Electrical Power Generation, Storage and Usage

8/10/2019 Electrical Power Generation, Storage and Usage

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Electrical Power Generation, Storage and Usage_______________________________________________________________________________________

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Electrical Power

Generation, Storageand UsageHand Out

Ir ing T. de Lange30-1-2011

Version 001Source: A.C.C. de Langen/HR&O


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Contents

1 Introduction and important facts 32 Definitions with time dependent functions 7

2.1 Voltage and Current 72.1.1 Time functions with series circuit 82.1.2 Time functions with parallel circuit 9

2.2 Period time T and frequency f 112.3 The mean value of a periodic function 11

2.3.1 Definition and calculation method 112.3.2 Special properties 13

2.4 The effective value: Root Mean Square 132.4.1 Definition 142.4.2 Special properties 15

2.5 The form factor 162.6 Overview and overview of periodic signals 162.7 Electric Energy and Power 17

2.8 Power and energy in ideal network elements 192.8.1 Power and energy in an ideal resistor 192.8.2 Energy and power in an ideal inductor 202.8.3 Energy and power in an ideal capacitor 21

3 Sine wave voltage and current 223.1 Sine wave description and display 223.2 Voltage and current with impedances 25

3.2.1 Behavior of resistors 253.2.2 Behavior of inductors 263.2.3 Behavior of capacitors 273.2.4 Behavior of R-L-C series circuits 283.2.5 Behavior of R-L-C parallel circuits 303.2.6 Summary 33

3.3 Types of power in alternating current 333.3.1 Active power with a resistor 343.3.2 Reactive power with an inductor 353.3.3 Reactive power with a capacitor 35

3.4 Power with an R-L-C circuit 363.5 P, Q and S with more elements in a circuit 383.6 Power triangles vs impedance triangles 383.7 Determination of R and L or C from nameplate data 39

4 Calculations with complex numbers and functions 404.1 The Complex Plane 41

4.1.1 The definition of a complex number 424.1.2 Carthesian versus polar co-ordinates 424.1.3 The transformation carthesian polar 43

4.2 Arithmatic actions with complex numbers 434.2.1 Adding and subtracting 444.2.2 Multiplying and dividing 44

4.3 A special complex time function 464.3.1 Adding and subtracting 47


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1 Introduction

Understanding electric systems is found difficult by most people. This can beunderstood, since it cannot be seen what is happening, like in mechanical systems. It isnot easy to see electrons flowing through a conductor with ones bear eyes!

But it is not necessary to see them flowing. You can get grip on electronics when youstudy the basic theory that is derived from physic research. When you accept thephysics rules in electronics, youre able to understand the basic behaviour of a lot ofelectronic components.

That is the point that will be emphasized in this lecture. Basic theory will be firstrepeated and extended in this hand-out, before using the book: Electrical Machines,Drives and Power Systems from Theodore Wildi, ISBN: 0130986372.

One should bear in mind that understanding electronics in the base gives you a lot ofpossibilities to create sophisticated drive systems, fun applications and can save you lots

of money when you are advised by third party, drive system suppliers.

In this lecture an effort in systematical thinking in magnetics theory, complex numbertheory, resistor/capacitor/inductor theory and inverter theory is made. In understandingthese theories, you are basically able to understand every topology of electricalmachine, inverter and transformer.

In this hand-out, Resistor, Capacitor and Inductor Theory will be discussed. AlsoComplex Number Theory will be dealt with.

In the book, electrical machines (motors, generators), power-electronics (how do youmake alternating current from a battery?) and batteries will be dealt with.

The first classes of the lecture will deal with basic theory, step by step, in order to give aclear introduction to the goal of this lecture. After that, it is easier to understand thebehavior of electrical machines. Then the fun part will start: understanding howmachines work, seeing them in real life, looking at demonstrations in class, tricks andtips, running machines in parallel, loadsharing, advanced options etc.

The lecture material will be elaborate; make an effort every week to study it andprepare possible questions.

This reader was edited by Dr. ir. ing. H.T. Grimmelius.


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P: Both voltage and current sources are energy sources. The none-ideal power sourcecan be seen as voltage source or as current source. The conversion is according to U T=IN.R with R = RT= RN.

Impedances

Z: the factor that gives the relation between voltage and current. Z is determined byresistors, coefficients of self-induction for inductors, capacity values of capacitors andthe time functions of current and voltage.

R: The relation between voltage and current with resistors and other none-reversibleenergy converters. The factor can always be used where electric energy is permanentlyconverted into heat or work. A resistor can be compared with a mechanical damper. Thequicker one moves the damper (read: the more one applies more voltage across theresistor), the more force it will require to move it (read: more current will flow throughthe resistor). So voltage can be compared with mechanical speed and current can becompared with mechanical force.

