ee2092_1_2011_fundamentals.doc by prof j r lucas

7/30/2019 EE2092_1_2011_fundamentals.doc by prof j r lucas

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Course Outline

UEE 2092 Theory of ElectricityB.Sc. Engineering Degree

Semester 2/3 2011

Credits: 2Lecturer: Prof. J. Rohan Lucas

Lectures: 2 hours per week.

Duration: 14 weeksAssignments/Tests: In-class

EE2092 Theory of Electricity May 2011 J R Lucas Page 1 of 92


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Web site

Department

http://www.elect.mrt.ac.lk

Noteshttp://www.elect.mrt.ac.lk/pdf_notes.htm

Past Question Papers & Answers

http://www.elect.mrt.ac.lk/pdf_qpapers.htm



3/92

Learning Objectives

To develop electrical analysis tools.

To provide a foundation in electrical

fundamentals through network theorems.

To apply dc and ac principles to solve circuits.

To apply matrix analysis to solve circuits To analyse circuits and waveforms.



4/92

Outline Syllabus

1. Fundamentals (6 hrs)

Review of electric circuits.

Circuit Transient Analysis Differential equation solution

Review of AC theory.



5/92

2. Coupled circuits & Dependent sources (4 hrs)

Series and parallel resonance.

Electromagnetic coupling in circuits

Mutual inductance

Analysis of coupled circuits Transformer as a coupled circuit.

Dependent sources

solving circuits with dependent sources.



6/92

3. Network theorems (4 hrs)

Superposition theorem,

Thevenin's theorem,

Norton's theorem,

Millmans theorem,

Reciprocity theorem,

Maximum power transfer theorem,

Nodal-mesh transformation theoremCompensation theorem.



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4. Matrix Analysis (4 hrs)

Network topology

Nodal and mesh analysis.

Two-port theory

Impedance parameters, admittance parameters, hybrid parameters ABCD parameters.



8/92

5. Three-phase Analysis (6 hrs)

Introduction to Three Phase

Analysis of three phase balanced circuits.

Single line equivalent circuits.

Analysis of three phase unbalanced circuits.

Symmetrical component Analysis.



9/92

6. Non-sinusoidal waveforms (4 hrs)

Periodic Waveforms

Waveform parameters: mean, rms, peak,rectified average etc.

Power, power factor

Harmonics, Fourier analysis.

Non-Periodic Waveforms

Laplace transform Transient analysis using Laplace transform.

EE2092 Theor of Electricit Ma 2011 J R L cas Page 9 of 92


10/92

Learning OutcomesAt the end of this module you should be able to

1. solve coupled circuits involving mutual impedance2. solve circuits with dependant sources

3. analyse the resonance of circuits

4. solve circuits using network theorems5. apply matrix analysis to solve large circuits

6. solve three phase circuits: balanced and unbalanced

7. analyse periodic waveforms: harmonics content8. analyse circuits with non-sinusoidal sources

9. obtain Laplace transforms and analyse transients.

EE2092 Th f El t i it M 2011 J R L P 10 f 92


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Continuous Assessment

30% of overall mark for moduleAt least 35% of allocated mark required to be

considered for a grade point

Components of Continuous Assessment

unannounced in-class tests (around 7) attendance at lectures tutorials/assignments may be given



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End of Semester Assessment

70% of overall mark for module

at least 40% of allocated marks required to beconsidered for a grade point

closed book examination of duration 2 hours all questions must be answered

will usually consist of 5 questions questions not necessarily of equal weightage



13/92

Recommended Texts

Electric Circuits, E.A.Edminster, SchaumOutline Series, McGraw Hill

Theory and Problems of Basic ElectricalEngineering, D P Kothari, I J Kothari, Prentice

Hall of India, New Delhi

Electrical Engineering Fundamentals, VincentDel Toro, Prentice Hall of India, New Delhi



