Subject: EE 213 ELECTRIC CIRCUIT I
(Week 2-3)
Chapter 1: DC/AC Sources and Electrical Circuit Components, Voltage and
Current Laws
Introduction
An electrical circuit is a complete conductive path through which electrons
flow from the source to the load and back to the source. The direction and magnitude
of the electrons flow however depend on the kind of source. In Electrical
Engineering, there are basically two types of voltage or current (Electrical Energy)
source which defines the kind of circuit and they are; Alternating Current (or voltage)
and Direct Current.
Learning Outcomes
1. Apply DC/AC Sources and Electrical Circuit Components
2. Understand Ohm’s Law
3. Solve problems in electric charge and electric current.
Learning Content
Electricity: Basic Principles
Structure of Matter






Matter- anything that occupies space and has mass.
Element- a substance that cannot be decomposed any further by chemical
reaction.
Compound- a combination of two or more elements.
Molecule- smallest particle that a compound can be reduced to before it breaks
down into its element.
Atom- smallest part that an element can be reduced to and still keeping the
properties if the element.
Electrons – negatively charge particles that revolve around the nucleus of an atom.
Electron is one of the lightest particles with a known mass. The mass of the
electron is about 9.11 x 10-31 kg.
N= 2n2
Where;
N = total number of electrons on a given shell
n = nth shell of the atom.






Protons – positively charged particles that stays in the nucleus of an atom. Proton
is very small but it is fairly massive compare to the other particles that
makes up matter. The mass of one proton is 1.675 x 10-27 kg.
Neutrons – particles having no charge. Neutrons are about the same size as
proton but their mass is slightly greater 1.675 x 10-27 kg.
Nucleus – the central part of the atom where the protons and neutrons of an atom.
Atomic number – represent the number of the electrons or protons of an atom.
Atomic mass – represent the sum of protons and neutrons of an atom.
Valence electrons – electrons found in the outermost shell or orbit
of an atom.
Electric Charge
Coulomb (C) – unit of electric charg , which is equivalent to 6.25x1018
electrons or protons . Named after the French physicist, Charles A.
coulomb (1736 – 1806).
remember; 1 electron or proton has a charge of 1.6 x 10-19 coulomb.
Potential Difference


Potential – the capability of doing work .
Volt (V) – unit of potential difference, which is equal to one joule of work done per
one coulomb of charge. Named after the Italian physicist, Alessandro C. Volta
(1754 – 1827) who invented the first electric battery.
𝑾
𝑬=
𝑸
Where;
E = potential difference of voltage (volt)
W = work done (joule)
Q = charge (coulomb)

Sample problem:
Determine the magnitude of the voltage required to accelerate an electron to
kinetic
energy of 8 𝑥 10−15 joules.
Note that one electron has a charge of 1.6 x 10−19 coulomb
Solution:
𝐸=
𝑊
𝑄
8 𝑋 10−15
= 1.6 𝑋 10−19 = 50 000 volts
Electric Current
When a potential difference between two charges forces a third charge to move,
the charge in motion is called an electric current.
Ampere (A) – unit of charge flow equal to one coulomb of charge past a given point
in one second. Named after the French mathematician, Andre M. Ampere (1775 –
1836).
𝑸
𝑰=
𝑻
Where;
I = current (ampere)
Q = charge (coulomb)
T = time (second)

Sample problem:
A car battery supplies a current of 50 ampere to the starter motor. How much
charge passes through the starter in 1⁄2 minute?
Solution:
60 𝑠
Q = It = 50 (0.5 min. x 1 𝑚𝑖𝑛) = 𝟏𝟓𝟎𝟎 𝒄
Resistance

The fact that a wire carrying a current can became hot, it is evident that the work
done by the applied force in producing the current must be accomplish against
some opposition or resistance.

Ohms(Ω) – practical unit of resistance , named after the german physicist , George
S. ohms (1787 – 1854)
𝑹=
𝑷𝑳 𝑷𝑽 𝑷𝑳𝟐
= 𝟐 =
𝑨
𝑨
𝑽
Where;
R = Resistance
A = cross sectional area
P = resistivity
L = length
V = volume


