3.1 A
Rcoil = ?
208 V
60 Hz
R=?
Lcoil =?
IR = 2.25 A
Icoil = 1.5 A
The equivalent components of impedance coil are resistance and inductance.
Let ⃑⃑⃑⃑⃑
𝑉𝑇 = 208 V0
IR = 2.25 A
circuit
coil
VT (coil)
coil
equal to Icoil
Icoil = 1. 5 A
IT = 3.1 A
To find circuit , let us consider this triangle
IR = 2.25 A
circuit
IT = 3.1 A
Icoil = 1.5 A
Using cosine law,
(1.5)2 = (2.25)2 + (3.1)2 – 2(2.25)(3.1)coscircuit
circuit = 27.06
250
(a) circuit = 27.06; pf = cos circuit = cos 27.06 = 0.89 lagging
(b) Let us find S: S = VTIT = (208)(3.1) = 644.8 VA
The figure below is the corresponding power triangle of the circuit since the current is
lagging.
S = 644.8 VA
QL = ?
 = 27.06
cos  =
𝑃
𝑆
P = ?(0.89) = 573.87 watts
and P = S cos = Spf = 644.8
(c) To find coil, consider again this triangle
IR = 2.25 A
Using cosine law,

coil
IT = 3.1 A
Icoil = 1.5 A
(3.1)2 = (2.25)2 + (1.5)2 - 2(2.25)(1.5)cos
 = 109.9
Now coil = 180 -  = 180 - 109.9 = 70.1
pfcoil = coscoil = cos 70.1 = 0.34
(d) To find VRcoil and VLcoil , let us consider this triangle
Vcoil = 208 V
VL coil
coil = 70.1
VR coil
251
𝑉𝑅𝑐𝑜𝑖𝑙 = 𝑉𝑐o𝑖𝑙 𝑐𝑜𝑠𝑐𝑜𝑖𝑙
𝑉𝐿𝑐𝑜𝑖𝑙
𝑅𝑐𝑜𝑖𝑙 =
𝑋𝐿 𝑐𝑜𝑖𝑙 =
and 𝐿 =
𝑉𝐿𝑐𝑜𝑖𝑙
𝐼𝑐𝑜𝑖𝑙
𝑋𝐿
2𝑓
𝑉𝑅𝑐𝑜𝑖𝑙
𝐼𝑐𝑜𝑖𝑙
=
=
= 208 𝑐𝑜𝑠 70.1 = 70.8 𝑉
= 𝑉𝑐𝑜𝑖𝑙 𝑠𝑖𝑛𝑐𝑜𝑖𝑙 = 208 𝑠𝑖𝑛 70.1 = 195.58 𝑉
=
70.8 𝑉
1.5 𝐴
195.58 𝑉
1.5 𝐴
130.39
2(60)
= 47.2 
= 130.39 
= 0.3459 𝐻
Example 7.2 A unity power-factor load of 1,794 watts is connected in parallel with a
load of 1,656 watts operating at a lagging power factor of 0.6. If the line voltage is 115,
calculate (a) the current in each load, (b) the load current, (c) the total power, (d) the
overall power factor, (e) the reactive volt-amperes.
Note: Let
I1 = 15.6 A
1 = 53.13°
I2 = 24 A
Solution:
Load 2
Load 1
Vs = 115 V
VT (ref)
IT = ?
252
Load 1: unity power factor – as defined a load with unity power factor is a purely
resistive load.
P1 = IR1 VR1
where P1 = 1,794 watts
VR1 = VS = 115 V, since resistance is the only component in load 1
IR = ?
𝐼1 =
𝑃1
𝑉1
=
1,794 𝑊
115 𝑉
= 15.6 𝐴
Load 2: lagging power factor ; the components are resistor and inductor.
P2 = 1,656 watts
pf = 0.6, with this 2 = cos-1 0.6 = 53.13
The power triangle for load 2
QL2 = ?
S2 = ?
 = 53.13
P = 1,656 W
𝑆2 =
To find S2:
𝑃2
𝑐𝑜𝑠 2
=
1,656 𝑊
𝑐𝑜𝑠 53.13
= 2,760 𝑉𝐴
S2 = I2V2
𝐼2 =
𝑆2
𝑉2
=
2,760 𝑉𝐴
115 𝑉
= 24 𝐴
I2 = 24 A-53.13 (negative because this current is lagging the total voltage)
QL2 = S2 sin 2
= 2760 sin 53.13 = 2,208 VAR
For the total current
IT = I1 + I2
= 15.60 + 24 A-53.13
= (15.6 + j0) + (14.4 – j19.2)
= 30 – j19.2
253
= 35.62-32.62
The equivalent power triangle of the circuit:
QL = 2,208 VAR
T = ?
P1 = 1,794 watts
P2 = 1,656 watts
PT = 3,450 watts
PT = 3,450 W
T = tan-1
2,208
3,450
= 32.62
QL2 = QT = 2,208 VARs
⃑⃑⃑⃑⃑1 = 50 + j75, ⃑⃑⃑⃑⃑
Example 7.3 In the circuit shown below, let 𝑍
𝑍2 = 80 + j25, ⃑⃑⃑⃑⃑
𝑍3 = 100
+ j70, determine :
a.
b.
c.
d.
e.
f.
the total impedance
the total current
the total power
the total reactive power
apparent power
the current in each impedance
75 ohms
50 ohms
I1
80 ohms
I3
100 ohms
120 V
60 Hz
I2
25 ohms
70 ohms
254
a.
⃑⃑⃑⃑⃑
𝑍𝑇 = ⃑⃑⃑⃑⃑
𝑍1 +
⃑⃑⃑⃑⃑
𝑍2 ⃑⃑⃑⃑⃑
𝑍3
⃑⃑⃑⃑⃑
⃑⃑⃑⃑⃑
𝑍
2+ 𝑍
3
(80+𝑗25)(100−𝑗70)
= (50 + 𝑗 75) + (80+𝑗25)+ (100−𝑗70) = 105.03 + 𝑗71.536 
= 127.0775 34.2588
⃑⃑⃑⃑⃑⃑
𝑉𝑇
⃑⃑⃑⃑⃑⃑
𝑍𝑇
1200
b.
𝐼⃑⃑⃑⃑𝑇 =
c.
𝑃 = 𝐼𝑇 2 𝑅𝑇 = (0.9443)2 (105.03) = 93.656 𝑊
d.
𝑸 = 𝐼𝑇 2 𝑋𝐶 = (0.9443)2 (71.536) = 63.789 𝑘𝑉𝐴𝑅 (㄰𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒)
e.
𝑺 = 𝐼𝑇 𝑉𝑇 = (0.9443)(120) = 113.315 VAR
f.
⃑⃑⃑
𝐼1 = 0.9443 𝐴 − 34.2588
=
127.0775 34.2588
= 0.9443 𝐴 − 34.2588
By current divider
100−𝑗70
⃑⃑⃑⃑
) = 0.6212 𝐴  − 55.2146
𝐼2 = (0.9443 𝐴 − 34.2588) (
180−𝑗45
80+𝑗25
⃑⃑⃑⃑
) = 0.4265 𝐴  − 2.8686
𝐼3 = (0.9443 𝐴 − 34.2588) (
180−𝑗45
.
255
Assessment No. 18
COMBINATION CIRCUITS
Name:
1.
Score: _________ Rating: ______
Aviso, Myrell Jud A.
In order to measure the power taken by a small 120-V 60-Hz single-phase induction
motor, it is connected in parallel with a non-inductive resistor across 120-V 60-Hz
mains. The currents measured are as follows: resistor current, 3.0 amp; motor
current, 4.2 amp; total current, 6.7 amp. Determine (a) power factor of circuit; (b)
power factor of motor; (c) power to motor; (d) total power to circuit.
IT = 6.7 A
R motor = ?
R=?
120 V
60 Hz
L motor
Im =
IR =
4.2 A
3A
256
2.
A parallel circuit consisting of a resistor, an impedance coil, and a capacitor of
negligible loss is connected across a 100-volt 25-Hz supply and takes a current of
3.1 amp. The current to the resistor is 2.5 amp, that of the capacitor 2.0 amp and
that to the impedance coil, 2.8 amp, Determine (a) power-factor angle and power
of entire circuit; (b) power factor and power-factor angle of impedance coil; (c)
resistance of impedance coil; (d) reactance and inductance of impedance coil.
257
3.
A parallel circuit consisting of a non-inductive resistor, an impedance coil, and a
capacitor of negligible loss is connected across 120-V 60-Hz mains and takes 6 amp
lagging current at a power factor of 0.936. The resistor takes 4 amp., and the
capacitor 3 amp. Determine (a) power to the impedance coil; (b) current to
impedance coil; (c) power-factor angle and power factor of impedance coil.
258
4.
The following information is given in connection with three loads connected to a
220-V source: load A = 12 A at unity power factor; load B is 40 amp at a power
factor of 0.5 lagging; load C is 20 amp at a leading power factor of 0.9. Calculate (a)
the total current, (b) the total power, (c) the reactive volt-amperes, (d) the
overall power factor.
259
5.
With a secondary load, the primary of a telephone induction coil, having an effective
resistance of 240 ohms and an inductance of 0.01583 henry, is in parallel with a 2F capacitor of negligible resistance. With 50 volts at a frequency of 1,000 Hz across
the primary, determine (a) current to primary; (b) current to capacitor; (c) line, or
total, current; (d) power to entire circuit; (e) phase angle between line current and
voltage.
260
6.
The currents at the junction point in a circuit have the following values I1 = (16 –
j4) A and (8 + j22) A. Calculate (a) I1 + I2, (b) I1 – I2, and (c) I2 – I1.
7.
Calculate the equivalent impedance of a circuit in which a coil of wire having a
value of 5 53.2 is connected parallel with a capacitive reactance of 6.25 ohms.
8.
A coil of wire having a value of (5 + j8) ohms is connected in series with a
capacitive reactance XC, and this series combination is then connected in parallel
with a resistor R. If the equivalent impedance of the circuit is 4 0, calculate the
values of XC and R.
.
261
Objective Test No. 6
ALTERNATING CURRENT CIRCUITS
1. A sine wave voltage is applied across an inductor. When the frequency of the voltage
is increased, the current
A. decreases
B. increases
C. does not change
D. momentarily goes to zero
2. An inductor and a resistor are in series with a sine wave voltage source. The
frequency is set so that the inductive reactance is equal to the resistance. If the
frequency is increased then
A. VR > VL
B. VL < VR
C. VL = VR
D. VL > VR
3. A sine wave voltage is applied across a capacitor. When the frequency of the voltage
is increased, the current
A. increases
B. decreases
C. remains constant
D. ceases
4. A capacitor and a resistor are connected in series to a sine wave generator. The
frequency is set so that the capacitive reactance is equal to the resistance and, thus,
an equal amount of voltage appears across each component. If the frequency is
decreases,
A. VR > VL
B. VL < VR
C. VL = VR
D. VL > VR
5. In an ac circuit, the voltage _____________.
A. leads the current
B. lags the current
C. is in phase with the current
D. is any of the above, depending on the component
6. The voltage lags behind the current by ¼ cycle in a ___________ circuit.
A. pure capacitor
B. pure inductor
C. pure resistor
D. capacitance with inductor
7. The unit of inductive reactance is the _________________.
A. henry
262
B. Tesla
C. farad
D. ohm
8. When voltage and current are in phase in an ac circuit, the ____________.
A. impedance is zero
B. resistance is zero
C. resistance is zero
D. phase angle is 90
9. The impedance of a circuit does not depend on ____________.
A. current
B. frequency
C. resistance
D. capacitance
10. The power dissipated as heated in an ac circuit depends on its ____________.
A. resistance
B. capacitive reactance
C. inductive reactance
D. impedance
11. The power factor of a circuit in which XL = XC ____________.
A. is zero
B. is one
C. depends on the ratio XC/XL
D. depends on the value of R
12. The inductive reactance of a 1-mH coil in a 5-kHz circuit is ____________.
A. 3.1 
B. 6.3 
C. 10 
D. 31 
13. The capacitive reactance of a 5 F capacitor in a 20 kHz circuit is __________.
A. 0.63 
B. 1.6 
C. 5 
D. 16 
14. A 2-F capacitor is connected to a 50-V, 400-Hz power supply. The current that
flows is
A. 0.20 mA
B. 0.25 A
C. 0.30 A
263
D. 0.40 A
15. In a series ac circuit R = 10 , XL = 8 , and XC = 6  when the frequency is f. The
impedance at this frequency is __________.
A. 10.2 
B. 12 
C. 24 
D. 104 
16. The total voltage in a series RL circuit ________ the current by an angle _____.
A. lags, of 90
B. lags between 0 to 90
C. leads, between 0 to 90
D. leads, between 90 and 180
17. In a series RL circuit, the inductor current _____ the resistor current.
A. lags
B. leads
C. is equal
D. is negative
18. The impedance triangle is similar to the __________ triangle with the resistance phasor
in place of the ____________.
