Siemens STEP 2000 Course

Basics of
Electricity

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Table of Contents

Introduction ..............................................................................2
Electron Theory .........................................................................4
Conductors, Insulators and Semiconductors ............................5
Electric Charges ........................................................................7
Current......................................................................................9
Voltage .................................................................................... 11
Resistance .............................................................................. 13
Simple Electric Circuit ............................................................. 15
Ohm’s Law ............................................................................. 16
DC Series Circuit .................................................................... 18
DC Parallel Circuit ...................................................................23
Series-Parallel Circuits ............................................................30
Power......................................................................................34
Magnetism .............................................................................37
Electromagnetism ..................................................................39
Introduction to AC ...................................................................42
AC Generators ........................................................................44
Frequency ...............................................................................47
Voltage and Current ................................................................48
Inductance ..............................................................................51
Capacitance ............................................................................56
Inductive and Capacitive Reactance .......................................61
Series R-L-C Circuit .................................................................67
Parallel R-L-C Circuit ................................................................69
Power and Power Factor in an AC Circuit ................................ 71
Transformers ...........................................................................75
Three-Phase Transformers ......................................................80
Review Answers .....................................................................83
Final Exam ..............................................................................84

1
 Introduction

Welcome to the first course in the STEP series,

Siemens Technical Education Program designed to prepare
our distributors to sell Siemens Energy & Automation products
more effectively. This course covers Basics of Electricity and is
designed to prepare you for subsequent courses on Siemens
Energy & Automation products.

Upon completion of Basics of Electricity you will be able to:

• Explain the difference between conductors and insulators

• Use Ohm’s Law to calculate current, voltage, and

resistance

• Calculate equivalent resistance for series, parallel, or

series-parallel circuits

• Calculate voltage drop across a resistor

• Calculate power given other basic values

• Identify factors that determine the strength and polarity of

a current-carrying coil’s magnetic field

• Determine peak, instantaneous, and effective values of an

AC sine wave

• Identify factors that effect inductive reactance and

capacitive reactance in an AC circuit

• Calculate total impedance of an AC circuit

• Explain the difference between real power and apparent

power in an AC circuit

• Calculate primary and secondary voltages of single-phase

and three-phase transformers

• Calculate kVA of a transformer

2
The objectives listed above may sound strange to you. You may
also wonder why you would need to know these things to sell
Siemens Energy & Automation products. Developing a basic
knowledge of electrical concepts, however, will help you to
better understand customer applications. In addition, you will be
better able to describe products to customers and determine
important differences between products.

If you are an employee of a Siemens Energy & Automation

authorized distributor, fill out the final exam tear-out card and
mail in the card. We will mail you a certificate of completion if
you score a passing grade. Good luck with your efforts.

3
 Electron Theory

Elements of an Atom All matter is composed of molecules which are made up of a

combination of atoms. Atoms have a nucleus with electrons
orbiting around it. The nucleus is composed of protons and
neutrons (not shown). Most atoms have an equal number of
electrons and protons. Electrons have a negative charge (-).
Protons have a positive charge (+). Neutrons are neutral. The
negative charge of the electrons is balanced by the positive
charge of the protons. Electrons are bound in their orbit by
the attraction of the protons. These are referred to as bound
electrons.

Electron

Proton

Nucleus

Free Electrons Electrons in the outer band can become free of their orbit
by the application of some external force such as movement
through a magnetic field, friction, or chemical action. These are
referred to as free electrons. A free electron leaves a void which
can be filled by an electron forced out of orbit from another
atom. As free electrons move from one atom to the next an
electron flow is produced. This is the basis of electricity.

4
 Conductors, Insulators and Semiconductors

Conductors An electric current is produced when free electrons move from

one atom to the next. Materials that permit many electrons to
move freely are called conductors. Copper, silver, aluminum,
zinc, brass, and iron are considered good contors. Copper is the
most common maal used for contors and is relatively insive.

Insulators Materials that allow few free electrons are called insulators.
Materials such as plastic, rubber, glass, mica, and ceramic are
good insulators.

An electric cable is one example of how conductors and

insulators are used. Electrons flow along a copper conductor to
provide energy to an electric device such as a radio, lamp, or a
motor. An insulator around the outside of the copper conductor
is provided to keep electrons in the conductor.
Rubber Insulator
Copper Conductor

5
Semiconductors Semiconductor materials, such as silicon, can be used
to manufacture devices that have characteristics of both
conductors and insulators. Many semiconductor devices will
act like a conductor when an external force is applied in one
direction. When the external force is applied in the opposite
direction, the semiconductor device will act like an insulator.
This principle is the basis for transitors, diodes, and other solid-
state electronic devices.

Transistor Diode

Review 1
1. List the three basic elements of an atom and state the
charge of each (positive, negative, or neutral).

Element Charge

____________ ____________

____________ ____________

____________ ____________

2. An electron forced out of orbit by an external force is

called a ____________ ____________ .

3. Conductors allow ____________ free electrons to flow

when an external electric force is applied.

4. Which of the following materials are good conductors?

a. copper e. aluminum
b. plastic f. glass
c. silver g. iron
d. rubber h. mica

5. Semiconductor devices can be manufactured to allow

____________ electrons to flow in one direction and ___
_________ electrons to flow in the opposite direction.

6
 Electric Charges

Neutral State of an Atom Elements are often identified by the number of electrons in
orbit around the nucleus of the atoms making up the element
and by the number of protons in the nucleus. A hydrogen
atom, for example, has only one electron and one proton. An
aluminum atom (illustrated) has 13 electrons and 13 protons. An
atom with an equal number of electrons and protons is said to
be electrically neutral.

Outer Band

Positive and Electrons in the outer band of an atom are easily displaced by
Negative Charges the application of some external force. Electrons which are
forced out of their orbits can result in a lack of electrons where
they leave and an excess of electrons where they come to rest.
The lack of electrons is called a positive charge because there
are more protons than electrons. The excess of electrons has a
negative charge. A positive or negative charge is caused by an
absence or excess of electrons. The number of protons remains
constant.

Neutral Charge Negative Charge Positive Charge

7
Attraction and Repulsion of The old saying, “opposites attract,” is true when dealing with
Electric Charges electric charges. Charged bodies have an invisible electric
field around them. When two like-charged bodies are brought
together, their electric field will work to repel them. When two
unlike-charged bodies are brought together, their electric field
will work to attract them. The electric field around a charged
body is represented by invisible lines of force. The invisible
lines of force represent an invisible electrical field that causes
the attraction and repulsion. Lines of force are shown leaving a
body with a positive charge and entering a body with a negative
charge.
Unlike Charges Attract Like Charges Repel

Coulomb’s Law During the 18th century a French scientist, Charles A. Coulomb,
studied fields of force that surround charged bodies. Coulomb
discovered that charged bodies attract or repel each other
with a force that is directly proportional to the product of the
charges, and inversely proportional to the square of the distance
between them. Today we call this Coulomb’s Law of Charges.
Simply put, the force of attraction or repulsion depends on
the strength of the charged bodies, and the distance between
them.

8
 Current

Electricity is the flow of free electrons in a conductor from

one atom to the next atom in the same general direction. This
flow of electrons is referred to as current and is designated
by the symbol “I”. Electrons move through a conductor at
different rates and electric current has different values. Current
is determined by the number of electrons that pass through
a cross-section of a conductor in one second. We must
remember that atoms are very small. It takes about
1,000,000,000,000,000,000,000,000 atoms to fill one cubic
centimeter of a copper conductor. This number can be simplified
using mathematical exponents. Instead of writing 24 zeros after
the number 1, write 1024. Trying to measure even small values
of current would result in unimaginably large numbers. For this
reason current is measured in amperes which is abbreviated
“amps”. The letter “A” is the symbol for amps. A current of one
amp means that in one second about 6.24 x 1018 electrons
move through a cross-section of conductor. These numbers are
given for information only and you do not need to be concerned
with them. It is important, however, that the concept of current
flow be unstood.

Units of Measurement The following chart reflects special prefixes that are used when
dealing with very small or large values of current:

Prefix Symbol Decimal

1 kiloampere 1 kA 1000 A
1 milliampere 1 mA 1/1000 A
1 microampere 1 mA 1/1,000,000 A

9
Direction of Current Flow Some authorities distinguish between electron flow and
current flow. Conventional current flow theory ignores the
flow of electrons and states that current flows from positive
to negative. To avoid confusion, this book will use the electron
flow concept which states that electrons flow from negative to
positive.

_ + _ +
Electron Flow Conventional
Current Flow

10
Voltage

Electricity can be compared with water flowing through a pipe.

A force is required to get water to flow through a pipe. This
force comes from either a water pump or gravity. Voltage is the
force that is applied to a conductor that causes electric current
to flow.
Water Flow Through a Pipe

Current Flow Through a Conductor

Electrons are negative and are attracted by positive charges.

They will always be attracted from a source having an excess
of electrons, thus having a negative charge, to a source having
a deficiency of electrons which has a positive charge. The force
required to make electicity flow through a conductor is called
a difference in potential, electromotive force (emf), or more
simply referred to as voltage. voltage is designated by the letter
“E”, or the letter “V”. The unit of measurement for voltage is
volts which is also designated by the letter “V”.

11
Voltage Sources An electrical voltage can be generated in various ways. A
battery uses an electrochemical process. A car’s alternator and
a power plant generator utilizes a magnetic induction process.
All voltage sources share the characteristic of an excess of
electrons at one terminal and a shortage at the other terminal.
This results in a difference of potential between the two
terminals.
Shortage of Electrons

Excess of Electrons

+ _

Batter y

Voltage Circuit Symbol The terminals of a battery are indicated symbolically on an

electrical drawing by two lines. The longer line indicates the
positive terminal. The shorter line indicates the negative
terminal.

