002 Chapter 1 Basics of Electrical Eng

7/29/2019 002 Chapter 1 Basics of Electrical Eng

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SKEU 1003 SEM I 2011/12

BASIC ELECTRICALENGINEERING

CHAPTER 1


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1.1 Introduction to Electrical Engineering

1.2 Systems of Units.

1.3 Electric Charge.1.4 Current.

1.5 Voltage.

1.6 Power and Energy.1.7 Circuit Elements.

2

CONTENTS


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What is

electrical

engineering?

Engineering

the discipline dealing with the art or science of

applying scientific knowledge to practical problems

Science Aims to understand the why and how of nature

Technology A product of engineering and science, the study ofthe natural world

Engineering discipline that deals with the

study and application of electricity and

electromagnetismn

Power, control systems, electronics and

telecommunications

BASIC ELECTRICAL ENGINEERING


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Branches of electrical

engineering

Communications:

sending information

through electrical signal

Control: controlling of

physical process using

electrical energy and

signal

Power system:generation,

tranmission and distribution

electric energy

Signal Processing:

process the signal

sense by sensor to

obtain requiredinformation


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Branches of electrical

engineering

Computer systems:normally found in our

daily live: washing

machine, car, personal

computerElectromagnetics: branch which

involve magnetic and electricfield, example magnetron in

microwave oven

Mechatronics: branch

combine mechanical

system and electronic

system


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Electric circuit theory &

electromagnetic theory

Fundamental theories in

electrical engineering


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7

Fakult i Kejurut eraan Elektrik


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8



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Quantity Basic unit Symbol

Length meter m

Mass kilogram Kg

Time second s

Electric current ampere A

Thermodynamic

temperature

kelvin K

Luminous intensity candela cd

Six basic units


9


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Electric quantity, unit& symbol

Quantity Symbol Unit

Charge Q Coulomb

Voltage V; v

Current I; i

Energy W

Power P; p

Impedance Z

Resistance R


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The derived units commonly used in electric circuit theory

Decimal multiples and

submultiples of SI units


1

1


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A simple electric circuit

batterylamp

current

What is an electrical circuit?

Circuit that consists of various types of elements/components

connected in closed paths by conductors


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Charge

an electrical property of an atomic particlesmeasured in coulombs ( C ).

Q (Coulomb) = I (ampere) x t (second)

1 ampere-hour = 3600 Coulomb.

The flow (motion) of electric charges createselectric current

I

e

conductor


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Law of conservation of charge

Charge can neither be created nor destroyed

Electric charge or electricity is mobile:

Can be transferred from one place to

another where it can be converted to

another form of energy

Transfer of charge transfer of energy

The energy is transferred and transform (convert) into

useful form as heat, sound, light


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Electric current the time rate of change of charge, measured in

amperes (A)

the relationship between current i, charge q andtime tis given by:

i = dq/dt ori(t) = dq(t)/dt

1 Ampere = 1 Coulomb/second

q time varying charges Q constant charges

The flow of electrons; can be produced by chemical action

or by a generator.


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Ifi = dq/dt,

q = ?


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Direct current (dc)

Conclusion from the graph?

t (sec)

i (amp)

A dc is a current that remain constant with time


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Alternating current (AC)?

Conclusion from the graph

i(amp)

t(sec)

An AC is a current that varies sinusoidally with time


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(i)(ii)

-3 A

3 A

Conventional current flow:(i) positive current flow,(ii) negative current flow


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Example 1Given q = (10 10e-2t)mC and t= 0.5 s.

Calculate the current, i.

i = dq/dt = d/dt(10 10e-2t) = 20e-2t

At t= 0.5 s, i= 20e-2x0.5 =


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Voltage

Also known as potential difference the energy or work required to move a unit

charge through an element (from one point to

another point). vab = dw/dq

Unit : volts (V) orjoule/coulomb

Voltage acrossan element is actually measuredbetween two points ( a and b), the plus (+) and

minus (-) signs are used to define reference

direction or voltage polarity.

1 volt = 1 joule/coulomb=1 newton-meter/coulomb


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Voltage polarity , vab

vab

+a

-b

9 V

+ a

- b

- 9 V

+ b

- a

(i) (ii) (iii)


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It must be remembered that the voltage has direction given by itspolarities (+, -) mark on the terminal. Polarity of voltage can also be

identified by using double subscript notation such Vab. The voltage Vab

can be interpreted in two ways: (1) point a is at a potential of Vab volts

higher than point b, or (2) the potential at point a with respect to point bis Vab.