A. current, resistor current
B. current, resistor voltage
C. voltage, impedance
D. voltage, resistor voltage
19. In the impedance triangle the inductive reactance and impedance phasor are
analogous to the ______ and _______ phasor respectively in the voltage triangle.
A. inductive voltage, total voltage
B. inductive current, total current
C. inductive voltage, resistive voltage
D. inductive current, resistive current
20. In a series RL circuit phasor diagram, total voltage may be represented by the _____
phasor and the resistor voltage may be represented the ___________ phasor.
A. current, voltage
B. current resistance, current
C. impedance, resistance
D. impedance, inductance
21. The phase angle of a series RL circuit is the angle between the ________ phasor and the
_________ phasor.
A. resistance, inductive reactance
264
B. resistance, impedance
C. inductive reactance, impedance
D. none of these
22. The phase angle of series RL circuit may be computed as ______ or ____ or ___________.
A. cos-1R/XL, sin-1XL/R, tan-1R/Z
B. cos-1R/Z, sin-1XL/R, tan-1R/XL
C. cos-1Z/XL, sin-1R/Z, tan-1XL/R
D. cos-1R/Z, sin-1XL/Z, tan-1 XL/R
23. A(n) ____ stores and returns energy to a circuit while a(n) _________ dissipates energy.
A. resistor, impedance
B. resistor, inductor
C. inductor, resistor
D. inductor, reactance
24. For an RL circuit, the power factor cannot be less than ______ or greater than ________.
A. 0, 1
B. 1,0
C. 0, -1
D. -1, 0
25. The voltage across the capacitor _________ the current through it by ______.
A. lags, 45
B. lags, 90
C. leads, 0
D. leads, 90
26. If the resistance in a series RC circuit is increased the magnitude of the phase angle
__.
A. increases
B. remains the same
C. decreases
D. changes in an indeterminate manner
27. In a series RC circuit, the current _______ the total voltage by an angle _______.
A. lags, of 45
B. lags, of 0
C. leads, between 0 to 90
D. leads, of 90
28. The resistance phasor of a series circuit points to the right. The capacitor reactance
phasor points ______ while the diagonal of the rectangle having there two phasors as
sided represents their ________.
A. up, impedance
265
B. down, impedance
C. left, current
D. up, total voltage
29. The phase angle for a series RC circuit is defined as the angle between the _______ and
the _______ phasors.
A. current, resistance voltage
B. current, total voltage
C. resistance voltage, capacitor voltage
D. R, XC
30. The phase angle for a series RC circuit may be computed as the angle between the
_______ and _________ phasors.
A. resistance, impedance
B. reactance, impedance
C. reactance, impedance
D. none of these
31. The power dissipated in a series RC circuit with R = 10 ohms and XC = 10 ohms
carrying an effective current of 3 amperes is ________ watts.
A. 30
B. 30/2
C. 90
D. 90/2
32. The magnitude of the power factor of an RC circuit with R = 10 ohms and XC = 10
ohms, 1.2 amp effective is
A. 1
B. 0.5
C. 0.707
D. 0
33. The magnitude of the power factor of an RC circuit with R = 30 ohms, XC = 40 ohms
and E = 100 V effective is
A. 60
B. 80
C. 100
D. 120
34. The net reactance in an RLC circuit is
A. 𝑋𝐿
B. 𝑋𝑐
C. 𝑋𝑐 + 𝑋𝐿
D. 𝑋𝐿 – 𝑋𝑐
35. The impedance of a series RLC circuit is
266
A. √𝑅 2 + 𝑋𝐿 2 + 𝑋𝑐 2
B. √𝑅 2 + 𝑋𝐿 2 + 𝑋𝑐 2
C. √𝑅 2 + (𝑋𝐿 2 − 𝑋𝑐 2 )
D. √𝑅 2 + (𝑋𝐿 − 𝑋𝑐 )2
.
2
267
7.2
Network Theorems
To recall Kirchhoff’s laws solve for the current in each impedance in the circuit below
so that you will have a basis for the answers in the network theorems. Remember that
whatever network theorem you use will arrive at the same result.
I1
75 ohms
a
50 ohms
I3
80 ohms
100 ohms
I
120 V
60 Hz
II
IB
I2
IA
25 ohms
70 ohms
⃑⃑⃑⃑𝑆 = 120 V0
Let 𝑉
By KVL at loop I
120 0 − (50 + 𝑗75) ⃑⃑⃑
𝐼1 − (80 + 𝑗25) ⃑⃑𝐼⃑2⃑ = 0
(50 + 𝑗75)𝐼⃑⃑⃑1 + (80 + 𝑗25) ⃑⃑𝐼⃑2⃑ = 120
equation 1
By KVL at loop 2
(80 + 𝑗25) ⃑⃑⃑⃑
𝐼2 − (100 − 𝑗70) ⃑⃑𝐼⃑3⃑ = 0
equation 2
By KCL at node a
⃑⃑⃑
𝐼1 − ⃑⃑𝐼⃑2⃑ − ⃑⃑⃑⃑
𝐼3 = 0
equation 3
Equating (1) , (2) and (3)
(50 + 𝑗75)𝐼⃑⃑⃑1 + (80 + 𝑗25) ⃑⃑𝐼⃑2⃑
= 120
(80 + 𝑗25) ⃑⃑𝐼⃑2⃑ − (100 − 𝑗70) ⃑⃑⃑⃑
𝐼3 = 0
⃑⃑⃑
𝐼1
Then using determinant method
− 𝐼⃑⃑⃑⃑2
− ⃑⃑𝐼⃑3⃑ = 0
268
(50 + 𝑗75)
𝐷= [
0
1
120
[ 0
0
⃑⃑⃑
𝐼1 =
(80 + 𝑗25)
(80 + 𝑗25)
−1
0
− (100 − 𝑗70)] = −22,125 − 𝑗8,150
−1
(80 + 𝑗25)
0
(80 + 𝑗25) − (100 − 𝑗70)]
−21,600 + 𝑗5,400
−1
−1
=
=
−22,125 − 𝑗8,150
−22,125 − 𝑗8,150
= 0.7805 − 𝑗0.5316 = 0.9443 − 34.26
(50 + 𝑗75) 120
0
[
0
0
− (100 − 𝑗70)]
−12,000 + 𝑗8,400
1
0
−1
⃑⃑𝐼⃑2⃑ =
=
=
−22,125 − 𝑗8,150
−22,125 − 𝑗8,150
= 0.3544 − 𝑗0.5102 = 0.6214 − 55.21
[
⃑⃑𝐼⃑3⃑ =
(50 + 𝑗75) (80 + 𝑗25)
0
(80 + 𝑗25)
1
−1
−22,125 − 𝑗8,150
120
0 ]
0 = −21,600 + 𝑗5,400 =
−22,125 − 𝑗8,150
= 0.4265 𝐴 − 2.86
To find the apparent power,, reactive power and true power of the circuit the formula
below is used
⃑⃑T ⃑IT̅
S⃑⃑T = V
This is called the conjugate method of finding power.
where:
⃑S⃑T = the total apparent power in volt-ampere (VA)
⃑V⃑T = the total voltage
⃑IT̅ = the conjugate of the total current
In the circuit above ⃑⃑⃑⃑⃑
VT = 120 V0,
⃑IT̅ = 0.944334.26
conjugate is
⃑IT = ⃑I1 = 0.9443 − 34.26 and its
269
⃑⃑ST = ⃑V⃑T ⃑IT̅
= (120 V0)(0.944334.26)
= 113.32 34.26
In rectangular form,
⃑⃑T ⃑IT̅
S⃑⃑T = V
= 93.66 + j63.79
Therefore S = 113.32 VA, P = 93.66 W, Q = 63.79 VAR (inductive).
7.2.1 Maxwell’s Mesh Equations
Example 7.3 In the circuit below, determine the current I1, I2, and I3. Determine also
the
Solution:
75 ohms
50 ohms
80 ohms
100 ohms
I
120 V
60 Hz
II
IB
25 ohms
70 ohms
IA
Let
⃑⃑⃑⃑𝑆 = 120 V0
𝑉
At mesh I
1200 − ⃑⃑⃑⃑
𝐼𝐴 (50 + 𝑗75)– ⃑⃑⃑⃑
𝐼𝐴 (80 + 𝑗25) + ⃑⃑⃑⃑
𝐼𝐵 (80 + 𝑗25) = 0
270
120 + 𝑗0 − ⃑⃑⃑⃑
𝐼𝐴 [(50 + 𝑗75) + (80 + 𝑗25)] + ⃑⃑⃑⃑
𝐼𝐵 (80 + 𝑗25) = 0
120 – ⃑⃑⃑⃑
𝐼𝐴 (130 + 𝑗100) + ⃑⃑⃑⃑
𝐼𝐵 (80 + 𝑗25) = 0
⃑⃑⃑⃑
(130 + 𝑗100)𝐼⃑⃑⃑⃑
𝐴 − (80 + 𝑗25)𝐼𝐵 = 120 equation 1
At mesh II
− ⃑⃑⃑⃑
𝐼𝐵 (80 + 𝑗25) + ⃑⃑⃑⃑
𝐼𝐴 (80 + 𝑗25) − ⃑⃑⃑⃑
𝐼𝐵 (100 – 𝑗70) = 0
− ⃑⃑⃑⃑
𝐼𝐵 [(80 + 𝑗25) + (100 – 𝑗70)] + ⃑⃑⃑⃑
𝐼𝐴 (80 + 𝑗25) = 0
− ⃑⃑⃑⃑
𝐼𝐵 (180 – 𝑗45) + ⃑⃑⃑⃑
𝐼𝐴 (80 + 𝑗25) = 0
⃑⃑⃑⃑
(80 + 𝑗25)𝐼⃑⃑⃑⃑
𝐴 − (180 – 𝑗45)𝐼𝐵 = 0
equation 2
Equate equations 1 and 2
⃑⃑⃑⃑
(130 + 𝑗100)𝐼⃑⃑⃑⃑
𝐴 − (80 + 𝑗25)𝐼𝐵 = 120
⃑⃑⃑⃑
(80 + 𝑗25)𝐼⃑⃑⃑⃑
𝐴 − (180 – 𝑗45)𝐼𝐵 = 0
Using determinant method
𝐷= [
(130 + 𝑗100)
(80 + 𝑗25)
− (80 + 𝑗25)
] = −22,125 − 𝑗8150
− (180 – 𝑗45)
120 − (80 + 𝑗25)
[
]
−21,600 + 𝑗5,400
0
−
(180
–
𝑗45)
⃑⃑⃑⃑
𝐼𝐴 =
=
= 0.7805 − 𝑗0.5316 𝐴
−22,125 − 𝑗8150
−22,125 − 𝑗8150
= 0.9443 𝐴 − 34.2589
(130 + 𝑗100) 120
]
−9,600 − 𝑗3,000
(80 + 𝑗25)
0
⃑⃑⃑⃑
𝐼𝐵 =
=
= 0.426 − 𝑗0.0213 𝐴
−22,125 − 𝑗8150
−22,125 − 𝑗8,150
[
= 0.4265 𝐴 − 2.862
To find 𝐼1 , 𝐼2 , and 𝐼3
⃑⃑⃑
𝐼1 = ⃑⃑⃑⃑
𝐼𝐴 = 0.9443 𝐴 − 34.2589
⃑⃑𝐼⃑2⃑ = ⃑⃑⃑⃑
𝐼𝐴 − ⃑⃑⃑⃑
𝐼𝐵 = (0.7805 − 𝑗0.5316) − (0.426 − 𝑗0.0213 𝐴) = 0.3545 − 𝑗0.5103
271
= 0.6214 − 55.2127
𝐼⃑⃑⃑⃑3 = 0.4265 𝐴 − 2.862
For self-assessment no. 19, solve the load current in each network using the network
theorem being required to apply.
R1
L1
150mH
10Ω
E
L
O
A
D
120 Vrms
60 Hz
0°
R
50Ω
R3
25Ω
L
C1
30µF
250mH
(a )
R1
10Ω
E
120 Vrms
60 Hz
30°
L1
150mH
L
O
A
D
R
50Ω
L
250mH
(b)
R3
C1
25Ω
30µF
E1
150 Vrms
60 Hz
-45°
272
Assessment No. 19
NETWORK THEOREMS
Name:
Aviso, Myrell Jud A.
7.2.2 Mesh Analysis
Score: _________ Rating: ______
273
7.2.3 Nodal Analysis
274
7.2.4 Thevenin’s Theorem
275
7.2.5 Norton’s Theorem
276
7.2.6 Superposition Theorem
277
7.2.7 Millman’s Theorem
.
278
Unit 8
Power Factor Correction or Improvement
LEARNING OUTCOMES
At the end of the lesson,
you are expected to:
1. discuss power factor correction or improvement.
2. determine sizes of capacitors to improve or correct the
power factor of a system.
