EE21

EE21
Basic Electrical Engineering

Prepared by:
ENGR. IMMA CONCEPCION C. LLANERA, REE., RME.
Faculty, EE Department
College of Engineering, Architecture & Technology
Palawan State University
 EE21

Prepared by:
ENGR. IMMA CONCEPCION C. LLANERA, REE., RME.
Faculty, EE Department
College of Engineering, Architecture & Technology
Palawan State University
This module has the following sections and corresponding icons:

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 Course Overview
COURSE DESCRIPTION:

• This course provides the students a sound background in the theory and
concepts of the fundamental and basic laws of electricity and magnetism.
Practical applications such as electrical equipment, electrical safety, blueprint
reading, house and commercial building wiring, and lighting are introduced.

COURSE OUTCOMES:
At the end of the course, students will be able to:
• Explain the fundamental concepts of Electrical Engineering and its practical
implementation;
• Explain the characteristics, uses and applications of DC and AC circuit
elements/devices and their parameters;
• Analyze DC and AC circuit by applying fundamental circuit laws, theorems
and techniques;
• Evaluate laws and theorem used in circuit analysis through experiment and
simulations; and
• Explain the composition of basic circuit and the use of circuit measuring
devises.

TIME FRAME:
• This course is intended for 18 weeks.

3
 Palawan State Univeristy
College of Engineering, Architecture and Technology

Department of Electrical Engineering

EE21: Basic Electrical Engineering

Module 1: Basic Concepts

Prepared by:
ENGR. IMMA CONCEPCION C. LLANERA, REE., RME.
Instructor 1
 Overview

COURSE OUTCOME -1:

Explain the fundamental concepts of Electrical Engineering and its
practical implementation.

INTENDED LEARNING OUTCOMES:

At the end of this module, you will be able to:
Discuss the important parameters of a circuit including symbols of circuits
elements.
Explain the characteristics of resistance, conductance, conductors,
insulators and semiconductors.
Identify the types of resistance.
Solve the resistance of a given material considering its temperature,
resistivity, cross-sectional area and length.
Identify the value of resistance of a carbon composition resistor using color
code.
Distinguish the relationship between voltage, current & resistance.
Calculate power and energy of a circuit using Ohm’s Law
Solve electric power consumption in given condition

TIME FRAME:
This module is intended for 5 weeks.

Electric circuit theory and electromagnetic theory are the two fundamental
theories upon which all branches of electrical engineering are built. Many branches of
electrical engineering, such as power plant, electrical machines, control systems,
electronics, communications, and instrumentation, are based on electric circuit theory.
Therefore, the basic electric circuit theory course is the most important course for an
electrical engineering student, and always an excellent starting point for a beginning
student in electrical engineering education. Circuit theory is also valuable to students
specializing in other branches of the physical sciences because circuits are a good model
for the study of energy systems in general, and because of the applied mathematics,
physics, and topology involved.
To actually determine the values of these variables in a given circuit requires that
we understand some fundamental laws that govern electric circuits. These laws, known
as Ohm’s law and Kirchhoff’s laws, form the foundation upon which electric circuit
analysis is built, and other laws and theorems.
In this module we shall also solve problems in evaluating resistive circuits using
different techniques commonly applied in circuit design and analysis. These techniques
include combining resistors in series or parallel, voltage division, current division, and
delta-to-wye and wye-to-delta transformations. The application of these laws and
techniques will be restricted to resistive circuits in this module. We will finally apply the
laws and techniques to real-life problems of electrical lighting.
 Lecture
NATURE OF ELECTRICITY
1.0
STRUCTURE OF THE ATOM

FIG. 1.1 ELECTRONS AND NUCLEUS OF AN ATOM

Matter is anything that has mass and occupies space. Matter is composed of very
small particles called atoms. All matter can be classified into either one of two groups:
elements or compounds. In an element, all the atoms are the same. Examples of
elements are aluminum, copper, carbon, germanium, and silicon. A compound is a
combination of elements. Water, for example, is a compound consisting of the elements
hydrogen and oxygen. The smallest particle of any compound that retains the original
characteristics of that compound is called a molecule.
Atoms are composed of subatomic particles of electrons, protons, and neutrons in
various combinations. The electron is the fundamental negative charge (-) of electricity.
Electrons revolve about the nucleus or center of the atom in paths of concentric “shells,”
or orbits (Fig. 1.1). The proton is the fundamental positive (+) charge of electricity. Protons
are found in the nucleus. The number of protons within the nucleus of any particular atom
specifies the atomic number of that atom. For example, the silicon atom has 14 protons
in its nucleus so the atomic number of silicon is 14. The neutron, which is the fundamental
neutral charge of electricity, is also found in the nucleus.

THE ELECTRIC CHARGE

Since some atoms can lose electrons and other atoms can gain electrons, it is
possible to cause a transfer of electrons from one object to another. When this takes
place, the equal distribution of the positive and negative charges in each object no longer
exists. Therefore, one object will contain an excess number of electrons and its charge
must have a negative, or minus (-), electric polarity. The other object will contain an
excess number of protons and its charge must have a positive, or plus (+), polarity
 When a pair of objects contains the same charge, that is, both positive (+) or both
negative (-), the objects are said to have like charges. When a pair of bodies contains
different charges, that is, one body is positive (+) while the other body is negative (-), they
are said to have unlike or opposite charges. The law of electric charges may be stated as
follows: (FIG 1.2)

FIG. 1.2 LIKE CHARGES REPEL EACH OTHER; UNLIKE CHARGES ATTRACT EACH OTHER

THE COULOMB

The magnitude of electric charge a body possesses is determined by the number

of electrons compared with the number of protons within the body. The symbol for the
magnitude of the electric charge is Q, expressed in units of coulombs (C). One electron
or proton has a charge of 1.6 x 10-19 coulomb.
Note: Copper is used as a universal reference conductor. 1 cubic centimeters of copper
has 8.5 x 1022 free electrons.

Free electrons – are electrons which are not attached to a specific atom and its free to
move. It is sometime called as conduction electrons.

FIG. 1.2.1 1 CUBIC CENTIMETERS OF COPPER HAS 8.5 X 1022 FREE ELECTRONS
Example1: Solve the number of free electrons in a copper conductor having a diameter
of 1.6 mm and a length of 10m.

