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I. BASIC CONCEPTS
A. ELECTRICAL UNITS
BASE UNITS in S.I
Length meter (m)Mass Kilogram (kg)
Time seconds (k)
Temperature Kelvin (k)
Current Ampere (A)
S.I. Unit Prefixes
Exa 1018 Atto 10-18
Peta 1015 Femto 10-15
Tera 1012 Pico 10-12
Giga 109 Nano 10-9
Mega 106 Micro 10-6
Kilo 103 Milli 10-3
Hecto 102 Centi 10-2
Deca 101 deci 10-1
Scientific Notation
It is the shifting of the decimal point either to the left or to the right of the given
number until there is only one significant digit to the left of the decimal point and
then multiplying the number with the appropriate power of 10 to retain its original
value.
A way of expressing a number in terms of the power of 10
Ex. 58,000 m = 58 x 104 m
Engineering Notation
It is an exponential format of specifying format on specifying numbers in which
the powers of 10 are limited to the multiples of three so that it corresponds to an
S.I. prefix.
Ex. 58,000 m = 58 km
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B. DEFINITION OF TERMS
Electric Charge
• The quantity of electric energy stored in battery, capacitors, elementary particles or
any insulated materials.
• Q (constant quantity)
• q (instantaneous quantity)
instantaneous – a varying quantity at a particular instant, expressed in Coulomb
1 Coulomb = 6.25 x 1018 electrons
1 electron = 1.602 x 10-19 coulomb
Electric Current
• A net flow of positive or negative charges that passes to a given point or a
specified period of time.
• A rate of transfer of electricity from one point to another.
• I or i
I = Q/t (constant quantity)
I = dq/dt ((instantaneous quantity)
1 Ampere= 1 coulomb / sec
Where:
I = Current (Ampere), A
Q = Charge (coulomb), C
t = Time (second), s
Voltage or Potential Difference
• It is the potential energy difference that exists between two points.
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• It is the amount of work done per unit charge.
V= W/Q
Where:
V= voltage (volts), V
W = work or energy ( Joules), J
Q = charge (coulomb), C
+ V -
“The charge will not move unless you apply the potential difference”
+ V - + V -
Voltage Drop Voltage Rise
C. CIRCUIT COMPONENTS
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1. Active Elements
• Elements capable of supplying energy.
• Components capable of controlling voltages to produce gain & switching action
in a circuit.
a. Voltage Source
• Independent current source
• Dependent / Controlled source
Independent Sources
V I
Capable of delivering voltage or current regardless of the network connection
Dependent Sources
+
-
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V I
Supplies voltage or current controlled by variable connected in some other part of the
network
2. Passive Elements
• Are elements capable of absorbing energy
• Are elements capable of storing energy but does not supply energy
a. Resistor – absorbs energy
b. Inductor – stores energy
c. Capacitor – stores energy
Resistor
• Its function is to limit the amount of current or divide the voltage in a circuit.
• It is also used to convert electrical energy into another form of energy like heat energy.
Unit: Ohm
Capacitor
+
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• Its basic function is to concentrate the electric field of voltage applied across the
dielectric. A capacitor is constructed of two conductor plates separated by an insulator
(dielectric)
Unit: Farad (F)
Inductor
• Its main function is to concentrate the magnetic field of electric current in a coil.
• An induced voltage is generated when the current changes its value or direction.
Unit: Henry (H)
A. Nature of Resistance
Resistance
• The property of a material or circuit element to oppose the flow of electrons.
Factors affecting the resistance of a conductor:
1. Length
2. Cross- sectional area
3. Nature of the material
∆
L
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R α L/A
• R = but V = A L
• L = V / A ; A = V / L
• R = =
Resistivity of Copper at 200C:
• Standard Annealed Copper
ρ= 1.7241 x 10-8 Ω-m
= 1.7241 x 10-6 Ω-cm
= 10.37 Ω-cmil / ft
• Hard – Drawn Copper
ρ= 1.77 x 108 Ω-m
= 1.77 x 106 Ω-m
= 10.65 Ω-cmil / ft
Mil (mil)
• A unit of length equivalent to one thousandth of an inch.
