By

ABDUL HAKIM HJ ABU BAKAR

Topic 5:

FUNDAMENTAL OF FUNDAMENTAL OF

ELECTRICAL & ELECTRONICSELECTRICAL & ELECTRONICS

CLB 10402CLB 10402

Objective

After this lesson, student should be able to :

Understand the concept of peak, rms,

and average value

Understand the concept of shape factor

and peak factor

Objective

Peak value

2PP PV V

2PP PI I

peakV

peak to peakV

Root-mean-square (r.m.s) value

0.707rms PV V

0.7071rms PI I

• Average heating effect over half – cycle :

2

2/ 2 ( ) 1

( )( ) 2

P

P

I R watt radianI R watt

rad

• If I is the value of dc voltage through the same resistance to produce the same heating effect :

therefore

2 21

2rms PI R I R

0.70712

Prms P

II I

Average value

0.637avg PV V

0.637avg PI I

Sinusoidal current

2 ( )

( )

Pave

I amp radI

rad

Average value of current

over a half-cycle :

0.637ave PI I amp

The total area under the half-cycle curve divided by the distance in radian of the curve along the horizontal axis

The voltage across the resistor is directly proportional to the current It follows that the relationship derived for current

also apply to voltage So, in general, average value of sinusoidal current or voltage is

0.637ave PI I

0.637 x maximum value

Shape factor

r.m.s value

Average value

0.707 x maximum value

0.637 x maximum value

11.1fk

Peak factor

Peak or maximum value

r.m.s value

maximum value

0.707 x maximum value

414.1pk

Conclusion

11.1fk

414.1pk

Shape factor

Peak factor

What is power factor?

• Is the ratio of true power to

apparent power in an AC circuit.

• PF = PT/ PA

Another way...

• PF = EI Cos / EI

• PF = Cos

• Power factor is the cosine of the angle.

Also…another way

Voltage against current

I

V

Cos

Conclusion

Purely resistive Purely reactive Equal

PF = 1 PF = 0 PF = 0.7071

100% consumed No power consumed 70% consumed

Objectives

• Understand the use of resistance,

inductance and capacitance in circuit

• Understand the concept of R-L-C circuit

in series and parallel

• Determine the total impedance, Z and

current, I in a circuit

Introduction

Resistance

Consume energy

Reduces current

Inductance

Stores energy in a magnetic field

WL = ½ LI2

Capacitance

Store energy in electric circuit

Wc = ½ CV2

Inductive AC circuit Capacitance AC circuit

Types

AC circuit

Resistance Capacitance Inductance

XL

XC R

Total Impedance

fLX L 2fC

X c2

1

2 2( )L CZ R X X

Total Circuit Current

I = E/Z

R L C

Given: Voltage = 150 V, 60 hertz.

Find I, if R = 1100Ώ

L = 8 mH

C = 2µF

What is RLC Circuit ?What is RLC Circuit ? What is RLC Circuit ?What is RLC Circuit ?

What we want to know?What we want to know? What we want to know?What we want to know?

• To know the value of each components

• To calculate Xc and XL

• To find current of each components

• To calculate total current

• To calculate the impedance

PropertiesProperties PropertiesProperties

• Voltage is constant

• Different current flow in each branch

Step by stepStep by step Step by stepStep by step

1. Value of each components

Resistance Ohm

Inductance Henry, H

Capacitance Farad, F

2. To find Xc and XL

fcX c

2

1

fLX L 2

If Xc = XL, it will be a resonance,

IC and IL will have a same amplitude but,

differs 180 in phase.

3. To find current

C

C

L

L

R

X

EI

X

EI

R

EI

4. Total current

5. Impedance

22 )( LCRT IIII

T

TT

I

EZ

OBJECTIVES

After this lesson, you should be able to:

Define true power

Define apparent power

POWER

Electrical power is the rate at which

electrical energy is converted to another

form, such as motion, heat, or an

electromagnetic field

Power, P (Watt)

APPARENT POWER

A measure of alternating current power that

is computed by multiplying the root-mean-

square current by the root-mean-square

voltage.

EIP

Measured in Volt-amps(VA)

TRUE POWER

In a resistive circuit, power consumed

by the resistor is dissipated into heat

and not returned to the source.

cosEIP

Measured in Watts,W

RESISTIVE CIRCUIT

Apparent Power(VA)

True Power(W)

Reacti

ve P

ow

er

(VA

R)

REACTIVE CIRCUITS

Apparent Power(VA)

True Power(W)

Reacti

ve P

ow

er

(VA

R)

APPLICATION

Electric Utility Company

- supply both true and apparent power

- apparent power is larger than true power thus

current supplied will be larger

- cost will arise

- Utility company will raise its raise for customers

operating at low power factors

Single-Phase Circuit Connection

Three-Phase to Single-Phase

Connection

3-Wire Edison Scheme

Summary

If a circuit is purely resistive, the current is in phase with the voltage. If it is purely inductive, the current lags the voltage by 90°. If the circuit is purely capacitive, the current leads the voltage by 90°.

If a circuit contains both resistance and inductance, the current lags the voltage by an angle less than 90o but the angle is greater than 0°.

If a circuit contains resistance and capacitance, the currents leads the voltage by an angle less than 90°, but the angle is greater than 0°

REFERENCES

• Edward Huges, “Electrical

Technology,” 8th Edition, Prentice Hall

• Theodore Wildi, “Electrical Machines,

Drives and Power Systems” 5th Edition,

Prentice Hall

• Marizan Sulaiman, “Analisis Rangkaian

Sistem Kuasa Elektrik Moden” Utusan

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