EE2092!1!2011 Fundamentals

EE2092 Theory of Electricity, May 2011 J R Lucas Page 1 of 84

Course Outline

UEE 2092 Theory of Electricity B.Sc. Engineering Degree

Semester 2/3 2011

Credits: 2

Lecturer: Prof. J. Rohan Lucas

Lectures: 2 hours per week.

Duration: 14 weeks

Assignments/Tests: In-class


Web site

Department

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

Notes

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

Past Question Papers & Answers

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


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.


Outline Syllabus

1. Fundamentals (6 hrs)

Review of electric circuits.

Circuit Transient Analysis

Differential equation solution

Review of AC theory.


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.


3. Network theorems (4 hrs)

Superposition theorem,

Thevenin's theorem,

Norton's theorem,

Millmans theorem,

Reciprocity theorem,

Maximum power transfer theorem,

Nodal-mesh transformation theorem

Compensation theorem.


4. Matrix Analysis (4 hrs)

Network topology

Nodal and mesh analysis.

Two-port theory

Impedance parameters,

admittance parameters,

hybrid parameters

ABCD parameters.


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.


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.


Learning Outcomes At the end of this module you should be able to

1. solve coupled circuits involving mutual impedance

2. solve circuits with dependant sources

3. analyse the resonance of circuits

4. solve circuits using network theorems

5. apply matrix analysis to solve large circuits

6. solve three phase circuits: balanced and unbalanced

7. analyse periodic waveforms: harmonics content

8. analyse circuits with non-sinusoidal sources

9. obtain Laplace transforms and analyse transients.


Continuous Assessment

30% of overall mark for module

At 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


End of Semester Assessment

70% of overall mark for module

at least 40% of allocated marks required to be considered 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


Recommended Texts

Electric Circuits, E.A.Edminster, Schaum

Outline Series, McGraw Hill

Theory and Problems of Basic Electrical

Engineering, D P Kothari, I J Kothari, Prentice

Hall of India, New Delhi

Electrical Engineering Fundamentals, Vincent

Del Toro, Prentice Hall of India, New Delhi


1. Fundamentals

1.1. Review of electric circuits

Resistance (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)


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 = dtidtdi

L= L.i.di = L.i2

L L

Figure 2 Circuit symbols for Inductance (a)

(b) (c)

i(t)

v(t)


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 = dtdtdv

Cv= C.v.dv = C.V 2

C C

Figure 3 Circuit symbols for Capacitance

(a) (b)

i(t)

v(t)


Impedance and Admittance

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

where Z(p) impedance operator,

Y(p) admittance operator. Impedances and Admittances

either linear or non-linear

Figure (a) Linear System Figure (b) Non-Linear System


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


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, behaviour would also depend on forcing frequency.

To determine the natural behaviour

forcing function must not have its own

specific frequency

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

(t) = 0 , t > 0

Unit Step

H(t)

1

t

Unit Impulse

(t)

t


Some properties of Unit Impulse (t)

Area under curve is unity.

1)( dtt, which gives

0

0

1)( dtt

Also

)0()()( fdtttf

and

)()()( fdtttf

t=0


Transient Analysis

Particular integral steady state solution

Complementary 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/R

Complementary function L p i + R i = 0

R

L

es(t)

i(t)

Series R-L circuit


Solution has the form

R

EeAti

tL

R

.)(

A is obtained from initial conditions

At t = 0, i = 0

R

EA

tL

R

eR

Eti 1)(

Unit impulse derivative of unit step

Response to unit impulse

derivative of response to unit step

t

i(t)

Step Response

es(t)

t

E

es(t)

t

E

t

d es(t) dt


With impulse excitation e(t) = E.(t)

Complementary function is same as before.

Particular integral is now different and equal to 0.

0.)( t

L

R

eAti

New A is obtained from new initial conditions.

tL

R

eL

Eti

)(

Or from derivative of unit step response as

tL

Rt

L

R

eL

Ee

R

E

dt

dti

1)(


Additional Problems 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)(.)( and t

L

R

L eEdt

tidLtv

.)(

.)(

VL

(t)

time

VR

(t)

time

E E


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 t

RCeAti

1

0)(

R

C E

)(.

1)(.)(

1ti

pCtiRdtti

C

01

pC

R


Constant A 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/R

01

RCeAR

E

giving A = E/R.

t

RCeR

Eti

1

)(

Voltage drops across elements can now be obtained using

Ohms law.