L: the relation between resulting magnetic flux and the current that causes this flux inan inductor. The mathematical relation is: u = L.di/dt or i = (1/L).udt. For the build upof the magnetic field energy has to be stored in the inductor. The energy can beregained when the magnetic field is broken down. One can speak of borrowing electricenergy for the build up of the magnetic field.An inductor can also be compared with a mechanical component: the spring. The moreone squeezes a spring, the more force it will require to squeeze it. Or: the longer oneapplies a voltage across the inductor, the higher the current will rise. Compare also thebehavior: only after movement, a force will develop. Or: only after applying voltage, acurrent will develop.

C: The relation between charge and voltage with a capacitor. The mathematicaldescription is: u(t) = (1/C).i(t)dt or I = C.du/dt. For the build up of the electric field(that is proportional to the voltage) electric energy has to be supplied to the capacitor.


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But this energy is regained when the electric field is broken down. Just as with theinductor, one can speak of borrowing electric energy. The capacitor can be comparedwith a mechanical mass. The more one applies force to a mass, the faster the mass willmove. Or: The more one applies current through a capacitor, the more voltage willdevelop across the terminals.

Kirchhoffs laws

Voltage law: Sum of the voltages in a mesh is zero.

Or : Usupply= in GAS= Uload= Uin UAS

Current law: Sum of the currents to a node is zero.

Or: Iin= Iout


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2 Definitions with time dependent functions

2.1 Voltage and current

Electric situations can be displayed in circuits of sources, resistors, inductors andcapacitors.

Due to all kinds of causes the sources could result in time varying voltages and currents.Under these circumstances, we have to calculate with Kirchhoffs laws, Ohms law(resistors) and the differential equations between voltage and current for inductors andcapacitors. So:

Making calculations on these kinds of situations is only useful when the time function isknown. In practice, this means repetitions in time T. One can speak of periodic timefunctions.

uR= R . iR iR= (1/R) uRuL= L . diL/dt iL= (1/L) uLdtuC= (1/C) iCdt iC= C . duC/dt


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Examples of time functions are:

When the signal is sine wave shaped, one speaks of alternating current or voltage (ac).

2.1.1 Time functions with series circuit

The current in each component is the same in a series R-L-C circuit.The voltage across the resistor is: uR= R.iThe voltage across the inductor is: uL= L.di/dtThe voltage across the capacitor is: uC= (1/C).i.dt

The total voltage is u = uR+ uL+ uC

So:

When the current is known, the voltage can easily be determined. This situation below isdone with R = 10, L = 10mH and C = 100F.

i.dtC

1+

dt

idL.+i*R=u

t

t

tot

0


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It can be seen that a step change in the current requires an infinite peak voltage across

the inductor. In practice, this is not feasible; the change in current will always require acertain amount of time. When the total voltage is known, the solving of the current willbe difficult on paper. But the solving can easily be done with numerical integrationcomputer programs (for example in Matlab or Matlab/Simulink).

2.1.2 Time functions with parallel circuit

The voltage across each component is the same in a parallel R-L-C circuit.The current through the resistor is: iR= (1/R).u

The current through the inductor is: iL= (1/L). ).u.dtThe current through the capacitor is: iC= C.du/dt.

The total current is i = iR+ iL+ iC


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This yields:

When the voltage is known, the current can easily be determined. Solving the voltage atknown total current requires numerical integration.

This situation below is done with R = 10, L = 10mH and C = 100F.

Now, one can see that a step change in the voltage needs an infinite current peakthrough the capacitor. This is not possible in practice either; changes in voltage across acapacitor will always take a certain amount of time.

totaal

t

t

i =1

R* u +

1

Lu.dt + C.

d u

dt0


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2.2 Period time T and frequency f

The shortest signal repetition time is called the period time. The value f = 1/T is therepetition frequency or the base frequency of the periodic signal. In figure x a periodicfunction is displayed.

Via a Fourier analysis, every arbitrary periodic signal can be seen as a sum of a puresine wave signal with same base frequency and a number (again sine wave) higherharmonics (i.e. with the frequency a multiple of the base frequency). Due to this, onecan also say angle frequency = 2..f = 2. /T

2.3 The mean value of a periodic function

When using periodic signals in direct current circuits, the mean value is often important.The rotational speed of a direct current motor is for example proportional to the meanvalue of the voltage.

The mean is indexed by m or by a dash above the symbol. So: um= .

2.3.1 Definition and calculation method

The mean value of a periodic function is defined as:

The mean value of function f is equal to the surface A under the function curve in thefunctiontime plane (in axis units).

F =F =1

T f(t) dtgem

0

T


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Some examples are given below:

( ) sin( ) 0f t f t F= =

( )2

f ff t t F

T= =

Both are determined with the surface under the curve.

Single side sineBy means of mathematics:

[ ]

0

0

180

0

1( ) sin( ). sin( ).

1 cos( )

( 2 )

ff t f t dt t dt

T T

f f

T

T

= =

= + =

=

1 fControlled half sine

By means of mathematics:

[ ]

1( ) sin( ). sin( ).