Specific resistance or resistivity – the amount of change of resistance in a
material per unit change in temperature.
Circular mil(CM) – area of circle having a diameter (d) of one mil.
𝑨 = 𝒅𝟐
remember: 1 inch = 1000 mil
1MCM = 1000 CM
Resistance
Resistivity
Length
Area
Ohm
Ohm-m
M
Sq. m.
Ohm
Ohm-cm
Cm
Sq.cm.
Ohm
Ohm-cm/ft
Ft.
CM
Sample problem:
 How many circular mils does a round copper rod of 0.25 inch diameter has?
Solution:
D = 0.25 in. x
𝟏𝟎𝟎𝟎 𝒎𝒊𝒍𝒔
𝟏𝒊𝒏.
=250 mils
A = 𝑑2 = (250)2 = 62,500 CM
Sample problem:
 What is the size in square millimeter (mm2 ) is the cable of 250 MCM size
Solution:
250 MCM = 250,000 CM
A = d2 = d = √A d= √250,000 = 500 mils
1 in.
D = 500 mils x 1000 mils x
π
π(12.7)2
A =D2 4 =
4
25.4 mm
1 in.
= 12.7 mm
= 𝟏𝟐𝟔. 𝟔𝟕mm2
Sample problem:

Determine the resistance of a bus bar of a copper if the length is 10 meters
long and the cross section is 4x4 cm. the resistivity is 1.724 µΩ-cm.
L = 10 in. x
R=
𝑝𝑙
𝑎
=
100 𝑐𝑚
= 1,000 𝑐𝑚
1 𝑖𝑛.
(1.742𝑥10−6 )(1,000)
16
𝟒
R = 1.078 x𝟏𝟎 𝒐𝒉𝒎
Conductor Undergoing a Drawing Process
In the process, the waste of the material is assumed negligible (efficiency i100%),
thus keeping the volume to be constant all throughout the process,

With the volume of the material constant, resistance varies directly as the
square of the length.
R2
𝑳𝟐 𝟐
=( )
𝑳𝟏
R1

With the volume of the material constant, resistance varies inversely as to
the fourth power of the diameter.
R2
𝒅𝟏 𝟒
=( )
𝒅𝟐
R1
Where;
Subscript 1: condition before the drawing process
Subscript 2: condition after the drawing process
Sample problem:
A kilometer of wire having a diameter of 11. 7 mm and a resistance of 0.031 ohm
is drawn so that its diameter is 5.0mm. What does its resistance become?
Solution:
R2
𝑫
= (𝑫𝟏 )
R1
𝟒
or
𝟐
𝟏𝟏𝟕
R2 = 0.031( 𝟓 )4
𝑫
R2= R1 (𝑫𝟏 )
𝟒
𝟐
= 0.93Ω
Effect of temperature in resistance
Experiments have shown that the resistance of all wires generally used
in practice electrical systems, increases as the temperature increases.
R2
R1
R2
R1
=
𝑻 + t2
𝑇 + t1
= 𝟏 + 𝒂t1(t2- t1)
𝒂 t1 =
𝟏
T+t1
Where
R1=initial resistance (ohm)
R2= final resistance (ohm)
T= interfered absolute temperature
T1= initial temperature
T2= final temperature
𝒂 = temperature coefficient of resistance

Temperature coefficient of resistance (𝒂)- Ohmic change per degree at
some specified temperature
Material
Silver
Copper
Aluminum
ρ(µCm/ft.)
9.9
10.37
17
T(degree
celcius)
𝒂 @ 20 degree
celcius
243
234.5
236
0.0038
0.00393
0.0039
Tungsten
Zinc
33
36
202
250
0.0045
0.0037
Sample problem:
The resistance of the tertiary winding of a power transformer is o.125 ohm at 25℃
and the temperature coefficient of resistance at 25℃ is 0.00393. what is the
resistance at 65℃.
Solution:
R2 = R1 ( 1
+ a∆𝒕)
R2 = 0.125[1 + 0.00393(65 – 25)] = 0.14465 Ω
Sample problem:
The resistance of a transformer winding is 0.25 ohm at 25℃.when operating at full
load , the temperature of the winding is 75℃. The temperature coefficient of
resistance of cooper at 0℃ is 4.27 x 10−3 per degree centigrade. What is the
winding resistance at full load.
Solution:
1
1
A0 = 𝑇 or T= a
0
1
T=4.27 𝑋 10−3= 234.192℃
𝑇−T2
R2 = R1(𝑇+T )
1
234.192+75
R2 = 0.25( 234.92+25 ) = 𝟎. 𝟐𝟗𝟖 𝒐𝒉𝒎
Insulation Resistance of Cable
Where:
R = Insulation resistance of the cable (ohm)
P = resistivity of the insulating material (ohm-meter)
L = length of cable (meter)
r1 = radius of the conductor
r2= radius of the outer surface of the insulation
Sample problem:
A certain wire has a conductor with a diameter of 0.5 in. and 0.15 inch thick
insulation wrapped around the conductor. If the specific resistance of the insulation
is known to be 1x1014 ohm-cm, what is the insulation resistance per 1000ft length
of this wire?
Solution:
100 𝑐𝑚
L = 1000 ft. x3.281𝑓𝑡 = 30,478.51 𝑐𝑚
𝑑 0.50
R1 = =
2
2
= 0.25
R2 = r1 + t = 0.25 + 0.15 = 0.40 inch
𝑷
r2
R = 𝟐𝝅𝑳=ln(r )
1
1𝑋1014
0.40
R = 2𝜋(30,478.51) ln (0.25) = 𝟐. 𝟒𝟓𝐱𝟏𝟎𝟖Ω
DC ELECTRIC CIRCUIT
Ohm’s Law
● It states that at constant temperature, the current flowing in a electric circuit is
directly proportional to the impress emf applied to the circuit and inversely
proportional to the equivalent resistance of the said circuit.
 Named after the German physicist, George S. Ohm (1787-1854)
𝑬
𝑰=
𝑹
Where:
E = impressed voltage (volt)
I = current drawn (ampere)
R = resistance (ohm)
Current (I): The unit of current is Ampere (A) and is defined as the quantity of current
forced by a pressure of one volt through a resistance of one Ohm.
Voltage (V): The unit of voltage is Volt (V) and is defined as amount of pressure
required to force a current of one ampere through a resistance of one Ohm.
While talking of voltages we should be careful about the following two terms:


Electromotive Force (E): It is the voltage that exists across the terminals of a
battery or dynamo which is not connected to any external circuit.
Potential Difference (U): It is the voltage that exists across the terminals of a
battery or dynamo that is connected to an external circuit. It may be calculated
using theABOVE given formula:
U = (E – Internal resistance of battery or dynamo) × I
Resistance (Ω): The unit of resistance is Ohm (Ω) and is defined as the amount of
resistance offered by a circuit so that a current of one ampere is allowed to flow through
it at a pressure of one volt.
Sample problem:
A circuit has a resistance of 8 ohm. If a voltmeter connected across its terminal
reads 10 V, how much current is flowing through the circuit?
Solution:
Using Ohm’s law:
𝐸
10
𝐼=
=
= 𝟏. 𝟐𝟓 𝑨
𝑅
8
Electrical Power


Power is the rate of energy transfer.
Watt – unit of electrical energy equal to one joule of energy consumed in
one second. Named after the British engineer and inventor James Watt
(1736-1819)
𝑬𝟐
𝟐
𝑷 = 𝑬𝑰 = 𝑰 𝑹 =
𝑹
Where:
P = electrical power (watt)
E = voltage (volt)
I = current (ampere)
R = resistance (ohm)
Sample problem:
A 200-V lamp has a hot resistance of 400 ohms. The power rating in watts of the
lamp is?
Solution:
𝑃=
𝐸2
2002
=
= 𝟏𝟎𝟎 𝑾
𝑅
400
Sample problem:
An electric motor drives a mechanical load, taking 18.8 A from a 230 V source.
Calculate the power input of the motor.
Solution:
P = EI = 230(18.8) = 𝟒𝟑𝟐𝟒 𝐖
ELECTRICAL ENERGY
 Energy is the capacity to do work.
𝑾 = 𝑷𝒕
Where:
W= electrical energy (joule)
P= electrical power (watt)
t= time (second)
Remember:
1calorie = 4.186 joule
1 BTU = 252 calories
1kWh = 3600 kJ =3413 BTU = 860 kcal
1 joule = 1 𝑥 107 ergs
1 day = 24 hours
1 month = 30 days = 720 hours
1 year = 365 days = 8760 hours
Sample problem:
A 10 hp motor runs at rated load for 5 hours. How many kWh is consumed?
Solution:
0.746 𝑘𝑊
𝑊 = 𝑃𝑡 = [10 ℎ𝑝 𝑥
] 5 = 𝟑𝟕. 𝟑 𝒌𝑾𝒉
ℎ𝑝
Sample problem:
A residential house has a lighting load of 1000 W and a small appliance load of
2000 W. If they are used at the same time, what will be the monthly bill at an energy
cost of P 0.40 per kilowatt-hour?
Solution: W = Pt = (1kW + 2kW)(720 hours) = 𝟐𝟏𝟔𝟎 𝐤𝐖𝐡
total cost = 2160 kWh x
P 0.40
kWh
= 𝐏 𝟖𝟔𝟒
Sample problem:
Ten kW is equal to a how many calories per second.
Solution:
𝑃 = 10,000
𝐽
𝑠
𝑥
1 𝑐𝑎𝑙
4.186 𝐽
= 𝟐𝟑𝟖𝟖. 𝟗𝟏𝟓
𝒄𝒂𝒍
𝒔
Series Connected Resistors