Solution:
Note: 1.6mm = 0.16 cm ; 10 m = 1000 cm
Volume = Area x Length
𝜋
Area of a circle = 4 𝑑 2
𝜋 𝜋
V = AL = (4 𝑑 2 )L = 4 (0.16)2 (1000) = 20.1 cm3

n = number of free electrons

8.5 𝑥 1022 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
n= 𝑥 20.1 𝑐𝑚3
𝑐𝑚3

n = 1.709 x 1024 electrons

THE ELECTROSTATIC FIELD

The fundamental characteristic of an electric charge is its ability to exert a force.

This force is present within the electrostatic field surrounding every charged object. When
two objects of opposite polarity are brought near each other, the electrostatic field is
concentrated in the area between them (Fig. 1.3). The electric field is indicated by lines
of force drawn between the two objects.
If an electron is released at point A in this field, it will be repelled by the negative
charge and will be attracted to the positive one. Thus both charges will tend to move the
electron in the direction of the lines of force between the two objects. The arrowheads in
Fig. 1.3 indicate the direction of motion that would be taken by the electron if it were in
different areas of the electrostatic field.

FIG. 1.3 THE ELECTROSTATIC FIELD BETWEEN TWO CHARGES OF OPPOSITE POLARITY

POTENTIAL DIFFERENCE / VOLTAGE

Because of the force of its electrostatic field, an electric charge has the ability to
do the work of moving another charge by attraction or repulsion. The ability of a charge
to do work is called its potential. When one charge is different from the other, there must
be a difference in potential between them.0
 The sum of the differences of potential of all the charges in the electrostatic field
is referred to as electromotive force (emf).

The basic unit of potential difference is the volt (V). The symbol for potential
difference is V, indicating the ability to do the work of forcing electrons to move. Because
the volt unit is used, potential difference is called voltage.

CURRENT

The movement or the flow of electrons is called current. To produce current, the
electrons must be moved by a potential difference. Current is represented by the letter
symbol I. The basic unit in which current is measured is the ampere (A). One ampere of
current is defined as the movement of one coulomb past any point of a conductor during
one second of time.
The definition of current can be expressed as an equation:

where

I = current, A
Q = charge, C
T = time, s

Example2: If a current of 2A flows through a point in a wire for 30s, how many
coulombs pass through the point in the wire?

Solution:
𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠
Q = IT = 2 x 30 sec
𝑠𝑒𝑐

Q = 60 coulombs

CURRENT FLOW

In a conductor, such as copper wire, the free electrons are charges that can be
forced to move with relative ease by a potential difference. If a potential difference is
connected across two ends of a copper wire (Fig. 1.4), the applied voltage (1.5 V) forces
the free electrons to move.
This current is a drift of electrons from the point of negative charge, -Q, at one end
of the wire, moving through the wire, and returning to the positive charge, +Q, at the other
end. The direction of the electron drift is from the negative side of the battery, through the
wire, and back to the positive side of the battery.
The direction of electron flow is from a point of negative potential to a point of
positive potential. The solid arrow (Fig. 1.4) indicates the direction of current in terms of
electron flow. The direction of moving positive charges, opposite from electron flow, is
considered the conventional flow of current and is indicated by the dashed arrow (Fig.
1.4).

FIG. 1.3 POTENTIAL DIFFERENCE ACROSS TWO ENDS OF A WIRE CONDUCTOR CAUSES ELECTRIC CURRENT

SOURCES OF ELECTRICITY

Chemical Battery
A voltaic chemical cell is a combination of materials which are used for
converting chemical energy into electric energy. A battery is formed when two or
more cells are connected. A chemical reaction produces opposite charges on two
dissimilar metals, which serve as the negative and positive terminals. The metals
are in contact with an electrolyte.
Generator
The generator is a machine in which electromagnetic inductance is used to
produce a voltage by rotating coils of wire through a stationary magnetic field or by
rotating a magnetic field through stationary coils of wire. Today more than 95
percent of the world's energy is produced by generators.

Thermal Energy
The production of most electric energy begins with the formation of heat
energy. Coal, oil, or natural gas can be burned to release large quantities of heat.
Once heat energy is available, conversion to mechanical energy is the next step.
Water is heated to produce steam, which is then used to turn the turbines that drive
the electric generators. A direct conversion from heat energy to electric energy will
increase efficiency and reduce thermal pollution of water resources and the
atmosphere
Magnetohydrodynamic (MHD) Conversion
In an MHD converter, gases are ionized by very high temperatures,
approximately 3000 degrees Fahrenheit (3000°F), or 1650 degrees Celsius
(1650°C). The hot gases pass through a strong magnetic field with current
resulting. The exhausted gases are then moved back to the heat source to form a
complete cycle. MHD converters have no mechanical moving parts.
 Thermionic Emission
The thermionic energy converter is a device that consists of two electrodes
in a vacuum. The emitter electrode is heated and produces free electrons. The
collector electrode is maintained at a much lower temperature and receives the
electrons released at the emitter.

Solar Cells
Solar cells convert light energy directly into electric energy. They consist of
semiconductor material like silicon and are used in large arrays in spacecraft to
recharge batteries. Solar cells are also used in home heating.
Piezoelectric Effect
Certain crystals, such as quartz and Rochelle salts, generate a voltage
when they are vibrated mechanically. This action is known as the piezoelectric
effect. One example is the crystal phonograph cartridge, which contains a Rochelle
salt crystal to which a needle is fastened. As the needle moves in the grooves of a
record, it swings from side to side. This mechanical motion is applied to the crystal,
and a voltage is then generated.
Photoelectric Effect

Some materials, such as zinc, potassium, and cesium oxide, emit electrons
when light strikes their surfaces. This action is known as the photoelectric effect.
Common applications of photoelectricity are television camera tubes and
photoelectric cells.
Thermocouples

If wires of two different metals, such as iron and copper, are welded together
and the joint is heated, the difference in electron activity in the two metals produces
an emf across the joint. Thermocouple junctions can be used to measure the
amount of current because current acts to heat the junction.

DIRECT AND ALTERNATING CURRENTS AND VOLTAGES

Direct current (dc) is current that moves through a conductor or circuit in one
direction only (Fig. 1.4a). The reason for the unidirectional current is that voltage sources

FIG. 1.4. WAVEFORMS OF A CONSTANT DC CURRENT AND DC VOLTAGE

 such as cells and batteries maintain the same polarity of output voltage (Fig. 1.4b).
The voltage supplied by these sources is called direct-current voltage, or simply dc
voltage. A dc voltage source can change the amount of its output voltage, but if the same
polarity is maintained, direct current will flow in one direction only.