279
Introduction
Most industries use a large number of electric motors; therefore, industrial plants
represent highly inductive loads. This means that industrial power systems operate at a
power factor of less than unity (1.0). However, it is undesirable for an industry to operate
at a low-power factor, since the electrical power system will have to supply more power
to the industry than is actually used.
A given value of volt-amperes (voltage x current) is supplied to an industry by the
electrical power system. If the power factor (pf) of the industry is low, the current must
be higher, since the power converted by the total industrial load equals VA x pf. The value
of the power factor decreases as the reactive power (unused power) drawn by the
industry increases. This is shown in Figure 8.1. We will assume a constant value of true
power, in order to see the effect of increases in reactive power drawn by a load. . The
smallest reactive power shown (VAR1) results in the volt-ampere value of VA1. As reactive
power is increased, as shown by the VAR2 and VAR3 values, more volt-amperes (VA2 and
VA3) must be drawn from the source. This is true sine the voltage component of the
supplied volt-amperes remains constant. This example represents the same effect as a
decrease in the power factor, since pf = W/VA, and, as VA increases, the pf will decrease
if W remains constant.
Utility companies usually charge industries for operating at power factors below a
specified level. It is desirable for industries to “correct” their power factor to avoid such
charges to make more economical use of electrical energy. Two methods may be used to
cause the power factor to increase: (1) power-factor corrective capacitors, and (2) threephase synchronous motors. Since the effect of capacitive reactance is opposite to that of
inductive reactance, their reactive effects will counteract one another. Either powerfactor-corrective capacitors, or three-phase synchronous motors, may be used to add the
effect of capacitance to an AC power line.
Q3 (vars)
S3
Q2 (vars)
3
S2
2
S1
Q1 (vars)
1
PT (watts)
Figure 8.1. Effect of increases in reactive power (VAR) on apparent power (VA).
280
An example of power factor of power factor correction
is shown in Figure 8-2. We will assume from the
example that both true power and inductive reactive power remain constant at values of
10 kW and 10 kVAR. In Figure 8-2(a), the formulas show that the power factor equals 70
per cent. However, if 5-kVAR capacitive reactive power is introduced into the electrical
power system, the net reactive power becomes 5 kVAR (10 kVAR inductive minus 5 kVAR
capacitive), a shown in Figure 8-2(b). With the addition of 5 kVAR capacitive to the
system, the power factor is increased to 89 per cent. Now, in Figure 8-2(c), if 10-kVAR
capacitive is added to the power system, the total reactive power (kVAR) becomes zero.
The true power is now equal to the apparent power; therefore, the power factor is 1.0, or
100 per cent, which is characteristic of a purely resistive circuit. The effect of the
increased capacitive reactive power in the system is to increase or “correct” the power
factor and, thus, to reduce the current drawn from the power distribution lines that
supply the loads. In many cases, it is beneficial for industries to invest in either powerfactor-corrective capacitors, or three-phase synchronous motors to correct their power
factor.
Capacitors for Power Factor Correction
Static capacitors are used for power factor correction in the system. They are constructed
similarly to the smaller capacitors used in electrical equipment, which have metal-foil
plates separated by paper insulation. Ordinarily, static capacitors are housed in metal
tanks, so that the plates can be immersed in an insulating oil to improve high-voltage
operation. The usual operating voltages of static capacitors are from 230 volts to 13.8
kilovolts. These units are connected in parallel with power lines, usually at the industrial
plants, to increase the system power factor. Their primary disadvantage is that their
capacitance cannot be adjusted to compensate or changing power factors.
Power factor correction can also be accomplished by using synchronous capacitors
connected across the power lines. (Three-phase synchronous motors are also called
synchronous capacitors). The advantage of synchronous capacitors over static capacitors
is that their capacitive effect can be adjusted as the system power factor increases or
decreases. The capacitive effect of a synchronous capacitor is easily changed by varying
the DC excitation voltage applied to the rotor of the machine. Industries considering the
installation of either static or synchronous capacitors should first compare he initial
equipment cost and the operating cost against the savings brought about by an increased
system power factor.
281
𝑄𝐿
𝑆
𝑆 = √𝑃2 + 𝑄𝐿 2 = 14.14 kVA
𝑝𝑓 =
true power
apparent power
𝑝𝑓 =
10 kW
14.14 kVA
𝑃
=
𝑃
𝑆
= 0.707 or 70.7 %
𝑆 = √𝑃2 + 𝑄𝑛𝑒𝑡 2 = √102 + 52 = 11.18 𝑘𝑉𝐴
𝑄𝐿
where 𝑄𝑛𝑒𝑡 = 𝑄𝐿 − 𝑄𝐶 = 10 − 5 = 5 𝑘𝑉𝐴𝑅
𝑆
𝑝𝑓 =
10 kW
11.18 kVA
= 0.8945 or 89.45 %
𝑃
𝑄𝐶
𝑄𝐿
𝑆 = √𝑃2 + 𝑄𝑛𝑒𝑡 2 = √102 + 02 = 10 𝑘𝑉𝐴
where 𝑄𝑛𝑒𝑡 = 𝑄𝐿 − 𝑄𝐶 = 10 − 10 = 5 𝑘𝑉𝐴𝑅
牜= 𝑃
𝑝𝑓 =
10 kW
10 kVA
= 1 or 100%
𝑄𝐶
Figure 8.2. Illustration of the effect of capacitive reactance on an inductive circuit:
(a) Reactive power = 10 kVAR inductive, (b) Reactive power = 10 kVAR inductive, and 5
kVAR capacitive, (c) Reactive power = 10 kVAR inductive, and 10 kVAR capacitive.
282
What is Power Factor Correction or Improvement?
Power factor correction or improvement consists of adding a capacitive reactive power
to an ac circuit in such a manner that the apparent power drawn from the source is
reduced without altering the current through or the voltage across the load itself.
In commercial use, the power factor should be close to unity for efficient
distribution of electric power. However, the inductive load of motors may result in a
power factor of 0.70, as an example, for the phase angle of 45. To correct for this
lagging inductive component of the current in the main line, a capacitor can be
connected across the line to draw leading current from the source. To bring the power
factor up to 1.0, that is unity pf, the value of capacitance is calculated to take the same
amount of volt-amperes as the VARS of the load.
What is the purpose of correcting power factor?
1. To make use of electrical energy more economical. This can be proven by
analyzing Figure 8.3 and Figure 8.4(b). In Figure 8.3 without the capacitor for
power factor correction the current is 4.55 A. If a capacitor is connected across the
load the current becomes 3.13 A. Lowering the current means lowering the energy
used by the load.
2. To reduce the cost of materials since cost of wires and cables depend on their sizes.
In a high current application large size of wire should be used. Small current uses
small sizes of wires.
3. It is desirable for industries to “correct” their power factor to avoid such charges.
Electrical utility companies impose penalties to the users in which electrical loads
have very low power factor.
What is the proper connection of the power factor devices? Series or Parallel
To find out, let us consider this problem: In the circuit below, an induction motor is
connected across a 220-V 60-Hz source, find the current through the motor or the
current in the line.
220 V
60 Hz
Induction Motor
700 watts, 70 % lagging
power factor
Figure 8-3. An induction motor in which the power factor is to be corrected.
283
The power triangle is shown below
QL
S
 = 45.57
P = 700 W
To find 
𝑝𝑓 = 𝑐𝑜𝑠  = 0.70 ;
 = c𝑜𝑠 −1 0.70 = 45.57
To find the current,
𝑝𝑓 = 𝑐𝑜𝑠  =
𝑃
=𝐼
𝑆
𝑃
𝑇 𝑉𝑇
𝑠𝑜 𝑡ℎ𝑎𝑡 𝐼𝑇 =
𝑃
𝑉𝑇 𝑝𝑓
=
700 𝑊
(220 𝑉)(0.70)
= 4.55 𝐴
To find the resistance and inductance of the motor,
XL
XL
Z
 = 45.57
R
Note: The motor is composed of resistance and inductance since the current is lagging
by 70%.
To find R,
𝑃
Using the formula 𝑃 = 𝐼 2 𝑅 𝑎𝑛𝑑 𝑅 = 𝐼2
𝑃
𝑅 = 𝐼2 =
700 𝑊
(4.55 𝐴)2
= 33.81 𝐴
For the inductance,
𝑋𝐿 = 𝑅 𝑡𝑎𝑛  = 33.81 𝑡𝑎𝑛 45.57 = 34.49 
284
𝐿 =
𝑋𝐿
2𝑓
=
34.49 
2(60)
= 91.49 𝑚𝐻
The equivalent circuit for the motor would be a series RL of resistance 33.81  and
inductance 91.49 mH.
R = 33.81 ohms
91.49 mH
Now connect a 32-F in series with the motor across the same source (see Figure 8-4a).
To find the current.
𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋2 )2
where 𝑅 = 33.81 
𝑋𝐿 = 34.49  because the frequency has not changed
𝑋𝐶 =
1
2(60)(32𝑥10−6 )
= 82.89 
𝑍 = √(33.81)2 + (34.49 – 82.89)2 = 59.04 
Then
𝐼=
𝑉𝑇
𝑍
=
220 𝑉
59.04 
= 3.73 𝐴
As you can see, the current is lowered but this current is not the rated current in the
motor. The load current is altered in this case, thus, this is not the correct connection of
the capacitor.
3.73 A
3.13 A
R = 33.81 ohms
R = 33.81 ohms
91.49 mH
220 V
60 Hz
C = 88 µF
220 V
60 Hz
91.49 mH
C = 88 µF
(a)
(b)
285
Figure 8.4. (a) A capacitor connected in series with the motor, (b) A capacitor connected
in parallel with the motor. This is the correct connection of the power factor correction
device because in this circuit the current in the load is not altered and the line current is
lowered.
Now let us connect the capacitor in parallel with the motor (Figure 8-4b) The
corresponding power triangle is shown below.
QL = 714.07 VARS
S
 = 45.57
P = 700 W
𝑄𝐿 = 𝑃 𝑡𝑎𝑛  = 700 𝑡𝑎𝑛 45.57 = 714.07 VARS
The capacitive reactive power is calculated by using the formula QC = VC2/XC
𝑄𝐶 =
(220)2
82.89
= 583.91 VARS
QL
QC
Qnet = 130.16 VARS
Snew
 = 45.57
P = 700 W
286
𝑄𝑛𝑒𝑡 = 𝑄𝐿 – 𝑄𝐶 = 714.07 − 583.91 = 130.16 VARS
 = 𝑡𝑎𝑛−1
130.16
= 10.53
700
pf = cos  = cos10.53 = 0.983 or 98.3%, the power factor is increased from 70%
to 98.3%.
For the total current
𝑆 = 𝑃 𝑐𝑜𝑠  = 700 𝑐𝑜𝑠 10.53 = 688.21 VA
𝐼𝑇 =
𝑆
걼𝑇
=
688.21 𝑉𝐴
220 𝑉
= 3.13 𝐴
In this circuit the current is lowered without altering the current in the motor and the
power factor is increased. Therefore the correction of capacitor and other power factor
correction devices is parallel with the load or across the power lines.
How to Obtain a Unity Power Factor
In order to obtain a unity power factor QL = QC so that the P = S
QL
S
QC - QL = 0
P
Example 8.1 An electric motor has an inductance of 0.045 H and a resistance of 30
ohms.
a) Calculate the current though the motor when connected to a 24-V 60-Hz power
source;
b) Calculate the power used by the motor;
c) Calculate the power factor of the load;
287
d) A capacitor is to be connected in across the load so that its power factor becomes
96% lagging. Calculate the capacitance of the capacitor.
e) Repeat (c) for 96% leading.
f) Repeat (c) for unity power factor.
Solution:
(a)
𝑋𝐿 = 2𝑓𝐿 = 2(60)(0.045) = 16.96 
𝒁 = √𝑅 2 + 𝑋𝐿 2 = √(30)2 + (16.96)2 = 34.36 
R = 30 ohms
24 V
60 Hz
L = 0.045 H
𝑰=
𝑽𝑻
𝒁
24 𝑉
= 34.36  = 0.6985 𝐴
(b) 𝑃 = 𝐼 2 𝑅 = (0.6985)2 (30) = 14.64 W
(c) 𝑝𝑓 =
𝑃
𝑆
=𝐼
𝑃
14.64
𝑇 𝑉𝑇
= (0.6985)(24) = 0.8732 𝑜𝑟 87.32%
or
 = 𝑡𝑎𝑛−1
𝑋𝐿
𝑅
= 𝑡𝑎𝑛−1
16.96
30
= 29.48
then 𝑝𝑓 = 𝑐𝑜𝑠  = 𝑐𝑜𝑠 29.48 = 0.87
(d)
The power triangle of the load,
𝑄𝐿 = 𝑃 tan 𝑜𝑙𝑑
= 14.64 tan 29.48
= 8.276 𝑉𝐴𝑅
𝑆𝑜𝑙𝑑
old = 29.48
𝑃 = 14.64 𝑊
288
By connecting a capacitor so that its power factor becomes 0.96 lagging,
For a 0.96 pf:  = 𝑐𝑜𝑠 −1 0.96 = 16.26
QL
Sold
old
Qnet = 8.276 - QC
new= 16.26
QC = ?