Vab = 5 V

Vba = - 5 V

Vab = ? V

Vba = ? V

V1 = 2 V

V2 = ? V

+

5V

-

a

b

Element,

X

a-

5V

+

b

Element,

X

+

V1

-

-

V2

+

b

a

Element,

X


2

3

Voltage have Polari t ies


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Power

Definition? Unit ?

Poweris the energy supplied or absorbed

per unit time

p = dw/dt

= (dw/dq).(dq/dt)

= v.i

p is time-varying quantity and is called

the instantaneous power

Powerabsorbed orsupply by an element is the

product of the voltage across the element and

the current through it


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W

p

t

v

i


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v

i

p

t


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Passive sign conventionpassive sign convention is satisfied when

the current enters through the positive

terminal of an element ,p = +vi. If the current enters through negative

terminal,p = -vi


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v

+

-

i

v

+

-

i

p = + vi,

absorbing power

p = -vi, supplying

power

Conclusion : +Power absorbed = -Power supplied

F k lt i K j t El kt ik


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Polarity references and the expression or power.


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3A

4 V

+

-

p= ?

Observe the directions of voltage and current for both

figure. Give conclusion

3A

4 V

+

-

p= ?

(a) (b)

Example 2


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3A

4 V

+

-

p = ?

(a)

3A

4 V

+

-

p = ?

(b)

Observe the directions of voltage and current for both

figure. Give conclusion

Example 3


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Power: Absorbed or supplied

Va

+

-

Ia

Va = 12 V

Ia = 2 A

Pa = ?

Vb

+

-

Ib

Vb = 12 V

Ib = 1 APb = ?

Vb

-

+

Ib

Vb = 12 V

Ib = -1 A

Pb = ?

Example 4


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Example 5

Determine P and the direction of power transfer

for the following set of current and voltage

(a)I = 15A, V = 20 V b) I = 4 A, V = -50 V

c) I = -5A, V = 100V d) I = -16A, V = -25V

I

A

B+

V

-


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Energy The capacity to do work

Measured in joules, (J) or watt.second

dt

dwp

Energy (w), absorbed or supplied by an element

from time t0to tis:

t

t

dtvidtpwt

t 00


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Note

Electrical energy is expressed in terms of

kWh (kilowatt-hours)

1 kWh =1 kW x 1 hour=1000 watt-hours

(= 1000 x 60 x 60 watt-sec)


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Example 6 How much energy does a 100 Watt electric bulb

consume in two hours?

W=pt=100Watt * 2 hour=200Watt.hour= 200(Watt.hour)*60(min/h)=1200 Watt.minute

= 12000 Watt.min*60(s/min)=720,000 Watt.sec

= 720,000 Joule=720Kj


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CIRCUIT ELEMENTSAn electric circuit is an

interconnection of circuit elements

linked together in a closed path sothat an electric current may flow

continuously.

Active elements and passiveelements


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Active

elements

Has ability to supply

energy

Eg: voltage sourse

or current source,

battery,generator

Pasive

elementsCannot supply power

Eg. Resistor, inductor,

capacitor

Resistor absorbs power(energy)

Inductor & capacitor has ability

to store energy and release

energy through the circuit



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Active Elements Passive Elements

Independent

sources

Dependant

sources

A dependent source is an active element in

which the source quantity is controlled by

another voltage or current.

They have four different types: VCVS, CCVS,

VCCS, CCCS. Keep in minds the signs of

dependent sources.


3

9


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-0.5 A

a b

c de+

4 V

-

- 1 V + + (-5 V) -

2.5 A

3 A

+

10 V-


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Active elements

dependent sources

Current sourceVoltage source

independent sources

Voltage source Current source

Voltage control Current control Voltage control Current control


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List of dependent sourcesVoltage controlled voltage source (VCVS)

Current controlled voltage source (CCVS)

Voltage controlled current source (VCCS)

Current controlled current source (CCCS)


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ACTIVE ELEMENTS:Independent Sources

Independent Voltage Source (IVS)

-An ideal (IVS) is an active

element that providesa specific voltage which

is completely

independent of current

flowing through it.

The current flow through it is dependent

on the circuit element connected to it

+

-Vs


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Independent Current Source (ICS) An ideal ICS is an active element that provides

a specific current which is completely independent

of voltage across it.