Sketches of voltage and current can now be done as

earlier.


Example 3 R C circuit supplied from a step voltage source, but with C initially charged to Vo Governing differential equation

oVdttiC

tiRte )(1

)(.)(

oVipC

iR .

1.

Complementary function remains same as earlier

except that the value of constant will be different.

The particular solution would again be zero.

R

C E

Vo


since voltage across capacitor cannot change suddenly

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

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

01

RCo eAR

VE

giving RVE

A o

tRCo e

R

VEti

1

)(

voltage across

t

RCoC eVEEtv

1

)(

v

C(t

)

time

E

Vo


Example 4 R L C circuit supplied from a step voltage

)(.

1)(..)(.

)(1)(

)(.)(

tipC

tipLtiR

dttiCtd

tidLtiRte

)(

1)(..)(..)(. 2 ti

CtipLtipRtep

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.

R

L

E

C

i(t)


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

complementary function is

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

giving 8006001060060062 jp

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

giving a solution of the form

A.e-600t

.ej800t

+ B.e-600t

.e-j

800t

,

or C.e-600t

.cos(800t+)


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 800t

Also since i(0)=0, vR(0) = 0.

vL(0) = 120 = giving at t=0

giving C = 300/800 = 0.375

td

id4.0 300

td

id

3000800cos800.0800sin.600. 06000600 eeCtd

id


i(t) = 0.375 e-600t

.sin 800t A

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 V

vC(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


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

0.4 p2 + 800 p + 4105 = 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 00.02

0.04

0.06

0.08

0.1

0.12

0 0.002 0.004 0.006 0.008 0.01

i(t)


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

0.4 p2 +1000 p +4105 = 0 or p2 + 2500 p + 106 = 0

giving (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

0.02

0.03

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)


1.3. Review of AC theory

Sinusoidal waveform has equation

v(t) = Vm sin( t + )

Period is T. Angular frequency = 2/T

Mean value = 0

v(t)

t T

T


Average Value (rectified)

Average of full-wave rectified waveform

average value for sinusoidal wave = (2/) Am

vrect (t)

t T

T


Effective value

Effective value = rms value

Rms value for sinusoidal wave = Am/2

rms value is always specified for ac waveforms.

Tt

t

eff

o

o

dttaT

A )(1 2

v2(t)

t T

T


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


Phasor Representation of Sinusoids

e j

= cos + j sin or e jt = cos t + j sin t

Sinusoidal waveform is the projection, on a particular

direction, of the complex exponential ejt.

a(t)

t

T

Amsint

t

O O

P

X


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 diagram

Ay

Ax


Usual to draw Phasor diagram using rms value A,

rather than with the peak value Am.

The phasor A is characterised by its magnitude A

and its phase angle .

Polar co-ordinates of phasor A written as A.

Cartesian co-ordinates Ax and Ay of phasor A, usually

written A = Ax + j Ay.


Phase difference

Consider Am sin (t+1) and Bm sin (t+2) as shown

Can be represented by rotating phasors Am e j(t+) and

Bm ej (t+) with peak amplitudes Am and Bm,

Or on normal phasor diagram with complex A and B with polar co-ordinates A and B.

O

y(t)

t

Amsin (t+1)

Bmsin (t+2)

T

Am

Bm


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)

= 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


Currents and voltages in simple circuit elements

(1) Resistor

for a sinusoid,

Let i(t) = Im cos (t+orReal part of [Im e(jt+ ]

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

(jt+

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

R

v (t)

i (t)

v(t) = R .i(t)

O t

Vmcos t

Imcos t

T


Vm = R.Im

and Vm/2 = R.Im/2

i.e. V = R . I

Note: V and I are rms values of voltage and current. no phase angle change has occurred in the resistor.

Note also that power dissipated in resistor R is

R . I 2 = V . I

V I O

Phasor diagram

R

V

I


0 t

Vmsin t

Imcos t

T

(2) Inductor

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

v(t) = Real [L.d

dt Im e(jt+ ]

= Real [L. j. Im ej(t+ ] = Real [j Vm e

j(t+ ]

L

v (t)

i (t)

dt

tidLtv

)()(


or v(t) = L.Im cos (t+ =L..Im sin (t+

L.Im cos (t++/2

= Vm cos (t++/2

Vm = L.Im

and Vm/2 = L.Im/2 Vrms = L.Irms Voltage waveform leads current waveform by 90o or /2 radians or current waveform lags voltage waveform by 90o for the inductor.