14/92

1. Fundamentals

1.1. Review of electric circuitsResistance (unit: ohm,; letter symbol: R, r)

v = R i, i = G v

p(t) = v(t) . i(t) = R . i2(t) = G . v 2(t)

w(t) = v(t) . i(t).dt = R . i2(t). dt = G. v 2(t).dt


R R

Figure 1 Circuit symbols for

Resistance

(a)(b

)

(c)

i(t)

v(t)


15/92

Inductance(unit: henry, H;letter symbol: L , l)

v = Lp i, i = v/Lp

Current through inductor cannot change suddenly.

p(t) = v(t) . i(t)

w(t) =v(t).i(t).dt = dtidtdiL = L.i.di = L.i2


L L

Figure 2 Circuit symbols for

Inductance

(a)(b

)

(c)

i(t)

v(t)


16/92

Capacitance (unit:farad, F; letter symbol: C , c )

v = (1/Cp) i, i = Cp v

Voltage across capacitor cannot change suddenly.

p(t) = v(t) . i(t)w(t)=v(t).i(t).dt = dtdtdvCv =C.v.dv = C.V2


C C

Figure 3 Circuit symbolsfor

Capacitance

(a) (b)

i(t)

v(t)


17/92

Impedance and Admittance

v = Z(p) i , i = Y(p) v

whereZ(p) impedance operator,

Y(p) admittance operator.

Impedances and Admittances

either linear or non-linear


Figure (a) Linear

System

Figure (b) Non-Linear

System


18/92

Active Circuit Elements

component capable of producing energy

sources of energy (or sources)

o voltage sources

o current sources.

Producing energy

non-electrical energy electrical energy



19/92

1.2. Natural Behaviour of R-L-C Circuits

Does not depend on the external forcing functions

Depends on internal properties of the system.

Give a pendulum an initial swing, let go

behaviour depends on natural frequency

if we push at some other rate, behaviourwould also depend on forcing frequency.

To determine the natural behaviour

forcing function must not have its ownspecific frequency



20/92

o step function, impulse function.

Unit Step Function H(t)

similar to a staircase step

H(t) = 0, t < 0

H(t) = 1, t > 0

Unit Impulse Function (t)

(t) = 0 , t < 0(t) = , t = 0


Unit Step

H(t)1

t

Unit Impulse

(t

) t


21/92

(t) = 0 , t > 0



22/92

Some properties of Unit Impulse (t)Area under curve is unity.

= 1)( dtt

, which gives +

=

0

0

1)( dtt Also

= )0()()( fdtttf

and

= )()()( fdtttf

EE2092 Th f El i i M 2011 J R L P 22 f 92

t=0


23/92

Transient Analysis

Particular integral steady state solutionComplementary function transient solution

Series R-L circuit

With step excitation

es(t) = E.H(t)

governing behaviour is)()( tHEteiR

dt

diL

s==+

Particular integral E/RComplementary function L p i+ R i = 0

EE2092 Th f El i i M 2011 J R L P 23 f 92

R

L

es(t

)

i(t)

Series R-L

circuit


24/92

Solution has the form

R

EeAti

tL

R

+=

.)(

A is obtained from initial conditionsAt t = 0, i = 0R

EA =

=

tL

R

eR

Eti 1)(

Unit impulse derivative of unit step

Response to unit impulse

derivative of response to unit step

EE2092 Theory of Electricity, May 2011 J R Lucas Page 24 of 92

t

i(t)

Step Response

es(t

)

t

E

es(t

) t

E

t

d

es(t)dt


25/92

With impulse excitation e(t) = E.(t)Complementary function is same as before.

Particular integral is now different and equal to 0.0.)( +=

tL

R

eAti

NewA is obtained from new initial conditions.t

L

R

eL

Eti

=)(

Or from derivative of unit step response ast

L

Rt

L

R

eL

Ee

R

E

dt

dti

=

= 1)(



26/92

AdditionalProblems on Circuit Transients

Example 1: Simple R C circuit supplied from

step voltage

Continuing from earlier solution,

voltage drops across each element is obtained as

==

tL

R

R

eEtiRtv 1)(.)(

andt

L

R

L eEdt

tid

Ltv

== .

