chapter1 -alternating voltage & current :)

8/13/2019 Chapter1 -Alternating Voltage & Current :)

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11/29/2013211/29/2013 Alternating Voltage & Current

Objectives

At the end of the chapter, student should be able

to:

describe the principle of ac voltage and currentgeneration.

calculate RMS and Average values for voltage

and current. explain and analyze the phasor representation.


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1. INTRODUCTION

Why AC is required? Whereas we already

have DCDC cannot be transmitted as economically as AC

transmission systems

Circuit Theory 1 & Engineering Math- importantElectrical unit, Ohms Law, Kirchoffs Law etc.

Abbreviations & SymbolsAC, DC

AC can be step-up and down

Nowadays, many electrical appliances are using AC

system


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What is AC system??

The magnitude of the voltage/current vary

in a repetitive manner. Eg: sinusoidal

wave, square wave, triangular wave

What is the characteristic of the AC

current/voltage?

It flows first in one direction and then in theother. The cycle of variations is repeated

exactly for each direction.

1. INTRODUCTION


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How to turn thecoils?

Hydro

WindGeothermal

Tidal


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Magnitude of the resulting voltage is proportional to the

rate at which flux lines are cut (Faraday's Law)


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Generation of an alternating e.m.f

E.m.f is induced when the flux

is being cut

No e.m.f is being generated in

the loop , if the loop AB are

moving parallelto the direction

of the magnetic flux.- no flux is

being cut. [(a)&(d)]


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Sine wave of induced e.m.f


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Formula of Induced EMF

Angular velocity = AL

AM = AL sin = X sin

Assumption: l= length of conductor of one side (m),

B= flux density (T)

So; total e.m.f. generated = 2 B l X sin (V)

X is also b n b is breadth / width of the loop in meters,

n is the speed of rotation in rpse = 2 B lb n sin (V)

For N numbers of coil, e = 2 B NAn sin (V)

Peak value Ep= 2B(NA)n (V)

A=area of loop


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p n

The generator shown before, has two poles or one pair of poles.

Machines can have two or more pairs of poles. If an a.c generator has

pairs of poles and if its speed is revolutions per second, thenfrequency no. of cycles per second

= no. of cycles per revolution x no. of revolutions per second.

f pn= Hz

Repeat the previous example if the generator uses six poles of magnet.

=f


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3. Waveform Terms & Definitions

Terms:

1. Period,Ttime taken to complete one cycle (s)

2. Peak value, Vmor Vp, Imor Ip

3. Peak-to-peak value, Vp-por Ip-p

4. Frequency, fNumber of cycles that occur in 1 second (Hz)

5. Amplitudedistance from its average to its peak

fT

1=


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Example:

Question:

i- Period

ii- Peak valueiii- Peak to peak value

iv- Frequency

v- Amplitude

Answer:

i- 10ms

ii- 10Viii- 20V

iv- 100Hz

v- 10V

Answer:

i- 10ms

ii- 5Viii-10V

iv-100Hz

v- 5V


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Example:

Find

i- Period

ii- Peak value

iii- Peak to peak value

iv- Frequency

The waveform at the left was

obtained from an oscilloscope with

the knobs turned at the following

positions:

Vertical axis 5V/div

Horizontal axis 10ms/div


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Example:

Answer:

(i) If Em=100V, determine the coil voltage at: (a) 30

(b) 330

(ii) If the coil rotates at = 3000/s, how long does it take to complete 1

revolution?

(i) (a) E30 = 50Volt

(b) E330= -50Volt

(ii) 1.2 sec


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Phase Shift/difference:

Phase shift occur when there are L or/and C exist in the

circuit3 important situations


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Phase lead or lag caused by C and L components


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5. Average & R.M.S. Values

Average

= sefengthcurvenderre

For a complete sinusoidal waveform, average value = 0

since it is symmetrical.