+
_

Units of Measurement The following chart reflects special prefixes that are used when
dealing with very small or large values of voltage:

Prefix Symbol Decimal

1 kilovolt 1 kV 1000 V
1 millivolt 1 mV 1/1000 V
1 microvolt 1 mV 1/1,000,000 V

12
 Resistance

A third factor that plays a role in an electrical circuit is

resistance. All material impedes the flow of electrical current
to some extent. The amount of resistance depends upon
composition, length, cross-section and temperature of the
resistive material. As a rule of thumb, resistance of a conductor
increases with an increase of length or a decrease of cross-
section. Resistance is designated by the symbol “R”. The unit of
measurement for resistance is ohms (Ω).

Resistance Circuit Symbols Resistance is usually indicated symbolically on an electrical

drawing by one of two ways. An unfilled rectangle is commonly
used. A zigzag line may also be used.

Resistance can be in the form of various components. A

resistor may be placed in the circuit, or the circuit might contain
other devices that have resistance.

Units of Measurement The following chart reflects special prefixes that are commonly
used when dealing with values of resistance:

Prefix Symbol Decimal

1 kilohm 1 kΩ 1000 Ω
1 megohm 1 MΩ 1,000,000 Ω

13
Review 2
1. Elements are identified by the number of ____________
in orbit around the nucleus.

2. A material that has an excess of electrons is said to

have a ____________ charge.

3. A material that has a deficiency of electrons is said to

have a ____________ charge.

4. Like charges ____________ and unlike charges

____________ .

5. The force that is applied to a conductor to cause current

flow is ____________ .

6. Electrons move from ____________ .

a. positive to negative
b. negative to positive

7. With an increase of length or a decrease of cross-

section of a conductor, resistance will ____________ .

a. increase
b. decrease

14
 Simple Electric Circuit

An Electric Circuit A fundamental relationship exists between current, voltage, and

resistance. A simple electric circuit consists of a voltage source,
some type of load, and a conductor to allow electrons to flow
between the voltage source and the load. In the following
circuit a battery provides the voltage source, electrical wire is
used for the conductor, and a light provides the resistance. An
additional component has been added to this circuit, a switch.
There must be a complete path for current to flow. If the switch
is open, the path is incomplete and the light will not illuminate.
Closing the switch completes the path, allowing electrons to
leave the negative terminal and flow through the light to the
positive terminal.

Switch _ _
+ +

An Electrical Circuit The following schematic is a representation of an electrical

Schematic circuit, consisting of a battery, a resistor, a voltmeter and an
ammeter. The ammeter, connected in series with the circuit,
will show how much current flows in the circuit. The voltmeter,
connected across the voltage source, will show the value of
voltage supplied from the battery. Before an analysis can be
made of a circuit, we need to understand Ohm’s Law.

+ _
A
+
+
_ V R
_

15
 Ohm’s Law

George Simon Ohm The relationship between current, voltage and resistance was
and Ohm’s Law studied by the 19th century German mathematician, George
Simon Ohm. Ohm formulated a law which states that current
varies directly with voltage and inversely with resistance. From
this law the following formula is derived:

E Voltage
I= or Current =
R Resistance

Ohm’s Law is the basic formula used in all electrical circuits.

Electrical designers must decide how much voltage is needed
for a given load, such as computers, clocks, lamps and motors.
Decisions must be made concerning the relationship of current,
voltage and resistance. All electrical design and analysis begins
with Ohm’s Law. There are three mathematical ways to express
Ohm’s Law. Which of the formulas is used depends on what
facts are known before starting and what facts need to be
known.

E E
I= E=IxR R=
R I

Ohm’s Law Triangle There is an easy way to remember which formula to use. By
arranging current, voltage and resistance in a triangle, one can
quickly determine the correct formula.

16
Using the Triangle To use the triangle, cover the value you want to calculate. The
remaining letters make up the formula.

E E
I= E=IxR R=
R I

Ohm’s Law can only give the correct answer when the correct
values are used. Remember the following three rules:

• Current is always expressed in amperes or amps

• Voltage is always expressed in volts
• Resistance is always expressed in ohms

Examples of Solving Using the simple circuit below, assume that the voltage
Ohm’s Law supplied by the battery is 10 volts, and the resistance is 5 Ω.

+ _
A
+
+
_ V R
_

To find how much current is flowing through the circuit, cover

the “I” in the triangle and use the resulting equation.

E 10 Volts
I= I= I = 2 Amps
R 5Ω

Using the same circuit, assume the ammeter reads 200 mA

and the resistance is known to be 10 Ω. To solve for voltage,
cover the “E” in the triangle and use the resulting equation.

E=IxR E = 0.2 x 10 E = 2 Volts

Remember to use the correct decimal equivalent when dealing

with numbers that are preceded with milli (m), micro (µ) or kilo
(k). In this example had 200 been used instead of converting
the value to 0.2, the wrong answer of 2000 volts would have
been calculated.

17
 DC Series Circuit

Resistance in a A series circuit is formed when any number of resistors are

Series Circuit connected end-to-end so that there is only one path for current
to flow. The resistors can be actual resistors or other devices
that have resistance. The following illustration shows four
resistors connected end-to-end. There is one path of current
flow from the negative terminal of the battery through R4, R3,
R2, R1 returning to the positive terminal.

R1 R2 R3 R4
+
_

Formula for Series The values of resistance add in a series circuit. If a 4 Ω

Resistance resistor is placed in series with a 6 Ω resistor, the total value
will be 10 Ω. This is true when other types of resistive devices
are placed in series. The mathematical formula for resistance in
series is:

Rt = R1 + R2 + R3 + R4 + R5

11 KΩ 2 KΩ 2 KΩ 100 Ω 1 KΩ

R1 R2 R3 R4 R5
+
_

Rt = R1 + R2 + R3 + R4 + R5
Rt = 11,000 + 2,000 + 2,000 + 100 + 1,000
Rt = 16,100 Ω

18
Current in a Series Circuit The equation for total resistance in a series circuit allows us to
simplify a circuit. Using Ohm’s Law, the value of current can be
calculated. Current is the same anywhere it is measured in a
series circuit.

E
I=
R
12
I=
10
I = 1.2 Amps

5Ω 1Ω 2Ω 2Ω 10 Ω

R1 R2 R3 R4 Rt

+ +
_ _
12 Volts 12 Volts

Original Circuit Equivalent

Circuit

Voltage in a Series Circuit Voltage can be measured across each of the resistors in a
circuit. The voltage across a resistor is referred to as a volt
age drop. A German physicist, Kirchhoff, formulated a law which
states the sum of the voltage drops across the resistances of
a closed circuit equals the total voltage applied to the circuit. In
the following illustration, four equal value resistors of 1.5 Ω each
have been placed in series with a 12 volt battery. Ohm’s Law
can be applied to show that each resistor will “drop” an equal
amount of voltage.

12 V

3V 3V 3V 3V

1.5 Ω 1.5 Ω 1.5 Ω 1.5 Ω

R1 R2 R3 R4

12 Volt Battery
_
+

19
 First, solve for total resistance:

Rt = R1 + R2 + R3 + R4
Rt = 1.5 + 1.5 + 1.5 + 1.5
Rt = 6 Ω

Second, solve for current:

I= E
R

I = 12
6
I = 2 Amps

Third, solve for voltage across any resistor:

E=IxR
E = 2 x 1.5
E = 3 Volts

If voltage were measured across any single resistor, the

meter would read three volts. If voltage were read across a
combination of R3 and R4 the meter would read six volts. If
voltage were read across a combination of R2, R3, and R4 the
meter would read nine volts. If the voltage drops of all four
resistors were added together the sum would be 12 volts, the
original supply voltage of the battery.

Voltage Division in a It is often desirable to use a voltage potential that is lower than
Series Circuit the supply voltage. To do this, a voltage divider, similar to the
one illustrated, can be used. The battery represents Ein which in
this case is 50 volts. The desired voltage is represented by Eout,
which mathematically works out to be 40 volts. To calculate this
voltage, first solve for total resistance.

Rt = R1 + R2
Rt = 5 + 20
Rt = 25 Ω

20
Second, solve for current:

Ein
I=
Rt
50
I=
25
I = 2 Amps

Finally, solve for voltage:

Eout = I x R2
Eout = 2 x 20
Eout = 40 Volts

R1 5Ω
+
_
Ein
Eout
R2 20 Ω 40 Volts

21
Review 3
1. The basic Ohm’s Law formula is ____________ .

2. When solving circuit problems; current must always be

expressed in ____________ , voltage must always be
expressed in ____________ and resistance must always
be expressed in ____________ .

3. The total current of a simple circuit with a voltage

supply of 12 volts and a resistance of 24 Ω is
____________ amps.

4. What is the total resistance of a series circuit with the

following values: R1=10 Ω, R2=15 Ω, and R3=20 Ω?
____________ Ω.

5. What is total current of a series circuit that has a 120

volt supply and 60 Ω resistance?

6. In the following circuit the voltage dropped across R1 is

____________ volts and R2 is ____________ volts.

1.5 Ω 1.5 Ω

R1 R2

+
_ 12 Volts

7. In the following circuit voltage dropped across R1 is

____________ volts, and R2 is ____________ volts.

5Ω 20 Ω

R1 R2

+
_ 100 Volts

22
 DC Parallel Circuit

Resistance in a A parallel circuit is formed when two or more resistances are

Parallel Circuit placed in a circuit side-by-side so that current can flow through
more than one path. The illustration shows two resistors placed
side-by-side. There are two paths of current flow. One path is
from the negative terminal of the battery through R1 returning
to the positive terminal. The second path is from the negative
terminal of the battery through R2 returning to the positive
terminal of the battery.

+
_ R1 R2

Formula for Equal To determine the total resistance when resistors are of equal
Value Resistors in a value in a parallel circuit, use the following formula:
Parallel Circuit
Value of any one Resistor
Rt =
Number of Resistors

In the following illustration there are three 15 Ω resistors. The

total resistance is:

Value of any one Resistor

Rt =
Number of Resistor

15
Rt =
3
Rt = 5 Ω

+ R1 R1 R1
_
15 Ω 15 Ω 15 Ω

23
Formula for Unequal There are two formulas to determine total resistance for
Resistors in a Parallel Circuit unequal value resistors in a parallel circuit. The first formula is
used when there are three or more resistors. The formula can
be extended for any number of resistors.