An alternating-current voltage (ac voltage) source periodically reverses or

alternates in polarity (Fig. 1.5a). Therefore, the resulting alternating current also
periodically reverses direction (Fig. 1.5b). In terms of conventional flow, the current flows
from the positive terminal of the voltage source, through the circuit, and back to the
negative terminal, but when the generator alternates in polarity, the current must reverse
its direction. The ac power line used in most homes is a common example. The voltage
and current direction go through many reversals each second in these systems.

FIG. 1.5. WAVEFORMS OF AC VOLTAGE AND AC CURRENT

 Lecture
Electrical Standards and
1.1 Conventions

INTRODUCTION
The international metric system of units or dimensions, commonly called SI, is used in
electricity. The abbreviation SI stands for systeme internationale. The seven base units of SI are
length, mass, time, electric current, thermodynamic temperature, light intensity, and amount of
substance (Table 1.1). Formerly the MKS metric system was used, where M stands for meter
(length), K for kilogram (mass), and S for seconds (time).

TABLE 1.1. BASE UNITS OF THE INTERNATIONAL METRIC SYSTEM

METRIC PREFIXES

TABLE 1.2. DERIVED SI UNITS

 Multiplier Prefix Symbol
1024 Yotta- Y
1021 Zetta- Z
1018 Eta- E
1015 Peta- P
1012 Tera- T
109 Giga- G
106 Mega- M
103 Kilo- k
102 hecto- h
101 deka- da
10-1 deci- d
10-2 centi- c
10-3 milli- m
10-6 micro- μ
10-9 nano- n
10-12 pico- p
10-15 femto- f
10-18 atto- a
10-21 zepto- z
10-24 yocto- y

TABLE 1.3. PREFIXES

Example3. A resistor has a value of 10M stamped on its case. How many ohms of resistance
does this resistor have?
Answer: The letter M denotes mega. Thus the resistor has a value of 10 megohms or
10,000,000 ohms.
Example4. A power station has a capacity of delivering 500000 watts (W). What is the capacity in
kilowatts (kw)?
Answer: Kilo stands for 1000. Thus, 500 OOO W = 500 kW
 Lecture
Graphical Symbols and
1.2 Electrical Diagrams

SCHEMATIC DIAGRAM
A simple electric circuit is shown in pictorial form in Fig. 1.6a. The same circuit is drawn in
schematic form in Fig. 1.6b. The schematic diagram is a shorthand way to draw an electric circuit,
and circuits usually are represented in this way. In addition to the connecting wires, three
components are shown symbolically in Fig. 2-lb: the dry cell, the switch, and the lamp.

FIG 1.6. A SIMPLE LAMP CIRCUIT

Note the positive (+) and the negative (-) markings in both pictorial and schematic
representations of the dry cell. The schematic components represent the pictorial components in
a simplified manner. A schematic diagram then is one that shows by means of graphic symbols
the electrical connections and the functions of the different parts of a circuit.

ONE-LINE DIAGRAM
A one-line, or single-line, diagram shows the component parts of a circuit by means of
single lines and appropriate graphical symbols. The single lines represent the two or more

FIG 1.7. A ONE-LINE DIAGRAM OF A SUBSTATION

conductors that are connected between the components in the actual circuit. The one-line
diagram shows the necessary basic information about the sequence of a circuit but does not give
the detailed information that is found in a schematic diagram. One-line diagrams are generally
used to show complex electrical systems without the individual conductors to the various loads.
Figure 1.7 is an example of a one-line diagram.

BLOCK DIAGRAM
The block diagram is used to show the relationship between the various component
groups or stages in the operation of a circuit. It shows in block form the path of a signal through
a circuit from input to output (Fig. 1.8). The blocks are drawn in the shape of squares or rectangles
that are joined by single

FIG 1.8. BLOCK DIAGRAM OF A TYPICAL TRANSISTOR RADIO RECEIVER CIRCUIT

lines. Arrowheads placed at the terminal ends of the lines show the direction of the signal path
from input to output as the diagram is read from left to right. As a general rule, the necessary
information to describe the components or stages is placed within the block. On some block
diagrams, devices such as antennas and loudspeakers are shown by standard symbols instead
of by blocks (Fig. 1.8).

WIRING DIAGRAM
The wiring, or connection, diagram is used to show wiring connections in a simple, easy-
to-follow way. They are very commonly used with home appliances and electrical systems of
automobiles (Fig. 1.9). The typical wiring diagram shows the components of a circuit in a pictorial
manner. The components are identified by name. Such a diagram also often shows the relative
location of the components within a given space. A color-coding scheme may be used to identify
certain wires or leads (Fig. 1.9).

FIG 1.9. WIRING DIAGRAM OF AN AUTOMOBILE STARTING CIRCUIT

ELECTRICAL PLAN
An integral part of any set of drawings for the construction of a building is the wiring plan
or layout. Architects, electrical designers, and contractors use floor-plan diagrams to locate
components of the building’s electrical system, such as receptacle outlets, switches, lighting
fixtures, and other wiring devices.

FIG 1.10. FLOOR PLAN OF A ROOM WITH FIXTURES, OUTLETS, AND SWITCHES.

STANDARD WIRING SYMBOLS ARE SHOWN.

These devices and wiring arrangements are represented by means of symbols (Fig. 1.10).
The living-room plan (Fig. 1.10) show two three-way switching arrangements. In one arrangement
a ceiling outlet is switched from two door locations. Similarly, two receptacle outlets on the north
wall are switched from two separate locations. The connection between switch and ceiling outlet
is drawn with a medium-weight solid line, indicating that the connection (conduit or cable) is to be
concealed in the walls or ceiling above. The cross lines indicate the conductors in the conduit or
cable. If cross lines are omitted, two wires are understood to be in the connection.
 Lecture
Resistance and Resistivity
1.3
Materials in general have a characteristic behavior of resisting the flow of electric
charge. This physical property, or ability to resist current, is known as resistance and is
represented by the symbol R.
Various factors affect the resistance of a material:
1. Temperature. The resistance of all substance changes to some degree with
temperature. In the case of pure metals, the resistance increases rapidly with a
rise in temperature. The extreme case of this is superconductivity. When materials
reach a sufficiently low temperature, their resistance can completely vanish;
electrical currents can run through them without any resistance.
2. Length. Resistance of a uniform conductor is directly proportional to its length.
When length increases, resistance increases.
3. Cross-Sectional Area. The resistance of a uniform conductor is inversely
proportional to its cross-sectional area. When cross-sectional area increases,
resistance decreases.
4. Material Properties. The resistance of a given conductor depends on the material
from which it is made. A numerical value called the resistivity is assigned to
materials based on how well they conduct electricity.