𝑄𝑛𝑒𝑡 = 𝑄𝐿 − 𝑄𝐶 = 8.276 − 𝑄𝐶
𝑡𝑎𝑛 𝑛𝑒𝑤 =
𝑄𝐿 − 𝑄𝐶
𝑃
𝑡𝑎𝑛 16.26 =
8.276 − 𝑄𝐶
14.64
𝑄𝐶 = 8.276 – 14.64 𝑡𝑎𝑛 16.26 = 4.006 VARS
𝑉𝐶 2
𝑋𝐶 =
𝑄𝐶
𝐶 =
=
(24)2
4.006
1
2(60)(143.78)
= 143.78 
= 18.45 F
(e) new = 16.26 (still 16.26 because the power factor is 0.96)
QL
Sold
old
new= 16.26 Qnet = QC - 8.276
Snew
QC
289
𝑡𝑎𝑛 𝑛𝑒𝑤 =
𝑡𝑎𝑛 𝑛𝑒𝑤 =
𝑡𝑎𝑛 16.26 =
𝑄𝑛𝑒𝑡
𝑃
𝑄𝐶 − 𝑄𝐿
𝑃
𝑄𝐶 − 8.276
14.64
𝑄𝐶 = 12.55 VARS
𝑋𝐶 =
𝐶 =
(f)
𝑄𝐶
=
𝑉𝐶 2
12.55
1
2(60)( 45.9)
= 45.9 
= 57.79 F
To make the power factor unity, QL = QC. Since QL = 8.276 VARS, QC is also
equal to 8.276 VARS.
QC = 8.276 VARS
𝑋𝐶 =
𝐶 =
.
𝑉𝐶 2
𝑉𝐶 2
=
𝑄愲
(24)2
8.276
1
2(60)( 69.6)
= 69.6 
= 38.11 F
290
Assessment No. 20
POWER CORRECTION OR IMPROVEMENT
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. A 5-kW AC motor has a power factor of 65% lagging when it is connected to a
200-V 50-Hz mains. What effective current does the motor draw?
2. In no. 1 what is the current through the motor when a capacitor of 80 F is
connected in parallel to the motor?
3. What is the power factor in the case of number 2?
291
4. In ac circuit takes a load of 160 kVA at a lagging power factor of 0.75 when
connected to a 460-V 60-Hz source. What value of capacitance that must be
connected across the load to make the power factor 0.89 lagging?
5. A small ac motor used in a washing is, in effect, an RL circuit. If the machine takes
311 watts and 4.5 amp from a 115-V source when operating normally, calculate
its power factor.
6. An induction motor draws 6.0 A at 0.8 lagging power factor from 208-V 60-Hz
source.
(a) What value of capacitance must be place in parallel with the motor to raise
the overall power factor to unity?
(b) What are the magnitudes of the final motor current, capacitor current, and
line current?
292
7. What value of capacitance is required to produce an overall power factor of 0.96
lagging with the motor of Problem 6?
8. A synchronous motor capable of operating with a leading power factor draws 15
kW from a distribution transformer while driving an air compressor. The
remainder of the load on the transformer is 80 kW at 0.85 lagging power factor.
(a) How many kilovars of capacitive reactive power must the synchronous motor
produce to raise the overall power factor to 0.96 lagging?
(b) What is the reactive factor of the synchronous motor operating in this manner?
293
9. The power factor of a load on a 120-V 60-Hz source is raised from 0.707 lagging
to 0.866 lagging by connecting a 53-F capacitor across the load. What is the
active power of the load?
10. The power factor of a load on a 120-V 60-Hz is raised from 0.866 lagging to 0.966
leading by connecting a 110 ½ - F capacitor in parallel with the load. What is
the rms load current?
.
294
Objective Test No. 7
POWER FACTOR CORRECTION OR IMPROVEMENT
1. A power factor of 1 indicates that the circuit phase angle is
A. 90 
B. 45
C. 180
D. 0
2. Energy sources are normally rated in
A. watts
B. volt-amperes
C. volt-amperes reactive
3. The magnitude of the power factor of an RC circuit with R = 10 , XC = 10  and
12 A effective is
A. 1
B. 0.5
C. 0.707
D. 0
4. The current in a series inductive circuit is 7.5 A at 25 Hz. The circuit takes 425 W,
and the power factor is 0.47. The resistance of the circuit is
A. 7.32 
B. 7.56 
C. 7.86 
D. 7.98 
5. A circuit with a resistor and capacitor in series takes 200 W at a power factor of
0.40 from 200-V 50-Hz supply. The capacitance of the circuit is
A. 32 F
B. 43.4 F
C. 66.4 F
D. 80 F
6. In a pure reactive circuit, the power factor is
A. lagging
B. zero
C. leading
D. unity
7. Power factor is defined as the ration of
A. volt-amperes to watts
B. watts to volt-amperes
C. volt-ampere reactive to watts
295
D. watts to volt-ampere reactive
8. In a series circuit consisting of resistance and reactance, power factor is defined
as the ratio of
A. resistance to impedance
B. resistance to reactance
C. reactance to impedance
9. For a parallel circuit consisting of resistance and reactance the value of power
factor is the ratio of
A. impedance to reactance
B. reactance to impedance
C. resistance to impedance
D. impedance to resistance
10. It is not easy to find the value of impedance in parallel circuit but power factor
can be easily obtained as the ratio of
A. active current to line current
B. reactive current to line current
C. line current to active current
11. The power factor of ac circuit containing both a resistor and a capacitor is
A. more than unity
B. leading by 90 degrees
C. leading power factor
12. In an ac circuit, a low value of reactive volt-ampere compared and with watts
indicates
A. high power factor
B. unity power factor
C. leading power factor
13. In a given ac circuit when power factor is unity the reactive power is
A. maximum
B. equal to I2R
C. zero
14. The capacitor for power factor correction are rated in terms of
A. voltage
B. VA
C. kW
D. kVAR
.
296
Unit 9
RESONANCE
LEARNING OUTCOMES
At the end of the lesson,
you are expected to:
1.
2.
3.
4.
5.
define resonance.
discuss the characteristics of a series resonance.
discuss the characteristics of a parallel resonance.
find the frequency for resonance.
solve the parameters of a circuit for resonance.
297
9.1 Resonant Effect
Inductive reactance increases as the frequency is increased, but capacitive reactance
decreases with higher frequencies. Because of these opposite characteristics, for any LC
combination there must be a frequency at which the XL equals the XC, as one increases
while the other decreases. This case of equal and opposite reactances is called
resonance, and the ac circuit is then a resonant circuit.
Any LC circuit can be resonant. It all depends on the frequency. At the resonant
frequency, an LC combination provides the resonance effect. Off the resonant frequency,
either below or above, the LC combination is just another ac circuit.
The frequency at which the opposite reactances are equal is the resonant
frequency. This frequency can be calculated as fr = 1/(2LC) where L is the inductance
in henrys, C is the capacitance in farads, and fr is the resonant frequency in hertz that
makes XL = XC.
In general, we can say that large values of L and C provide a relatively low resonant
frequency. Smaller values of L and C allow higher values for fr. The resonance effect is
most useful for radio frequencies, where the required values of microhenrys for L and
picofarads for C are easily obtained.
The most common applications of resonance in RF circuits is called tuning. In this
use, the LC circuit provides maximum voltage output at the resonant frequency,
compared with the amount of output at any other frequency either below or above
resonance. This idea is illustrated in Figure 9-1, where the LC circuit resonant at 1000
kHz magnifies the effect of this particular frequency. The result is maximum output at
1000 kHz, compared with lower or higher frequencies.
Resonant
LC circuit
fr = .
(a)
1
.
(b)
298
Figure 9.1. Applications of resonance. (a) LC circuit resonant at fr of 1000 kHz to provide
maximum output at this frequency. (b) Wavemeter as an example of tuning an LC circuit
for resonance at different frequencies.
For the wavemeter in Figure 9-1(b), note that the capacitance C can be varies to
provide resonance at different frequencies. The wavemeter can be tuned to any one
frequency in any range depending on the LC combination.
Tuning in radio and television are applications of resonance. When you tune a
radio to one station, the LC circuits are tuned to resonance for that particular carrier
frequency. Also, when you tune a television receiver to a particular channel, the LC
circuits are tuned to resonance for that station. There are almost unlimited uses for
resonance in ac circuits.
From Wikipedia
Resonance of a circuit involving capacitors and inductors occurs because the collapsing
magnetic field of the inductor generates an electric current in its windings that charges
the capacitor, and then the discharging capacitor provides an electric current that builds
the magnetic field in the inductor. This process is repeated continually. An analogy is a
mechanical pendulum.
At resonance, the series impedance of the two elements is at a minimum and the parallel
impedance is at maximum. Resonance is used for tuning and filtering, because it occurs
at a particular frequency for given values of inductance and capacitance. It can be
detrimental to the operation of communications circuits by causing unwanted sustained
and transient oscillations that may cause noise, signal distortion, and damage to circuit
elements.
Parallel resonance or near-to-resonance circuits can be used to prevent the waste of
electrical energy, which would otherwise occur while the inductor built its field or the
capacitor charged and discharged. As an example, asynchronous motors waste inductive
current while synchronous ones waste capacitive current. The use of the two types in
parallel makes the inductor feed the capacitor, and vice versa, maintaining the same
resonant current in the circuit, and converting all the current into useful work.
Since the inductive reactance and the capacitive reactance are of equal magnitude, ωL =
1/ωC, so:
where ω = 2πf, in which f is the resonance frequency in hertz, L is the inductance in
henries, and C is the capacitance in farads when standard SI units are used.
299
The quality of the resonance (how long it will ring when excited) is determined by its Q
factor, which is a function of resistance. A true LC circuit would have infinite Q, but all
real circuits have some resistance and smaller Q and are usually approximated more
accurately by an RLC circuit.
9.2 Series Resonance
10 
L = 239 µH
XL = 1500 ohms
300 µV
C = 106 pF
XC = 1500 ohms
(a)
XL, 
XL = 2fL
XL = XC at fr
fr
(b)
frequency
XC = 1/(2fC)
X C, 
(b)
Figure 9.2. Series Resonance. (a) Schematic diagram of series RLC, (b) Graph to show
reactances XC and XL are equal and opposite at the resonant frequency fr.
300
Highest current
value at fr
(a)
I, A
Small current values
30
current
CurrentSmall
maximum
at values
fr
below fr
above fr
20
(b)
Figure10
9.3. Graphs showing maximum current at resonance for the series
circuit shown in Figure 9-2. (a) Amplitudes of individual cycles. (b) response
curve to show amount of I below and above resonance. Values of I are in
Table 9.1.
600
800
1000
1200
Frequency, kHz
1400
301
Table 9.1
Series-Resonance calculations for the Circuit in Figure 9-2.
f
XL =
kHz
2fL
XC =
Net Reactance, 
1/(2fL) XC - XL
Z, 
XL – XC
I=
VL =
VC =
VT/Z
IXL
IXC
A
V
V
600
900
2500
1600
1600
0.19
171
475
800
1200
1875
675
675
0.44
528
825
1000
1500
1500
0
0
10
30
45,000
45,000
1200
1800
1250
550
550
0.55
990
688
1400
2100
1070
1030
1030
0.29
609
310
9.3 Characteristics of a Series Resonance:
a.
b.
c.
d.
The inductive reactance and the capacitive reactance of the circuit are equal.
Therefore, the impedance is minimum and is equal to the resistance of the
circuit.
Hence, the current in the circuit is maximum and is equal to E/R.
The current through the circuit is in phase with the applied voltage.
This condition is known as resonance. The series RLC circuit if Figure 9-2(a) is
called a resonant circuit, and the frequency at which resonance occurs is called the
resonant frequency of the circuit.
The letter symbol for resonant frequency is fr.
Since at resonance XL = XC, then 2fL = 1/2fC, for which
𝑓𝑟 =
1
2√𝐿𝐶
where fr is in hertz, L is in henrys, and C is in farads.
302
9.4 Q-factor of a Series Circuit
In the case of an RLC circuit it is defined as equal to the voltage magnification in the
circuit at resonance.
𝑄 − 𝑓𝑎𝑐𝑡𝑜𝑟 =
where R = resistance;
1
𝑅
𝐿
√
𝐶
L = inductance;
C = capacitance
In the case of series resonance, the higher quality factor, i.e., Q factor means not
only higher voltage magnification but also higher selectivity of the tuning coil.
Example 9.1 A coil in a tined circuit in a radio receiver has an inductance of 300 H and
a resistance of 15 . What value of capacitance must be connected in series with the
coil for the circuit to be series resonant at 840 kHz?