)(

.)(


VL(t)

time

VR(t)

time

EE


27/92

Example 2 Simple R C circuit supplied from

step voltage (switch closed onto a battery)

Governing differential equation is

e(t) = R . i(t) +

For complementary function,

, giving p = 1/RC

No current through capacitor at steady state.

particular solution steady state solution i(t) = 0 solution is in the form tRCeAti 10)( +=


R

C

E

)(.

1)(.)(

1ti

pCtiRdtti

C+=

01

=+pC

R


28/92

ConstantA can be obtained from initial conditions:

Voltage across a capacitor cannot change suddenly

at t=0, vc(t) = 0, vR(t) = E and i(t) = E/R0

1

= RCeAR

E giving A = E/R.

tRCeREti1

)(

=

Voltage drops across elements can now be obtained using

Ohms law.

Sketches of voltage and current can now be done as

earlier.



29/92

Example 3 R C circuit supplied from astep voltage source, but with C initially

charged to VoGoverning differential equation

oVdttiC

tiRte ++= )(1

)(.)(

oVipC

iR ++=.

1.

Complementary function remainssame as earlier except that the value

of constant will be different.

The particular solution would again be zero.

since voltage across capacitor cannot change suddenly

At t=0, vc(t) = Vo.


R

CE

Vo


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vR(t) = E Vo and i(t) = (E Vo)/R0

1

=

RCo eAR

VE giving RVEA o= t

RCo eR

VEti

1

)(

=

voltage across( )

tRC

oCeVEEtv

1

)(

=


vC(t)

time

E

Vo


31/92

Example 4 R L C circuit supplied from astep voltage

)(.

1)(..)(.

)(1)(

)(.)(

tipC

tipLtiR

dtti

Ctd

tidLtiRte

++=

++=

)(1

)(..)(..)(. 2 tiC

tipLtipRtep ++=

Complementary function is

R.p + L.p2 + 1/C = 0

Solution of equation can have real or

complex roots dependant on the values of

components.(a) R = 480 , L = 0.4 H, C = 2.5 F, E = 120 V


R

L

E

C

i(t)


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complementary function is

0.4 p2 + 480 p + 4105 = 0 or p2 + 1200 p + 106 = 0

giving 80060010600600 62 jp ==

Particular solution would be i(t)=0 at t=

giving a solution of the formA.e-600t.ej800t + B.e-600t.e-j800t,

or C.e-600t.cos(800t+ )



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Using initial conditions [2 energy storing elements]

at t=0, i(t) = 0 and vC(t) = 0

0 = C.e-6000.cos(8000+)gives = 90o as C cannot be zero [trivial solution]

giving the solution i(t) = C. e-600t.sin 800tAlso since i(0)=0, vR(0) = 0.

vL(0) = 120 = giving at t=0giving C = 300/800 = 0.375

i(t) = 0.375 e-600t.sin 800tA


td

id4.0 300=

td

id

[ ] 3000800cos800.0800sin.600. 06000600 =+= eeCtd

id


34/92

Using Ohms law,

vR(t) = 480 i(t) = 180 e-600t.sin 800t V

vL(t)

=

0.4

p.i(t) = 90

e-600t

.sin 800t

+

120

e-600t

.cos

800t VvC(t) = 120 - 90 e

-600t.sin 800t - 120 e-600t.cos 800t V


i(t) Vc(t)

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0 0.002 0.004 0.006 0.008 0.01

0

20

40

60

80

100

120

140

0 0.002 0.004 0.006 0.008 0.01


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(b) R = 800 , L = 0.4 H, C = 2.5 F, E = 120 Vcomplementary function is

0.4 p2 + 800 p + 4

105 = 0 or p2 + 2000 p + 106 = 0

giving (p +1000)2 = 0 or p = 1000 (repeated roots)

In this case the solution is of the form

i(t) = A.t e-1000t + 0

At t=0, i(t) = 0, [automatically satisfied]and di(t)/dt = 300

300.01000 0100001000 == eAeAdt

di

giving A = 300

i(t) = 300 t e-1000t

A


0

0.02

0.04

0.06

0.08

0.1

0.12

0 0.002 0.004 0.006 0.008 0.01

i(t)