V = jL.I or V = L.IImpedance 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.

j L

V

I

dt

d


(3) Capacitor

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

v(t) = Real [ dteI

Cv tjm .

1 )( ]

= Real [1

C j. . Im e(jt+ ] = Real [

1

j Vm e(jt+ ]

C

v (t)

i (t)

0 t

Vmsin t

Imcos t

T

dtiCv .

1


or v(t) = 1

CI t dtm cos( ).

1

C .Im sin (t+

1

C .Im cos (t+/2=Vm cos (t+/2

Vm = 1

C .Im and Vm/2 =1

C .Im/2

i.e. Vrms = 1

C .Irms

Voltage waveform lags current waveform by 90o or /2 radians or the current waveform leads the voltage waveform by 90o for a capacitor.

Thus V =1

j C .I or V =1

C .I

Impedance Z =1

j C , and V = Z . I Power dissipation in a pure capacitor is zero, but product V.I not zero.

1

j C

V

I

V

I

O Phasor diagram


Impedance and Admittance in an a.c. circuit

Impedance Z is a complex quantity.

Relates complex rms voltage to complex rms current.

V = Z . I

Admittance Y is inverse of impedance Z.

I = Y . V

In cartesian form

Z = R + j X and Y = G + j B

Real part of Z is resistive, usually denoted by R,

Imaginary part of Z is called a reactance denoted by X.

ZY

1


A pure inductor and a pure capacitor has a reactance

only and not a resistive part, while a pure resistor has

only a resistive part and not a reactive part.

Z = R + j 0 for a resistor,

Z = 0 + jL for an inductor,

and Z = = 0 j for a capacitor.

Real part of Y is a conductance, denoted by G,

Imaginary part of Y is a susceptance, denoted by B.

C

1

Cj

1


Relationships between components of Z and of Y

G j B YZ R j X

R j X

R X

1 12 2

so that G

R

R XB

X

R X

2 2 2 2, and

It must be remembered that, in a complex circuit, G

does not correspond to the inverse of the resistance R,

but its effective value is influenced by X as well.


V VL

VR I

Phasor diagram

VL

Simple Series Circuits

R-L series circuit

Consider I as reference

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

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

so that total series impedance is

Z = R + jL Phasor diagram has been drawn with I as reference. [i.e. I is drawn along the x-axis direction].

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

By phasor addition, V = VR + VL.

R jL I

VR VL

V


If V is taken as reference,

I is seen to lag voltage by the

same angle that the voltage was 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.

Note: Power dissipation can only

occur in the resistive part of the

circuit and is equal to R.I 2.

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

Phasor diagram

V VC

VR I

Phasor diagram

V

VL

VR I


R-C series circuit

VR = R.I, VC = 1

j C . I

and V = VR + VC

V = (R + Cj1

).I

so 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 to R.I 2, not equal to product V . I.

R 1

j C I

VR VC

V

V

VC

VR

I


L-C series circuit

VL = jL.I,

VC = 1

j C .I = j1

C

and V = VL + VC

V = (jL +1

j C ).I

so that total series impedance

is Z = jL +1

j C = jL j1

C

Total impedance is purely reactive, and all voltages in

the circuit are inphase but perpendicular to current I.

Phasor diagram

V

VC

VL

I or

VC

VL

I V

1

j C

jL I

VC VL

V


Resultant voltage corresponds to algebraic difference

of voltages VL and VC and direction could be either up

or down depending on which voltage is more.

When L =1

C , total impedance of circuit is zero.

Circuit current I, for a given V, would be very large (only limited by the internal impedance of the source). This condition is known as series resonance.

In L-C circuit, current either lags or leads the voltage

by an angle equal to 90o and the resultant circuit is

either purely inductive or capacitive. No power dissipation can occur in the circuit and but the

product V . I for the circuit is non zero.