1 mil = 1 x 10 -3 in
Square mil (mil2
)
• A cross-sectional area of a square whose side is equivalent to 1 mil.
Circular mil
• A cross-sectional area of circle whose diameter is equivalent to 1 mil.
1 cmil = / 4 sq.mil
Where:
p = resistivity or specific resistance of a
given
material at a certain temperature.
(ohm – m)
L = length (m)
A = cross-sectional area (m2)
V = volume (m3)
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Area in cmil = D2 = D12 x 106
Where:
D = Diameter in mils
D1 = Diameter in inch
Table 1. Resistivity Constant of some common electrical material at 200C.
Material ρ at 200C (Ω-m)
Silver 1.64x 10-8
Copper 1.72 x 10-8
Aluminum 2.83 x 10-8
Tungsten 5.50 x 10-8
Nickel 7.80 x 10-8
Iron 12.0 x 10-8
Constantan 49.0 x 10-8
Nichrome 11- x 10-8
Conductance
• The property of the material that allows easily the flow of current.
G = 1 / R =
G = 1 / ρ x A / L
G = σ A / L
Where:
G = conductance in mho or Siemens (S)
σ = conductivity constant in S/ m
Percent Conductivity
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_DE_ = _AO + OD_
BC AO+ OC
_R2_ = _/Tx/ + T2_
R1 /Tx/ + T1
Therefore:
R2 = R1 [/Tx/ + T2]
[/Tx/ + T2]
Where:
R1 = resistance at temperature T1
R2 = resistance at temperature T2
T1 = Initial temperature
T2 = Initial temperature
Tx = inferred zero resistance temperature
= inferred absolute zero temperature
Inferred Zero Resistance Temperature
• The temperature in which the material inhibits zero resistance or super
conductivity.
Or R2 = R1 + EF
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∆ABC = ∆BEF
_EF_ = _OD – OC_
BC AO + OC
_EF_ = _T2 – T1 _
R1 /Tx / + T1
EF_ = __∆t_ _
R1 /Tx / + T1
Let: α1 = ___1____
/Tx / + T1
Therefore:
R2 = R1 + R1 α1 ∆t
R2 = R1 (1+ α1 ∆t)
Where:
α1 = temp. Coefficient of resistance @T1 in /°C
∆t = change in temperature
Table 2. Inferred zero resistance temperature and temperature coefficient of resistance
of some common electrical material.
Material Tx (0
C) (/0
C)Silver 243 0.0038
Copper 234.5 0.00393
Aluminum 236 0.0039
Tungsten 202 0.0045
Nickel 147 0.006
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Iron 180 0.0055
Nichrome 6250 0.00016
Constantan 0.000008
Carbon -0.0005
RESISTOR
Digit Multiplier Tolerance
Resistors Color Coding
Color Digit Multiplier Tolerance
Black 0 x 100
-Brown 1 x 101 -Red 2 x 102 -Orange 3 x 103 -Yellow 4 x 104 -Green 5 x 105 -Blue 6 x 106 -Violet 7 x 107 -Gray 8 x 108 -White 9 x 109 -Gold - x 0.1 +/- 5%Silver - x 0.01 +/- 10%
No color - - +/- 20%
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SAMPLE PROBLEMS:
1. What are the values of the following resistors? Tolerance?
a. Brown Red Brown Silver
b. Green Blue Black
c. Brown Gray Yellow Gold
d. Red Red Gold Gold
e. Brown Black Green
2. A certain resistance was measured to be 30Ω at 200C and 40Ω at 950C. Find the
temperature coefficient at 00
C and 250
C and the inferred zero resistance temperature.
GIVEN:
R 1 = 30 Ω T 1 = 20 oC
R 2 = 40 Ω T 2 = 95 oC
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REQUIRED:
α oc , α 25 , T x
SOLUTION:
A
B
3. What is the resistance at normal room temperature of 60m of copper wire having a
diameter of 0.64mm?(ρ copper = 1.72 x 10-8 Ω-m at 200C)
SOLUTION:
40(Tx + 20 ) = 30(Tx +95 )
Tx = 205
R = 3.21Ω
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4. What is the conductance of 100ft of no. 14 AWG iron wire which has a diameter of
64mils at a temperature of 200C? (ρ iron = 1,2 x 10-6Ω-cm)
SOLUTION:
5. A platinum coil has a resistance of 3.146Ω at 400C and 3.767Ω at 1000C. Find the
resistance at 00C and the temperature coefficient of resistance at 400C.