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(c) R = 1000 , L = 0.4 H, C = 2.5 F, E = 120 Vcomplementary function is

0.4 p2

+1000 p +4105

= 0 or p2

+ 2500 p + 106

= 0giving (p +500)(p+2000) = 0 or p = 500 or 2000

In this case the solution is of the form

i(t) = A e-500t + B e-2000t

At t=0, i(t) = 0 = A + B

At t=0, vC(t) = 0 ,

gives di(t)/dt = 300.

i.e. 500A 2000B = 300,gives A = 0.2 = B

i(t) = 0.2 (e-500t e-2000t) A


0

0.01

0.020.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0 0.002 0.004 0.006 0.008 0.01

i(t)


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1.3. Review of AC theory

Sinusoidal waveform has equation

v(t) = Vm sin( t + )Period is T. Angular frequency = 2/TEE2092 Theory of Electricity, May 2011 J R Lucas Page 37 of 92

v(t)

tT

T


38/92

Mean value = 0



39/92

Average Value (rectified)

Average of full-wave rectified waveform

average value for sinusoidal wave = (2/) Am


vrect

(t)

tT

T


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Effective value

Effective value =

rms value

Rms value for sinusoidal wave = Am/2


+

=Tt

t

eff

o

o

dttaT

A )(1 2

v2(t)

t

T

T


41/92

rms value is always specified for ac waveforms.

Form Factor and Peak Factor

Form Factor = rms value/average value

= 1.1107 1.111 for sinusoidal

Peak Factor = peak value/rms value

= 2 = 1.4142 for sinusoidal



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Phasor Representation of Sinusoids

ej = cos + j sin or ejt = cos t + j sin t


a(t

)

t

T

Amsint

t

O O

P

X


43/92

Sinusoidal waveform is the projection, on a particular

direction, of the complex exponential ejt.

A reference direction is chosen, usually horizontal.


R

P

0

Am

a(t)

t

T

Amsin

( t+)

t

0 0

R

X

P

Rotating Phasor

diagram

R

P

0

Am

reference

direction

0

A=Am

2

Phasor dia ram

Ay

Ax


44/92

Usual to drawPhasor diagram using rms value A,

rather than with thepeak value Am.

The phasorA is characterised by its magnitude Aand its phase angle .Polar co-ordinates of phasorA written as A.Cartesian co-ordinates Ax and Ayof phasorA, usually

written A = Ax + j Ay.



45/92

Phase difference

ConsiderAm sin (t+1) and Bm sin (t+2) asshown

Can be represented by rotating phasorsAm e

j(t+1

)

andBm ej (t+2)with peak amplitudesAmand Bm,


1 O

2

y(t

)

t

Amsin

(t+1)

Bmsin

(t+2)

T

2

1

2

1

Am

Bm


46/92

Or on normal phasor diagram with complex A and B

with polar co-ordinates A1 and B2.



47/92

Example 1

Find the addition and subtraction of the 2 complex

numbers given by 1030o

and 25 48o

.Addition = 10 30o + 25 48o

= 10(0.8660 + j0.5000) + 25(0.6691 +

j0.7431)= (8.660 + 16.728) + j (5.000 + 18.577)

= 25.388 + j 23.577 = 34.647 42.9o

Subtraction = 1030o

2548o

= (8.660 16.728) + j (5.000 18.577)



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= 8.068 j 13.577 = 15.793239.3o

Example 2

Find the multiplication and the division of the two

complex numbers given by 1030o and 25 48o.

Multiplication = 10 30o * 2548o = 250

78o

Division = 10 30o 2548o = 0.4 18o



49/92

Currents and voltages in simple

circuit elements

(1) Resistor

for a sinusoid,


R

v

(t)

i

(t)

v(t) =

O t

Vmcos

t

Imcos

t

T


50/92

Let i(t) = Im cos (t+) or Real part of[Ime(jt+) ]

v(t) = Real [R.Im e(jt+)] = Real [Vm.e(jt+)]

or v(t)= R . Im cos(t+) = Vm cos (t+ )

Vm = R.Imand Vm/2 = R.Im/2

i.e. V = R . I


VIO

Phasor diagram

R

V

I


51/92

Note: V and I are rms values of voltageand current.

no phase angle change has occurred in theresistor.