Solution:
𝑋𝐿 = 2𝑓𝐿 = 2 𝑥 (840 𝑘𝐻𝑧) 𝑥 (300 𝐻) = 1583 
At resonance,
𝑋𝐶 = 𝑋𝐿 = 1583 
C=
1
2 𝑥 (840 𝑘𝐻𝑧) 𝑥 (1583 )
= 120 𝑝𝐹
9.5 Parallel Resonance
120 Vrms
65 Hz
0°
E
R
100Ω
L
200mH
C
30µF
At 65 IL = IC and IT = IR. This condition is called parallel resonance. Other
characteristics of parallel resonance are:
303
1. The admittance of the circuit is minimum and is equal to the conductance of the
circuit.
2. The current drawn is minimum.
3. The phase angle between the current and voltage is zero, the power factor is
unity.
4. The resonant frequency is given by 𝑓𝑟 =
1
2√𝐿𝐶
if the resistance in the
inductance and capacitance branches is negligible.
9.6 Half-Power Frequencies
The half-power frequencies are those frequencies at which the power dissipation in the
circuit is half of the power dissipation at resonant frequency fr. They are the
corresponding frequencies f1 and f2 at the value of current 𝐼 =
current at resonance in RLC series circuit.
I0
current
0.707 I0
f1
fr
f2
10
frequency
Hence power Po drawn by the circuit at resonance is
𝑃0 = 𝐼0 2 𝑅
600
800
1000
1200
Frequency, kHz
1400
𝐼𝑜
√2
; where Io is the
304
𝐼
2
1
𝐼
2 𝑜
2
𝑅 (=half the power at resonance)
2
1
2
𝑅 (=half the power at resonance)
Power in the circuit at f1 = ( 𝑜 ) 𝑅 =
√2
𝐼
Power in the circuit at f2 = ( 𝑜 ) 𝑅 =
√2
𝐼
2 𝑜
9.7 Bandwidth and Selectivity
The difference (f2 –f1) is called the bandwidth of the resonant network. The ratio of the
bandwidth to the resonant frequency is defined as the selectivity of the circuit.
When frequency is varied in RLC circuit, the selectivity becomes
𝑓2 − 𝑓1
1
=
𝑓𝑟
𝑄0
Example 9.2 A parallel resonant circuit having L = 100 H and C = 10 pF. If Q of the
circuit is 50, what is the bandwidth of the resonant circuit?
Solution:
𝑓− =
=
1
2√𝐿𝐶
1
2√(100 𝑥 10−6 )(10 𝑥 10−12 )
= 503 kHz
.
305
Assessment No. 21
RESONANCE
Name:
Aviso, Myrell Jud A.
Score: _________ Rating: ______
1. A series circuit with a 12-ohm resistor and a 32-ohm inductive reactance is
connected across 240-volt 60-Hz mains. Determine (a) capacitive reactance that
will make circuit resonant; (b) current; (c) circuit power; (d) value of capacitance
and inductance.
2. Prob. 1 determine for resonance (a) value of capacitance C with resistance R,
inductance L, and frequency as given; (b) value of inductance L with C and f as given;
(c) value of f with L and C as given; (d) voltage across inductor and capacitor in (a) ,
(b) , (c); (e) current and power in (a), (b), (c).
306
3. In a series circuit the resistance is 1,000 ohms and the inductance 0.008 henry and
the capacitance is adjustable. Determine (a) value of capacitance to give resonance
at 1,000 Hz; (b) current if emf is 40 volts; (c) voltage across inductance and across
capacitance; (d) power.
.
307
Objective Test No. 8
RESONANCE
1. There will ________ be a frequency called the ______ frequency, at which _______.
A. sometimes, natural, XL = XC
B. always, natural, R = 0
C. always, resonate, XL = XC
D. sometimes, resonant, R = 0
2. For the series RLC circuit at resonance the current amplitude is _____ for a fixed
voltage amplitude and power factor.
A. minimum, zero
B. minimum, unity
C. maximum, zero
D. maximum, unity
3. In an RLC circuit, the current at resonance is
A. maximum
B. minimum
C. infinity
D. zero
4. In RLC circuits, the current at resonance is
A.
B.
C.
D.
minimum in parallel circuit and minimum in series circuit.
maximum in parallel circuit and minimum in series circuit.
maximum in either parallel or series circuit.
minimum in either parallel or series circuit.
5. A series resonance is capacitive at f = 100 Hz. The circuit will be inductive
somewhere at
A. f < 100 Hz
B. f > 100 Hz
C. f = 100 Hz by increasing the value of the resistance
D. none of these
6. At frequency less than the resonant frequency
A. series circuit is capacitive and parallel circuit is inductive
B. series circuit is inductive and parallel circuit is capacitive
C. both circuits are inductive
D. both circuits are capacitive.
308
7. The value of current at resonance in a series RLC circuit affected by the value of
A. R
B. C
C. L
D. all of these
8. In resonant circuit, the power factor at resonance is
A. zero
B. 1
C. 0.5
D. 0.707
9. Which of the following statement is true for a series RLC circuit tuned at resonant
frequency
A. the voltage across C > applied
B. the voltage across L > applied voltage
C. the voltage across L and C > applied voltage
D. the voltage across both L and C < applied voltage
10. The power factor at resonance in RLC parallel circuit is
A. zero
B. 0.08 lagging
C. 0.8 lagging
D. unity
11. The quality factor of RLC circuit will increase if
A. R increases
B. R decreases
C. impedance increases
D. voltage increases
12. Higher the Q of a series circuit,
A. broader it resonance curve
B. b. narrower its pass band
C. greater its bandwidth
D. sharper its resonance
13. Selectivities of different resonant circuits are compared in terms of their
A. impedances
B. reactances
C. frequencies
D. bandwidths
309
SUPPLEMENTARY PROBLEMS
1. The equation of the emf on an inductive circuit is e = 400 sin 377t and the current
is i = 40 sin (377t - 60°). What is the inductance? [0.023 H]
2. An impedance coil takes 25 A when connected across 220-V 25 Hz mains and 5.4
A from a 110-V 60-Hz mains. What is the inductance? [0.054 H]
3. Across what direct emf must an impedance coil having a resistance of 20 ohms and
an inductance of 0.05 H be connected in order that it may take the same power that
it takes from a 220-V, 60-Hz mains? [160 V]
4. A reactor whose resistance is 24 ohms and a power factor of 0.60 is connected in
series with a 50- µF capacitor across a 220-V 60-Hz supply. What is the voltage
across the reactor? [276 V]
5. The maximum of voltage and current outputs of an alternator are 300 volts and 20
amperes, respectively. What is the power output in watts of the voltage leads the
current by 30°? [2598 W]
6. A 50-µF and 100-µF capacitors are connected in series across a 100sin (t + 30°)
voltage. What is the equation of the current? [i = 1.26 sin (t + 120°)]
7. The ohmic resistance of a large contactor is measured to be 20 ohms. a voltage of
230 volts is impressed on the contactor and the current taken is 3.2 A. Neglecting
core loss, determine the inductance of the contactor. [183 mH]
8. A load of (20 + j35) ohms is connected across 220-volt source. Determine the
power factor and VARS. [49.6%, 1042 VARS]
9. A single-phase motor is rated 5 HP, 75% power factor and 220 volts. What is the
approximate size of capacitor is necessary to raise the power factor to about 95%?
[2 KVAR]
10. A coil of 50-ohm resistance and 150-mH inductance is connected in parallel with a
550-µF capacitor. The source voltage is 100 sin (t + 30°) volts. What is the
equation of the line current? [1.25 sin (t + 75.5°)]
11. A coil of 40-ohm resistance and 30-ohm reactance are connected in parallel with a
capacitance C to produce resonance. The source voltage is 100 volts. What is the
power drawn by the circuit? [120 W]
12. An impedance load consisting of 12-ohm resistance and 16-ohm inductive
reactance is connected across a 60-Hz, 100-V source. Find the capacitance of a
condenser which may be paralleled with this load to bring the power factor to 0.80
lagging. [46.42 µF]
13. A voltage of 100 volts is impressed across a circuit consisting of an inductive
reactance XL of 20 ohms in series with a parallel connection of a resistance R and
a capacitance C. the voltage across L is 100 volts and that across C is also 100 volts.
What is the real power supplied by the source? [433 W]
310
14. A single phase AC source is impressed on a three-branch parallel circuit with a 20ohm resistance in one branch, a 20-ohm capacitive reactance in another branch,
and in the third branch is an RL section in which the resistance RL is fixed and an
inductive reactance which is variable over a wide range. For what value of
resistance RL is it possible to adjust the circuit so that there will be only one
condition of unity power factor as XL is varied? [10 ohms]
15. A single-phase 240-V induction motor is connected in parallel with a 160-ohm
resistor. The motor takes 2 A and the total current is 3 A. determine the total
power taken by the circuit. [580 W]
16. When 1 A is passed through three impedance coils, A, B, and C in series the voltage
drops are respectively, 6, 3, and 8 volts on direct current and 7, 5, and 10 V,
respectively on alternating current. when alternating current flows through in the
circuit, what is its overall power factor? [0.78]
17. Two impedance coils which have resistance of 2.6 and 3.7 ohms take 5 and 10
amperes, respectively, from a 220-V, 60-Hz supply. What current will they take
when connected in series across 220-V, 60-Hz mains? [3.39 A]
18. A non-inductive resistance of 15 ohms in series with a condenser takes 5 A from a
220-V mains. What current will this circuit take from 220-V, 25-Hz supply? [2.19
A]
19. A single-phase load on 200 volts takes 5 kW at 0.60 lagging power factor. Find the
kVAR size of condenser which may be paralleled with this motor to bring the
resultant power factor to 0.90 lagging? [4.25 kVAR]
20. The values of the instantaneous currents in the branches of a parallel circuit are as
follows :
𝑖1 = 5 sin 346 t ; 𝑖2 = 10 sin (346t + π/4); 𝑖3 = 7.5 sin (346 t + π/2); 𝑖4 = 8 sin
(346 t − π/3)
Express the resultant line current in the same form as the original expression and
determine the r.m.s.value and the frequency of this current. [12.5 A; 55 Hz]
21. Four coils are connected in series. Each has induced in it a sinusoidal e.m.f. of 100
V, 50 Hz and there is a phase difference of 14 electrical degrees between one coil
and the next. What is the total e.m.f. generated in the circuit ? [384 V]
22. The instantaneous voltage across each of the four coils connected in series is given
by
𝑣1 = 100 sin 471 t; 𝑣2 = 250 cos 471 t; 𝑣3 = 150 sin (471 t + π/6); 𝑣4 = 200 sin (471
t − π/4)
Determine the total p.d. expressed in similar form to those given. What will be the
resultant p.d. If v2 is reversed in sign ? [v = 414 sin (471 t + 26.5°); v = 486 sin (471 t
− 40°)]
23. An alternating voltage of v = 100 sin 376.8 t is applied to a circuit consisting of a
coil having a resistance of 6Ω and an inductance of 21.22 mH.
a. Express the current flowing in the circuit in the form i = Im sin (376.8 t  )
b. If a moving-iron voltmeter, a wattmeter and a frequency meter are connected
in the circuit, what would be the respective readings on the instruments ? [i
= 10 sin (376.8 t − 53.1°); 70.7 V; 300 W; 60 Hz]
311
24. Three circuits A, B and C are connected in series across a 200-V supply. The voltage
across circuit A is 50 V lagging the supply voltage by 45° and the voltage across
circuit C is 100 V leading the supply voltage by 30°. Determine graphically or by
calculation, the voltage across circuit B and its phase displacement from the supply
voltage. [79.4 V ; 10° 38′ lagging]
25. Three alternating currents are given by
𝑖1 = 141 sin (ωt + π/4) ; 𝑖2 = 30 sin (ωt + π/2); 𝑖3 = 20 sin (ωt − π/6)
and are fed into a common conductor. Find graphically or otherwise the equation of
the resultant current and its r.m.s. value. [i = 167.4 sin (ωt + 0.797), Irms = 118.4 A]
26. Four e.m.fs 𝑒1 = 100 sin ωt, 𝑒2 = 80 sin (ωt − π/6), 𝑒3 = 120 sin (ωt + π/4) and
𝑒4 = 100 sin (ωt − 2π/3) are induced in four coils connected in series so that the
vector sum of four e.m.fs. is obtained. Find graphically or by calculation the
resultant e.m.f. and its phase difference with (a) 𝑒1 and (b) 𝑒2 . If the connections
to the coil in which the e.m.f. e2 is induced are reversed, find the new resultant
e.m.f . [208 sin (ωt – 0.202) (a) 11°34′ lag (b) 18′ 26′ lead; 76 sin (ωt + 0.528)]
27. Draw to scale a vector diagram showing the following voltages :
v1 = 100 sin 500t; v2 = 200 sin (500t + π/3); v3 = − 50 cos 500t; v4 = 150 sin (500 t
− π/4)
Obtain graphically or otherwise, their vector sum and express this in the form Vm sin
(500 t  ) using v1 as the reference vector. Give the r.m.s. value and frequency of the
resultant voltage. [360.5 sin (500 t + 0.056); 217 V; 79.6 Hz]
28. The voltage applied to a coil having R = 200 Ω, L = 638 mH is represented by e =
20 sin 100 π t. Finda corresponding expression for the current and calculate the
average value of the power taken by the coil. [i = 0.707 sin (100 πt − π/4); 50 W]
(I.E.E. London)
29. The coil having a resistance of 10 Ω and an inductance of 0.2 H is connected to a
100-V, 50-Hz supply. Calculate (a) the impedance of the coil (b) the reactance of
the coil (c) the current taken and (d) the phase difference between the current and
the applied voltage. [(a) 63.5 Ω (b) 62.8 Ω (c) 1.575 A (d) 80°57′ ]
30. An inductive coil having a resistance of 15 Ω takes a current of 4 A when connected
to a 100-V, 60 Hz supply. If the coil is connected to a 100 -V, 50 Hz supply, calculate
(a) the current (b) the power (c) the power factor. Draw to scale the vector
diagram for the 50-Hz conditions, showing the component voltages. [(a) 4.46 A (b)
298 W (c) 0.669]
31. When supplied with current at 240-V, single-phase at 50 Hz, a certain inductive
coil takes 13.62 A. If the frequency of supply is changed to 40 Hz, the current
increases to 16.12 A. Calculate the resistance and inductance of the coil. [17.2 W,