To remember trends in electrical resistance, it is helpful to think of water flowing in

a pipe. If the length of the pipe increases, the resistance increases. Think of drinking
through a very long straw -- It would take a lot of effort!
On the other hand, increasing the diameter (and so the cross-sectional area)
decreases the resistance. A straw with a bigger diameter is easier to use, especially when
drinking those thick fast-food store milkshakes.
The resistance of any material with a uniform cross-sectional area A depends on
A and its length ℓ, as shown in Fig. 1.11(a). You can represent resistance (as measured
in the laboratory), in mathematical form,
𝜌ℓ
𝑅 =
𝐴
Ω−𝐶𝑀
where ρ is known as the resistivity of the material in Ω − 𝑚𝑒𝑡𝑒𝑟 or in .
𝑓𝑡

Good conductors, such as copper and aluminum, have low resistivities, while
insulators, such as mica and paper, have high resistivities. Table 1.3 presents the values
of ρ for some common materials and shows which materials are used for conductors. The
circuit element used to model the current-resisting behavior of a material is the resistor.
For constructing circuits, resistors are usually made from metallic alloys and carbon
compounds. The circuit symbol for the resistor is shown in Fig. 1.11(b), where R stands
for the resistance of the resistor. The resistor is the simplest passive element.
FIG 1.11. RESISTOR (A) CROSS SECTION AND (B) CIRCUIT SYMBOL FOR RESISTANCE

Ohm defined the constant of proportionality for a resistor to be the resistance, R.

(The resistance is a material property which can change if the internal or external
conditions of the element are altered, e.g., if there are changes in the temperature.) Thus,
the resistance R of an element denotes its ability to resist the flow of electric current; it is
measured in ohms (Ω).
Type of Material Resistivity at 20 0C
(Elements or Alloys) 𝑖𝑛 𝑥10−8 Ω − 𝑚 Ω − 𝐶𝑀
𝑖𝑛
𝑓𝑡
Silver 1.6458 9.9
Copper, annealed 1.7239 10.37
Copper, hard-drawn 1.7705 10.65
Gold, pure 2.3274 14
Aluminum 2.8261 17
Magnesium 4.6548 28
Tungsten 5.4860 33
Zinc 5.9847 36
Nickel 7.8134 47
Iron, cast 8.9771 54
Platinum 9.9745 60
Iron, commercial 12.4682 75
Lead 12.9440 132
Mercury 95.9220 577
Brass 6.9822 42
German Silver 33.0823 199
Manganin 44.0543 265
Lucero 46.5479 280
Advance 48.8753 294
Constantan 50.2053 302
Excello 91.4334 550
Nichrome 99.7456 600
Nichrome II 109.7201 660
Chromel 620-655
Ω−𝐶𝑀
Table 1.4. RESISTIVITY BOTH 𝑖𝑛 Ω − 𝑚 & 𝑖𝑛 𝑓𝑡
Note: If not specified, a copper wire is assumed an annealed copper wire and at 20 0C
conditions.

Circular Mil (CM) – it is an old unit in specifying the cross section are of a wire. It is equal
to the are of a circle having a diameter of 1mil.

FIG 1.11.1 CIRCULAR MIL

A = cross section area of the wire, in cmils (CM)

d = diameter of wire, in mils

Note: 1 inch = 1000 mils; 1 MCM = 1000 CM

Example5: Using the given particulars calculate the resistances of the following
conductors at 200C: (a) material copper, = 1,000 ft, CM = 3,220 cir mils; (b) material
= aluminum, length 4 miles, diameter 162 mils; (c) material Advance, length 486
in., diam eter 0.0159 in.
Solution:
𝜌ℓ 10.37𝑥1000
a. 𝑅 = = = 3.2205 Ω
𝐴 3220
𝜌ℓ 17𝑥4𝑥5280
b. 𝑅 = = = 13.6806 Ω
𝐴 162

Note: 1 mile = 5280 ft and when the diameter is given in mils, the area in circular
mils will only be the square of the diameter.
486
𝜌ℓ 294𝑥
12
c. 𝑅 = =( = 47.0986 Ω
𝐴 0.0159𝑥1000)2

Note: 12 in= 1 ft and 1inch = 1000 mils

TEMPERATURE-RESISTANCE EFFECTS
Resistance of an electric conductor also depends upon the temperature. The
resistance of all conductors will increase as the temperature of the conductor increases.

𝑅2 𝑇 + 𝑡2
=
𝑅1 𝑇 + 𝑡1

R1 = resistance at temperature t1
R2 = resistance at temperature t2
T = inferred absoluter zero temperature
 1 𝑅2
Let : 𝛼 = 𝑇+𝑡1 , then 𝑅1 = 1 + 𝛼(𝑡2 − 𝑡1)

𝛼 = temperature coefficient of resistance at t1

Note: For annealed copper wire, the inferred absolute zero temperature is 234.50C

Example6: A copper winding has a resistance of 0.25 ohm at a temperature of

180C. Calculate the temperature in the winding when after a period of operation,
the resistance increases to 0.31 ohm.
Solution:
𝑅2 𝑇+𝑡2
𝑅1
= 𝑇+𝑡1
0.31 234.5+𝑡2
=
0.25 234.5+18

t2 = 78.60C

COLOR CODE FOR CARBON COMPOSITION RESISTORS

FIG 1.12. 4 BAND COLOR CODE RESISTOR CHART

Since carbon resistor is physically quite small, it is more convenient to used a color code
indicating the resistance value than to imprint the numerical value on the casing. The
color is read from left to right.

