introduction of basic electricity (1)

8/6/2019 Introduction of Basic Electricity (1)

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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



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.



• 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



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

+

-



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

+



• 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



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)



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



_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



∆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



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%



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



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



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



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.



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



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



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



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



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



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

introduction of basic electricity (1)

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