CHAPTER 1: ALTERNATING VOLTAGE & CURRENT 1. 2. 3. 4.

5. 6. 11/29/2013

Introduction Generating AC Voltages Waveform & Definitions Voltages & Currents as function of time Average & R.M.S. Values Phasor Representation

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Objectives At the end of the chapter, student should be able to:  describe the principle of ac voltage and current generation.  calculate RMS and Average values for voltage and current.  explain and analyze the phasor representation.

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1. INTRODUCTION Circuit Theory 1 & Engineering Math- important •Electrical unit, Ohm’s Law, Kirchoff’s Law etc. •Abbreviations & Symbols – AC, DC

Why AC is required? Whereas we already have DC •DC cannot be transmitted as economically as AC transmission systems • AC can be step-up and down

• Nowadays, many electrical appliances are using AC system

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1. INTRODUCTION Most alternating systems operate on a sinusoidal basis.  Sinusoidal waveform are tricky things to draw. By using phasor, it can be represent as straight lines.  By ing up such lines, we can undertake apparently difficult additions and subtractions and this simplifies the later analyses which will be considered. 

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1. INTRODUCTION 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 the other. The cycle of variations is repeated exactly for each direction. 

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2. GENERATING AC VOLTAGES

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Hydro Tidal

How to turn the coils?

Wind

Geothermal 11/29/2013

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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 parallel to the direction of the magnetic flux.- no flux is being cut. [(a)&(d)] 11/29/2013

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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 θ = X sin θ Assumption: l = length of conductor of one side (m), B= flux density (T) So; total e.m.f. generated = 2 B l X sin θ (V) X is also π b n

b is breadth / width of the loop in meters,

e = 2 B l b π n sin θ (V)

n is the speed of rotation in rps For N numbers of coil, e = 2 π B N A n sin θ (V)

A=area of loop

Peak value Ep = 2πB(NA)n (V) 11/29/2013

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Example: A coil of 160 turns is rotated at 1500 rpm in a magnetic field having a uniform density of 0.15 T. The mean area per turn is 50 cm2. Calculate a) the frequency; b) the period; c) the maximum value of the induced emf; d) the value of the induced emf when the coil has rotated through 30 from the position of zero emf. Solution:

a) Frequency, f = 1500/60 = 25 Hz;

b) Period, T = 1/f = 1/25 = 0.04s; c) Em = 2 π B N A n = 2 π (0.15) (160) (50x10-4) (1500/60) = 18.84 V; d) e = 2 π B N A n sin θ = 18.84 x sin 300 = 9.42 V

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 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 p pairs of poles and if its speed is n revolutions per second, then frequency f = 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.

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3. Waveform & Definitions

: 1. Period,T – time taken to complete one cycle (s)

2. Peak value, Vm or Vp, Im or Ip 3. Peak-to-peak value, Vp-p or Ip-p 4. Frequency, f – Number of cycles that occur in 1 second (Hz) 5. Amplitude – distance from its average to its peak

T = 11/29/2013

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Example:

Question: i- Period ii- Peak value iii- Peak to peak value iv- Frequency v- Amplitude

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Answer: i- 10ms ii- 10V iii- 20V iv- 100Hz v- 10V

Alternating Voltage & Current

Answer: i- 10ms ii- 5V iii-10V iv-100Hz v- 5V

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Example:

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 Find i- Period ii- Peak value iii- Peak to peak value iv- Frequency

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4. Voltages & Currents as function of time Previously we’ve seen that: e = 2 π B N A n sin θ (V) & Ep = 2 π B N A n (V) » e = Ep sin θ (V) The angular velocity, ω =

 Rad/s t

The angular velocity, ω = 2 π f Rad/s » e = Ep sin ω t (V) At the end, alternating voltage and current can be formed as function of time as:

» e (t) = Ep sin ω t (V) » i (t) = Ip sin ω t (A)

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Example: (i) If Em=100V, determine the coil voltage at: (a) 30° (b) 330° (ii) If the coil rotates at ω = 3000/s, how long does it take to complete 1 revolution?