1 cos( )2

ff t f t dt t dt

T T

f

= =

= +


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2.3.2 Special properties

In many cases the true periodic function can be seen as a superposition of more, simpleperiodic functions. When looking at the previous definition, the following will apply:

1

n

sum i

i

F F

=

=

In other words: the sum of the mean values of the superposed partial functions is equalto the mean value of the sum function. Proof:

( )1 2 1 21

1 1 1... ...

n

sum i

i

F F F dt F dt F dt F T T T =

= + + = + + =

N.B.: The mean value is independent on the period time T and frequency f.

2.4 The effective value: Root Mean Square (R.M.S)

The effective value is important when one has to look at energy. It is the value that isconverted to direct current value in energetic view. This will receive more attention lateron, but a short introduction will follow now. The heat developing in a resistor, usingdirect current is: P = I2.R. With alternating current this is: 1/T.i2.Rdt = R/T.i2dt. Thesquare of the effective current value is: 1/T.i2dt. The effective value of a sine wavealternating current is .2..

N.B.: different signals have different values. The effective value is indexed by eff or bythe use of the capital for voltage and/or current. So u effor U.

2.4.1 Definition

The effective value of a periodic function is defined as:

In other words: The square of the effective value is equal to the mean value of thesquared function. N.B.: The effective value is always independent of the period time Tand the frequency f.

One can determine the effective value by means of mathematical calculation orgraphical determination of the surface under the squared curve. Some examples aregiven below:

eff1T

0

T

2F = f (t)dt


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By means of the surface: A1 =A2

2 2

2

2

2 2

12

2eff

f T fF

T

F F f

= =

= =

By means of mathematics:

( 2 2

2 2 1( ) sin( ) ( ) sin ( ) 1 cos(22

f t f t f t f t f = = =

Other examples:

Saw tooth: By means of mathematics:

2

21 1( ) ( ) 33 3

eff

f ff t t F f t dt f

T T= = = =

Saw tooth likeSee previous example:

2

13

3 3eff

fF f= =

Square wave:By means of surface:

2 2

1 1 2 2eff

f t f tF

T

+=

f.2

1=dt

2

t)2-(1f

T

1=f=F:So

2T

0

effcos


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2.4.2. Special properties

When a true periodic function can be seen as a superposition of more, simple functions,then there are sometimes possibilities to determine the effective value from the partialeffective values. In this situation this will apply:

fe2= (1/T) {f1

2+ f22+...+ 2.f1.f2+ ....} dt

fe2= (1/T)f1

2 .dt + (1/T)f22.dt + (1/T)2.f1.f2.dt + ....

fe2= fe1

2+ fe22 + mix-term1-2+ ......

If the integral over the mix terms yields zero, the square of the effective value is equalto the sum of the effective values of the partial functions.The mix term contributions are zero when:a) the partial functions are not zero at the same time So, if f1 0 then f2 = 0 and viceversa

b) one function is constant and the mean value of the other function is zero.

This last situation can be a direct current with noise or ripple. This is a periodic signalwith known mean value. For this applies:

F2eff= f2m+ r

2eff with r the noise or ripple.

2.5 The form factor

The mean value and the effective value of a periodic function are proportional to thepeak value of the function. The proportionality is different per function. Because ofalternating use of the effective and mean value, one defines a form factor. This factor isdefined as the effective value by the mean value.

The value of the form factor can vary between 1 and infinite. Also the form factor isindependent of period time and frequency.

2.6 Overview and resume of periodic signals

For periodic signal there are the next understandings:

feffectiveThe form factor F = ---------

Fmean


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a) The mean value

_ 1 Tf = fmean = --- f(t) dt

T 0

_ _With special case f = fi

b) The effective value

With special case f2eff= f2gem + r

2eff, only when: rgem=0.

c) The form factor

feff.F = -----

Fmean.

Function Mean Value Eff Value Ff.sin t

Sine 0 .2.f

f.sin tRectificationboth sides

2/.f .2.f .2.

f.sin tRectificationsingle side

1/.f .f .

f.t/TSaw tooth .f 3/3.f 2/3.3

(t)dtf=F2

T

0

T1

eff


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2.7 Electric energy (W) and power (P)

Electromechanical energy exchange can based on Lorentz force, Coulomb force andinduction voltages. Electrically seen, the energy is determined by voltage and current.This will be highlighted now.

The voltage u = uABand the current i. The current i in the time t equals an amount ofcharge Q = i. t Coulomb, while the voltage is the same as the amount of energy perCoulomb. In time t an amount energy W is converted, where W = u.Q = u.i. t.Over a certain time period it is needed to add up all these contributions. So with u and ias the momentary values of voltage and current there will apply: W = W = u.idt.

t1

W = u(t).i(t) dt (in Joule or watt.seconds)0

The amount of energy converted is thus determined by factors voltage, current andtime. The amount of energy is most of the time important due to economical reasons.

For the construction of electromechanical installations, the current and voltage areimportant. Durage of usage is often less important. As energy determining quantity onedoes not use the amount of energy itself, but the amount of energy per second. Onespeaks in this case of power

The momentary power:

d Wp(t) = --- = u(t).i(t)

dt


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For the usual frequencies the momentary power is a time function. The mean value is agood indicator of power consumption:

W = pdt t.pm

This mean power is indicated by the capital P. The concerned unity has the dimension ofvolt-ampere, or Watt (W). 1 Watt is 1 Joule/second or 1 volt-ampere/second. For directcurrent situations, the momentary power P=U.I is constant. So here, the momentary

power equals the mean power.