1 = 1 + 1 + 1
Rt R1 R2 R3

In the following illustration there are three resistors, each of

different value. The total resistance is:

1 = 1 + 1 + 1
Rt R1 R2 R3

1 = 1 + 1 + 1
Insert Value of the Resistors
Rt 5 10 20

1 = 4 + 2 + 1
Find Lowest Common Denominator
Rt 20 20 20
1 = 7 Add the Numerators
Rt 20
Rt = 20 Invert Both Sides of the Equation
1 7
Rt = 2.86 Ω Divide

+
R1 R2 R3
_
5Ω 10 Ω 20 Ω

24
 The second formula is used when there are only two resistors.

R1 x R2
Rt =
R1 + R2

In the following illustration there are two resistors, each of

different value. The total resistance is:

R1 x R2
Rt =
R1 + R2
5 x 10
Rt =
5 + 10
50
Rt =
15
Rt = 3.33 Ω

+
R1 R2
_
5Ω 10 Ω

Voltage in a When resistors are placed in parallel across a voltage source,

Parallel Circuit the voltage is the same across each resistor. In the following
illustration three resistors are placed in parallel across a 12 volt
battery. Each resistor has 12 volts available to it.

12 Volt
Battery
+
_ R1 12 V R2 12 V R3 12 V

25
Current in a Current flowing through a parallel circuit divides and flows
Parallel Circuit through each branch of the circuit.

It

+
_ R1 R2 R3

I1 I2 I3

It

Total current in a parallel circuit is equal to the sum of the

current in each branch. The following formula applies to current
in a parallel circuit.

It = I 1 + I 2 + I 3

Current Flow with Equal When equal resistances are placed in a parallel circuit,
Value Resistors in a opposition to current flow is the same in each branch. In the
Parallel Circuit following circuit R1 and R2 are of equal value. If total current (It)
is 10 amps, then 5 amps would flow through R1 and 5 amps
would flow through R2.

It = 10 Amps

+
_ R1 R2

I1 = I2 =
5 Amps 5 Amps
It = 10 Amps

It = I 1 + I 2
It = 5 Amps + 5 Amps
It = 10 Amps

26
Current Flow with Unequal When unequal value resistors are placed in a parallel circuit,
Value Resistors in a opposition to current flow is not the same in every circuit
Parallel Circuit branch. Current is greater through the path of least resistance.
In the following circuit R1 is 40 Ω and R2 is 20 Ω. Small values
of resistance means less opposition to current flow. More
current will flow through R2 than R1.

12 Volts
+
_ R1 R2
40 Ω 20 Ω
I1 = I2 =
0.3 Amps 0.6 Amps
It = 0.9 Amps

Using Ohm’s Law, the total current for each circuit can be
calculated.

I1 = E
R1

I1 = 120 Volts
40 Ω
I1 = 0.3 Amps

I2 = E
R2

I2 = 120 Volts
20 Ω
I2 = 0.6 Amps

It = I 1 + I 2
It = 0.3 Amps + 0.6 Amps
It = 0.9 Amps

27
 Total current can also be calculated by first calculating total
resistance, then applying the formula for Ohm’s Law.

R1 x R2
Rt =
R1 + R2
40 Ω x 20 Ω
Rt =
40 Ω + 20 Ω
800 Ω
Rt =
60 Ω

Rt = 13.333 Ω

It = E
Rt

It = 12 Volts
13.333 Ω

It = 0.9 Amps

28
Review 4
1. The total resistance of a parallel circuit that has four
20 Ω resistors is ____________ Ω.

2. Rt for the following circuit is ____________ Ω.

+ R1 R2 R3
_
10 Ω 20 Ω 30 Ω

3. Rt for the following circuit is ____________ Ω.

+ R1 R2
_
5Ω 10 Ω

4. Voltage available at R2 in the following circuit is

____________ volts.

12 Volts
+ R1 R2
_
5Ω 10 Ω

5. In a parallel circuit with two resistors of equal value and

a total current flow of 12 amps, the value of current
through each resistor is ____________ amps.

6. In the following circuit current flow through R1 is _____

_______ amps, and R2 is ____________ amps.

24 Volts
+ R1 R2
_
10 Ω 10 Ω

29
 Series-Parallel Circuits

Series-parallel circuits are also known as compound circuits. At

least three resistors are required to form a series-parallel circuit.
The following illustrations show two ways a series-parallel
combination could be found.

Parallel Branches

+
_

Parallel Branches

Devices in Series

+
_

Simplifying a Series-Parallel The formulas required for solving current, voltage and resistance
problems have already been defined. To solve a series-parallel
circuit, reduce the compound circuits to equivalent simple
circuits. In the following illustration R1 and R2 are parallel with
each other. R3 is in series with the parallel circuit of R1 and R2.

R1 10 Ω
R3 10 Ω

+
_ R2 10 Ω

30
 First, use the formula to determine total resistance of a parallel
circuit to find the total resistance of R1 and R2. When the
resistors in a parallel circuit are equal, the following formula is
used:

Value of any One Resistor

R =
Number of Resistors

R = 10 Ω
2

R = 5Ω

Second, redraw the circuit showing the equivalent values. The

result is a simple series circuit which uses already learned
equations and methods of problem solving.

R3 10Ω R3 5Ω

+
_

Simplifying a In the following illustration R1 and R2 are in series with each

Series-Parallel Circuit other. R3 is in parallel with the series circuit of R1 and R2.
to a Parallel Circuit

R1 10Ω
+
_ R3 20Ω

R2 10Ω

First, use the formula to determine total resistance of a series

circuit to find the total resistance of R1 and R2. The following
formula is used:

R = R1 + R 2
R = 10 Ω + 10 Ω
R = 20 Ω

31
 Second, redraw the circuit showing the equivalent values. The
result is a simple parallel circuit which uses already learned
equations and methods of problem solving.

+
_
R = 20 Ω R3 = 20 Ω

+
_ Rt = 10 Ω

32
Review 5
1. Calculate equivalent resistance for R1 and R2 and total
resistance for the entire circuit.

R1 20 Ω
R3 10 Ω

+
_ R2 30 Ω

R1/R2 equivalent resistance = ____________ Ω

Total resistance = ____________ Ω

2. Calculate equivalent resistance for R1 and R2 and total

resistance for the entire circuit.

R1 30 Ω
+
_ R3 20 Ω

R2 10 Ω

R1/R2 equivalent resistance = ____________ Ω

Total resistance = ____________ Ω

33
 Power

Work Whenever a force of any kind causes motion, work is

accomplished. In the illustration below work is done when a
mechanical force is used to lift a weight. If a force were exerted
without causing motion, then no work is done.

Electric Power In an electrical circuit, voltage applied to a conductor will cause

electrons to flow. Voltage is the force and electron flow is the
motion. The rate at which work is done is called power and is
represented by the symbol “P”. Power is measured in watts
and is represented by the symbol “W”. The watt is defined as
the rate work is done in a circuit when 1 amp flows with 1 volt
applied.

Power Formulas Power consumed in a resistor depends on the amount of

current that passes through the resistor for a given voltage. This
is expressed as voltage times current.

P=ExI

or

P = EI

Power can also be calculated by substituting other components

of Ohm’s Law.

P = I2R

and
E2
P=
R

34
Solving a Power Problem In the following illustration power can be calculated using any of
the power formulas.

I = 2 Amps

+
_ R=6Ω
12 Volts

P = EI
P = 12 Volts x 2 Amps
P = 24 Watts

P = I2 R
P = (2 Amps)2 x 6 Ω
P = 24 Watts

E2
P=
R
(12 Volts)2
P=
6Ω
144
P=
6

P = 24 Watts

Power Rating of Equipment Electrical equipment is rated in watts. This rating is an indication
of the rate at which electrical equipment converts electrical
energy into other forms of energy, such as heat or light. A
common household lamp may be rated for 120 volts and 100
watts. Using Ohm’s Law, the rated value of resistance of the
lamp can be calculated.

E2 which can be transposed to R = E

2
P=
R P
(120 Volts)2
R=
100 Watts

R = 144 Ω

35
 Using the basic Ohm’s Law formula, the amount of current flow
for the 120 volt, 100 watt lamp can be calculated.

E
I=
R
120 Volts
I=
144 Ω

I = 0.833 Amps

A lamp rated for 120 volts and 75 watts has a resistance of

192 Ω and a current of 0.625 amps would flow if the lamp had
the rated voltage applied to it.

E2
R=
P
(120 Volts)2
R=
75 Watts

R = 192 Ω

E2
I=
P
120 Volts
I=
192 Ω

I = 0.625 Amps

It can be seen that the 100 watt lamp converts energy faster
than the 75 watt lamp. The 100 watt lamp will give off more
light and heat.

Heat Current flow through a resistive material causes heat. An

electrical component can be damaged if the temperature is
too high. For this reason, electrical equipment is often rated for
a maximum wattage. The higher the wattage rating, the more
heat the equipment can dissipate.

36
 Magnetism

The principles of magnetism are an integral part of electricity.

Electromagnets are used in some direct current circuits.
Alternating current cannot be understood without first
understanding magnetism.

Types of Magnets The three most common forms of magnets are the horse-shoe,
bar and compass needle.

All magnets have two characteristics. They attract and hold

iron. If free to move, like the compass needle, the magnet will
assume roughly a north-south position.

Magnetic Lines of Flux Every magnet has two poles, one north pole and one south
pole. Invisible magnetic lines of flux leave the north pole and
enter the south pole. While the lines of flux are invisible, the
effects of magnetic fields can be made visible. When a sheet
of paper is placed on a magnet and iron filings loosely scattered
over it, the filings will arrange themselves along the invisible
lines of flux.