0.05 H] (London Univ.)
32. A voltage v (t) = 141.4 sin (314 t + 10°) is applied to a circuit and a steady current
given by i (t) = 14.4 sin (314 t − 20°) is found to flow through it. Determine (i) the
p.f. of the circuit and (ii) the power delivered to the circuit. [0.866 (lag); 866 W]
33. A circuit takes a current of 8 A at 100 V, the current lagging by 30° behind the
applied voltage. Calculate the values of equivalent resistance and reactance of the
circuit. [10.81 Ω ; 6.25 Ω]
34. Two inductive impedance A and B are connected in series. A has R = 5 Ω, L = 0.01
H; B has R = 3 Ω , L = 0.02 H. If a sinusoidal voltage of 230 V at 50 Hz is applied to
the whole circuit calculate (a) the current (b) the power factor (c) the voltage
312
drops. Draw a complete vector diagram for the circuit. [(a) 18.6 (b) 0.648 (c) VA =
109.5 V, VB = 129.5 V] (I.E.E. London)
35. A coil has an inductance of 0.1 H and a resistance of 30 Ω at 20°C. Calculate (i) the
current and (ii) the power taken from 100-V, 50-Hz mains when the temperature
of the coil is 60° C, assuming the temperature coefficient of resistance to be 0.4%
per°C from a basic temperature of 20°C. [(i) 2.13 A (ii) 158.5 W] (London Univ.)
36. An air-cored choking coil takes a current of 2 A and dissipates 200 W when
connected to a 200-V, 50-Hz mains. In other coil, the current taken is 3 A and the
power 270 W under the same conditions. Calculate the current taken and the total
power consumed when the coils are in series and connected to the same supply.
[1.2, 115 W] (City and Guilds, London)
37. A circuit consists of a pure resistance and a coil in series. The power dissipated in
the resistance is 500 W and the drop across it is 100 V. The power dissipated in
the coil is 100 W and the drop across it is 50 V. Find the reactance and resistance
of the coil and the supply voltage. [9.168Ω; 4Ω; 128.5V]
38. A choking coil carries a current of 15 A when supplied from a 50-Hz, 230-V supply.
The power in the circuit is measured by a wattmeter and is found to be 1300 watt.
Estimate the phase difference between the current and p.d. in the circuit. [0.3768]
(I.E.E. London)
39. An ohmic resistance is connected in series with a coil across 230-V, 50-Hz supply.
The current is 1.8 A and p.ds. across the resistance and coil are 80 V and 170 V
respectively. Calculate the resistance and inductance of the coil and the phase
difference between the current and the supply voltage. [61.1 Ω, 0.229 H, 34°20′ ]
(App. Elect. London Univ.)
40. A coil takes a current of 4 A when 24 V d.c. are applied and for the same power on
a 50-Hz a.c. supply, the applied voltage is 40. Explain the reason for the difference
in the applied voltage. Determine (a) the reactance (b) the inductance (c) the
angle between the applied p.d. and current (d) the power in watts. [(a) 8 Ω (b)
0.0255 H (c) 53°7′ (d) 96 W]
41. An inductive coil and a non-inductive resistance R ohms are connected in series
across an a.c. supply. Derive expressions for the power taken by the coil and its
power factor in terms of the voltage across the coil, the resistance and the supply
respectively. If R = 12 Ω and the three voltages are in order, 110 V, 180 V and 240
V, calculate the power and the power factor of the coil. [546 W; 0.331]
42. Two coils are connected in series. With 2 A d.c. through the circuit, the p.ds. across
the coils are 20 and 30 V respectively. With 2 A a.c. at 40 Hz, the p.ds. across the
coils are 140 and 100 V respectively. If the two coils in series are connected to a
230-V, 50-Hz supply, calculate (a) the current (b) the power (c) the power factor.
[(a) 1.55 A (b) 60 W (c) 0.1684]
43. It is desired to run a bank of ten 100-W, 10-V lamps in parallel from a 230-V, 50Hz supply by inserting a choke coil in series with the bank of lamps. If the choke
coil has a power factor of 0.2, find its resistance, reactance and inductance. [R =
4.144 Ω, X = 20.35 Ω , L = 0.065 H] (London Univ.)
44. At a frequency for which ω = 796, an e.m.f. of 6 V sends a current of 100 mA
through a certain circuit. When the frequency is raised so that ω = 2866, the same
voltage sends only 50 mA through the same circuit. Of what does the circuit consist
? [R = 52 Ω, L = 0.038 H in series] (I.E.E. London)
313
45. A capacitor having a capacitance of 20 µF is connected in series with a noninductive resistance of 120 Ω across a 100-V, 50-Hz supply, Calculate (a) voltage
(b) the phase difference between the current and the supply voltage (c) the power.
Also draw the vector diagram. [(a) 0.501A (b) 52.9° (c) 30.2 W]
46. A capacitor and resistor are connected in series to an a.c. supply of 50 V and 50 Hz.
The current is 2 A and the power dissipated in the circuit is 80 W. Calculate the
resistance of the resistor and the capacitance of the capacitor. [20 Ω ; 212 µF]
47. A voltage of 125 V at 50 Hz is applied to a series combination of non-inductive
resistor and a lossless capacitor of 50 µF. The current is 1.25 A. Find (i) the value
of the resistor (ii) power drawn by the network (iii) the power factor of the
network. Draw the phasor diagram for the network. [(i) 77.3 Ω (ii) 121 W (iii)
0.773 (lead)] (Electrical Technology-1, Osmania Univ.)
48. A black box contains a two-element series circuit. A voltage (40 − j30) drives a
current of (40 − j3) A in the circuit. What are the values of the elements ? Supply
frequency is 50 Hz. [R = 1.05 ; C = 4750 µF] (Elect. Engg. and Electronics Bangalore
Univ.)
49. Following readings were obtained from a series circuit containing resistance and
capacitance : V = 150 V ; I = 2.5 A; P = 37.5 W, f = 60 Hz. Calculate (i) Power
factor (ii) effective resistance (iii) capacitive reactance and (iv) capacitance. [(i)
0.1 (ii) 6 Ω (iii) 59.7 Ω (iv) 44.4 µF]
50. An e.m.f. represented by e = 100 sin 100 π t is impressed across a circuit consisting
of 40-Ω resistor in series with a 40-µF capacitor and a 0.25 H indicator. Determine
(i) the r.m.s. value of the current (ii) the power supplied (iii) the power factor. [(i)
1.77 A (ii) 125 W (iii) 1.0] (London Univ.)
51. A series circuit with a resistor of 100 Ω capacitor of 25 µF and inductance of 0.15
H is connected across 220-V, 60-Hz supply. Calculate (i) current (ii) power and
(iii) power factor in the circuit. [(i) 1.97 A; (ii) 390 W (iii) 0.9 (lead)] (Elect. Engg.
and Electronics Bangalore Univ.)
52. A series circuit with R = 10 Ω, L = 50 mH and C = 100 µF is supplied with 200 V/50
Hz. Find (i) the impedance (ii) current (iii) power (iv) power factor. [(i) 18.94 Ω
(ii) 18.55 A (iii) 1966 W (iv) 0.53 (leading)] (Elect. Engg. & Electronics Bangalore
Univ.)
53. A coil of resistance 10 Ω and inductance 0.1 H is connected in series with a 150-µF
capacitor across a 200-V, 50-Hz supply. Calculate (a) the inductive reactance, (b)
the capacitive reactance, (c) the impedance (d) the current, (e) the power factor
(f) the voltage across the coil and the capacitor respectively. [(a) 31.4 Ω (b) 21.2 Ω
(c) 14.3 Ω (d) 14 A (e) 0.7 lag (f) 460 V, 297 V]
54. A circuit is made up of 10 Ω resistance, 12 mH inductance and 281.5 µF capacitance
in series. The supply voltage is 100 V (constant). Calculate the value of the current
when the supply frequency is (a) 50 Hz and (b) 150 Hz. [8 A leading; 8 A lagging]
55. A coil having a resistance of 10 Ω and an inductance of 0.2 H is connected in series
with a capacitor of 59.7 µF. The circuit is connected across a 100-V, 50-Hz a.c.
supply. Calculate (a) the current flowing (b) the voltage across the capacitor (c)
the voltage across the coil. Draw a vector diagram to scale. [(a) 10 A (b) 628 V (c)
635 V]
56. A coil is in series with a 20 µF capacitor across a 230-V , 50-Hz supply. The current
taken by the circuit is 8 A and the power consumed is 200 W. Calculate the
inductance of the coil if the power factor of the circuit is (a) leading and (b) lagging.
314
Sketch a vector diagram for each condition and calculate the coil power factor in each
case. [0.415 H; 0.597 H; 0.0238 ; 0.0166]
57. A circuit takes a current of 3 A at a power factor of 0.6 lagging when connected to
a 115-V, 50-Hz supply. Another circuit takes a current, of 5 A at a power factor of
0.707 leading when connected to the same supply. If the two circuits are connected
in series across a 230-V, 50Hz supply, calculate (a) the current (b) the power
consumed and (c) the power factor. [(a) 5.5 A (b) 1.188 kW (c) 0.939 lag]
58. A coil of insulated wire of resistance 8 ohms and inductance 0.03 H is connected to
an a.c. supply at 240 V, 50-Hz. Calculate (a) the current, the power and power
factor (b) the value of a capacitance which, when connected in series with the
above coil, causes no change in the values of current and power taken from the
supply. [(a) 19.4 A, 3012 W, 0.65 lag (b) 168.7 µF] (London Univ.)
59. A series circuit, having a resistance of 10 Ω, an inductance of 0.025 H and a variable
capacitance is connected to a 100-V, 25-Hz single-phase supply. Calculate the
capacitance when the value of the current is 8 A. At this value of capacitance, also
calculate (a) the circuit impedance (b) the circuit power factor and (c) the power
consumed. [556 µF (a) 1.5 Ω (b) 0.8 leading (c) 640 W]
60. An alternating voltage is applied to a series circuit consisting of a resistor and ironcored inductor and a capacitor. The current in the circuit is 0.5 A and the voltages
measured are 30 V across the resistor, 48 V across the inductor, 60 V across the
resistor and inductor and 90 V across the capacitor. Find (a) the combined copper
and iron losses in the inductor (b) the applied voltage. [(a) 3.3 W (b) 56 V] (City &
Guilds, London)
61. When an inductive coil is connected across a 250-V, 50-Hz supply, the current is
found to be 10 A and the power absorbed 1.25 kW. Calculate the impedance, the
resistance and the inductance of the coil. A capacitor which has a reactance twice
that of the coil, is now connected in series with the coil across the same supply.
Calculate the p.d. across the capacitor. [25 Ω; 12.5 Ω; 68.7 mH; 433 V]
62. A voltage of 200 V is applied to a series circuit consisting on a resistor, an inductor
and a capacitor. The respective voltages across these components are 170, 150 and
100 V and the current is 4 A. Find the power factor of the inductor and of the circuit.
[0.16; 0.97]
63. A pure resistance R, a choke coil and a pure capacitor of 50µ F are connected in
series across a supply of V volts, and carry a current of 1.57 A. Voltage across R is
30 V, across choke coil 50 V and across capacitor 100 V. The voltage across the
combination of R and choke coil is 60 volt. Find the supply voltage V, the power
loss in the choke, frequency of the supply and power factor of the complete circuit.
Draw the phasor diagram. [60.7 V; 6.5 W; 0.562 lead] (F.E. Pune Univ.)