Overview electric energy and electric powerGeneral Direct Current

Momentary Power p(t) = u(t).i(t) = dW/dt P =U.I = W/tMean Power P = 1/T.p(t)dt P = U.IEnergy W = p(t)dt = u(t).i(t) W = P.t = U.I.t

These functions are applicable, no matter what the time function of the periodic signal is

(so applicable for dc, ac, saw tooth, etc).

In daily life, humanity has to deal with large amounts of electric energy. For example:the energy consumption of an electric heater of 2kW. The energy consumption for onehour will then be: W = P.t = 2000.3600 = 7.2x106J. If one has to deliver this energy ona bicycle, one has to paddle for about a 12 hours period, non stop, on moderate power.

To keep numbers reasonable, one often uses kilowatt hours (kWh). This is the energythat is consumed by a device of 1 kilowatt in one hour.So 1 kWh = 1000.3600 = 3.6x106J.

2.8 Power and energy in ideal network elements

2.8.1 Power and energy in an ideal resistor

An electric heating element can be seen as an ideal resistor. In this resistor electricenergy is directly converted into heat and is directly lost. This partially the case inelectric motors. Here electric energy, through magnetism and current carrying windings,

The mean power:

1 TP = pgem= --- p(t) dt

T 0


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is also directly converted into work (thus finally heat). So an electric motor can also bepartially displayed as an ideal resistor.

The relation between voltage and current for an ideal resistor is: u(t) = R.i(T). Voltageand current have the same time function. For the momentary power applies:

P(t) = R.i2(t) = u2(t)/R

For the mean power applies:

1 T 1 T URP = --- R.i dt = R . I = --- u/R dt = --- = UR.IR

T 0 T 0 R

The active power across a resistor equals:

The resistive value times the square of the effective value of the current through theresistor: P=I2.R

The amount of electric energy that is converted in the resistor equals the integral ofpower over time.

2.8.2 Energy and power in an ideal inductor

For an ideal inductor applies: u = L.di/dt

For the momentary power applies:

d i(t) dp(t) = L ------ . i(t) = -- { L.i(t) }

dt dt

For the mean power, one can find PL= 0:

1 T 1P = --- p(t) dt = --- { L.i(T) - L.i(0) } = 0 watt

T 0 T

For the amount of stored energy in the inductor can be found: W(t) = .L.i2(t0). Thisfollows from:

1W = W1 - W0= p(t) dt = L.i(t1 ) - L.i(t0)

0


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The inductor needs a supply of energy when the current increases and its energy isgiven away again when the current decreases. The mean power however is zero Watt.The energy is stored in the build up of the magnetic field and releases again when themagnetic field is reduced. The amount of stored energy is WL= .L.i

2.

WL = L.i en PL= 0 watt.

2.8.3 Energy and power in an ideal capacitor

The relation for an ideal capacitor is: I = C.du/dt

This gives for the momentary power p(t):

d u(t) dp(t) = C ------ . u(t) = --- { C . u(t) }

dt dt

For the mean value one can find:

1 T 1P = --- p(t) dt = --- { C.u(T) - C.u(0) } = 0 watt

T 0 T

The amount of stored energy is WC= .C.u2(t)

1W = W1- W0= p(t) dt = C.u

2(t1) - C.u2(t0)

0

A capacitor requires additional energy when the voltage increases, which will bereleased again when the voltage decreases. The mean power is, just as with theinductor, zero Watt. The electric energy is stored in the build up of the electric field andrelease again when the field is reduced.

WC = C .u and PC= 0 watt.


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3 Sine wave voltage and current

In practice, electric energy is delivered as sine wave alternating current to the publicgrid. It is produced by electricity plants. In Europe, the voltage is 230Vac and thefrequency is 50 Hz. In Japan the voltage is 100Vac and the frequency 60 Hz. In Americaone uses 120V and 60 Hz. In Australasia and other British colonies one uses 240V 50Hz.

Also other periodic time functions can be seen as a sum of sine wave voltages/currentswith frequencies equal to a multiple of the repetition frequency (Fourier analysis).

The calculation on the basis of a pure sine wave signal is therefore important basicknowledge for the consideration of electric situations.

3.1 Sine wave description and display

A general description of a sine wave signal can be u = .sin(t+) with the peakvalue, the angle frequency and the phase shift on tom t = 0 s. Due to similarity ofsine and cosine one can also use u = .cos(t+). In connection with future use of thecomplex alternating current theory, the cosine will be used as description of signals.

SINE SIGNAL u(t) = cos( t + )


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Features of the description are:a) The modulus in peak value or effective valueb) The phase shift = shift of the peakc) The frequency .