Area for half cycle of sine wave is

0

Ip

ppp IIdI 2cossin 00

==


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R.M.S. (Root Mean Square)Also known as effective value

Value to do useful workIt is an equivalent dc valueE.g: 240Vac capable of producing the

same average power as 240 volts of steady dc

E R

i P

t

Pdc = P ave

Pdc= Pave= I2R

For dc

For acE R

i P

t

Pave=

P(t)

2

2

RIm


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P(t) = i2R = (Imsin t)2R = Im

2R sin2t

tRIm 2cos12

12

= tRIRI mm

2cos22

22

=

To get the average value of P(t), the term cos 2 t will bezero, thus

Pave=2

2

RIm

Paveragedc= Paverage acI2R =

2

2

RIm

I2

2

2

mI

=

mm

rms I7070

2

III .===


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==T

rms di

TII

0

21

m

m II

I 707.02==

So, the r.m.s value of a sinusoidal current is

or,

Likewise for the average voltage and rms voltage.


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av

rms

value

value=

Form factor

rmsvalue

valuemax

=Peak or crest factor

Form factor and peak factor

Form Factor is the ratio between the average value and the RMS value.

Crest Factor is the ratio between the R.M.S. value and the Peak value

of the waveform .


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6. Phasor Representation

Representation of an alternating quantity by a phasor


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Phasor representation of analternating quantity for the

first half cycle


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Phasor Representation

PhasorMagnitude & Angle It is actually a complex number

Generally written as e = Epand I = Ipin phasor form

In complex form, e = Epcos + iEpsin

I = Ipcos + iIpsin

e (t) = Ep sin t e = Ep0 (V)

i (t) = Ip sin t I = Ip 0 (A)

Phasor diagram for above:

EpIp


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i = Imsin(t + ) = Im

i = Imsin(t - ) = Im -

Phasordiagram


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Example:

Phase different between Voltage and current waveforms are 40, and voltage lags.

Using current as the reference, sketch the phasor diagram and the corresponding

waveform.

v(t)= Vmsin(t - 40)

Im

Vm

j

40

40

i(t)= Imsin t

Repeat the above if the voltage leads by the same angle.


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Example:

Given v = 20 sin (t + 30) and I =18 sin (t - 40), draw the phasor diagram,

determine phase relationships, and sketch the waveforms.

V leads I by 700 or I lags V by 700


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Addition & Subtraction of Phasors

Add i t ion of phasors Substract ion o f phasors

C=A+B D=A-B


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Addition & Subtraction of Phasors

Vector Breakup Method

The instantaneous values of two alternating voltages are represented respectively

by v1=60 sin (V) and V2= 40 sin (- /3) (V). Derive an expression for the

instantaneous value of :

a) the sum;

b) the difference of these voltage (V1V2)

V1= 60V

V2= 40V

V1+ V2

First assume up positive, down negative

Right positive and left negative


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V1= 60V

V2= 40V V1+ V2

Horizontal component:

For V1= + 60V

For V2= cos 600x 40 = + 20V

Total Horizontal component = +80V (Means 80V to the right)

Vertical component:

For V1= 0V

For V2= - 40 sin 600= - 34.64V

Total Vertical component = - 34.64V (Means 34.64V down)

80V34.64V

V1+ V2

V1+ V2 =22

643480 .

= 87.2V

?


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= 87.2V

80V34.64V

V1+ V2

= tan-134.64/80

= - 23.410

V1+ V2=87.2 sin (- 23.410) (V)

b) V1 - V2 = V1+ (- V2)

V1= 60V

V2= 40V

V1+ V2

-V2= 40V V1 + (- V2)

600

Apply the same steps as in

previous

Try by yourselves

Next, try to sketch the waveform

for both


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11/29/2013

The instantaneous values of two alternating voltages are representedrespectively by v1=40 sin (V) and V2= 60 sin (+ /3) (V). Derive an

expression for the instantaneous value of :

a) the sum;

b) the difference of these voltage

Exercise

chapter1 -alternating voltage & current :)

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