64. An a.c. series circuit has a resistance of 10 Ω, an inductance of 0.2 H and a
capacitance of 60 µF. Calculate (a) the resonant frequency (b) the current and (c)
the power at resonance, given that the applied voltage is 200 V. [46 Hz; 20 A; 4
kW]
65. A circuit consists of an inductor which has a resistance of 10 Ω and an inductance
of 0.3 H, in series with a capacitor of 30 µF capacitance. Calculate (a) the
impedance of the circuit to currents of 40 Hz (b) the resonant frequency (c) the
peak value of stored energy in joules when the applied voltage is 200 V at the
resonant frequency.
[58.31 Ω; 53 Hz; 120 J]
315
66. A resistor and a capacitor are connected in series with a variable inductor. When
the circuit is connected to a 240-V, 50-Hz supply, the maximum current given by
varying the inductance is 0.5 A. At this current, the voltage across the capacitor is
250 V. Calculate the values of (a) the resistance (b) the capacitance (c) the
inductance. [480 Ω, 6.36 µF; 1.59 H]. Neglect the resistance of the inductor
67. A circuit consisting of a coil of resistance 12Ω and inductance 0.15 H in series with
a capacitor of 12µF is connected to a variable frequency supply which has a
constant voltage of 24 V. Calculate (a) the resonant frequency (b) the current in
the circuit at resonance (c) the voltage across the capacitor and the coil at
resonance. [(a) 153 Hz (b) 2 A (c) 224 V]
68. A resistance, a capacitor and a variable inductance are connected in series across
a 200-V, 50-Hz supply. The maximum current which can be obtained by varying
the inductance is 314 mA and the voltage across the capacitor is then 300 V.
Calculate the capacitance of the capacitor and the values of the inductance and
resistance. [3.33 µF, 3.04 H, 637 Ω] (I.E.E. London)
69. A 250-V circuit, consisting of a resistor, an inductor and a capacitor in series,
resonates at 50 Hz. The current is then 1A and the p.d. across the capacitor is 500
V. Calculate (i) the resistance (ii) the inductance and (iii) the capacitance. Draw
the vector diagram for this condition and sketch a graph showing how the current
would vary in a circuit of this kind if the frequency were varied over a wide range,
the applied voltage remaining constant. [(i) 250 Ω (ii) 0.798 H (iii) 12.72 µ F] (City
& Guilds, London)
70. A resistance of 24 Ω, a capacitance of 150 µF and an inductance of 0.16 H are
connected in series with each other. A supply at 240 V, 50 Hz is applied to the ends
of the combination. Calculate (a) the current in the circuit (b) the potential
differences across each element of the circuit (c) the frequency to which the supply
would need to be changed so that the current would be at unity power-factor and
find the current at this frequency. [(a) 6.37 A (b) VR = 152.9 V, VC = 320 V, VC =
123.3 V (c) 32 Hz; 10 A] (London Univ.)
71. A series circuit consists of a resistance of 10 Ω, an inductance of 8 mH and a
capacitance of 500 µµF. A sinusoidal E.M.F. of constant amplitude 5 V is introduced
into the circuit and its frequency varied over a range including the resonant
frequency. At what frequencies will current be (a) a maximum (b) one-half themaximum ? [(a) 79.6 kHz (b) 79.872 kHz, 79.528 kHz] (App. Elect. London Univ.)
72. A circuit consists of a resistance of 12 ohms, a capacitance of 320 µF and an
inductance of 0.08 H, all in series. A supply of 240 V, 50 Hz is applied to the ends of
the circuit. Calculate :
c. the current in the coil.
d. the potential differences across each element of the circuit.
e. the frequency at which the current would have unity power-factor.
[(a) 12.4 A (b) 149 V, 311 V (c) 32 Hz] (London Univ.)
73. A series circuit consists of a reactor of 0.1 henry inductance and 5 ohms resistance
and a capacitor of 25.5 µ F capacitance. Find the resonance frequency and the
percentage change in the current for a divergence of 1 percent from the resonance
frequency. [100 Hz, 1.96% at 99 Hz; 4.2% at 101 Hz] (City and Guilds, London)
74. A capacitor of 50 µF capacitance is connected in parallel with a reactor of 22 Ω
resistance and 0.07 henry inductance across 200-V, 50-Hz mains. Calculate the
316
total current taken. Draw the vector diagram in explanation. [4.76 A lagging, 17º12′
] (City & Guilds, London)
75. A non-inductive resistor is connected in series with a capacitor of 100 µF
capacitance across 200-V, 50-Hz mains. The p.d. measured across the resistor is
150 V. Find the value of resistance and the value of current taken from the mains
if the resistor were connected in parallel-with the capacitor instead of in series. [R
= 36.1 Ω; 8.37 Ω] (City & Guilds, London)
76. An impedance of (10 + j15) Ω is connected in parallel with an impedance of (6 −
j8)Ω. The total current is 15 A. Calculate the total power. [2036 W] (City & Guilds,
London)
77. The load on a 250-V supply system is : 12 A at 0.8 power factor lagging ; 10 A at 0.5
power factor lagging ; 15 A at unity power factor ; 20 A at 0.6 power factor leading.
Find (i) the total lead in kVA and (ii) its power factor. [(i) 10.4 kVA (ii) 1.0] (City &
Guilds, London)
78. A voltage having frequency of 50 Hz and expressed by V = 200 + j100 is applied to
a circuit consisting of an impedance of 50 ∠30º Ω in parallel with a capacitance of
10 µF. Find (a) the reading on a ammeter connected in the supply circuit (b) the
phase difference between the current and the voltage. [(a) 4.52 (b) 26.6º lag]
(London University)
79. A voltage of 200º ∠30º V is applied to two circuits A and B connected in parallel.
The current in A is 20 ∠60º A and that in B is 40 ∠−30º A. Find the kVA and kW in
each branch circuit and the main circuit. Express the current in the main circuit in
the form A + jB. [kVAA = 4, kVAB = 8, kVA = 12, kWA = 3.46, kWB = 4, kW = 7.46,
I = 44.64 − j 2.68] (City & Guilds, London)
80. A coil having an impedance of (8 + j6) Ω is connected across a 200-V supply.
Express the current in the coil in (i) polar and (ii) rectangular co-ordinate forms.
If a capacitor having a susceptance of 0.1 S is placed in parallel with the coil, find
(iii) the magnitude of the current taken from the supply. [(i) 20 ∠ 36.8ºA (ii) 16 −
j12 A (iii) 17.9 A] (City & Guilds, London)
81. A coil-A of inductance 80 mH and resistance 120 Ω, is connected to a 230-V, 50 Hz
single-phase supply. In parallel with it in a 16 µF capacitor in series with a 40 Ω
non-inductive resistor B. Determine (i) the power factor of the combined circuit
and (ii) the total power taken from the supply. [(i) 0.945 lead (ii) 473 W] (London
University)
82. A choking coil of inductance 0.08 H and resistance 12 ohm, is connected in parallel
with a capacitor of 120 µF. The combination is connected to a supply at 240 V, 50
Hz Determine the total current from the supply and its power factor. Illustrate your
answers with a phasor diagram. [3.94 A, 0.943 lag] (London University)
83. A choking coil having a resistance of 20 Ω and an inductance of 0.07 henry is
connected with a capacitor of 60 µF capacitance which is in series with a resistor
of 50 Ω. Calculate the total current and the phase angle when this arrangement is
connected to 200-V, 50 Hz mains. [7.15 A, 24º39′ lag] (City & Guilds, London)
84. A coil of resistance 15 Ω and inductance 0.05 H is connected in parallel with a noninductive resistance of 20 Ω. Find (a) the current in each branch (b) the total
current (c) the phase angle of whole arrangement for an applied voltage of 200 V
at 50 Hz. [9.22 A ; 10A ; 22.1º]
85. A sinusoidal 50-Hz voltage of 200 V (r.m.s) supplies the following three circuits
which are in parallel : (a) a coil of inductance 0.03 H and resistance 3 Ω (b) a
317
capacitor of 400 µF in series with a resistance of 100 Ω (c) a coil of inductance 0.02
H and resistance 7 Ω in series with a 300 µF capacitor. Find the total current
supplied and draw a complete vector diagram. [29.4 A] (Sheffield Univ. U.K.)
86. A 50-Hz, 250-V single-phase power line has the following loads placed across it in
parallel : 4 kW at a p.f. of 0.8 lagging ; 6 kVA at a p.f. of 0.6 lagging; 5 kVA which
includes 1.2 kVAR leading. Determine the overall p.f. of the system and the
capacitance of the capacitor which, if connected across the mains would restore
the power factor to unity. [0.844 lag ; 336 µF]
87. Define the terms admittance, conductance and susceptance with reference to
alternating current circuits. Calculate their respective values for a circuit
consisting of resistance of 20Ω, in series with an inductance of 0.07 H when the
frequency is 50 Hz. [0.336 S, 0.0226 S, 0.0248 S] (City & Guilds, London)
88. Explain the terms admittance, conductance, susceptance as applied to a.c. circuits.
One branch A, of a parallel circuit consists of a coil, the resistance and inductance
of which are 30 Ω and 0.1 H respectively. The other branch B, consists of a 100 µF
capacitor in series with a 20 Ω resistor. If the combination is connected 240-V, Hz
mains, calculate (i) the line current and (ii) the power. Draw to scale a vector
diagram of the supply current and the branch-circuit currents. [(i) 7.38 A (ii) 1740
W] (City & Guilds, London)
89. Find the value of capacitance which when placed in parallel with a coil of resistance
22 Ω and inductance of 0.07 H, will make it resonate on a 50-Hz circuit. [72.33 µF]
(City & Guilds, London)
90. A parallel circuit has two branches. Branch A consists of a coil of inductance 0.2 H
and a resistance of 15 Ω ; branch B consists of a 30 mF capacitor in series with a 10
Ω resistor. The circuit so formed is connected to a 230-V, 50-Hz supply. Calculate
(a) current in each branch (b) line current and its power factor (c) the constants
of the simplest series circuit which will take the same current at the same power
factor as taken by the two branches in parallel. [3.57 A, 2.16 A ; 1.67 A, 0.616 lag,
8.48 Ω, 0.345 H]
91. A 3.73 kW, 1-phase, 200-V motor runs at an efficiency of 75% with a power factor
of 0.7 lagging. Find (a) the real input power (b) the kVA taken (c) the reactive
power and (d) the current. With the aid of a vector diagram, calculate the
capacitance required in parallel with the motor to improve the power factor to 0.9
lagging. The frequency is 50 Hz. [4.97 kW ; 7.1 kVA ; 5.07 kVAR ; 35.5 A ; 212µF]
92. The impedances of two parallel circuits can be represented by (20 + j15) and (1 −
j60) Ω respectively. If the supply frequency is 50 Hz, find the resistance and the
inductance or capacitance of each circuit. Also derive a symbolic expression for
the admittance of the combined circuit and then find the phase angle between the
applied voltage and the resultant current. State whether this current is leading or
lagging relatively to the voltage. [20 Ω; 0.0478 H; 10 Ω; 53 µF; (0.0347 − j
0.00778)S; 12º38′ lag]
93. One branch A of a parallel circuit consists of a 60-µF capacitor. The other branch B
consists of a 30 Ω resistor in series with a coil of inductance 0.2 H and negligible
resistance. A 140 Ω resistor is connected in parallel with the coil. Sketch the circuit
diagram and calculate (i) the current in the 30 Ω resistor and (ii) the line current
if supply voltage is 230-V and the frequency 50 Hz. [(i) 3.1 ∠− 44º (ii) 3.1 ∠ 45º A]
94. A coil having a resistance of 45 Ω and an inductance of 0.4 H is connected in parallel
with a capacitor having a capacitance of 20 µF across a 230-V, 50-Hz system.
318
Calculate (a) the current taken from the supply (b) the power factor of the
combination and (c) the total energy absorbed in 3 hours. [(a) 0.615 (b) 0.951 (c)
0.402 kWh] (London University)
95. A series circuit consists of a resistance of 10 Ω and reactance of 5 Ω. Find the
equivalent value of conductance and susceptance in parallel. [0.08 S, 0.04 S]
96. An alternating current passes through a non-inductive resistance R and an
inductance L in series. Find the value of the non-inductive resistance which can be
shunted across the inductance without altering the value of the main current.
[ω2L2/2R] (Elec. Meas. London Univ.)
97. A p.d. of 200 V at 50 Hz is maintained across the terminals of a series-parallel
circuit, of which the series branch consists of an inductor having an inductance of
0.15 H and a resistance of 30 Ω, one parallel branch consists of 100-µF capacitor
and the other consists of a 40-Ω resistor. Calculate (a) the current taken by the
capacitor (b) the p.d. across the inductor and (c) the phase difference of each of
these quantities relative to the supply voltage. Draw a vector diagram representing
the various voltage and currents. [(a) 29.5 A (b) 210 V (c) 7.25º, 26.25º] (City &
Guilds, London)
98. A coil (A) having an inductance of 0.2 H and resistance of 3.5 Ω is connected in
parallel with another coil (B) having an inductance of 0.01 H and a resistance of 5
Ω. Calculate (i) the current and (ii) the power which these coils would take from a
100-V supply system having a frequency of 50-Hz. Calculate also (iii) the resistance
and (iv) the inductance of a single coil which would take the same current and
power. [(i) 29.9 A (ii) 2116 W (ii) 2.365 Ω (iv) 0.00752 H] (London Univ.)