Remarks:a) The height of the voltage or current is called the modulus value. This value is veryimportant. Dependent on the application one uses the peak value or the effective value.This is possible since = 2.u resp. = 2.i. In energy applications one always usesthe effective value. In signal processing, one uses most of the time the peak value.

b) The phase angle gives the shift of the peak compared to t = 0 s. To the left ispositive, to the right is negative. In many cases the phase shift is also important dat. Itcan be seen from the sine wave form that sin(t) = cos(t-90). Usually, the phase

angle is given in degrees, however the calculations are to be made using radians.

c) With energy applications in Europe, the frequency is 50Hz, so = 100rad/s.

In many cases with known frequency, only modulus value and phase angle areimportant. In stead of a value versus time curve a so-called phasor diagram can beused. Although erroneously called, it is sometimes called a vector diagram. The phasordiagram can be seen as a polar display of current and voltage. A sine wave signal isdisplayed by an arrow with length that is equal to the modulus value and an angle thatequals the phase angle compared to the horizontal reference axis.


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For example: in the above phasor diagram are displayed:

U1= 15 cos(t+60) Vdus 15 < 60 V

U2= 10 sin(t+60) = 10 cos(t-30) Vdus 10


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3.2 Voltage and current with impedances

An impedance is defined as: Z = / = U/I. Z gives the relation between the modulusvalue of the voltage and the current. Next to the influence on the modulus valuesimpedances have also influence on the phase shift between voltage and current. For thisphase shift the angle z = u i. As ideal impedances a resistor, inductor andcapacitor are used. A general impedance can be seen as a series and/or parallel circuitsof these components.

3.2.1 Behavior of resistors

i(t) = cos(t+i)

u(t) = R.i(t)= R. cos(t+i)= cos(t+u)

For the modulus value can be seen: = R..

As a result of the resistor a phase shift of z= u i= 0 occurs. Voltage and currentare thus in phase. This can be seen in the phasor diagram.


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3.2.2 Behavior with inductors

i(t) = cos(t+i)

u(t) = L di/dt= -L. sin(t+i)= L. cos(t+i+90)= cos(t+u)

For the modulus value applies:

= L . = XL.

As a result of the property of the inductor there is a phase shift between voltage andcurrent:

L= u- i= +90

The sine wave voltage across the inductor leads the sine wave current wit 90 or ..This can be seen in the phasor diagram.

When looking at the modulus values, L replaces the resistive value R. The value L iscalled the reactance of the inductor and is assigned by X L. The ideal inductor does not


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have a resistive component, but an inductive reactance en causes a phase shift L=+90.

3.2.3 Behavior of capacitors

i(t) = cos(t+_i)

u(t) = 1/C i dt= +1/C. sin(t+i)= 1/C. cos(t+i-90)= cos(t+u)

For the modulus value applies:

= /(C) = XC .

As a result of the property of a capacitor, the voltage lags the current by 90. This canbe seen in the phasor diagram.


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Now, 1/(C) replaces the value of R. The value 1/(C) is called the reactance of thecapacitor and is assigned by XC. The ideal capacitor does not have a resistivecomponent, but a pure capacitive reactance. This causes a phase shift between voltageand current of C= -90.

Summary:

Kind of Impedance Resistor Reactance Inductor Reactance CapacitorImpedance Z R XL= L XC= 1/CPhase shift 0 +90 -90

3.2.4 Behavior of R-C-L series circuit

From the previous sections can be concluded that an impedance can be a composite ofa resistor and a reactance. The voltage across the resistive is in phase with the current,the voltage across the reactances leads resp. lags the current. The inductive reactance

has a positive phase shift of 90 whereas the capacitive reactance has a negative phaseshift. Both reactances can neutralize each other. The simplest series circuit is given:

Through each component flows the current:

i(t)=.cos(t+i)

For the three partial voltages applies:

uR(t) = RS. cos(t+i)uL(t) = L. cos(t+i+90) = XL. cos(t+i+90)uC(t) = 1/C. cos(t+i-90) = -XC. cos(t+i+90)

The voltage across the reactance (inductor and capacitor together) is:

uX(t) = (XL-XC). cos(t+i+90) = XS. cos(t+i+90) = -XS. sin(t+i).

For the total voltage yields:

u(t) = Rcos- Xsin with R=RS., X=XS. en =.t+i


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Goniometry says: cos(+) = cos().cos() sin().sin(). When tg() = X/R= XS/RSone can construct a triangle, composed of the phasor diagrams of the voltages acrossthe resistor and reactances:

In the equations for the voltage this gives:

u(t)/ = (RS/cos).(cos.cos-sin.sin) = (RS/cos).cos(+) -> Z=R/cos

In the triangle one can see that (R/cos()) is equal to the hypotenuse.

The voltage across the resistor, the voltage across the reactances and the total voltageform a right-angled triangle. This triangle is called the voltage-triangle for a seriescircuit. Here the voltage phasor of the resistor has the same direction as the currentphasor. The phase shift between the resulting voltage and current is equal to

For an RCL circuit applies:

The vector sum of the phasor for voltage across theresistor and the phasor for voltage across the

reactance gives the phasor for the total voltage


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arctg(X/R). The voltage phasors follow from the current phasor by means of theresistor (R with angle 0) and the reactance (X with angle 90). Just as with the voltagetriangle one can define an impedance triangle, existing of RS, XSand Z.