R-L-C series circuit

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

j C . I

and V = VR +VL + VC

V = (R + jL +1

j C ).I

so that the total series impedance is

Z = R +jL +1

j C = R + j(L 1

C )

R jL

I

VR VL

V 1

j C

VC


Z R LC

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 to R . I 2 and that

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


Simple Parallel Circuits

R-L parallel Circuit

Consider V as reference

V = R.IR,

V = jL.IL,

and I = IR + IL

IV

R

V

j L

total shunt admittance = 1 1

R j L

Phasor diagram

I IL

IR V

IL

R

jL

IR

V

IL

I


R-C parallel Circuit


V = R.IR,

IC = jC.V,

and I = IR + IL

I

V

RV j C .

total shunt admittance = 1

Rj C

Phasor diagram

I IC

IR V

IC

R

1

j C

IR

V

IC

I


R-L-C parallel Circuit


V = R.IR, V = jL.IL, IC = jC.V

and I = IR + IL + IC

IV

R

V

j LV j C

.

total shunt admittance

= 1 1

R j Lj C

Shunt resonance will occur when 1

LC giving minimum value of shunt admittance.

R

1

j C

IR

V

IL I

IC

jL

Phasor diagram

I IL+IC

IR V

IC

IL


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 . I

purely 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 P of circuit.


apparent power S = V . I

Since a difference exists between apparent power

and the active power, we define a new term reactive

power Q for the reactance X.

Instantaneous value of power is given by

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

If v(t) = Vm cos t and i(t) = Im cos (t ), where voltage has been taken as reference and the current

lags voltage by a phase angle hen p(t) = Vm cos t . Im cos (t ) = Vm Im .. 2 cos t . cos (t ) = 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 be

given by the constant value Vm Im cos

P = Vm Im cos cos.

2.

2

mm IV

V . I cos

i(t)

current lagging voltage by angle inphase quadrature

Vm Im cos

p(t)

t T

v(t)

t

p(t)

p(t)


Power Factor

Power Factor = active power/apparent power

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 of resistor, inductor and capacitor,

P lies between V. I and 0

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


Reactive Power

Defined as product of voltage and current components

which are quadrature (90o out of phase).

reactive power Q = V. I sin For L and C, reactive power Q = V. I Unlike 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.

Inductive reactive power and capacitive reactive

power have opposite signs.


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


P1

Q1

1

Qc

Q2 2

For a load

lagging power factor cos 1

active power P1, reactive power Q1

If 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 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 = V

2.C

The value of C can be determined.


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 current and voltage laws,

and Ohms Law

with differential operator

p =

dt

d


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) = L p . iL(t), C p .vC(t) = ic(t)

eliminate other variables

L p . iL(t) = r . ir(t)

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

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

i.e. C p . e(t) = C p . r . ir(t) + i(t) + C p . R . i(t)

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

i.e. L.C.p2. e(t) = [L.C. p

2.r + 1 + C p . 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 operator p

In a similar way

any current and any voltage would be related by an admittance transfer function of differential operator p

any two voltages or any two currents would be related by a transfer gain of the differential operator p

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

be replaced using the relationship p = j.


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 3T E

TT

dtT

tE

Tdttf

T00

).3

2(1

).(1

2)

2

32()

2

32(

2

0

2 E

T

TT

T

E

T

tt

T

ET

t


rms value =

TT

dtT

tE

Tdttf

T0

22

0

2 .)3

2(1

).(1

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 = 2

2

E

E


Example 4

A certain 50 Hz, alternating

voltage source has an internal

emf of 250 V and an internal

inductance of 31.83 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 mH

250 V


Solution

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

= j 10.00

Current I = jXRj 10

250

, | V | = 230 V

active power P = | I |2 R =

RXR

])10([

25022

2

,

voltage V = (R+jX) . I

R+jX

If we simply differentiate and obtain the condition for maximum power,

0,0

X

Pand

R

P, 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

R2 + (X + 10)2 = 1.185 R2 + 1.1815 X2

i.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 = R

X

]2010[5104.6

2502

2

,0

dR

dP

gives the condition

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

or X+5 = R dRdX


also 20 dRdX

+ 0 = 0.1815(2R + 2X dRdX

)

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

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

j10

250 V 4.983-j4.584


I= 33.97A

- 47.38o

42.61o E=250V

VR=169.3 V

VC=155.7V

VL=339.7V

under given condition

I = 416.5983.4

250

584.4983.410

250

jjj

i.e. I = o38.47360.7

250

= 33.97-47.38o A

terminal voltage at source (load voltage)

= 33.97-47.38o 6.771-42.61o = 230.0-90o V

voltage across resistive part of load

= 4.98333.97-47.38o = 169.3-47.38o V

voltage 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 at

230 V ? and what is the Power ?

If parallel capacitance is connected

with load to achieve 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 kW

31.83 mH

250 V, 50 Hz

Load


The 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

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

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