99. Two coils, one (A) having R = 5 Ω, L = 0.031 H and the other (B) having R = 7 Ω ;
L = 0.023 H, are connected in parallel to an a.c. supply at 200 V, 50 Hz. Determine
(i) the current taken by each coil and also (ii) the resistance and (iii) the
inductance of a single coil which will take the same total current at the same power
factor as the two coils in parallel. [(i) IA = 18.28 A, IB = 19.9 A (ii) 3.12 Ω (iii) 0.0137
H] (London Univ.)
100.
Two coils are connected in parallel across 200-V, 50-Hz mains. One coil
takes 0.8 kW and 1.5 kVA and the other coil takes 1.0 kW and 0.6 kVAR. Calculate
(i) the resistance and (ii) the reactance of a single coil which would take the same
current and power as the original circuit. [(i) 10.65 Ω (ii) 11.08 Ω] (City & Guilds,
London)
101.
An a.c. circuit consists of two parallel branches, one (A) consisting of a coil,
for which R = 20 Ω and L = 0.1 H and the other (B) consisting of a 40-Ω noninductive resistor in series with 60-µF capacitor. Calculate (i) the current in each
branch (ii) the line current (iii) the power, when the circuit is connected to 230-V
mains having a frequency of 50 Hz. Calculate also (iv) the resistance and (b) the
inductance of a single coil which will take the same current and power from the
supply. [(i) 6.15 A, 3.46 A (ii) 5.89 (iii) 1235 W (iv) 35.7 Ω (b) 0.0509 H] (London
Univ.)
102.
One branch (A) of a parallel circuit, connected to 230-V, 50-Hz mains
consists of an inductive coil (L = 0.15 H, R = 40 Ω) and the other branch (B)
consists of a capacitor (C = 50 µF) in series with a 45 Ω resistor. Determine (i) the
power taken (ii) the resistance and (iii) the reactance of the equivalent series
circuit. [(i) 946 W (ii) 55.4 Ω (iii) 4.6 Ω] (London Univ.)
319
103.
A resistance of 20 W and a coil of inductance 31.8 mH and negligible
resistance are connected in parallel across 230 V, 50 Hz supply. Find (i) The line
current (ii) power factor and (iii) The power consumed by the circuit. [(i) 25.73 A
(ii) 0.44 T lag (iii) 246 W] (F. E. Pune Univ.)
104.
Two impedances Z1 = (150 + j157) ohm and Z2 = (100 + j 110) ohm are
connected in parallel across a 220-V, 50-Hz supply. Find the total current and its
power factor. [24 ∠ − 47ºA ; 0.68 (lag)] (Elect. Engg. & Electronics Bangalore Univ.)
105.
Two impedances (14 + j5)Ω and (18 + j10) Ω are connected in parallel
across a 200-V, 50-Hz supply. Determine (a) the admittance of each branch and of
the entire circuit ; (b) the total current, power, and power factor and (c) the
capacitance which when connected in parallel with the original circuit will make
the resultant power factor unity. [(a) (0.0634 − j0.0226), (0.0424 − j0.023)
(0.1058 − j0.0462 S) (b) 23.1 A, 4.232 kW, 0.915 (c) 147 µF]
106.
A parallel circuit consists of two branches A and B. Branch A has a resistance
of 10 Ω and an inductance of 0.1 H in series. Branch B has a resistance of 20 Ω and
a capacitance of 100 µF in series. The circuit is connected to a single-phase supply
of 250 V, 50 Hz. Calculate the magnitude and the phase angle of the current taken
from the supply. Verify your answer by measurement from a phasor diagram
drawn to scale. [6.05 ∠ − 15.2º] (F. E. Pune Univ.)
107.
Two circuits, the impedances of which are given by Z1 = (10 + j15) Ω and
Z2 = (6 − j8) Ω are connected in parallel. If the total current supplied is 15 A, what
is the power taken by each branch ? [737 W ; 1430 W] (Elect. Engg. A.M.A.E. S.I.)
108.
A voltage of 240 V is applied to a pure resistor, a pure capacitor, and an
inductor in parallel. The resultant current is 2.3 A, while the component currents
are 1.5, 2.0 and 1.1 A respectively. Find the resultant power factor and the power
factor of the inductor. [0.88 ; 0.5]
109.
Two parallel circuits comprise respectively (i) a coil of resistance 20 Ω and
inductance 0.07 H and (ii) a capacitance of 60 µF in series with a resistance of 50
Ω. Calculate the current in the mains and the power factor of the arrangement
when connected across a 200-V, 50-Hz supply. [7.05 A ; 0.907 lag] (Elect. Engg. &
Electronics, Bangalore Univ.)
110.
Two circuits having the same numerical ohmic impedances are joined in
parallel. The power factor of one circuit is 0.8 lag and that of other 0.6 lag. Find the
power factor of the whole circuit. [0.707] (Elect. Engg. Pune Univ.)
111.
How is a current of 10 A shared by three circuits in parallel, the impedances
of which are (2 − j5) Ω, (6 + j3)Ω and (3 + j4) Ω. [5.68 A ; 4.57 A, 6.12 A]
112.
A piece of equipment consumes 2,000 W when supplied with 110 V and
takes a lagging current of 25 Determine the equivalent series resistance and
reactance of the equipment. If a capacitor is connected in parallel with the
equipment to make the power factor unity, find its capacitance. The supply
frequency is 100 Hz. [3.2 Ω, 3.02 Ω, 248 µF] (Sheffield Univ. U.K.)
113.
A capacitor is placed in parallel with two inductive loads, one of 20 A at 30º
lag and one of 40º A at 60º lag. What must be current in the capacitor so that the
current from the external circuit shall be at unity power factor ? [44.5 A] (City &
Guilds, London)
114.
An air-cored choking coil is subjected to an alternating voltage of 100 V. The
current taken is 0.1 A and the power factor 0.2 when the frequency is 50 Hz. Find
the capacitance which, if placed in parallel with the coil, will cause the main current
320
to be a minimum. What will be the impedance of this parallel combination (a) for
currents of frequency 50 (b) for currents of frequency 40 ? [3.14 µF (a) 5000 Ω (b)
1940 Ω] (London Univ.)
115.
A circuit, consisting of a capacitor in series with a resistance of 10 Ω, is
connected in parallel with a coil having L = 55.2 mH and R = 10 Ω, to a 100-V, 50Hz supply. Calculate the value of the capacitance for which the current take from
the supply is in phase with the voltage. Show that for the particular values given,
the supply current is independent of the frequency. [153 µF] (London Univ.)
116.
In a series-parallel circuit, the two parallel branches A and B are in series
with C. The impedances are ZA = (10 − j8) Ω, ZB = (9 − j6) Ω and ZC = (100 + j0).
Find the currents IA and IB and the phase difference between them. Draw the
phasor diagram. [IA = 12.71 ∠−30º58′ IB = 15∠−35º56′ ; 4º58′ ] (Elect. Engg. &
Electronics Bangalore Univ.)
117.
Two loads are connected in parallel to an AC source and take currents of
32 and 53 amp, respectively. If the sinusoidal waves of the component currents
are out of phase of by 36, calculate the resultant current.
118.
Two equal voltages are out of phase with each other by 60. If their
geometric sum is 300 volts, calculate the rms value of each one.
119.
The following information is given in connection with an AC source that
delivers current to three loads in parallel: IA = 30 amp; IB = 20 A and lags behind
IA by 45; IC = 40 A and leads IA by 30. Using IB as the reference phasor,
determine the resultant current and the angle between it and IA.
120.
Three type of electric device A, B, and C are connected in series across a
115-V AC source, and voltage drops across the units are equal. If EA leads EB by
60 and EC lags EB by 60, what are the emfs across A, B, and C using EB as the
reference.
121.
The resultant value of two current that are out of phase by 30 is 100 amp.
If one of them is 80 amp, what is the other?
122.
Two electric devices A and B are connected in parallel, and the rms
current in A is 50 A. If the current in B lags behind A by 60 and the line current
is 80 A, determine the current in B.
123.
A lamp load and a single-phase motor are connected in parallel. The lamp
takes 6 amp and the motor takes 4 amp from a 120-V 60-Hz source. The current
to the lamp load is in phase with the voltage, and the current to the motor lags
50. Determine the total current.
124.
Two alternators are connected to the same 2,300-volt bus bars.
Alternator A delivers 150 amp, and alternator B delivers 200 amp. The current of
alternator A leads that of B by 20. Determine the total current to the load.
125.
The current in a pure capacitor circuit is given by the equation i = 4 cos
5,000t. If the rms value of the impressed emf is 110 volts, what are equations for
the voltage and power?
126.
The current in a pure inductor circuit is given by the equation i = 4 sin
5,000t. If the rms value of the impressed emf is 110 volts, what are equations for
the voltage and power?
127.
A small AC motor used in a washing machine is, in effect, an RL circuit. If
the machine takes 311 watts and 4.5 amp from a 115-V source when operating
normally, calculate its power factor.
321
128.
An impedance coil has a resistance of 20 ohms and an inductive reactance
of 40 ohms. For what values of resistance will be connected in series with the coil
to make the overall power factor of the circuit be 0.6?
129.
An impedance coil has a resistance of 20 ohms and an inductive reactance
of 40 ohms. What additional inductance should be inserted in series in the circuit
if the overall power factor is to be reduced to 0.5?
130.
A series circuit takes 371.2 watts at a power factor of 0.8 from a 116-volt
60-Hz source. What are the values of R and L?
131.
An RL circuit takes a current of 7 amp that lags behind the 231-volt source
by 35. Calculate the power factor, power, impedance, resistance, and inductive
reactance of the circuit.
132.
What should be the capacitance of a capacitor, in series with a 250-ohm
resistor, that will limit the current to 1.2 amp when the circuit is connected to a
600-V 60-Hz source.
133.
A series circuit consists of a 66.2-F and a variable resistor. For what two
values of resistance will the power taken by the circuit be 172.8 watts, if the
impressed 60-Hz emf is 120 V?
134.
A series RLC circuit consist of a 25-ohm resistor, a 0.221-henry inductor,
and a 66.3-F capacitor. If the series circuit is connected to a 125-V variablefrequency source, what will be the frequency at which the current is 5 A.
135.
The power input to an electric motor is 8.8 kW when operating normally
from 230-V AC source. If the current is 45 amp under this condition, calculate the
power factor, reactive factor, and reactive kilovolt-amperes.
136.
An AC circuit takes a load of 160 kVA at a lagging power factor of 0.75
when connected to a 460-V source. Calculate the current, power factor, power,
reactive kilovolt-amperes, and reactive factor.
137.
When an impedance coil is connected to a 114-V 60-Hz source, the current
is 3 amp. If the current rises to 4 amp when a 116-V 25-Hz source is impressed
across the same coil, determine the values of the resistance and inductance.
138.
An RLC circuit consist of a 55.4-ohm resistor, a 170-mH inductor, and an
82.8-F capacitor. If the potential drop across the resistor is 72 volts, calculate
the emf of the 60-Hz source.
139.
An impedance coil having a resistance of 18 ohms and an inductance of
0.122 henry is connected in series with a capacitor to a 117-V 60-Hz source. What
is the microfarad value of the capacitor if the circuit current is 6.5 amp?
140.
A telephone receiver has an impedance of 306 ohm at 800 Hz and a
resistance of 60 ohms. For what value of capacitance of a capacitor in series with
the coil will the power factor be unity at 1,000 Hz?
141.
A resistor and an inductor are connected in parallel to a 120-V 60-Hz
source. If the total current ad power in the circuit are, respectively, 17 amp and
1,800 watts, calculate the values of R and L.
142.
A resistor and an inductor are connected in parallel to a 120-V 60-Hz
source. The total current ad power in the circuit are, respectively, 17 amp and
1,800 watts. If a 177-F capacitor is connected in parallel with the circuit,
calculate the values of R and L.
References
“Electrical Power Systems Technology”. Stephen W. Fardo, Dale R. Patrick
322
“Electrical Engineering”, Chester Dawes
“Electric Circuits”, Charles Siskind
“Understanding AC Circuits”, Dale R. Patrick
”Electric Circuit Fundamentals”, Thomas L. Floyd
“1001 Solved Problems in Electrical Engineering”, Romeo Rojas
“Textbook-Reviewer in Electrical Engineering”, Marcialito M. Valenzona
“Electricity: Principles and Application”, Richard J. Fowier
“A Textbook in Electrical Technology”. Volume 1. Basic Electrical Engineering in SI
System of Units. B.L. Theraja, A.K. Theraja