Answer: (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 circuit

•3 important situations

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In phase: V and I are in phase since the angular displacement is zero.

Lead: V is leading i for certain amount of angular displacement, . Or Lag: i is lagging V for certain amount of angular displacement, . Lead: i is leading V for certain amount of angular displacement, . Or Lag: V is lagging i for certain amount of angular displacement, . 11/29/2013

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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 = area under curve length of base

•For a complete sinusoidal waveform, average value = 0 since it is symmetrical. •Area for half cycle of sine wave is 

 I p sin  d =  I p cos

Ip



 0

= 2I p

0 0

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

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•Therefore, average value for a full wave rectifier: Full wave average

I ave =

Ip Iave 0



22 I p  2

= 0.637 I p

2

•Furthermore, average value for half wave rectifier: Half- wave average

I ave =

Ip Iave

0

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2I p 2

= 0.318 I p

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R.M.S. (Root Mean Square) •Also known as effective value •Value to do useful work •It is an equivalent dc value – E.g: 240Vac capable of producing the same average power as 240 volts of steady dc i

P

Pdc = P ave E

R

For dc

t

Pdc = Pave = I2R

i

P

2

P(t) Pave=

For ac

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E

R

Im R 2

t

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P(t) = i2R = (Im sin ωt)2R = Im2R sin2 ω t

= Im

2

1 R  1  cos 2t 2

2



2

= I m R  I m R cos 2t 2 2

To get the average value of P(t), the term cos 2 ωt will be zero, thus 2 I Pave = m R 2

Paverage dc = Paverage ac

I = I rms =

Im 2

= 0.707I m

2

= Im R 2 2 2 I I = m 2

I2 R

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 So, the r.m.s value of a sinusoidal current is I = I rms

1 T 2 = 0 i d T or,

Im I= = 0.707 I m 2

Likewise for the average voltage and rms voltage. 11/29/2013

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

=

valuerms valueav

Crest Factor is the ratio between the R.M.S. value and the Peak value of the waveform . Peak or crest factor

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=

valuemax valuerms

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6. Phasor Representation Representation of an alternating quantity by a phasor

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Phasor representation of an alternating quantity for the first half cycle

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Phasor Representation • Phasor – Magnitude & Angle • It is actually a complex number e (t) = Ep sin t  e = Ep  0

(V)

i (t) = Ip sin t  I = Ip  0

(A)

Phasor diagram for above: Ip

Ep

• Generally written as e = Ep   and I = Ip   in phasor form

• In complex form, e = Ep cos  + iEp sin  I = Ip cos  + iIp sin 

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i = Im sin(t +  ) = Im  

Phasor diagram

i = Im sin(t -  ) = Im  -

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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.



j

v(t) = Vm sin(t - 40)

i(t) = Im sin t

Im

40 Vm 40

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

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or

I lags V by 700

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Addition & Subtraction of Phasors Addition of phasors

Substraction of phasors

C=A+B

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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 (V1 – V2) V1= 60V

V1+ V2

V2= 40V

First assume up positive, down negative Right positive and left negative 11/29/2013

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V1= 60V

V1+ V2

V2= 40V

Horizontal component: For V1 = + 60V For V2 = cos 600 x 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) θ

80V

34.64V V1+ V2

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V1+ V2 = 802  34.642 θ?

= 87.2V 38

θ

80V

34.64V V1+ V2 = 87.2V

θ = tan-1 34.64/80 = - 23.410 » V1+ V2 =87.2 sin ( - 23.410) (V)

b) V1 - V2 = V1 + (- V2) -V2= 40V

V1 + (- V2)

Apply the same steps as in previous

V1= 60V

600

Try by yourselves… Next, try to sketch the waveform for both

V2= 40V 11/29/2013

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Exercise The instantaneous values of two alternating voltages are represented

respectively 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

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