Series circuit – the resistors are connected end to end.
 The total resistance is equal to the sum of the individual resistances.
𝑹𝒕 = 𝑹𝟏 + 𝑹𝟐 + … + 𝑹𝒏
 The total voltage is equal to the sum of the voltage drop across each resistance.
𝑬𝒕 = 𝑬𝟏 + 𝑬𝟐 + … + 𝑬𝒏
 The current flowing in each resistances are the same
𝑰𝒕 = 𝑰𝟏 = 𝑰𝟐 = … = 𝑰𝒏
Parallel Connected Resistors
Parallel circuit – the resistors are connected across each other.
 The total resistance is equal to the reciprocal of the reciprocals of the individual
resistances.
𝟏
𝑹𝒕 =
𝟏
𝟏
𝟏
+ + …+
𝑹𝟏 𝑹𝟐
𝑹𝒏
 The voltage drop across each resistor is equal to the total voltage.
𝑬𝒕 = 𝑬𝟏 = 𝑬 𝟐 = … = 𝑬𝒏
 The total current is equal to the sum of the currents flowing in each resistance.
𝑰𝒕 = 𝑰𝟏 + 𝑰𝟐 + … + 𝑰𝒏
𝑬
𝑬
𝑬
𝑬𝒕 = 𝒕 + 𝒕 + … + 𝒕
𝑹𝟏
𝑹𝟐
𝑹𝒏
 The total conductance is equal to the sum of the individual conductance.
𝑮𝒕 =
𝟏
𝑹𝒕
+
𝟏
𝑹𝟏
+
𝟏
𝑹𝟐
+ …+
𝟏
𝑹𝒏
Sample problem:
A 5-ohm resistance is connected in parallel with a 10-ohm resistance. What is the
equivalent resistance?
Solution:
𝑹𝒕 =
𝟏
𝟏
𝟏
+
𝑹𝟏 𝑹𝟐
=
𝟏
𝟏 𝟏
+
𝟓 𝟏𝟎
𝑹𝒕 = 𝟑. 𝟑𝟑 𝜴
Sample problem:
Three resistor 𝑅1 , 𝑅2 𝑎𝑛𝑑 𝑅3 are connected in parallel and take a total current of
7.9 A from a dc source. The current through 𝑅1 is half of that through 𝑅2 . If 𝑅3 is
36 ohm and takes 2.5 A, determine the values of 𝑅1 𝑎𝑛𝑑 𝑅2 .
Solution:
It = I1 + I2 + I3
Et = E1 = E2 = E3
7.9 = 0.5I2 + I2 + 2.5
E3 = I3 R 3 = 2.5(36) = 90 V
I2 = 3.6 A
I1 = 0.5(3.6) = 1.8 A
R1 =
𝑹𝟐 =
E1
I1
=
𝑬𝟐
𝑰𝟐
=
90
1.8
= 50Ω
𝟗𝟎
𝟑.𝟖
= 𝟐𝟓𝜴
Series-parallel Connected Resistors
Series-parallel circuit – a combinational circuit which when simplified will result into a
series circuit.
𝑹𝒕 = 𝑹𝟏 +
𝟏
𝟏
𝟏
+
𝑹𝟐
𝑹𝟑
Parallel-series Connected Resistor
Parallel-series circuit – a combinational circuit which when simplified will result into a
parallel circuit.
𝑹𝒕 =
𝟏
𝟏
𝟏
𝑹𝟏 + 𝑹𝟐 + 𝑹 𝟑
Remember: Independent of the circuit connection, the total power drawn by the circuit is
equivalent to the sum of the power drawn by each resistances in the circuit.
𝑷𝒕 = 𝑷𝟏 + 𝑷𝟐 + … + 𝑷𝒏
𝑷𝒕 = 𝑰𝒕 𝟐 𝑹𝒕
Where:
𝑃𝑡 = total power drawn by the circuit
𝑃1 , 𝑃2 , 𝑃3 = power drawn by the resistors 1, 2 and 3 respectively
Sample problem:
Two resistance of 10 and 15 ohms each respectively are connected in parallel. The two
are then connected in series with a 5-ohm resistance. What is the equivalent
resistance?
Solution:
𝑹𝒕 = 𝟓 +
𝟏
𝟏
𝟏
+
𝟏𝟎 𝟏𝟓
𝑹𝒕 = 𝟏𝟏 𝒐𝒉𝒎𝒔
Sample problem:
A 5-ohm resistance is connected in parallel with a 10- ohm resistance. Another set, a 6ohm and an 8-ohm resistance are also connected in parallel. The two set are connected
in series. What is the equivalent resistance?
Solution
𝑹𝒕 =
𝟏
𝟏
𝟏
+
𝟓 𝟏𝟎
+
𝑹𝒕 = 6.76 𝜴
𝟏
𝟏 𝟏
+
𝟔 𝟖
Sample problem:
Two resistor of 5 and 10 ohm resistance are connected in parallel across a 12 V battery
with 0.2 internal resistance. What power is drawn by the two resistor?
Let: R p = equivalent resistance of the parallel resistors.
Solution: Type equation here. R p =
1
1 1
+
5 10
= 3.33Ω
R t = 3.33 + 0.2 = 3.53Ω
It =
Et
12
=
= 3.4 A
Rt
3.53
PP = It 2 R P = (3.4)2 (3.33) = 𝟑𝟖. 𝟓 𝐰𝐚𝐭𝐭𝐬