37
 By drawing lines the way the iron filings have arranged
themselves, the following picture is obtained. Broken lines
indicate the paths of magnetic flux lines. The field lines exist
outside and inside the magnet. The magnetic lines of flux
always form closed loops. Magnetic lines of flux leave the
north pole and enter the south pole, returning to the north pole
through the magnet.

Interaction between When two magnets are brought together, the magnetic flux
Two Magnets field around the magnet causes some form of interaction. Two
unlike poles brought together cause the magnets to attract each
other. Two like poles brought together cause the magnets to
repel each other.

38
 Electromagnetism

Left-Hand Rule for An electromagnetic field is a magnetic field generated by

Conductors current flow in a conductor. Whenever current flows a magnetic
field exists around the conductor. Every electric current
generates a magnetic field. A definite relationship exists
between the direction of current flow and the direction of the
magnetic field. The left-hand rule for conductors demonstrates
this relationship. If a current-carrying conductor is grasped
with the left hand with the thumb pointing in the direction
of electron flow, the fingers will point in the direction of the
magnetic lines of flux.

Current-Carrying Coil A coil of wire carrying a current, acts like a magnet. Individual
loops of wire act as small magnets. The individual fields add
together to form one magnet. The strength of the field can be
increased by adding more turns to the coil. The strength can
also be increased by increasing the current.

39
Left-Hand Rule for Coils A left-hand rule exists for coils to determine the direction of the
magnetic field. The fingers of the left hand are wrapped around
the coil in the direction of electron flow. The thumb points to the
north pole of the coil.

Electromagnets An electromagnet is composed of a coil of wire wound around

a core. The core is usually a soft iron which conducts magnetic
lines of force with relative ease. When current is passed
through the coil, the core becomes magnetized. The ability to
control the strength and direction of the magnetic force makes
electromagnets useful. As with permanent magnets, opposite
poles attract. An electromagnet can be made to control the
strength of its field which controls the strength of the magnetic
poles.

A large variety of electrical devices such as motors,

circuit breakers, contactors, relays and motor starters use
electromagnetic principles.

40
Review 6
1. The rate at which work is done is called
___________ .

2. The basic formula for power is ____________ .

3. In a circuit with a 12 volt supply and 4 Ω resistance the

power consumed is ____________ watts.

4. The two characteristics of all magnets are; they attract

and hold ____________ , and if free to move will assume
roughly a ____________ position.

5. Lines of flux always leave the ____________ pole and

enter the ____________ pole.

6. The left-hand rule for conductors states that when

the ___________ hand is placed on a current-carrying
conductor with the ____________ pointing in the
direction of electron flow, the fingers will point in the
direction of ____________ .

41
 Introduction to AC

The supply of current for electrical devices may come from a

direct current source (DC), or an alternating current source (AC).
In direct current electricity, electrons flow continuously in one
direction from the source of power through a conductor to a
load and back to the source of power. Voltage in direct current
remains constant. DC power sources include batteries and DC
generators. In allowing current an AC generator is used to make
electrons flow first in one direction then in another. Another
name for an AC generator is an alternator. The AC generator
reverses terminal polarity many times a second. Electrons will
flow through a conductor from the negative terminal to the
positive terminal, first in one direction then another.

42
AC Sine Wave Alternating voltage and current vary continuously. The graphic
representation for AC is a sine wave. A sine wave can represent
current or voltage. There are two axes. The vertical axis
represents the direction and magnitude of current or voltage.
The horizontal axis represents time.

+ Direction

0
Time

- Direction

When the waveform is above the time axis, current is flowing in

one direction. This is referred to as the positive direction. When
the waveform is below the time axis, current is flowing in the
opposite direction. This is referred to as the negative direction.
A sine wave moves through a complete rotation of 360
degrees, which is referred to as one cycle. Alternating current
goes through many of these cycles each second. The unit of
measurement of cycles per second is hertz. In the United
States alternating current is usually generated at 60 hertz.

Single-Phase and Alternating current is divided into single-phase and three-phase

Three-Phase AC Power types. Single-phase power is used for small electrical demands
such as found in the home. Three-phase power is used where
large blocks of power are required, such as found in commercial
applications and industrial plants. Single-phase power is shown
in the above illustration. Three-phase power, as shown in the
following illustration, is a continuous series of three overlapping
AC cycles. Each wave represents a phase, and is offset by 120
electrical degrees.

Phase 1 Phase 2 Phase 3

+

43
 AC Generators

Basic Generator A basic generator consists of a magnetic field, an armature, slip

rings, brushes and a resistive load. The magnetic field is usually
an electromagnet. An armature is any number of conductive
wires wound in loops which rotates through the magnetic field.
For simplicity, one loop is shown. When a conductor is moved
through a magnetic field, a voltage is induced in the conductor.
As the armature rotates through the magnetic field, a voltage
is generated in the armature which causes current to flow. Slip
rings are attached to the armature and rotate with it. Carbon
brushes ride against the slip rings to conduct current from the
armature to a resistive load.

Pole Piece

Magnetic Field

Armature

R1
Brush

Slip Ring

Basic Generator Operation An armature rotates through the magnetic field. At an initial
position of zero degrees, the armature conductors are moving
parallel to the magnetic field and not cutting through any
magnetic lines of flux. No voltage is induced.

R1

44
Generator Operation from The armature rotates from zero to 90 degrees. The conductors
Zero to 90 Degrees cut through more and more lines of flux, building up to a
maximum induced voltage in the positive direction.

90
Degrees

R1

Generator Operation from The armature continues to rotate from 90 to 180 degrees,
90 to 180 Degrees cutting less lines of flux. The induced voltage decreases from a
maximum positive value to zero.

180
Degrees
S

R1

Generator Operation from The armature continues to rotate from 180 degrees to 270
degrees. The conductors cut more and more lines of flux, but
in the opposite direction. voltage is induced in the negative
direction building up to a maximum at 270 degrees.

270
Degrees

R1

45
Generator Operation from The armature continues to rotate from 270 to 360 degrees.
270 to 360 Degrees Induced voltage decreases from a maximum negative value to
zero. This completes one cycle. The armature will continue to
rotate at a constant speed. The cycle will continuously repeat as
long as the armature rotates.

360
Degrees
S

One Revolution

R1

46
 Frequency

The number of cycles per second made by voltage induced in

the armature is the frequency of the generator. If the armature
rotates at a speed of 60 revolutions per second, the generated
voltage will be 60 cycles per second. The accepted term for
cycles per second is hertz. The standard frequency in the United
States is 60 hertz. The following illustration shows 15 cycles in
1/4 second which is equivalent to 60 cycles in one second.

1/4 Second

Four-Pole AC Generator The frequency is the same as the number of rotations per
second if the magnetic field is produced by only two poles.
An increase in the number of poles, would cause an increase
in the number of cycles completed in a revolution. A two-pole
generator would complete one cycle per revolution and a four-
pole generator would complete two cycles per revolution. An
AC generator produces one cycle per revolution for each pair of
poles.

One Revolution

R1

47
 Voltage and Current

Peak Value The sine wave illustrates how voltage and current in an AC
circuit rises and falls with time. The peak value of a sine wave
occurs twice each cycle, once at the positive maximum value
and once at the negative maximum value.
Peak Value
+

0
Time

- Peak Value

Peak-to-Peak Value The value of the voltage or current between the peak positive
and peak negative values is called the peak-to-peak value.

Peak-to-Peak
0 Value
Time

Instantaneous Value The instantaneous value is the value at any one particular time.
It can be in the range of anywhere from zero to the peak value.

+
Instantaneous Value

0
Time

48
Calculating Instantaneous The voltage waveform produced as the armature rotates
Voltage through 360 degrees rotation is called a sine wave because
instantaneous voltage is related to the trigonometric function
called sine (sin θ = sine of the angle). The sine curve represents
a graph of the following equation:

e = Epeak x sin θ

Instantaneous voltage is equal to the peak voltage times the

sine of the angle of the generator armature. The sine value is
obtained from trigonometric tables. The following table reflects
a few angles and their sine value.

Angle Sin θ Angle Sin θ

30 Degrees 0.5 210 Degrees -0.5

60 Degrees 0.866 240 Degrees -0.866

90 Degrees 1 270 Degrees -1

120 Degrees 0.866 300 Degrees -0.866

150 Degrees 0.5 330 Degrees -0.5

180 Degrees 0 360 Degrees 0

The following example illustrates instantaneous values at 90,

150, and 240 degrees. The peak voltage is equal to 100 volts.
By substituting the sine at the instantaneous angle value, the
instantaneous voltage can be calculated.

+ 90° = +100 Volts

150° = +50 Volts

240° = -86.6 Volts

-

Any instantaneous value can be calculated. For example:

240°
e = 100 x -0.866
e = -86.6 volts

49
Effective Value of an Alternating voltage and current are constantly changing values.
AC Sine Wave A method of translating the varying values into an equivalent
constant value is needed. The effective value of voltage and
current is the common method of expressing the value of AC.
This is also known as the RMS (root-mean-square) value. If the
voltage in the average home is said to be 120 volts, this is the
RMS value. The effective value figures out to be 0.707 times
the peak value.
+

Peak Value
169.7 Volts

169.7 Vpeak x 0.707 = 120 Vrms

The effective value of AC is defined in terms of an equivalent
heating effect when compared to DC. One RMS ampere of
current flowing through a resistance will produce heat at the
same rate as a DC ampere.

For purpose of circuit design, the peak value may also be

needed. For example, insulation must be designed to withstand
the peak value, not just the effective value. It may be that
only the effective value is known. To calculate the peak value,
multiply the effective value by 1.41. For example, if the effective
value is 100 volts, the peak value is 141 volts.

Review 7

1. The graphic representation of AC voltage or current

values over a period of time is a ____________
____________ .

2. Each phase of three phase AC power is offset by

____________ degrees.

3. An AC generator produces ____________ cycle per

revolution for each pair of poles.

4. What is the instantaneous voltage at 240 degrees for a

peak voltage of 150 volts?

5. What is the effective voltage for a peak voltage of 150

volts?
50
 Inductance

The circuits studied to this point have been resistive.