Example7:
Answer:
Yellow = 4
Green = 5
Orange = x 1000
Silver = 10% tolerance
The final expression of the resistance is 45,000 ± 10% ohms.
 Lecture
Ohm’s Law
1.4
So far, we have been introduced to both dependent and independent voltage and
current sources and were cautioned that they were idealized active elements that could
only be approximated in a real circuit. We are now ready to meet another idealized
element, the linear resistor. The resistor is the simplest passive element, and we begin
our discussion by considering the work of an obscure German physicist, Georg Simon
Ohm, who published a pamphlet in 1827 that described the results of one of the first
efforts to measure currents and voltages, and to describe and relate them mathematically.
One result was a statement of the fundamental relationship we now call Ohm’s law, even
though it has since been shown that this result was discovered 46 years earlier in England
by Henry Cavendish, a brilliant semirecluse.
Ohm’s law defines the relationship between current, voltage, and resistance. There are
three ways to express Ohm’s law mathematically.
1. The current in a circuit is equal to the voltage applied to the circuit divided by
the resistance of the circuit:

2. The resistance of a circuit is equal to the voltage applied to the circuit divided
by the current in the circuit:

3. The applied voltage to a circuit is equal to the product of the current and the
resistance of the circuit:

FIG 1.13. THE OHM’S LAW CIRCLE

Example8. An electric iron draws 4A at 120V. Find its resistance.
Solution:
𝑉 120
𝑅= = = 30 𝛺
𝐼 4
 Lecture
Electric Power
1.5
Power is the rate at which work is done, and in the mechanical system it is
generally expressed in horsepower, abbreviated hp; it is equivalent to the rate of 33,000
ft-lb of work per minute. This unit was originated by James Watt in 1782 because he found
it necessary to specify how much power his new steam engines would develop in terms
of the horses they were intended to replace. The practice then was to have the horses
push a crank as they walked into a 24-ft-diameter circle; this action resulted in the
operation of pumps that, raised water to elevated tanks. Since the average horse was
assumed to exert a force of about 175lbs and made 2.5 complete revolutions per minute,
it followed that 1 hp was equivalent to 175 X (24π) X 2.5, or 33,000 ft-lb per min. Later,
when the unit, of electrical power was adopted, it, was designated the watt in honor of the
inventor James Watt.
Following the fundamental definition of power, i.e., the time rate of doing work, the
unit of electrical power is defined in terms of the joule per second, the latter being the
work done when one coulomb of electricity is moved through a potential difference of one
volt in one second. Remembering that 1 coulomb per second is 1 amp from the equation
i=dq/dt or simply i=q/t, it follows that 1 joule per sec = 1 watt, and this is one product of V
(volts) and I (amperes) that is unity. In general, therefore, the power in watts is
𝑃 = 𝑉𝐼
Since V=IR, P may be written in terms of I and R; thus
𝑃 = (𝐼𝑅)𝐼 = 𝐼 2 𝑅
Also, V/R may be substituted for I, in which case we can also say that
𝑉 𝑉2
𝑃 = 𝑉 (𝑅) = 𝑅

Any of the above equations may be used to determine the circuit power if two
proper values of V, I and R, are known; they may also be employed to evaluate V, I or R
if two latter three terms and P are known.
Electrical power is frequently converted to mechanical power as, for example, in
the case of the electric motor; a precise relationship is therefore required between the
mechanical and electrical equivalents of power.
Accurate, measurements indicate that
1 hp = 746 watts
and, since 1,000 watts = 1 kilowatt (1 kw), 1 hp is sometimes taken to be
approximately 3/4 kw.
The electrical unit of power is the watt. In theory, the watt can be related to other
measures of power:
1 watt (W) = 3.143 Btu/hr
1 kilowatt (kW) = 1000 W
1 megawatt (MW) = 1000000 W
Example9. What is the resistance of a load that takes 2500 watts from a 115-volt
source?
Solution:
𝑉2 1152
𝑅 = = = 5.29 𝛺
𝑃 2500

Example10: The 560ohm resistor is connected to a circuit which causes a

current of 42.4 mA to flow through it. Calculate the voltage across the resistor and
the power it is dissipating.
Solution:
V = RI = (560)(0.0424) = 23.7 V – voltage across the resistor
Several ways to obtain the power dissipated:
P= Vi = (23.7)(0.0424) = 1.005 W; or
P= V2/R = (23.7)2/560 = 1.003 W; or
P = I2 R = (0.0424)2(560) = 1.007 W
 Lecture
Electric Energy
1.6
Energy and work are essentially the same and are expressed in identical units.
Power is different, however, because it is the time rate of doing work. With the watt unit
for power, one watt used during one second equals the work of one joule, or one watt is
one joule per second. The joule (J) is a basic practical unit of work or energy.
The kilowatthour (kWh) is a unit commonly used for large amounts of electric
energy or work. The amount of kilowatthours is calculated as the product of the power in
kilowatts (kW) and the time in hours (h) during which the power is used.

𝑉2
𝑊 = 𝑃𝑡 = 𝑉𝐼𝑡 = 𝐼 2 𝑅 𝑡 = ( 𝑅 ) 𝑡 in joules (J)

Note that a joule is equal to 1 watt-second where P is given in watts and t in

seconds.
Since the watt-second is ordinarily too small for practical computations a larger
unit, the kilowatt-hour, abbreviated as kwhr, is generally used. Thus,
𝑘𝑤ℎ𝑟 = (𝑃 𝑥 ℎ𝑟) / 1000

Example11. How much electric energy, in kilowatthours, is delivered to an electric

motor during on 8-hr period when operating from constant 230-volt source, if the
average currents are 34 amp for 2 hr, 38 amp for 1.5 hr, 26 amp for 4 hr, and 12
amp for 0.5 hr?
Solution:
𝐸𝑛𝑒𝑟𝑔𝑦 = [(230𝑥34𝑥2) + (230𝑥38𝑥1.5) + (230𝑥26𝑥4) + (230𝑥12 𝑥0.5)]𝑥1/1,000
= 54.05 𝑘𝑤ℎ𝑟

Example12. A water heater has a resistance of 5.3 ohms and takes 43.5 amp
when in operation. If it is in service on the average of 2 hrs per day, how many
kilowatthours of energy will be expended during a 30-day month?
Solution:
𝑊 = 𝑃𝑜𝑤𝑒𝑟 𝑥 𝑡𝑖𝑚𝑒 = 𝐼 2 𝑅 𝑡 = 43.52 𝑥5.3𝑥(2 𝑥 30)/1000 = 601.7355 𝑘𝑤ℎ𝑟

