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
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EE2092 Theory of Electricity May 2011 J R Lucas Page 1 of 92
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
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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.
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Outline Syllabus
1. Fundamentals (6 hrs)
Review of electric circuits.
Circuit Transient Analysis Differential equation solution
Review of AC theory.
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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.
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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.
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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.
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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.
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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.
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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
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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
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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
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R R
Figure 1 Circuit symbols for
Resistance
(a)(b
)
(c)
i(t)
v(t)
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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
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L L
Figure 2 Circuit symbols for
Inductance
(a)(b
)
(c)
i(t)
v(t)
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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
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C C
Figure 3 Circuit symbolsfor
Capacitance
(a) (b)
i(t)
v(t)
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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
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Figure (a) Linear
System
Figure (b) Non-Linear
System
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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
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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
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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
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Unit Step
H(t)1
t
Unit Impulse
(t
) t
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(t) = 0 , t > 0
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Some properties of Unit Impulse (t)Area under curve is unity.
= 1)( dtt
, which gives +
=
0
0
1)( dtt Also
= )0()()( fdtttf
and
= )()()( fdtttf
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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
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R
L
es(t
)
i(t)
Series R-L
circuit
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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
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t
i(t)
Step Response
es(t
)
t
E
es(t
) t
E
t
d
es(t)dt
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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)(
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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
== .
)(
.)(
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VL(t)
time
VR(t)
time
EE
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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)( +=
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R
C
E
)(.
1)(.)(
1ti
pCtiRdtti
C+=
01
=+pC
R
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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.
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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.
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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
)(
=
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vC(t)
time
E
Vo
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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
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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
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td
id4.0 300=
td
id
[ ] 3000800cos800.0800sin.600. 06000600 =+= eeCtd
id
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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
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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
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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
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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
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Mean value = 0
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Average Value (rectified)
Average of full-wave rectified waveform
average value for sinusoidal wave = (2/) Am
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vrect
(t)
tT
T
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Effective value
Effective value =
rms value
Rms value for sinusoidal wave = Am/2
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+
=Tt
t
eff
o
o
dttaT
A )(1 2
v2(t)
t
T
T
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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
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a(t
)
t
T
Amsint
t
O O
P
X
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Sinusoidal waveform is the projection, on a particular
direction, of the complex exponential ejt.
A reference direction is chosen, usually horizontal.
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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
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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.
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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,
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1 O
2
y(t
)
t
Amsin
(t+1)
Bmsin
(t+2)
T
2
1
2
1
Am
Bm
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Or on normal phasor diagram with complex A and B
with polar co-ordinates A1 and B2.
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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
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Currents and voltages in simple
circuit elements
(1) Resistor
for a sinusoid,
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R
v
(t)
i
(t)
v(t) =
O t
Vmcos
t
Imcos
t
T
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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
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VIO
Phasor diagram
R
V
I
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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
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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+) ]
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L
v (t)
i (t)
dt
tidLtv
)()( =
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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.
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(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+) ]
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C
v
(t)
i
(t) 0 t
Vmsin
t
Imcos t
T
/2= dtiCv .
1
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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.
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1
j C
V
I
V
I
O Phasordiagram
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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,
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ZY
1=
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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.
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C
1
Cj
1
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V
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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.
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R
jLI
VR VL
V
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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.
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Phasor diagram
V VC
VR
I
Phasor diagram
V
VLV
R
I
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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.
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C i i iV
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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.
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R1
j CI
VR VC
V
V
VC
VR
I
L C i i iV
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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.
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Phasor
diagram
V
VC
VL
Ior
VC
VL
IV
jLI
VC VL
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R L C i i i
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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
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R
jLI
VR
VL
V
VC
j1
j(1
)
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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.
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Si l P ll l Ci i
V
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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+
EE2092 Theory of Electricity, May 2011 J R Lucas Page 68 of 92
Phasor
diagram
IIL
IR
V
IL
R
jL
IR
IL
I
R C parallel Circuit V
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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+
EE2092 Theory of Electricity, May 2011 J R Lucas Page 69 of 92
Phasor
diagram
I IC
IR V
IC
R
1
j C
IR
V
IC
I
R L C parallel CircuitV
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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.
EE2092 Theory of Electricity, May 2011 J R Lucas Page 70 of 92
R
1
j C
IR
IL
I
IC
jL
Phasor
diagram
I IL+I
C
IR V
IC
IL
P
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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.
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V I [ (2 t ) +
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= 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 .
EE2092 Theory of Electricity, May 2011 J R Lucas Page 73 of 92
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
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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
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For inductor or capacitor V I exists although P = 0
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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
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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.
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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
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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.
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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 =
EE2092 Theory of Electricity, May 2011 J R Lucas Page 80 of 92
dt
d
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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)
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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
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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.
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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
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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
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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.
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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
EE2092 Theory of Electricity, May 2011 J R Lucas Page 87 of 92
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)
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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
EE2092 Theory of Electricity, May 2011 J R Lucas Page 89 of 92
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 ?
EE2092 Theory of Electricity, May 2011 J R Lucas Page 90 of 92
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
=
EE2092 Theory of Electricity, May 2011 J R Lucas Page 91 of 92
= 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