Resistance and voltage are not the only circuit properties that
effect current flow, however. Inductance is the property of an
electric circuit that opposes any change in electric current.
Resistance opposes current flow, inductance opposes change
in current flow. Inductance is designated by the letter “L”. . The
unit of measurement for inductance is the henry (h).

Current Flow and Current flow produces a magnetic field in a conductor. The
Field Strength amount of current determines the strength of the magnetic
field. As current flow increases, field strength increases, and as
current flow decreases, field strength decreases.
0 Degrees 30 Degrees 90 Degrees
No Current Increasing Maximum
Current Current

Any change in current causes a corresponding change in the

magnetic field surrounding the conductor. Current is constant
in DC, except when the circuit is turned on and off, or when
there is a load change. Current is constantly changing in AC,
so inductance is a continual factor. A change in the magnetic
field surrounding the conductor induces a voltage in the
conductor. This self-induced voltage opposes the change in
current. This is known as counter emf. This opposition causes
a delay in the time it takes current to attain its new steady
value. If current increases, inductance tries to hold it down. If
current decreases, inductance tries to hold it up. Inductance is
somewhat like mechanical inertia which must be overcome to
get a mechanical object moving or to stop a mechanical object
from moving. A vehicle, for example, takes a few moments to
accelerate to a desired speed, or decelerate to a stop.

51
Inductors Inductance is usually indicated symbolically on an electrical
drawing by one of two ways. A curled line or a filled rectangle
can be used.

Inductors are coils of wire. They may be wrapped around a core.

The inductance of a coil is determined by the number of turns
in the coil, the spacing between the turns, the coil diameter,
the core material, the number of layers of windings, the type of
winding, and the shape of the coil. Examples of inductors are
transformers, chokes, and motors.

Simple Inductive Circuit In a resistive circuit, current change is considered

instantaneous. If an inductor is used, the current does not
change as quickly. In the following circuit, initially the switch
is open and there is no current flow. When the switch is
closed, current will rise rapidly at first, then more slowly as the
maximum value is approached. For the purpose of explanation,
a DC circuit is used.

+
_ R1

L1

Inductive Time Constant The time required for the current to rise to its maximum value is
determined by the ratio of inductance (in henrys) to resistance
(in ohms). This ratio is called the time constant of the inductive
circuit. A time constant is the time (in seconds) required for the
circuit current to rise to 63.2% of its maximum value. When
the switch is closed in the previous circuit, current will begin to
flow. During the first time constant current rises to 63.2% of its
maximum value. During the second time constant, current rises
to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes
about five time constants for current to reach its maximum
value.

52
 100.0%
98.1%
94.9%
86.4%

63.2%

First Time Second Time Third Time Fourth Time Fifth Time
Constant Constant Constant Constant Constant

Similarly, when the switch is opened, it will take five time

constants for current to reach zero. It can be seen that
inductance is an important factor in AC circuits. If the frequency
is 60 hertz, current will rise and fall from its peak value to zero
120 times a second.
100.0%
First Time Second Time Third Time Fourth Time Fifth Time
Constant Constant Constant Constant Constant

36.8%

13.6%
5.1%
1.9%
0%

Calculating the Time The time constant is designated by the symbol “τ”. To
Constant of an determine the time constant of an inductive circuit use one of
Inductive Circuit the following formulas:

L (henrys)
τ (in seconds) =
R (ohms)
L (millihenrys)
τ (in milliseconds) =
R (ohms)
L (microhenrys)
τ (in microseconds) =
R (ohms)

53
 In the following illustration, L1 is equal to 15 millihenrys and
R1 is equal to 5 Ω. When the switch is closed, it will take
3 milliseconds for current to rise from zero to 63.2% of its
maximum value.

+
_ R1 5 Ω

L1 15 mh

15 mh
τ=
5Ω
τ = 3 milliseconds

Formula for Series Inductors The same rules for calculating total resistance can be applied. In
the following circuit, an AC generator is used to supply electrical
power to four inductors. There will always be some amount of
resitance and inductance in any circuit. The electrical wire used
in the circuit and the inductors both have some resistance and
inductance. Total inductance is calculated using the following
formula:

Lt = L 1 + L 2 + L 3

2 mh 2 mh 1 mh 1 mh

R1 L1 L2 L3 L4

AC Generator

Lt = L1 + L2 + L3 + L4
Lt = 2 mh + 2 mh + 1 mh + 1 mh
Lt = 6 mh

54
Formula for Parallel In the following circuit, an AC generator is used to supply
Inductors electrical power to three inductors. Total inductance is
calculated using the following formula:

1 = 1 + 1 + 1
Lt L1 L2 L3

R1

L1 L2 L3
5 mh 10 mh 20 mh

1 = 1 + 1 + 1
Lt 5 10 20
1 = 7
Lt 20

Lt = 2.86 mh

55
 Capacitance

Capacitance and Capacitors Capacitance is a measure of a circuit’s ability to store an elec

cal charge. A device manufactured to have a specific amount of
capacitance is called a capacitor. A capacitor is made up of a
pair of conductive plates separated by a thin layer of insulating
material. Another name for the insulating material is dielectric
material. When a voltage is applied to the plates, electrons are
forced onto one plate. That plate has an excess of electrons
while the other plate has a deficiency of electrons. The plate
with an excess of electrons is negatively charged. The plate
with a deficiency of electrons is positively charged.
Negative Plate
Dielectric Material
Positive Plate

Direct current cannot flow through the dielectric material

because it is an insulator; however it can be used to charge a
capacitor. Capacitors have a capacity to hold a specific quantity
of electrons. The capacitance of a capacitor depends on the area
of the plates, the distance between the plates, and the material
of the dielectric. The unit of measurement for capacitance is
farads (F). Capacitors usually are rated in µF (microfarads), or pF
(picofarads).

Capacitor Circuit Symbols Capacitance is usually indicated symbolically on an electrical

drawing by a combination of a straight line with a curved line, or
two straight lines.

56
Simple Capacitive Circuit In a resistive circuit, voltage change is considered
instantaneous. If a capacitor is used, the voltage across the
capacitor does not change as quickly. In the following circuit,
initially the switch is open and no voltage is applied to the
capacitor. When the switch is closed, voltage across the
capacitor will rise rapidly at first, then more slowly as the
maximum value is approached. For the purpose of explanation,
a DC circuit is used.

+
_ R1

C1

Capacitive Time Constant The time required for voltage to rise to its maximum value in a
circuit containing capacitance is determined by the product of
capacitance, in farads, times resistance, in ohms. This product is
the time constant of a capacitive circuit. The time constant gives
the time in seconds required for voltage across the capacitor to
reach 63.2% of its maximum value. When the switch is closed
in the previous circuit, voltage will be applied. During the first
time constant, voltage will rise to 63.2% of its maximum value.
During the second time constant, voltage will rise to 63.2% of
the remaining 36.8%, or a total of 86.4%. It takes about five
time constants for voltage across the capacitor to reach its
maximum value.
100.0%
98.1%
94.9%
86.4%

63.2%

First Time Second Time Third Time Fourth Time Fifth Time
Constant Constant Constant Constant Constant

57
 Similarly, during this same time, it will take five time constants
for current through the resistor to reach zero.
100.0%
First Time Second Time Third Time Fourth Time Fifth Time
Constant Constant Constant Constant Constant

36.8%

13.6%
5.1%
1.9%
0%

Calculating the Time To determine the time constant of a capacitive circuit, use one
Constant of a of the following formulas:
Capacitive Circuit
τ (in seconds) = R (megohms) x C (microfarads)
τ (in microseconds) = R (megohms) x C (picofarads)
τ (in microseconds) = R (ohms) x C (microfarads)

In the following illustration, C1 is equal to 2 µF, and R1 is equal

to 10 Ω. When the switch is closed, it will take 20 microseconds
for voltage across the capacitor to rise from zero to 63.2%
of its maximum value. It will take five time constants, 100
microseconds for this voltage to rise to its maximum value.

+
_ R1 10 Ω
C1 2µF

τ = RC
τ = 2µF x 10 Ω
τ = 20 microseconds

58
Formula for Connecting capacitors in series decreases total capacitance.
Series Capacitors The effect is like increasing the space between the plates. The
formula for series capacitors is similar to the formula for parallel
resistors. In the following circuit, an AC generator supplies
electrical power to three capacitors. Total capacotance is
calculated using the following formula:

1 = 1 + 1 + 1
Ct C1 C2 C3

5 µF 10 µF 20 µF

R1
C1 C2 C3

1 = 1 + 1 + 1
Ct 5 10 20
1 = 7
Ct 20

Ct = 2.86 µF

Formula for In the following circuit, an AC generator is used to supply

Parallel Capacitors electrical power to three capacitors. Total capacitance is
calculated using the following formula:

Ct = C1 + C2 + C3

R1

C1 C2 C3
5 µF 10 µF 20 µF

Ct = 5 µF + 10 µF + 20 µF
Ct = 35 µF

59
Review 8
1. The total inductance for this circuit is ___________ .

4 mh 2 mh 3 mh 1 mh

R1 L1 L2 L3 L4

2. The total inductance for this circuit is ____________ .

R1

L1 L2 L3
5 mh 10 mh 10 mh

3. The total capacitance for this circuit is ____________ .

5 µF 10 µF 10 µF

R1
C1 C2 C3

4. The total capacitance for this circuit is ____________ .

R1

C1 C2 C3

5 µF 10 µF 10 µF

60
 Inductive and Capacitive Reactance

In a purely resistive AC circuit, opposition to current flow is

called resistance. In an AC circuit containing only inductance,
capacitance, or both, opposition to current flow is called
reactance. Total opposition to current flow in an AC circuit that
contains both reactance and resistance is called impedance,
designated by the symbol “Z”. Reactance and impedance are
expressed in ohms.