Note also that power dissipated in resistorR is

R . I 2 = V . I



52/92

0 t

Vm

sin

t

Imcos

t

T

/2

(2) Inductor

Let i(t) = Im cos (t+) or Real part of[ Ime(jt+) ]

v(t) = Real [L. ddtIme(jt+) ]= Real [L. j .Imej(t+) ] = Real [j

Vmej(t+) ]


L

v (t)

i (t)

dt

tidLtv

)()( =


53/92


54/92

or current waveform lags voltage waveform by 90o for

the inductor.

V = jL.I or V = L.I90Impedance Z is thus defined as jL.V = Z.I corresponds to the generalised form of Ohms Law.

Power dissipation in a pure inductor is zero.

Energy is only stored but product V . I is not zero.



55/92

(3) Capacitor

Let i(t) = Im cos (t+) or Real part of [ Imej(t+) ]

v(t) = Real [ += dteICv tjm .1 )( ]

= Real [1

C j. .Ime(jt+) ] = Real [

1

jVm

e(jt+) ]


C

v

(t)

i

(t) 0 t

Vmsin

t

Imcos t

T

/2= dtiCv .

1


56/92

or v(t) = 1C I t dtm cos( ). + = 1C.Im sin (t+)= 1C .Im cos (t+/2) =Vm cos

(t+/2) Vm = 1C.Im and Vm/2 = 1C.Im/2i.e. Vrms=

1

C.IrmsVoltage waveform lags current waveform by 90o or /2radians or the current waveform leads the voltagewaveform by 90o for a capacitor.

Thus V = 1j C .I or V =1

C.I90ImpedanceZ=

1

j C

, and V = Z . IPower dissipation in a pure capacitor is zero, butproduct V.Inot zero.


1

j C

V

I

V

I

O Phasordiagram


57/92

Impedance and Admittance in an a.c.

circuit

ImpedanceZis a complex quantity.Relates complex rms voltage to complex rms current.

V = Z . I

Admittance Yis inverse of impedanceZ.I = Y . V

In cartesian form

Z = R + j X and Y = G + j BReal part ofZis resistive, usually denoted byR,


ZY

1=


58/92

Imaginary part ofZis called a reactance denoted byX.

A pure inductor and a pure capacitor has a reactance

only and not a resistive part, while a pure resistor hasonly a resistive part and not a reactive part.

Z = R + j 0 for a resistor,

Z = 0 + jL for an inductor,and Z = = 0 jfor a capacitor.

Real part ofY is a conductance, denoted by G,

Imaginary part ofY is asusceptance, denoted by B.


C

1

Cj

1


59/92

V


60/92

V VL

VRIPhasor

diagram

VL

Simple Series Circuits

R-L series circuit

ConsiderI as referenceVR = R.I, VL = jL.I,

and V = VR + VL = (R + jL).I

so that total series impedance isZ = R + jLPhasor diagram has been drawn with

Ias reference.[i.e.Iis drawn along the x-axis direction].

VR is in phase with I, VL is leading I by 90o.


R

jLI

VR VL

V


61/92

By phasor addition, V = VR +

VL.

If V is taken as reference,I is seen to lag voltage by the

same angle that the voltagewas leading the current earlier.

In R-L circuit, current I lags voltage

V by an angle less than 90o and the

circuit is said to be inductive.


Phasor diagram

V VC

VR

I

Phasor diagram

V

VLV

R

I


62/92

Note: Power dissipation can only occur in the resistive

part of the circuit and is equal toR.I2.

This is not equal to product V . I for the circuit.