3.2.5 Behavior of R-C-L parallel circuits

The simplest parallel circuit is given:

Across each component is the voltage:

u(t)=.cos(t+u).

For an impedance Z, existing of R and Xinseries applies:

= .Z with Z =(R+X)

z = u -i = arctg(X/R)


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For the three currents applies:

iR(t)= 1/RS. cos(t+u)iL(t)= 1/L. cos(t+u-90)

= -1/XL. cos(t+u+90)iC(t)= C. cos(t+i+90)

= +1/XC. cos(t+u+90)

The current through the reactance (inductor and capacitor together) is:

iX(t) =(-1/XL+1/XC). cos(t+_u+90)=1/XP. cos(t+_u+90)= -XP. sin(t+_u).

For the total current applies:

iX(t) =(-1/XL+1/XC). cos(t+u+90)=1/XP. cos(t+u+90)= -XP. sin(t+u)

When tg() = X/R= RP/XPthen the triangle below can be constructed. This triangle isconstructed with the phasor diagrams of the currents through the resistor andreactance.

In the equations for the current this gives:

i(t)=(/Z).cos(+)

Here /Z is the hypotenuse in the triangle of currents.


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The currents through the resistor, reactances and the total circuit form a right-angledtriangle. This triangle is called the current triangle for the parallel circuit. The resistorcurrent phasor has the same direction as the current phasor. The phase shift betweenthe resulting current and voltage is = arctg(X/R). Calculations are now done with 1/R,1/X and 1/Z. 1/Z is called the admittance of the circuit. Just as with the current triangleone can define an admittance triangle. This triangle consists of 1/RP, 1/XPand 1/Z.

For an RCL parallel circuit applies:

The vector sum of the phasor of the current throughthe resistor and the phasor of the current through thereactance gives the phasor for the total current.

For an impedance Z, existing of X and Rparallel applies:

= /Zwith 1/Z =((1/R)+(1/X))y=-z= i-z= arctg(R/X)


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3.2.6 Summary

Impedance is U/I. Impedance consists of resistance and reactance. For the resistancepart the current and voltage are in phase, whereas for the reactance part the voltageand current are shifted 90. Reactance is assigned by X, For an inductor, XL= L andfor a capacitor XC = 1/(C). In general the combination of resistance and reactance

gives the impedance Z, which is dependent on the partial values of R and X. The phaseshift between resulting voltage and current is determined by these partial values.

Circuit Impedance Z z= u- iR and X in series (R2+X2) atan(X/R)R and X in parallel ((1/R)2+(1/X)2) -atan(R/X)

3.3 Types of power in alternating current

In the general consideration of energy (W) and power (P) applies:

Element Active power P Energy WResistor I2.R = U2/R p.dt (converted)Inductor 0 .L.i2(t) (exchanged)Capacitor 0 .C.u2(t) (exchanged)

When one looks only at sine wave signals, there are some other types of power that canbe distinguished. Next to the active power P, there are two other types for alternatingcurrent: the reactive power Q and the complex power S. For pure sine wave signals onecan distinguish:


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a) P = active power= U.I (resistor) [Watt]= the mean of the momentary, consumed power (is zero for

inductor and capacitor).

b) Q = reactive power= the peak value of the borrowed momentary power for a reactance

(inductor and/or capacitor)

The unit is [VAr]. Q is 0 VAr for an ideal resistor

c) S = complex power= U.I (without distinction resistance/reactance)

The unit is [VA ]

For the explanation of P, Q and S some elementary network situations with inductor,capacitor and resistor will be discussed.

3.3.1 The active power with a resistor

The active power is equal to the mean value of the momentary power. Paragraph 2.8says that P is zero fr both an inductor and capacitor.

For a resistor in User Arrow System applies: u(t) = .sin(t) and i = .sin(t) and R =

U/I = u/i. Voltage and current are in phase. For the power applies: p(t) = ..sin2(.t) =...(1-cos(2t). This power is positive for a user under all circumstances. The limitsare between 0 and ., where the maximum is twice the mean power. The time functionhas double frequency. The mean value is the active power: P = ./2 = UR.IR= IR

2.R =UR

2/R.


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3.3.2 Reactive power with an inductor

For an inductor in User Arrow System applies: i(t) = .sin(t) and u = .cos(t). U =XL.I or u = XL.i. The voltage leads the current by 90. The power is: p(t) =..sin(t).cos(t) = ..sin(2t). This momentary power is alternating from positiveto negative. The mean value is 0 Watt. The reactive power Q is the peak value of themomentary power. It is equal to ./2 and the time function has twice the supplyfrequency.

In the figures above u(t), i(t), p(t) and Q are displayed.

3.3.3 Reactive power with a capacitor

For a capacitor in User Arrow System applies: i(t) = .sin(t) and u = -.cos(t). U =XC.I or u = XC.i.