Inductive Reactance Inductance only affects current flow when the current is
changing. Inductance produces a self-induced voltage (counter
emf) that opposes changes in current. In an AC circuit, current
is changing constantly. Inductance in an AC circuit, therefore,
causes a continual opposition. This opposition to current flow is
called inductive reactance and is designated by the symbol XL.

Inductive reactance is dependent on the amount of inductance

and frequency. If frequency is low, current has more time
to reach a higher value before the polarity of the sine wave
reverses. If frequency is high, current has less time to reach
a higher value. In the following illustration, voltage remains
constant. Current rises to a higher value at a lower frequency
than a higher frequency.

+ +

Current Current

0 0

_ _

Low Frequency High Frequency

The formula for inductive reactance is:

XL = 2πfL
XL = 2 x 3.14 x frequency x inductance

61
 In a 60 hertz, 10 volt circuit containing a 10 mh inductor, the
inductive reactance would be:

XL = 2πfL
XL = 2 x 3.14 x 60 x 0.10
XL = 3.768 Ω

Once inductive reactance is known, Ohm’s Law can be used to

calculate reactive current.

E
I=
Z
10
I=
3.768
I = 2.65 Amps

Phase Relationship Current does not rise at the same time as the source voltage
between Current and in an inductive circuit. Current is delayed depending on the
Voltage in an amount of inductance. In a purely resistive circuit, current and
Inductive Circuit voltage rise and fall at the same time. They are said to be “in
phase.” In this circuit there is no inductance. Resistance and
impedance are the same.

+
Voltage

Current

In a purely inductive circuit, current lags behind voltage by 90

degrees. Current and voltage are said to be “out of phase”. In
this circuit, impedance and inductive reactance are the same.
90 Degrees
+
Voltage
Current

62
 All inductive circuits have some amount of resistance. AC
current will lag somewhere between a purely resistive circuit,
and a purely inductive circuit. The exact amount of lag depends
on the ratio of resistance and inductive reactance. The more
resistive a circuit is, the closer it is to being in phase. The more
inductive a circuit is, the more out of phase it is. In the following
illustration, resistance and inductive reactance are equal.
Current lags voltage by 45 degrees.
45 Degrees
XL = 10 Ω
+
Voltage
Current

R = 10 Ω

Calculating Impedance in When working with a circuit containing elements of inductance,

an Inductive Circuit capacitance, and resistance, impedance must be calculated.
Because electrical concepts deal with trigonometric functions,
this is not a simple matter of subtraction and addition. The
following formula is used to calculate impedance in an inductive
circuit:

Z = R2 + XL2

In the circuit illustrated above, resistance and inductive

reactance are each 10 ohms. Impedance is 14.1421 ohms.
A simple application of Ohm’s Law can be used to find total
circuit current.

Z = 102 + 102
Z= 200
Z = 14.1421 Ω

Vectors Another way to represent this is with a vector. A vector is

a graphic represention of a quantity that has direction and
magnitude. A vector on a map might indicate that one city is
50 miles southwest from another. The magnitude is 50 miles
and the direction is southwest. Vectors are also used to show
electrical relationships. As mentioned earlier, impedance (Z) is
the total opposition to current flow in an AC circuit containing
reactance, inductance, and capacitance.

63
 The following vector illustrates the relationship between
reactance and inductive reactance of a circuit containing equal
values of each. The angle between the vectors is the phase
angle represented by the symbol θ. When inductive reactance
is equal to resistance the resultant angle is 45 degrees. It is this
angle that determines how much current will lag voltage.

Ω
1 2
.14
XL = 10 Ω

14
=
Z
θ
R = 10 Ω

Capacitive Reactance Capacitance also opposes AC current flow. Capacitive reactance

is designated by the symbol XC.The larger the capacitor, the
smaller the capacitive reactance. Current flow in a capacitive AC
circuit is also dependent on frequency. The following formula is
used to calculate capacitive reactance.

1
XC =
2πfC

Capacitive reactance is equal to 1 divided by 2 times pi, times

the frequency, times the capacitance. The capacitive reactance
for a 60 hertz circuit with a 10 microfarad capacitor is:

1
XC =
2πfC
1
XC =
2 x 3.14 x 60 x 0.000010

XC = 265.39 Ω

Once capacitive reactance is known, Ohm’s Law can be used to

calculate reactive current.

E
I=
Z
10
I=
265.39
I = 0.0376 Amps

64
Phase Relationship between The phase relationship between current and voltage are
Current and Voltage opposite to the phase relationship of an inductive circuit. In a
purely capacitive circuit, current leads voltage by 90 degrees.
90 Degrees
+
Voltage

Current

All capacitive circuits have some amount of resistance. AC

current will lead somewhere between a purely resistive circuit
and a purely capacitive circuit. The exact amount of lead
depends on the ratio of resistance and capacitive reactance. The
more resistive a circuit is, the closer it is to being in phase. The
more capacitive a circuit is, the more out of phase it is. In the
following illustration, resistance and capacitive reactance are
equal. Current leads voltage by 45 degrees.
45 Degrees XC = 10 Ω
+
Voltage

Current

R = 10 Ω

Calculating Impedance in The following formula is used to calculate impedance in a

a Capacitive Circuit capacitive circuit:

Z = R2 + XC2

In the circuit illustrated above, resistance and capacitive re

tance are each 10 ohms. Impedance is 14.1421 ohms.

Z = 102 + 102
Z= 200
Z = 14.1421 Ω

65
 The following vector illustrates the relationship between
resistance and capacitive reactance of a circuit containing equal
values of each. The angle between the vectors is the phase
angle represented by the symbol θ. When capacitive reactance
is equal to resistance the resultant angle is -45 degrees. It is
this angle that determines how much current will lead voltage.
R = 10 Ω

Z
=
XC = 10 Ω

14
.14
1 2
Ω
Review 9
1. In a circuit containing inductance, capacitance, or both,
opposition to current flow is called ____________ .

2. Total opposition to current flow in a circuit that contains

both reactance and resistance is called ____________ .

3. In a 50 hertz circuit, containing a 10 mh inductor, the

inductive reactance is ____________ ohms.

4. In a purely inductive circuit, ____________

a. current and voltage are in phase

b. current leads voltage by 90 degrees
c. current lags voltage by 90 degrees

5. In a purely capacitive circuit, ____________

a. current and voltage are in phase

b. current leads voltage by 90 degrees
c. current lags voltage by 90 degrees

6. In a 50 hertz circuit, containing a 10 microfarad

capacitor, the capacitive reactance is ____________
ohms.

7. In a circuit with 5 Ω resistance, and 10 Ω inductive

reactance, impedance is ____________ ohms.

8. In a circuit with 5 Ω resistance, and 4 Ω capacitive

reactance, impedance is ____________ ohms.

66
 Series R-L-C Circuit

Circuits often contain elements of resistance, inductance, and

capacitance. In an inductive AC circuit, current lags voltage by
90 degrees. In a AC capacitive circuit, current leads voltage by
90 degrees. It can be seen that inductance and capacitance
are 180 degrees apart. Since they are 180 degrees apart, one
element will cancel out all or part of the other element.

XL

R
XC

An AC circuit is:

• Resistive if XL and XC are equal

• Inductive if XL is greater than XC
• Capacitive if XC is greater than XL

Calculating Total The following formula is used to calculate total impedance

Impedance in a Series of a circuit containing resistance, capacitance, and inductance:
R-L-C Circuit
Z = R2 + (XL - XC)2

In the case where inductive reactance is greater than capacitive

reactance, subtracting XC from XL results in a positive number.
The positive phase angle is an indicator that the net circuit
reactance is inductive, and current lags voltage.

In the case where capacitive reactance is greater than inductive

reactance, subtracting XC from XL results in a negative number.
The negative phase angle is an indicator that the net circuit
reactance is capacitive and current leads voltage. In either case,
the value squared will result in positive number.

67
Calculating Reactance and In the following 120 volt, 60 hertz circuit, resistance is 1000 Ω,
Impedance in a Series inductance is 5 mh, and capacitance is 2 µF. To calculate
R-L-C Circuit total impedance, first calculate the value of XL and XC, then
impedance can be calculated.

R = 1000 Ω L = 5 mh C = 2 µF

XL = 2πfL
XL = 6.28 x 60 x 0.005
XL = 1.884 Ω

1
XC =
2πfC
1
XC =
6.28 x 60 x 0.000002

XC = 1,327 Ω

Z = R2 + (XL - XC)2

Z = 10002 + (1.884 - 1,327)2

Z = 1,000,000 + ( - 1,325.116)2

Z = 1,000,000 + 1,755,932.41

Z = 2,755,932.41

Z = 1,660.1 Ω

Calculating Circuit Current Ohm’s Law can be applied to calculate total circuit current.
in a Series R-L-C Circuit
I= E
Z

I= 120
1,660.1
I = 0.072 Amps

68
 Parallel R-L-C Circuit

Calculating Impedance in a Total impedance (Zt) can be calculated in a parallel R-L-C circuit
Parallel R-L-C Circuit if values of resistance and reactance are known. One method
of calculating impedance involves first calculating total current,
then using the following formula:

E
Zt = t
It
Total current is the vector sum of current flowing through the
resistance plus, the difference between inductive current and
capacitive current. This is expressed in the following formula:

It = IR2 + (IC - IL)2

In the following 120 volt, 60 hertz circuit, capacitive reactance

has been calculated to be 25 Ω and inductive reactance
50 Ω. Resistance is 1000 Ω. A simple application of Ohm’s Law
will find the branch currents. Remember, voltage is constant
throughout a parallel circuit.

R = 1000 Ω XL = 50 Ω XC = 25 Ω

E E E
IR = IL = IC =
R XL XC
120 120 120
IR = IL = IL =
1000 50 25
IR = 0.12 Amps IL = 2.4 Amps IL = 4.8 Amps

69
 Once the branch currents are known, total current can be
calculated.

It = IR2 + (IC - IL)2

It = 0.122 + (4.8 - 2.4)2

It = 0.0144 + 5.76

It = 5.7744

It = 2.4 Amps

Impedance is now found with an application of Ohm’s Law.