C i i iV


63/92

R-C series circuit

VR = R.I, VC =1

j C . I

and V = VR + VC V = (R + Cj1 ).Iso that Z = R +

1

j C

Phasor diagram has been drawn with V as reference.

In R-C circuit, current I leads voltage V by an angle less than

90o and the circuit is said to be capacitive.

Power dissipation can only occur in the resistive part of the

circuit and is equal toR.I2

, not equal to product V . I.


R1

j CI

VR VC

V

V

VC

VR

I

L C i i iV


64/92

L-C series circuit

VL = jL.I,

VC =1

j C

.I =j1

C

and V = VL + VC

V = (jL + 1j C ).Iso that total series impedanceis Z = jL + 1j C = jLj 1C

Total impedance is purely

reactive, and all voltages in the circuit are inphase butperpendicular to current I.


Phasor

diagram

V

VC

VL

Ior

VC

VL

IV

jLI

VC VL


65/92

R L C i i i


66/92

R-L-C series circuit

VR = R.I, VL = jL.I, VC =1

j C . I

and V = VR +VL + VC V = (R + jL + 1j C ).Iso that the total series impedance is


R

jLI

VR

VL

V

VC

j1

j(1

)


67/92

Z = R +jL + 1j C = R + j(L1C)Z R L

C= +

2

21

Magnitude has a minimum value at L =

1

C

This is the series resonance condition.

In an R-L-C circuit, the current can either lag or lead

the voltage, and the phase angle difference between

the current and the voltage can vary between 90o

and resultant circuit is either inductive or capacitive.

Note that the power dissipation can only occur in the

resistance in the circuit and is equal toR . I2

and thatthis is not equal to product V . I for the circuit.


Si l P ll l Ci i

V


68/92

Simple Parallel Circuits

R-L parallel Circuit

ConsiderV as referenceV = R.IR,

V = jL.IL,

and I = IR + IL

I VR Vj L= + total shunt admittance = 1 1R j L+


Phasor

diagram

IIL

IR

V

IL

R

jL

IR

IL

I

R C parallel Circuit V


69/92

R-C parallel Circuit

ConsiderV as reference

V = R.IR,

IC = jC.V,

and I = IR+ IL I VR V j C= + . total shunt admittance = 1R j C+


Phasor

diagram

I IC

IR V

IC

R

1

j C

IR

V

IC

I

R L C parallel CircuitV


70/92

R-L-C parallel Circuit

ConsiderV as reference

V = R.IR, V = jL.IL, IC = jC.Vand I = IR+ IL + IC

I VR Vj L V j C= + + . total shunt admittance

= 1 1R j L j C+ +

Shunt resonance will occur when1

LC=

giving minimum value of shuntadmittance.


R

1

j C

IR

IL

I

IC

jL

Phasor

diagram

I IL+I

C

IR V

IC

IL

P


71/92

Power

In an a.c. circuit,

power loss occurs only in resistive parts of circuit in general power loss is not equal to product V . Ipurely inductive parts and purely capacitive parts

of a circuit do not have any power loss.To account for this apparent discrepancy,

define product V . I as apparent power S of circuit.

Apparent power has the unit volt-ampere (VA).

watt (W) is used only for active power Pof circuit.



72/92

V I [ (2 t ) +


73/92

= Vm Im [cos (2 t ) + cos ]

Waveform of power p(t) has a sinusoidally varying

component and a constant component.

Average value of power (active power) P would begiven by the constant value Vm Im cos .


i(t)

current lagging voltage by angle inphase

Vm

Im

cos

p(t)

tT

v(t)

t

p(t)

p(t)

P V I IV V I


74/92

P = Vm Im cos = cos.2.2 mmIV = V . I cos

Power Factor

PowerFactor = activepower/apparentpower

For sinusoidal quantities, equal to term cos .

For a resistor, = 0o so that P = V . I

For an inductor, = 90o lagging, so that P = 0

For an capacitor, = 90o leading, so that P = 0

For combinations ofresistor, inductorand capacitor,P lies between V. I and 0


For inductor or capacitor V I exists although P = 0


75/92

For inductor or capacitor, V. I exists although P = 0.