The voltage lags the current by 90. Analogous with the consideration of the inductorfollows the momentary power of the capacitor. This momentary power is in contra-

phase with that of an inductor (compare the behavior of a mass-spring system). In thefigure u(t), i(t), p(t) and Q are given for the capacitor.


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The mean power is zero Watt. For the reactive power applies: Q = ./2 = UC.IC= IC2.XC

= UC2/XC.

Since the reactive power of the inductor and capacitor are in counter phase, the reactivepower of the inductor is said to be positive and the reactive power of the capacitornegative.

3.4 Power with an R-C-L circuit

By considering an arbitrary RCL circuit as a series circuit, one can say for the currentand voltage:

i(t) = sintu(t) = sin(t+) = [cos.sint + sin.cost]


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For the momentary power applies:

p(t) = . sint.[cos.sint + sin.cost]= . cos.sint + . sin.sint.cost

(with resistor) ( with reactance)

The term sin2(t) contributes to active power.

P = (1/T) ..cossint dt = ..cos= U.I.cos

The value cos() is called the power factor of the circuitry. This will also be dealt with inthe chapters on electric machines. The term sin(t).cos(t) determines the reactivepower Q. The peak value of sin(t).cos(t) is equal to , so: Q = ...sin() =U.I.sin(). With active and reactive power U.I is a determining factor, next to the phaseshift angle . Therefore it can be said that complex power is the product of current andvoltage: S = U.I. So:

Complex power S = U.IActive power P = S.cos()Reactive power Q = S.sin()

Since sin2() + cos2() = 1, S2= P2+ Q2. Due to this relation it is possible to display

the powers in a right-angled triangle: the power triangle. This triangle is displayedbelow. The triangle is similar to the impedance triangle of a circuit.

N.B.: in this case U is the total voltage across a circuit and I the total current throughthe circuit.


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3.5 P, Q and S with more elements in a circuit

The network elements resistor, inductor and capacitor can be implemented in series-parallel circuits. By taking the whole circuit as one, it is still possible to make calculationsfor P, Q and S. But it is also possible to take the powers of the separate elements.Looking at the sum of powers, one can say:

The total active power P is the sum of the active powers of the separate powers: P tot=Pparts

The total reactive power Q is the sum of the reactive powers of the separate powers:Qtot= Qparts= Qinductive parts- Qcapacitive parts.

The total complex power is S = (P2+ Q2). N.B.: adding the partial complex powersdoes not give the right solution. This always has to be done via calculation of the partialreactive and active powers!

3.6 power triangles versus impedance triangles

From the previous sections can be concluded:

P =UR.IR= IR2.R = UR

2/R = U.I.cos = S.cosQ =UX.IX= IX

2.X = UX2/X = U.I.sin = S.sin

S = (P2+ Q2) = IS2.Z = US

2/Z = U.I

These powers can be displayed in a power triangle for each situation. This triangle isright-angled, where P is the base, Q is the opposite side and S is the hypotenuse.

The impedance Z can be seen as an equivalent series circuit of RSand XS. Here Z2= RS2+ XS

2, which can be described in an impedance triangle. XS= XLXCis called positive inthe case of inductive behavior and negative in the case of capacitive behavior. It is alsopossible to define an admittance triangle for a parallel circuit in the same way. Here, thebase is 1/RP, the opposite side 1/XPand the hypotenuse 1/ZP. Now XP, obtained through1/XP= 1/XC 1/XL, is positive at capacitive behavior and negative at inductive behavior.


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4 Calculations with complex numbers and functions

Sine wave voltages and currents be displayed with phasor diagrams, instead ofcurrent/voltage curves.

For these vector displays the following rules are applicable:

The phasor length = modulus value (U and I or and )The angle = momentary phase angle (t + u) and (t + i)

Voltage and current are displayed in this polar display by corner speed rotatingarrows. The angle between the arrows of voltage and current is constant en isdetermined by the impedance.

The factor t determines both for voltage and current the rotational speed. The wholediagram can thus be displayed as static arrows in a rotating polar plane.

For the display applies:

Length phasor = modulus valueAngle = phase angle on t = 0s


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In the previous sections impedance is defined as a quantity that determines the ratioand phase shift between voltage and current. So an impedance can also be displayed ina polar plane by means of a phasor with length Z = U/I = / with an angle Zcompared to the reference axis. For a series impedance, this phasor consists of aresistive component (RS) in the same direction as the reference axis and a reactivecomponent (XS) perpendicular to this.

4.1 The complex plane

The complex plane is a display method for complex numbers. The numbers are definedin this plane by means of a rectangular co-ordinate system and/or polar co-ordinatesystem.

In a rectangular co-ordinate system the horizontal axis is the real axis (Re-axis) and thevertical axis the imaginary axis (Im-axis). In a polar co-ordinate system the Re-axis isalso the reference axis.

Complex numbers can be seen as vectors in a complex plane. Calculations with thesenumbers have many similarities compared to calculations with regular vectors. Due tothese similarities between complex and phasor display of voltage and current, it is thebasis for the most important calculation method for sine wave signals. For this, one usesEulers theorem with complex time function F.ejtthat is related to the cosine function.