E
Zt = t
It
120
Zt =
2.4
Zt = 50 Ω

70
 Power and Power Factor in an AC Circuit

Power consumed by a resistor is dissipated in heat and not

returned to the source. This is true power. True power is the rate
at which energy is used.

Current in an AC circuit rises to peak values and diminishes to

zero many times a second. The energy stored in the magnetic
field of an inductor, or plates of a capacitor, is returned to the
source when current changes direction.

Although reactive components do not consume energy, they do

increase the amount of energy that must be generated to do
the same amount of work. The rate at which this non-working
energy must be generated is called reactive power.

Power in an AC circuit is the vector sum of true power and

reactive power. This is called apparent power. True power is
equal to apparent power in a purely resistive circuit because
voltage and current are in phase. voltage and current are also in
phase in a circuit containing equal values of inductive reactance
and capacitive reactance. If voltage and current are 90 degrees
out of phase, as would be in a purely capacitive or purely
inductive circuit, the average value of true power is equal to
zero. There are high positive and negative peak values of power,
but when added together the result is zero.

True Power and The formula for apparent power is:

Apparent Power Formulas
P = EI

Apparent power is measured in volt-amps (VA).

True power is calculated from another trigonometric function,

the cosine of the phase angle (cos θ). The formula for true
power is:

P = EI cos θ

True power is measured in watts.

71
 In a purely resistive circuit, current and voltage are in phase.
There is a zero degree angle displacement between current
and voltage. The cosine of zero is one. Multiplying a value by
one does not change the value. In a purely resistive circuit the
cosine of the angle is ignored.

In a purely reactive circuit, either inductive or capacitive, current

and voltage are 90 degrees out of phase. The cosine of 90
degrees is zero. Multiplying a value times zero results in a zero
product. No power is consumed in a purely reactive circuit.

Calculating Apparent Power In the following 120 volt circuit, current is equal to 84.9
in a simple R-L-C Circuit milliamps. Inductive reactance is 100 Ω and capacitive
reactance is 1100 Ω. The phase angle is -45 degrees. By
referring to a trigonometric table, the cosine of -45 degrees is
found to be .7071.

R = 1000 Ω XL = 100 Ω XC = 1100 Ω

120 VAC

The apparent power consumed by the circuit is:

P = EI
P = 120 x 0.0849
P = 10.2 VA

The true power consumed by the circuit is:

P = EI cos θ
P = 120 x 0.0849 x 0.7071
P = 7.2 Watts

Another formula for true power is:

P = I2R
P = 0.08492 x 1000
P = 7.2 Watts

72
Power Factor Power factor is the ratio of true power to apparent power in an
AC circuit. Power factor is expressed in the following formula:

True Power
PF =
Apparent Power

Power factor can also be expressed using the formulas for true
power and apparent power. The value of EI cancels out because
it is the same in the numerator and denominator. Power factor
is the cosine of the angle.

EI cos θ
PF =
EI

PF = cos θ

In a purely resistive circuit, where current and voltage are in

phase, there is no angle of displacement between current and
voltage. The cosine of a zero degree angle is one. The power
factor is one. This means that all energy delivered by the source
is consumed by the circuit and dissipated in the form of heat.

In a purely reactive circuit, voltage and current are 90 degrees

apart. The cosine of a 90 degree angle is zero. The power factor
is zero. This means the circuit returns all energy it receives from
the source to the source.

In a circuit where reactance and resistance are equal, voltage

and current are displaced by 45 degrees. The cosine of a 45
degree angle is .7071. The power factor is .7071. This means the
circuit uses approximately 70% of the energy supplied by the
source and returns approximately 30%.

73
Review 10
1. An AC circuit is ____________ if inductive reactance and
capacitive reactance are equal.

2. A series AC circuit is ____________ if there is more

inductive reactance than capacitive reactance.

3. A series AC circuit is ____________ if there is more

capacitive reactance than inductive reactance.

4. In a 120 VAC, 60 hertz series circuit, with 1000 Ω

of resistance, 10 mh of inductance and 4 µF of
capacitance, impedance is ____________ Ω and current
is ____________ amps.

5. In the illustrated circuit,

R = 1000 Ω XL = 200 Ω XC = 1000 Ω

120 VAC
60 Hz

It is __________ amps, and impedance is __________ Ω.

6. True power is measured in ____________ .

7. A circuit with 0.2 amps flowing through 100 Ω of

resistance, is consuming ____________ watts.

74
 Transformers

Mutual Induction Transformers are electromagnetic devices that transfer electrical

energy from one circuit to another by mutual induction. Mutual
induction is the coupling of inductances by their mutual
magnetic fields. In a single-phase transformer there are two
coils, a primary and a secondary coil. The following circuit
illustrates mutual induction. The AC generator provides electrical
power to the primary coil. The magnetic field produced by
the primary induces a voltage into the secondary coil, which
supplies power to a load.

Primary Coil Secondary Coil

Transformers are used to step a voltage up to a higher level,

or down to a lower level. Transformers are used extensively
in power distribution systems, allowing power companies
to transfer electrical energy many miles. Power generators
typically generate high voltages. This voltage varies, depending
on the generator, but a typical voltage might be 15 KV. The
voltage is stepped up through a transformer to higher levels
for transmission to substations. Typical voltages range from
115 KV to 765 KV. The electrical power is received at substation
transformers many miles away where it is stepped down.
Typical voltage might be 34 KV or 69 KV. From here, electrical
power is fed to a distribution substation. It can also be fed
directly to factory locations. If the power is fed to a factory,
transformers at the factory site reduce the voltage to usable
levels. The power fed to a distribution substation is reduced by
transformers at the substation for factory and home use.

75
Coefficient of Coupling Mutual inductance between two coils depends on their flux
linkage. Maximum coupling occurs when all the lines of flux
from the primary coil cut through the secondary winding.
The amount of coupling which takes place is referred to as
coefficient of coupling. To maximize coefficient of coupling, both
coils are often wound on an iron core which is used to provide
a path for the lines of flux. The following discussion of step-up
and step-down transformers applies to transers with an iron
core.

Lines of Flux
Confined to
Iron Core

Lines of Flux
that don’t Couple

Voltage, Current, and the There is a direct relationship between voltage, impedance,
Number of Turns in a Coil current, and the number of coil turns in a transformer. This
relationship can be used to find either primary or secondary
voltage, current, and the number of turns in each coil. It is the
number of turns which determine if a transformer is a step up
or step down transformer. The following “rules-of-thumb” apply
to transformers:

• If the primary coil has fewer turns than the secondary coil,
it is a step-up transformer.

• If the primary coil has more turns than the secondary coil,
it is a step-down transformer.

When the number of turns on the primary and seconday coils of

a transformer are equal, input voltage, impedance, and current
are equal to output voltage, impedance, and current.

76
Step-Up Transformer A step-up transformer is used when it is desirable to step
voltage up in value. The following circuit illustrates a step-
up transformer. The primary coil has fewer turns than the
secondary coil. The number of turns in a transformer is given as
a ratio. When the primary has fewer turns than the secondary,
voltage and impedance are stepped up. In the circuit illustrated,
voltage is stepped up from 120 VAC to 240 VAC. Because
impedance is also stepped up, current is stepped down from 10
amps to 5 amps.

1:2

Primary Coil Secondary Coil

900 Turns 1800 Turns

120 VAC Supply 240 VAC

10 Amps 5 Amps

Step-Down Transformer A step-down transformer is used when it is desirable to

step voltage down in value. The following circuit illustrates a
step-down transformer. The primary coil has more turns than
the secondary coil. The step-down ratio is 2:1. voltage and
impedance are stepped down, current is stepped up.

2:1

Primary Coil Seconday Coil

1800 Turns 900 Turns

240 VAC Supply 120 VAC Out

5 Amps 10 Amps

77
Single-Phase Transformer 120 or 240 VAC single-phase transformers are used to supply
lighting, receptacle, and small appliance loads. A transformer
with a 240 VAC secondary can be used to supply 240 VAC to
larger appliances such as stoves, air conditioners and heaters.
A 240 VAC secondary can be tapped in the center to provide
two sources of 120 VAC power.

Primary Primary

120 VAC 120 VAC

120 VAC 240 VAC

Secondary Ground Secondary

Formulas for Calculating the There are a number of useful formulas for calculating, voltage,
Number of Primary and current, and the number of turns between the primary and
Secondary Turns of a secondary of a transformer. These formulas can be used with
Transformer either step-up or step-down transformers. The following legend
applies to the transformer formulas:

ES = secondary voltage
EP = primary voltage
IS = secondary current
IP = primary current
NS = turns in the secondary coil
NP = turns in the primary coil

To find voltage:

EP x IP ES x IS
ES = EP =
IS IP

To find current:

EP x IP ES x IS
IS = IP =
ES EP

To find the number of coil turns:

ES x NP EP x NS
NS = NP =
EP ES

78
 Using the values for the step-down transformer in the example
of the previous page, the secondary voltage can be verified.

EP x IP
ES =
IS
240 Volts x 5 Amps
ES =
10 Amps
1200
ES =
10
ES = 120 Volts

Transformer Ratings Transformers are rated in kVA (kilovolt-amps). This rating is

used rather than watts because loads are not purely resistive.
Only resistive loads are measured in watts. The kVA rating
determines the current a transformer can deliver to its load
without overheating. Given volts and amps, kVA can be
calculated. Given kVA and volts, amps can be calculated.

Volts x Amps
kVA =
1000
kVA x 1000
Amps =
Volts

Using the illustrated step-down transformer, the kVA rating can

be calculated. The kVA rating of a transformer is the same for
both the primary and the secondary.

240 x 5
Primary kVA =
1000

Primary kVA = 1.2 kVA

120 x 10
Secondary kVA =
1000

Secondary kVA = 1.2 kVA

Transformer Losses Most of the electrical energy provided to the primary of a

transformer is transferred to the secondary. Some energy,
however, is lost in heat in the wiring or the core. Some losses in
the core can be reduced by building the core of a number of flat
sections called laminations.