Reactive Power

Defined as product of voltage and current componentswhich are quadrature (90oout of phase).

reactive power Q = V. I sinFor L and C, = 90,reactivepowerQ = V. IUnlike inphase, where same direction means positive,

with quadrature, there is no natural positive direction.

Usual to define

Inductive reactive power when current lagging voltage Capacitive reactive power when current leading voltage.



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Power factor correction


77/92

Power factor correction

Although reactive power does not consume any

energy, it reduces the power factor below unity.When power factor is below unity,

for same power transfer P the current

required becomes larger and the

power losses in the circuit becomes

still larger (power loss I2).Thus supply authorities encourage industries to

improve their power factors to be close to unity.


P1

Q1

1

P

For a load


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P1

Q1

1

Qc

Q2

2

For a load

lagging power factor cos 1

active power P1, reactive power Q1If power factor is improved to a new value,

cos 2 (>cos 1)

leading reactive power Qc must be added. usually done by using capacitors.With pure capacitors, active power is unchanged at P1,

assuming supply voltage remains unchanged.Thus new reactive power Q2 = Q1 Qc


What is known is P cos and cos


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What is known is P1, cos 1 and cos 2.

Thus Q1 = P1 tan 1 and Q2 = P1 tan 2

so that Qc = Q1 Q2 = P1 tan 1 P1 tan 2

Reactive power supplied by a capacitor is dependant

on its capacitance C and the voltage across it V.

Thus Qc = P1 tan 1 P2 tan 2 = V2.Y = V2.C

The value of C can be determined.


Impedance Admittance and Transfer functions


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R

e

iC

r

L

ir

vr

iL

i

C

vC

vL

vR

Impedance, Admittance and Transfer functions

Each element is either governed by a constant,

differentiation or integration.Bilateral linear circuit governed by

differential equation.

written using Kirchoffs currentand voltage laws, and

Ohms Law

with differential operator

p =


dt

d


81/92

To find relationship between e(t) and ir(t).

e(t) = vr(t) + vC

(t) vR

(t), vr(t) = vL

(t)i(t) = ir(t) + iL(t) = iC(t)

vr(t) = r . ir(t), vL(t) = Lp . iL(t), Cp .vC(t) = ic(t)

eliminate other variablesLp . iL(t) = r . ir(t)

Lp . i(t) = Lp .[ ir(t) + iL(t)] = (Lp + r ). ir(t)

e(t) = r . ir(t) + vC(t) + R . i(t)

i.e. Cp . e(t) = Cp . r . ir(t) + i(t) + Cp . R . i(t)


L p C p e(t) = L p C p r i (t) + (1 + C p R) i (t)


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Lp . Cp . e(t) Lp . Cp . r . ir(t) + (1 + Cp . R) . ir(t)

i.e. L.C.p2. e(t) = [L.C. p2.r + 1 + Cp . R] . ir(t)

This is a differential equation involving terms up to the

second derivative of both e(t) and ir(t) of the form

f1(p). e(t) = f2(p). ir(t)

or e(t) = Zr(p). ir(t)

Zr(p) impedance transfer function of differential operatorp

In a similar way any current and any voltage would be related by an

admittance transfer function of differential operatorp any two voltages or any two currents would be related by

a transfer gain of the differential operatorp


In the case of sinsusoidal a c the differential operator p will


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In the case of sinsusoidal a.c., the differential operatorp will

be replaced using the relationship p = j.


Example 3


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Example 3

Determine mean value, average

value, peak value, rms value,

form factor and peak factor.

Solution

Mean value

[This result could have been written by inspection considering areas].

Average value = 65

]5[6

)])((2[1

31

21

32

21

ET

T

ETETE

T==

Peak value = 2E


2E

0 T 2T 3TE

==TT

dtT

tE

Tdttf

T00

).3

2(1

).(1

2)

232()

232(

2

0

2 ETTT

TE

Ttt

TE

T

t

====

rms value= =TT

dtT

tE

Tdttf

T

222 .)3

2(1

).(1


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rms value TTfT 00 )()(T

t

T

T

t

T

E

0

32

)3

(3

1)

32(

=

= EE

== ]81[9

)(2

form factor =2.1

5

6

6/5==

E

E

peak factor = 22

=E

E


Example 4


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Example 4

A certain 50 Hz, alternating

voltage source has an internalemf of 250 Vand an internal

inductance of31.83 mH.