Current Division Theorem (CDT)
𝐼1 =
𝐼𝑡 𝑅2
𝑅1 + 𝑅2
𝐼2 =
𝐼𝑡 𝑅1
𝑅1 + 𝑅2
Remember: If there are more than resistors in parallel, reduce first the circuit in two
resistors in parallel before applying the CDT
Sample problem:
A 10-ohm and a 20-ohm resistance are connected in parallel. Another resistance of 5ohm is connected in series with the two. If the supply voltage is 48 volts, what is the
current through the 10-ohm resistance?
Solution:
𝑅𝑡 = 5 +
𝐼𝑡 =
By CDT:
1
1
1
10 + 20
= 11.67 𝑜ℎ𝑚𝑠
𝐸𝑡
48
=
= 4.113 𝐴
𝑅𝑡
11.67
𝐼10 =
𝐼𝑡 (20)
4.113(20)
=
= 2.74 𝐴
10 + 20
30
Exercises
TEST I
1. A positively charged dielectric has a charged of 2 coulombs. If 12.5 X10 18 free
electrons are added to it, what will be the net charge on the said dielectric?
2. A cloud of 2.5 X1019 electrons move past a given point every 2 seconds. How
much is the intensity of the electron flow?
3. The current in an electric lamp is 5 amperes. What quantity of electricity flows
towards the filament in 6 minutes?
4. The substation bus bar is made up of 2-inches round copper bars 20ft.long.
What is the res istance of each bar resistivity is 1.724 x 10-6 ohm-cm
5. Determine the resistance of a bus bar made of copper if the length is 10 meters
long and the cross-section is a 4 x 4 cm2. Use 1.7241 micro-ohm-cm as the resistivi
ty.
6. A 500 MCM ACSR cable has 37 strands. Determine the diameter in mils of each
strand.
7. A given ire has a resistance of 17.5 ohms. If the length is 560m, how much
length must be cut-off from the wire in order to reduce its resistance to 12.5 ohms
8. A cylindrical rubber insulated cable has a diameter of 0.18 inch and an insulation
thickness of 0.25 inch. If the specific resistance of rubber is 10 14 ohm-cm,
determine the insulation resistance per 1000-ft length of the cable
9. The resistance of a copper wire at 30 degree Celsius is 50 ohms. If the
temperature coefficient of copper at 0 degree Celsius is 0.00427, what is the
resistance at 100 degree Celsius .
TEST II
1. 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?
2. A 120 DC motor draws a current of 100 A AND I LOCATED 1000 ft from the
supply source. If the diameter of the copper transmission line is o.45 inch, what
must be the voltage of the supply?
3. The hot resistance of an incandescent lamp is 10 ohms and the rated voltage is
50 V. Find the series resistance required to operate the lamp from an 80V supply.
4. Three resistors of 10, 15 and 20 ohms each are connected in parallel. What is
the equivalent resistance?
5. A 10-ohm and a 20-ohm resistance are connected in parallel. Another resistance
of 5 ohm is connected in series with the two. If the supply voltage is 48 volts, what
is the current through the 10 ohm resistance?
Flexible Teaching Learning Modality (FTLM) adopted
Example:
Online (synchronous) TelEducation moodle, Facebook messenger
Remote (asynchronous)
module, problems sets
References:
Complete Electrical Engineering Formulas and Principles, by: Romeo A.
Rojas Jr.
1001 Solved Problems in ELECTRICAL ENGINEERING, by: Romeo A.
Rojas
https://circuitdigest.com/tutorial/ac-circuit-theory