79
 Three-Phase Transformers

Delta Connections Three-phase transformers are used when three-phase power

is required for larger loads such as industrial motors. There
are two basic three-phase transformer connections, delta and
wye. Delta transformers are used where the distance from
the supply to the load is short. A delta is like three single-
phase transformers connected together. The secondary of a
delta transformer is illustrated below. For simplicity, only the
secondary of a three-phase transformer is shown. The voltages
shown on the illustration are secondary voltages available
to the load. Delta transformers are schematically drawn in a
triangle. The voltages across each winding of the delta triangle
represents one phase of a three phase system. The voltage is
always the same between any two wires. A single phase (L1 to
L2) can be used to supply single phase loads. All three phases
are used to supply three phase loads.
L1

480 Volts 480 Volts

480 Volts

L2

480 Volts

480 Volts

L3

L1 to L2 = 480 volts
L2 to L3 = 480 volts
L1 to L3 = 480 volts

80
Balanced Delta Current When current is the same in all three coils, it is said to be
balanced. In each phase, current has two paths to follow. For
example, current flowing from L1 to the connection point at
the top of the delta can flow down through one coil to L2, and
down through another coil to L3. When current is balanced, coil
current is 58% of the line current measured on each phase. If
the line current is 50 amps on each phase, coil current would be
29 amps.
50 Amps L1

29 Amps 29 Amps

50 Amps L2

29 Amps

50 Amps L3

Unbalanced Delta Current When current is different in all three coils, it is unbalanced. The
following diagram depicts an unbalanced system.
43.6 Amps L1

Coil A Coil B
30 Amps 20 Amps

26.4 Amps L2

Coil C
10 Amps

36 Amps L3

Though current is usually measured with an ammeter, line cur

rent of an unbalanced delta transformer can be calculated with
the following formulas:

IL1 = IA2 + IB2 + (IA x IB)

IL2 = IB2 + IC2 + (IB x IC)

IL3 = IA2 + IC2 + (IA x IC)

81
Wye Connections The wye connection is also known as a star connection. Three
transformers are connected to form a “Y” shape. The wye
transformer secondary, (shown below) has four leads, three
phase connectors, and one neutral. The voltage across any
phase (line-to-neutral) will always be less than the line-to-line
voltage. The line-to-line voltage is 1.732 times the line-to-neutral
voltage. In the circuit below, line-to-neutral voltage is 277 volts.
Line-to-line voltage will be 480 volts (277 x 1.732).
L1

480 Volts
277 Volts
L2

277 Volts
480 Volts

N
480 Volts

277 Volts

L3

Review 11
1. If the primary of a transformer has more turns than the
secondary, it is a ____________ transformer.

2. If the primary of a transformer has fewer turns than the

secondary, it is a ____________ transformer.

3. The secondary voltage of an iron-core transformer with

240 volts on the primary, 40 amps on the primary, and
20 amps on the secondary is ____________ volts.

4. A transformer with a 480 volt, 10 amp primary, and a

240 volt, 20 amp secondary will be rated for
____________ kVA.

5. A wye connected, three-phase transformer secondary,

with 240 volts line-to-line will have ____________ volts
line-to-neutral.

82
 Review Answers

Review 1 1) electron (-), proton (+), neutron (neutral); 2) free electrons;

3) many; 4) a, c, e, g; 5) many, few.

Review 2 1) electrons; 2) negative; 3) positive; 4) repel, attract; 5) voltage;

6) b; 7) a.

E
I=
Review 3 1) R ; 2) amps, volts, ohms; 3) .5 amps; 4) 45 Ω; 5) 2 amps;
6) 6 volts, 6 volts; 7) 20 volts, 80 volts.

Review 4 1) 5 Ω; 2) 5.45 Ω; 3) 3.33 Ω; 4) 12 volts; 5) 6 amps;

6) 2.4 amps, 1.6 amps.

Review 5 1) 12 Ω, 22 Ω; 2) 40 Ω, 13.33 Ω.

Review 6 1) power; 2) P = E x I; 3) 36 watts; 4) iron, north-south;

5)north, south; 6) left, thumb, lines of flux.

Review 7 1) sine wave; 2) 120 degrees; 3) one; 4) -129.9 volts;

5) 106.05 volts rms.

Review 8 1) 10 mh; 2) 2.5 mh; 3) 2.5 µF; 4) 25 µF.

Review 9 1) reactance; 2) impedance; 3) 3.14 Ω; 4) c; 5) b; 6) 318.5 Ω;

7) 11.18 Ω; 8) 6.4 Ω.

Review 10 1) resistive; 2) inductive; 3) capacitive; 4) 1198 Ω, .1 amp;

5) 84.9 milliamps,1414.2 Ω; 6) watts; 7) 4 watts;

Review 11 1) step-down; 2) step-up; 3) 480 volts; 4) 4.8 kVA;

5) 138.56 volts.

83
 Final Exam

The final exam is intended to be a learning tool. The book

may be used during the exam. A tear-out answer sheet is
provided. After completing the test, mail the answer sheet in for
grading. A grade of 70% or better is passing. Upon successful
completion of the test a certificate will be issued.

Questions 1. A material that is a good insulator is

a. copper c. silver
b. aluminum d. rubber

2. A material with more protons than electrons has a

a. negative charge c. neutral charge

b. positive charge d. no charge

3. In a simple electric circuit with a 12 volt supply, and a

24 Ω resistor, current is

a. 2 amps c. 0.2 amps

b. 5 amps d. 0.5 amps

4. The total resistance in a series circuit containing three,

10 Ω, resistors is

a. 10 Ω c. 3.33 Ω
b. 30 Ω d. 100 Ω

5. In a 12 volt series circuit where R1=10 Ω, R2=20 Ω,

and R3=10 Ω, current flow through R2 is

a. 0.3 amps c. 0.25 amps

b. 0.5 amps d. 3.33 amps

84
6. In a circuit containing three 30 Ω resistors in parallel,
the total resistance is

a. 30 Ω c. 10 Ω
b. 90 Ω d. 0.1 Ω

7. The rate at which work is done is called

a. energy c. efficiency
b. power d. power factor

8. Power in a simple 12 volt, 4 amp series circuit is

a. 3 amps c. 48 amps
b. 3 watts d. 48 watts

9. The instantaneous voltage at 150 degrees of an AC sine

wave whose peak voltage is 480 volts is

a. 415.7 volts c. 240 volts

b. 480 volts d. 0 volts

10. The effective voltage of an AC sine wave whose peak

voltage is 480 volts is

a. 415.7 volts c. 480 volts

b. 339.4 volts d. 679 volts

11. The time constant of a series circuit with a 10 mh

inductor, and a 5 Ω resistor is

a. 2 milliseconds c. 2 microseconds
b. 2 seconds d. .5 seconds

12. The total inductance of a series circuit containing

three inductors with values of 10mh, 20 mh, and 40 mh
is

a. 5.7 pF c. 70 Ω
b. 5.7 mh d. 70 mh

13. The time constant for a series circuit with a 20 Ω

resistor and a 4 µF capacitor is

a. 80 microseconds c. 5 microseconds
b. 80 milliseconds d. 5 milliseconds

85
 14. Total capacitance for a series circuit containing a 2 µF,
4 µF, and 8 µF capacitors is

a. 14 µF c. 1.14 µF
b. 0.875 µF d. 4 µF

15. Total opposition to current flow in an AC circuit that

contains both reactance and resistance is called

a. resistance c. impedance
b. reactance d. capacitance

16. In a 60 hertz circuit containing 5 millihenrys of

inductance, inductive reactance is

a. 1.884 Ω c. 0.0005 Ω
b. 1884 Ω d. 0.05 Ω

17. In a purely inductive circuit

a. current leads voltage by 90 degrees

b. current lags voltage by 90 degrees
c. current and voltage are in phase
d. current leads voltage by 30 degrees

18. In a series AC circuit with a 20 Ω resistor and a 10 Ω

inductive reactance, impedance is

a. 30 Ω c. 14.1 Ω
b. 10 Ω d. 22.4 Ω

19. A series AC circuit containing more capacitive reactance

than inductive reactance is

a. inductive c. capacitive
b. resistive d. in phase

20. An iron-core transformer with 120 volt, 10 amp primary,

and 5 amp secondary is a

a. step down transformer with a 60 volt secondary

b. step up transformer with a 240 volt secondary
c. step up transformer with a 480 volt secondary
d. step down transformer with a 30 volt secondary

86
87
 quickSTEP Online Courses

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downloaded in PDF format. These files contain the most recent
changes and updates to the STEP courses.

A unique feature of the quickSTEP site is our pictorial glossary.

The pictorial glossary can be accessed from anywhere within
a quickSTEP course. This enables the student to look up an
unfamiliar word without leaving the current work area.

88
Siemens STEP 2000
Test Answer Form

 Select the STEP 2000 course you are testing.

 Basics of Electricity  Basics of Sensors  Panelboards
 Basics of Electrical Products  General Motion Control  Power Monitoring and
 AC Motors  Motor Control Centers Management with ACCESS
 Basics of AC Drives  Busway  Safety Switches
 Basics of Control Components  Load Centers  Switchboards
 Basics of PLCs  Molded Case Circuit Breakers

 Tell us about you.

First Name Last Name

Company Name
Title

Mailing Address

City ST ZIP

Phone ext

Fax email

 Fill in your answers below.

a b c d a b c d
1.     11.    
2.     12.    
3.     13.    
4.     14.    
5.     15.    
6.     16.    
7.     17.    
8.     18.    
9.     19.    
10.     20.    

 Send this Test Answer Form to EandM for grading.

If you achieve a score of 70% or better and live in Northern California, we’ll send you a certificate of
completion! If you have any questions, contact EandM Training at 415.369.3535 or fax 707.473.3190 or
training@eandm.com.

Attn: Step 2000 Training Department

EandM
126 Mill St.
Healdsburg, CA 95448
707.473.3190 fax