If the terminal voltage is to be

maintained at 230 V,

determine the value of the maximum power that can

be delivered to a load (R + jX) and the values of

R and X under these conditions.Draw also the phasor diagram showing the voltages

and currents in the circuit under these conditions.


31.83

mH250

V

Solution R+jX


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Solution

at 50 Hz,xs = jL = j 25031.8310-3 = j 10.00

Current I =jXRj ++10

250

, | V | = 230 Vactive power P = | I |2 R = RXR

++

])10([

25022

2

,

voltage V = (R+jX) . I


If we simply differentiate and obtain the condition for maximumpower,

, we can show that these give the conditions

[R2 + (X + 10)2].1 R.2R = 0 and (X + 10) = 0 or X = 10 and R = 0

Then P = , V = This obviously is not the required solution.

| V |2 = 2302 = (R2+X2). | I |2 = ])10([250

)(22

222

++

+

XRXR


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| V | 230 (R X ). | I | ])10([ ++ XR R2 + (X + 10)2 = 1.185 R2 + 1.1815 X2i.e. 20 X + 102 = 0.1815 (R2 + X2)

or R2 + X2 = 5.5104 (102 + 20X)

Differentiate for maximum power keeping voltage constraint

i.e. P = RX + ]2010[5104.6250

2

2

,0=dRdP

gives the condition

(20X+100).1 R.20. 0=dRdX

or X+5 = RdRdX

also 20dR

dX+ 0 = 0.1815(2R + 2X

dR

dX

)

so that 20(X+5) = 0.363R2 +0.363X(X+5)


i.e. 20X + 100 = 0.363R2 +0.363X2 +1.815X


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i.e. 20X 100 0.363R 0.363X 1.815X

0.363(R2 + X2) = 18.185X + 100

but 20 X + 100 = 0.1815 (R2

+ X2

) = (9.0925X+50)i.e. 10.9075X =50

X = 4.584

giving R = 4.983 Substitution gives

Pmax = 5750 W,

under given condition

I = 416.5983.4250

584.4983.410

250

jjj +=

+

i.e. I = o38.47360.7250

= 33.97-47.38o A


j10

250

V

4.983-j4.584

terminal voltage at source (load voltage)


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I=

33.97A

-

47.38o

42.61oE=25

0V

VR=169.

3 V

VC=155.

7V

VL=339.

7V

g ( g )

=33.97-47.38o6.771-42.61o = 230.0-90o Vvoltage across resistive part of load

= 4.98333.97-47.38o = 169.3-47.38o Vvoltage across capacitive part

= 4.584-90o33.97-47.38o

= 155.7-137.38o

Example 5

If the load is purely resistive, what

would be its value for power transfer at230 V ? and what is the Power ?


31.83

mH

250 V, 50

Load

If parallel capacitance is connected with load to achieve


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p e c p c ce s co ec ed w o d o c eve

maximum power, what is the value of R, C and Pmax.

Solution

L = 25031.8310-3 = 10

250 = (R2 + 102)I, 230 = RI

230(R2

+ 102

) = 250R R2

+ 102

= 1.1815R2

i.e. R = 23.47 , P = 2302/23.47 = 2.253 kWThe solution of example 4 can be used, except that we

need to find the parallel equivalent of the load.G + j B = oj 38.47360.7

1

584.4983.4

1

=


= 0.092+j0.1000


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0.092+j0.1000

effective value of resistance = 1/0.0920 = 10.87

parallel capacitance = B/ =0.1000/(250) = 318.3 F

maximum power = 2302/10.87 = 4.867 kW

ee2092_1_2011_fundamentals.doc by prof j r lucas

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