The voltage across it is dependent on the

circuit element connected to it.

Is


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Practical sources (actual sources) have

power limit which it can supplied

p = vi.

Under normal condition, each source is

supplying power but under certain condition

the source can absorb powerexample

battery charging circuit : where a battery has

been charge using battery charger and thecurrent enter through positive (+)

terminal of battery.


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DEPENDENT SOURCEDependent source supply (produce)

voltage or current that is controlled by

the voltage or current at other part ofthe circuit. Dependent sources are

useful in modeling electronic circuits


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iv+

-

(a) (b)

Symbols for:

(a)Dependent voltage source

(b) dependent current source


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Question: State active elements, passive

element and the two dependent sources?

6 3 1

8VY

2

7 3 IX

IX

12 30 V

+ VY -


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Passive Element Electrical elements that absorbed or received or store

power.

Current enters the positive terminal and leaves thenegative terminal OR

Current flows through a passive element is in oppositedirection of the voltage across the element.

I

v-+

IA BPassive Element


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RESISTANCE

Definition?

Symbol?

Unit? a materials opposition to the flow of electric

current orelements ability to resist the flow of

electric current

R

Ohm


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Depend on geometry and type of material used R = (l/A)

resistivity of

material.

l

A

Difference material have difference value

of .


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Resistivity of material,

Material Resistivity of material, (m)

Conductors

Aluminum 2.8 x 10-8

Copper 1.72 x 10-8

Silver 1.64 x 10-8

Gold 2.45 x 10-8

Semiconductors

Carbon

4 x 10-5

Silicon 6.4 x 102

Insulators

Glass 1012

Teflon 3 x 1012


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Example (i-c) Calculate the resistance of

i) Copperii) Glass

if the radius and the length of materialsare 1.025x10-3 m and 10m respectively


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i) Copper

A = (1.025x10-3)2 = 3.3x10-6 m2

AlRcopper = Rc =

052.0103.31010x1.72

6

8-

x


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ii) Glass

A = (1.025x10-3)2 = 3.3x10-6 m2

Rglass = Rg = Al

Exxx 03.31003.3103.3

10101

18

6

12


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Resistor Resistor is a circuit element that uses to

resist the flow of current.

+ -V

i

OHMS LAW 1827 by George Simon

Ohm


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Ohms Law

The voltage, vacross a resistor is directlyproportional to the current i, flowing

through the resistor. i.e

v i

v = Ri

v

iR = v/i


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Rx = v/ix

Ry =v/iy

Rx >Ry

V

ix

iy

i

v


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R = 0

+

-

a

b

v = 0

Therefore v = 0

R = 0/i = 0

Short circuit

i

R = (infinity)

i = 0a

+v

-

b

Therefore i = 0

R = v/0 =

open circuit


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Example (i-d)

What is the resistance of copper conductor with aradius of 1.025 x 10-3 and length of 10 m.

A= (1.025x10-3)2 = 3.3x10-6 m2

A

L

052.0

103.3

1010x1.72

6

8-

xRcopper= Rc =


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For the glass with the same radius and length ascopper, the resistance for the glass will be.

A

L

Exxx 03.31003.3

103.310101 186

12

Rg=

CONCLUSION?


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Conductance Reciprocal of resistance

G

Measure of how well an element willconduct electric current

v

i

R

G 1 ((mho,

siemens (S)

or

SI unit


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The ability of an element to conductelectric current.

Compare:

i) If R = 100 , G = 1/100 = 0.01 S

ii) If R = 0.002 ,

G = 1/0.002 = 500 S

Conclusion?


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Power Resistor

passive sign convention, power absorb byresistor is p = vi

Usingohms law,

+ -V

i

v = iR

p = vi =(iR)i = i2R


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i = v/R

Rv

R

vvvip

2

v= iR

RiiiRvip2)(

Power

dissipatedby aresistor


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Power in term of GG = 1/R

GvvGvvip

GviG

iiRv

2)(

G

ii

G

ivip

2


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Resistor formula


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Example (i-e)

DC

+

-

v

15V

i

Determine the value R and iif power absorbs by R is 0.5 W.


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Practice problem 2.2 (page 34)

Calculate the voltage, the conductance Gand the powerp

2 mA

i

+

V-


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Type of resistors

Fixed resistor

Variable Resistor - potentiometer.

-

rheostat.

) P t ti t ) P t ti tb) Rh t t

002 Chapter 1 Basics of Electrical Eng

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