It must be understood, of course, that the primary basis for the cost of all electrical
service is energy, measured in kilowatthours. For the measurement of such energy all
installations are provided with watthour meters—integrating meters—usually at a point in
the circuit where the wires enter the building to be served. The energy cost is calculated
from meter readings, usually taken at monthly intervals. In addition, several extra charges
that take into account such factors as maximum demand, low power factor, variable costs
of generation fuel, generation cost, losses, surcharges etc.
 Example13. The difference between the two succeeding monthly meter readings
is 224 kwhr. Assume that the foregoing rate schedule is Php 9.40 per kwhr.
Neglecting other charges, calculate the cost service for electrical energy for the
monthly billing.
Solution:
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = (𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑) 𝑥 𝑅𝑎𝑡𝑒 = 224 𝑥 9.40 = 𝑃ℎ𝑝 2105.60
 Lecture
Basic Electrical Circuits
1.7
Basic Electrical Circuit
An electric circuit is a continuous path along which an electric current can flow. A simple
circuit is composed of a power source (e.g., battery or generator); the load, an electrical
component or group of components that consume electricity (e.g., a lamp or appliance);
and a set of conductors that carry current from the source to the load (e.g., wires). See
Figure 1.14. If the circuit is broken at any point, current will not flow.

a. Closed circuit b. Open circuit c. Switched circuit (open)

FIGURE 1.14 A closed circuit is an uninterrupted path that allows a continuous flow of
current through an electrical conductor. Shown in (a) is a closed lighting circuit in which
current flows from the battery to the lamp and back to the battery. The resistance to
current flow in the lamp's filament causes the lamp to illuminate. If the path of current flow
in a circuit is interrupted or opened (turned off), an open circuit result. The broken wire in
the second circuit (b) causes the circuit to open. A switch is installed in a circuit (c) to
allow the circuit to open or close to control operation of the lamp.
Closed Circuit
To keep current flowing in an electrical conductor, there must be a difference in charge
between the ends of the conductor. A closed circuit is an uninterrupted path that allows a
continuous flow of current through an electrical circuit. In a building electrical system, a
circuit is closed when a switch is turned on, allowing current to flow uninterrupted and the
lamp to light. The switch completes the conductor path, which allows current to flow from
the power source through the conductor (a wire) to the lamp. To complete the circuit, a
second conductor runs from the lamp back to the power source.
Open Circuit
If the path of current flow is interrupted such as if the switch in a circuit is opened (turned
off), an open circuit result. The switch breaks the conductor path, which prevents current
from flowing from the power source through the conductor to the lamp. An open circuit
prevents flow of current.

Short Circuit
If an inadvertent shortcut develops in a circuit that permits current flow through an
unintentional path, a short circuit is created. A short circuit occurs when current leaks out
of the intended conductor path such as out of a wire with damaged insulation. A short
circuit is a dangerous condition especially if it incorrectly energizes the metal cabinet or
housing of an appliance a hazard if a person touches the appliance.

A circuit may also have a control device and/or a protective device, but these are optional.
A control device either opens or closes the path of the circuit. Light switches, thermostats,
and time clocks are examples of common control devices found in circuits. An overcurrent
protection device is used to protect either the load and/or the conductors from excessive
heat from high amperage conditions. Most protective devices open the circuit, thereby
interrupting the path of current if excessive current is flowing in the circuit. Common
examples of protective devices include fuses and circuit breakers.
 Lecture
Direct-Current Series Circuit
1.8
A series circuit is a circuit in which there is only one path for current to flow along.
In the series circuit (Fig. 1.15), the current I is the same in all parts of the circuit. This
means that the current flowing through RI is the same as the current through R2, is the
same as the current through R3, and is the same as the current supplied by the battery.

FIG. 1.15 SERIES CIRCUIT

The equivalent resistance for a series connected circuit is the algebraic sum all the
resistors.
𝑅𝑡 = 𝑅1 + 𝑅2 + 𝑅3 + … + 𝑅𝑛
As stated before, in series circuit, only one current I passes all through the
elements of the circuit. Thus, we can say that,
𝐼𝑡 = 𝐼1 = 𝐼2 = 𝐼3 = … = 𝐼𝑛
meaning the current passing through the R1 is the same as the current passing
through R2.
However, the sum of voltage drops across each individual resistor are equal to
the total emf impressed into the circuit. Mathematically,
𝑉𝑡 = 𝑉1 + 𝑉2 + 𝑉3 + … + 𝑉𝑛

When n equal resistors are connected in series, Fig. 1.16, to source of emf or
volts, each one having a resistance of R ohms, the following conditions prevail:
1. The current I through all resistors is the same.
2. The total equivalent resistance of the circuit nR.
3. The voltage drops across the individual resistors are equal.
4. The voltage drop across each resistor ER = E/n = IR.
 FIG.1.16 EQUAL RESISTORS CONNECTED IN SERIES

Example14. Three resistors are connected in series across a 12-volt battery. R1=
1 ohm, R2= 2 ohms, and R3= 3 ohms. Find the total resistance the ohmmeter
registers when connected across the series connection?
Solution:
𝑅𝑡 = 𝑅1 + 𝑅2 + 𝑅3 = 1 + 2 + 3 = 6 𝑜ℎ𝑚𝑠
Example15.Seven identical 47-kilohm resistors are connected in series to a 120 V
dc source. Calculate the equivalent resistance and the current flowing through the
circuit.
Solution:
𝑅𝑡 = 𝑛𝑅 = 7 𝑥 47 𝑘𝛺 = 329 𝑘𝛺
𝐼𝑡 = 𝑉𝑡/𝑅𝑡 = 120 / 329,000 = 0.3647 𝑚𝐴

Consider the series circuit illustrated in Figure 1.15 with two resistors R1 and R2
connected with a fixed voltage source V carries a current of I.
Using the ohm’s Law that It = t / Rt, It is equal to I1 and I2, let us simplify it using
the generic I. 𝑅𝑡 = 𝑅1 + 𝑅2
𝑉𝑡
𝐼 =
𝑅1 + 𝑅2
𝑉2
Let us also apply Ohm’s Law in the resistor R2, we can say that 𝐼 = 𝑅2.

Since, both currents I are equal we can equate the two equations.

𝑉2 𝑉𝑡
= 𝑅1+𝑅2, manipulating the equation, we can say that
𝑅2
𝑅2
𝑉2 = 𝑉𝑡 𝑥 [𝑅1+𝑅2]
𝑅1
Likewise, 𝑉1 = 𝑉𝑡 𝑥 . Also, when more than two resistors are involved, we
𝑅1+𝑅2
can generalize the Voltage divider rule as,
𝑅𝑎
𝑉𝑎 = 𝑉𝑡 𝑥 𝑅𝑡
 Figure 1.17 Voltage Divider Circuit

Example14. Two resistors 10 ohms and 15 ohms are connected in series across
a 24V battery. Find the voltage drop across the 15-ohm resistor.
Solution:
Given:
Total Voltage = 24 volts
Resistor 1 = 10 ohms
Resistor 2 = 15 ohms
Required:
Voltage Drop at Resistor 2 = ?