II. Simple Resistive Circuits
A. Ohm’s Law
• Basic law of electricity
States that if the voltage is kept constant, less resistance results in high
current and more resistance results in less current.
• The voltage is directly proportional to the current.
R = 1.7627
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Basic Electric Circuit:
V α I
V = RI
Where:
V = voltage (volts, V)
I = current (ampere, A)
R = resistance (ohms, Ω)
Current Flow
A. Electron flow
• Actual flow of current since electrons are the moving charges therefore it moves
from the negative of the source going to the positive of the sources.
B. Conventional Current Flow
• The assumed direction of the flow of current which is opposite from the electron
flow, that is from the positive of the source to the negative of the source.
Power and Energy
Work
• Accomplish of motion against on opposing force.
• The amount of force multiplied by the distance travelled.
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Note: Work is performed whenever energy is converted from one form to another.
W = V x Q
W = F x d : expressed in joules (J)
Where:
V = voltage (Volts)
Q = charge (Coulombs)
F = force (Newton)
D = distance (meters)
Joule
• SI unit of energy and work
• Amount of energy required to raise one coulomb of electric charge through a
potential difference of one volt.
Power
• The rate at which energy and work.
• The rate of doing work.
• The amount of work done per unit time.
P = W / t = joules / sec = watts
Where:
P = power (watts)
W = work (joules)
t= Time (seconds)
Electric Power
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The energy is lost in the form of heat
Note: energy from the moving charges of electron is transferred through inter- atomic
collision. As the electrons collide with the atoms of the resistors energy is transferred
thereby causing heat.
P = W / t
But: W = V x Q (refer to voltage)
t = Q / I (refer to current)
P= _V x Q_
Q / I
Therefore electric power:
P = V I - - - - - - 1
By applying Ohm’s Law
Substitute,
V = I R
Therefore,
P = I2 R - - - - - - -2
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Substitute,
I = V / R
Therefore,
P = V2
- - - - - - -2
R
Where:
P = power (watts)
W = work (joules)
t = time (seconds)
V = voltage (volts)
I = current (ampere)
Watt
• SI unit of electric power
• Amount of power when joule of energy / work is consumed / done in one second.
Kilowatt- Hour (kw-hr)
Unit of electric energy or electric work.
Energy = power x time
Where:
P = power (watts)
W = work (joules)
t = time (seconds)
Horsepower (Hp)
• Mechanical output power.
Note: 1Hp = 746W = 0.746KW
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Efficiency
• The ratio of useful output power of a device with the total input power.
η= =
Cascaded System
Overall Efficiency
η= = x x
η== = η1 x η2 x η3
Sample Problems:
1. A certain appliance uses 300W. If it is allowed to operate continuously for 30 days, how
many kilowatt-hours of energy does it consume?
SOLUTION:
n
1
Po1
Pi2
n
2
Po2
Pi3
n
3
Po
3Pi1
E = 216kw-hrs
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2. A power supply for a transistor amplifier develops an output of 25V at 2.4A. If the overall
efficiency of the power supply is 80%. What input current will it draw from 120V source?
SOLUTION:
3. A battery is rated at 100Ah and has a lifetime of 25hr, what is its rated current? What
power can it deliver if is terminal voltage is 3V? What is the total amount of energystored in the battery?
SOLUTION:
4. How long a 75Ω heater should be supplied by a 120V source to produce 15KJ of heat
energy?
SOLUTION:
Iin = 0.625 A
Irated = 4 Amp
P = 12 watts
W = 300w-hr = 0.30kw-hr
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5. What electrical input power must be provided to an electric motor which develops
mechanical energy at the rate of 24HP with an efficiency of 85%?
SOLUTION:
6. A ½ HP motor draws a current of 4A from a 120V line. Calculate the electric power used
by the motor and the motor efficiency.
SOLUTION:
Pinput = 21.06 KW
P = 480watts
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