𝑅2 10
𝑉2 = 𝑉𝑡 𝑥 = 24𝑥 = 9.6 𝑉
𝑅1 + 𝑅2 25
 Lecture
Direct-Current Parallel Circuit
1.9
The second connection is the parallel circuit, the load resistances are connected
across one another so that the total current, entering one junction, divides to pass through
the individual ports in definite ratios and combines at the other junction to leave the latter.
In the parallel circuit the total current It divides into three parts I1, I2, and I3 to pass
through the respective resistors RI, R2, and R3.

Figure1.18 Parallel Circuit

When a number of resistors are connected in parallel, Fig. 1.18, to a source of E,
the following conditions prevail:
1. The same voltage E is impressed across all resistors. Meaning,
𝐸𝑡 = 𝐸1 = 𝐸2 = 𝐸3 = … = 𝐸𝑛
2. The total current is the algebraic sum of all the current passing each through
branch or resistor. Mathematically speaking,
𝐼𝑡 = 𝐼1 + 𝐼2 + 𝐼3 + … + 𝐼𝑛
3. The equivalent resistance of parallel connected resistors is the reciprocal of the
summation of the reciprocal of each resistor.
1
𝑅𝑡 = 1 1 1 1
( )+ ( )+ ( )+ … + ( )
𝑅1 𝑅2 𝑅3 𝑅𝑛

When n equal resistors are connected in parallel, Fig. 2.4, to source of emf of volts,
each one having a resistance of R ohms, the following conditions prevail:
1. The voltage impressed through all resistors is the same.
𝑅
2. The total equivalent resistance of the circuit 𝑅𝑡 = .
𝑛

3. The current across the individual resistors are equal.

𝐼𝑡
4. The current across each resistor 𝐼𝑟 = 𝑜𝑟 𝐼𝑡 = 𝑛(𝐼𝑟)
𝑛
 Figure1.19. Equal Resistors connected in Parallel
Example15. Two resistors are connected across a 9V dry cell. R1 = 6ohms and
R2= 9ohms. Find the equivalent resistance of the two.
Solution:
1 1
𝑅𝑡 = = = 3.6 Ω
1 1 1 1
𝑅1 + 𝑅2 6+9
Another way to solve parallel circuits using your calculators,
𝑅𝑡 = 𝑅1 || 𝑅2
Note: we can denote parallel connected resistors using the symbol of two parallel
lines “ || ”.
𝑅𝑡 = 𝑅1 | |𝑅2 = 6 | |9 = (6−1 + 9−1 )−1 = 3.6 Ω
We can expect the same results.

Example16. Ten resistors of which 120 ohms each are connected in parallel, find
the equivalent resistance.
Solution:
𝑅 120
𝑅𝑡 = = = 12 Ω
𝑛 10

Consider a circuit composed of two resistors R1 and R2 connected across a

voltage source E. From the circuit we will have
𝐼𝑡 = 𝐼1 + 𝐼2
From here we can say that I1 = It – I2 or I2 = It – I1.
As the same voltage E is present across each resistor, we can find the current flowing
across each resistors as
𝐸 𝐸
𝐼1 = 𝑎𝑛𝑑 𝐼2 =
𝑅1 𝑅2
Substituting this I1 and I2 in It, we have
𝐸 𝐸
𝐼𝑡 = + , we extract E on the expression, we have
𝑅1 𝑅2

1 1
𝐼𝑡 = 𝐸 ( + ), we add the two fraction bounded by the bracket
𝑅1 𝑅2
 𝑅1 + 𝑅2
𝐼𝑡 = 𝐸 𝑥 , manipulating further
𝑅1 𝑅2

𝑅1 𝑅2
𝐸 = 𝐼𝑡 𝑥
𝑅1 + 𝑅2

Now solving for I1 gives us:

𝐸 𝑅1 𝑅2
𝐼1 = , 𝑢𝑠𝑒 𝐸 = 𝐼𝑡 𝑥
𝑅1 𝑅1 + 𝑅2
𝑅 𝑅 1
𝐼1 = 𝐼𝑡 𝑥 𝑅 1+𝑅2 𝑥 𝑅 , dividing it by R1 we have
1 2 1

𝑅2
𝐼1 = 𝐼𝑡 𝑥 𝑅 (eq 2.13)
1 +𝑅2

Now solving for I2 gives us:

𝑅1
𝐼2 = 𝐼𝑡 𝑥 𝑅 (eq 2.13 a)
1 +𝑅2

When instructed to use current divider formula for more than two resistors, it is
generalized that,

𝑅𝑐
𝐼𝑎 = 𝐼𝑡 𝑥 𝑅 (eq 2.13 b)
𝑎 + 𝑅𝑐

Where Ia is the unknown current

Ra = is the resistance along the branch where Ia passes
Rc, equivalent resistance of the parallel connection excluding the resistance of Ia

Example17. Two resistors are connected in parallel to constant current source

of 10 Amps. R1= 4ohms and R2=6ohms. Find the current passing through the
6ohm resistor using current divider rule.
Solution:
𝑅1
𝐼2 = 𝐼𝑡 𝑥 = 4𝐴
𝑅1 + 𝑅2

Complex circuits are frequently found in practice to consist of two or more

interconnected sections, each of which may be a single resistor or two or more resistors
in series or parallel. The interconnection of the several sections may likewise be arranged
in series or parallel. Figure 1.20 illustrates two possible combinations designated series-
parallel and parallel-series circuits.
 Figure1.20 Combination of series and Parallel Connection

First, we discuss the Fig. 1.20a, the series-parallel connection. The major
connection of the circuit is in series because we can see that parallel connection where
connected in series to the voltage source E. We have two parallel connection connected
in series, first we have R1 || R2 and second R4 || R5 || R6.
So, for Fig. 1.20a, the equivalent resistance will be the sum of (R1 || R2), R3, and
(R4 || R5 || R6). Let us take an example by putting values on the figure.

Example1.18. In fig 2.5a, E= 120Vdc, R1=3, R2=5, R3=7, R4=9, R5=8 and,
R6=10. Find the total resistance.

Solution:
𝑅𝑡 = (𝑅1 || 𝑅2) + 𝑅3 + (𝑅4 || 𝑅5 || 𝑅6)
= (3 || 5) + 7 + (9 || 8 || 10) = 11.8502 Ω

This time we discuss the other figure shown in Fig.1.20b, the Parallel-Series circuit
connection. The major connection is parallel since the minor connection in each branch
connected across the voltage source, E, is resistors connected in series.
From the Fig. 1.20b, the equivalent resistance of the circuit is the parallel
connection of the three branches with series connected resistors in each branch. We can
put that to an equation, Rt = (R1 + R2) || R3 || (R4 + R5 + R6).
To fully appreciate the circuit, let’s consider an example by designating some
values in the circuit.

Example 2.19. In fig 2.5b, E= 120Vdc, R1=3, R2=5, R3=7, R4=9, R5=8 and,
R6=10. Find the total resistance.
Solution:
𝑅𝑡 = (𝑅1 + 𝑅2) || 𝑅3 || (𝑅4 + 𝑅5 + 𝑅6)
= (3 + 5) || 7 || (9 + 8 + 10)
= 3.2798 Ω
 PRACTICE PROBLEMS
1. How much charge is represented by 7,900 electrons?
2. How many electrons can be represented for an 10mC of charge?
3. An iron wire has a resistance of 62 ohms at a room temperature of 24 0C. (a) What
will be its resistance at 80 0C, (b) at -20 0C?
4. The tungsten filament in an incandescent lamp has a resistance of 9.81 ohms at a
room temperature of 22 0C and a resistance of 132 ohms at normal operating
temperature. Calculate the temperature of the heated filament.
5. An electric iron draws 4A at 120V. Find its resistance.
6. What is the resistance of a load that takes 2500 watts from a 115-volt source?
7. How much electric energy, in kilowatthours, is delivered to an electric motor during
on 8-hr period when operating from constant 230-volt source, if the average
currents are 34 amp for 2 hr, 38 amp for 1.5 hr, 26 amp for 4 hr, and 12 amp for
0.5 hr?
8. A water heater has a resistance of 5.3 ohms and takes 43.5 amp when in
operation. If it is in service on the average of 2 hrs per day, how many kilowatthours
of energy will be expended during a 30-day month?
9. Three resistors are connected in series across a 12-volt battery. R1= 1 ohm, R2=
2 ohms, and R3= 3 ohms. Find the total resistance the ohmmeter registers when
connected across the series connection?
10. Seven identical 47-kilohm resistors are connected in series to a 120 V dc source.
Calculate the equivalent resistance and the current flowing through the circuit.
11. Two resistors 8 ohms and 7 ohms are connected in series across a 30-V battery.
Find the voltage drop across the 8-ohm resistor.
12. Two resistors are connected across a 9V dry cell. R1 = 6ohms and R2= 9ohms.
Find the equivalent resistance of the two.
13. Ten resistors of which 120 ohms each are connected in parallel, find the equivalent
resistance.
14. Two resistors are connected in parallel to constant current source of 10 Amps. R1=
4ohms and R2=6ohms. Find the current passing through the 6ohm resistor using
current divider rule.

ASSESSMENT
1. In simple words, how do you describe the concept of current, charge, voltage, and
resistance. (5 pts each)
2. How much charge is represented by 7,900 electrons? (5pts)
3. How many electrons can be represented for a 10nC of charge? (5pts)
4. How much charge is represented by 7,900 charged particles? (5pts)
5. How many charged particles can be represented for a 10nC of charge? (5pts)
6. Calculate the resistance of an object with a square cross-section whose side is 0.1
inch with a length of 5m. Assume the material is copper. (5pts)
7. Calculate the resistance of a copper wire with cross-sectional area of 2.0 sq. mm
of length 10 ft. (5pts)
8. Find the resistance of a copper bar 1mx1mx1m. (5pts)
9. A coil of copper wire has a resistance of 78 ohms at a room temperature of 24 0C.
(a) What will be its resistance at 80 0C, (b) at -20 0C, (c) at 120 0F? (15pts)
10. Derive the value of resistivity in ohm-meter given the exact values from the ohm-
CM/ft value. Show your proper derivation with correct unit analysis. Just pick one
element or material. (5 pts)
11. Why is it that silver is not drawn to be used in wires? Explain your answer. (5pts)
 12. An electric iron draws 6A at 230V. If it is used for 2 hours. Find the power, energy
consumed and the electric bill assuming the prevailing schedule per kilowatt-hour
is 12.36 Php. (15 pts)
13. A load of 10 ohms was connected to a 12-volt battery. The current drawn was 1.18
amperes. What is the internal resistance of the battery? (5pts)
14. Three resistors are connected in series across a 12-volt battery. R1= 1 ohm, R2=
2 ohms, and R3= 3 ohms. Find the power dissipated in the 3-ohm resistor. (5pts)
15. A circuitry has five branches connected in parallel across a 9-V source. One of the
branches consists of four commercial resistors rated 30ohms-1/4watt, 20ohms-
1/4watt, 22ohm-1/4watt and 24ohm-1/4watt. Find the voltage across the 30ohm-
1/4watt resistor. (5pts)
16. Suppose you are working on a project, and you badly need a 5-volt source and
you do not have a regulated voltage source but you only have a 12V battery in
your home. Design a circuit that gives an output voltage of 5V from an input of 12
V using the voltage divider principle. Use commercially available resistors available
in the market. (Hint: Find the values of R1 and R2) (10pts)
17. Two resistors of 10 ohms and 12 ohms were connected in parallel. Another two
resistors with 20 and 25 ohmic value were connected in parallel. The two sets of
resistors are connected across each other. Find the equivalent resistance of the
two connected parallel connections. (5pts)
18. Four resistors, 5Ω, 15Ω, 10Ω and 8Ω, are connected across a 24-volt battery.
Calculate the current passing through the 15Ω resistor using current divider
formula. (5pts)

ANSWER KEY
Answer Key for this Module will be provided in the next Module.

References

Alexander, C. K., Sadiku, M. N., (2013). Fundamentals of Electric Circuits (6th ed.).
McGraw-Hill, Inc.
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Herman, S. L. (2009). Delmar’s Standard Handbook of Electricity (4th ed.). Cengage
Learning
Kang, J. S. (2018). Electric Circuits (1st ed.). Cengage Learning
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Rojas, R. A. (2012). 1001 Solved Problems in Electrical Engineering. Excel Review
Center.
Rojas, R. A. (2017). Electrical Engineering Pocket Reviewer (Problem Solving Type) (1st
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Theraja, B. L., Theraja, A. K. (2005). A Textbook of Electrical Technology – Volume I:
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