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For the use of this method it is important to master calculations with complex numbersand complex time functions.

4.1.1 The definition of a complex number

In mathematics, adding, subtracting, multiplying and dividing with real numbers is dealtwith. A special calculation is extraction of the root. This calculation can easily be donewith real, positive numbers. But with negative numbers problems occur. Real numberswhose squares are negative do not exist. These numbers are called imaginary numbers.An imaginary number is a real number multiplied by the factor (-1). This factor isassigned by j in electronics. In mathematics it is assigned by i. The reason is toprevent mix ups with the letter i for current.

The following applies in making calculations:j = (-1) j = -1 j3= - j en j4= 1

Some examples of imaginary numbers are: 3j and 2.15j.

A complex number is the sum of a real number and an imaginary number

Some examples of complex numbers are: (2+3j), (3.85-2.34j) and (-34+16j). Thenumbers are assigned by an underlined symbol. For example: G = A + B.j (A and B realnumbers).

4.1.2 Carthesian versus polar co-ordinates

Although hard to imagine, complex numbers can easily be graphically displayed bymeans of two perpendicular co-ordinate axes in the complex plane. As said before, thehorizontal axis is called the real axis and the vertical axis the imaginary axis. Whencomplex numbers are displayed in this way, one speaks of carthesian co-ordinates. Incarthesian co-ordinates a complex number G is displayed as : G = X+ j.Y.


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To define points in the complex plane, polar co-ordinates can be used too. With polardescription of complex numbers the real axis is the reference axis. The co-ordinates aredetermined by the length of the distance between the point and the origin (modulus)and the angle (argument) between connecting line and reference axis. For thedescription Eulers theorem is used: ej= cos() + j.sin(). So:

A complex number is displayed in polar co-ordinates as:

g = g


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For subtracting the same principle applies:

G1-G2= (a-c) + j.(b-d)

For adding and subtracting the carthesian method is used. The polar notation is notsuitable for adding and subtracting.

Adding and subtracting of complex numbers equals adding and subtracting of vectorsand takes place in Cartesian notation

4.2.2 Multiplying and dividing

When multiplying in the carthesian notation it can be seen that:

Gt = G1.G2 = (a+jb).(c+jd)= (ac+j.bd) + j(ad + bc)

Gt = (ac - bd) + j(ad + bc)

In the polar notation it can be seen that:

Gt = G1.G2 = G1ej1. G2e

j2

Gt = G1. G2ej(1 +2)


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In the polar calculation method the resulting modulus is the product of the two modulusvalues. The resulting angle is the sum of the two partial angles. This calculation can thusbe done with the two methods. In many cases the polar method will be preferable here,considering the amount of calculations.

Dividing can be done in a similar way:

Carthesian:

(a+jb) * (c-jd)Gt = G1/G2= (a+jb)/(c+jd) = -------------------

(c+jd) * (c-jd)

(ac+bd) + j(bc-ad)Gt = --------------------------

(c + d)

Polar:

G1Gt = ---- e

j(1 -2)G2

Multiplying and dividing can be done in both the carthesian and polar method. Polarmethod is preferred

Remark:

Multiplying by j -> 90 phase shift (counter clock wise)Dividing by j -> 90 phase shift (clock wise)


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4.3. A special complex time function

In relation with the so-called alternating current theory there is a function in thecomplex plane that is of great importance. This function is:

Where:

R = modulus value = corner frequencyt = time variable = phase angle at t = 0s

The display of this function in the complex plane results in a rotating vector (length R)whose endpoint is on a circle and whose center is in the origin. The real part of thisfunction equals the cosine function; the imaginary part represents the sine function).

According to Eulers theorem the function can also be written as an exponential functionwith complex power:

R(t) = R . ej.ejt= R . ejt

Here

R = R


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For further consideration two functions are taken:

X(t) = X ej( t+ x) = X.ejt and Y(t) = Y ej(t+y) = Y.ejt

The corner frequency is the same for both functions.

4.3.1 Adding and subtracting

When determining the sum of these two time functions in polar notation then it can beseen that:

X(t) + Y(t) = ( X ejx+ Y ejy). ejt= (X + Y).ejt= T . ejt

In other words: The result is obtained by taking the complex modulus values of the twotime functions together and to supply them with the time function determining part. Asame method applies with subtraction.

4.3.2 Multiplying and dividing

In polar notation multiplication gives:

X(t) . Y(t) = ( X ejx. Y ejy). ejt. ejt = (X . Y).ej2t= P . ej2t

The result of multiplication is a complex time function with double corner frequency anda complex modulus value equal to the product of the separate modulus values.

When dividing, this method gives:

X(t)/Y(t) = X/Y = D

The result is a complex number which has become time independent.

4.3.3 Integrating and differentiating

Integrating and differentiating are done with the inductor and capacitor. Differentiationof a complex time function gives:

d X(t) X(t)------ = lim -------dt t->0 t

In the complex plane it can be seen that X(t), with small enough t, is perpendicular tovalue X(t). The size is equal to the arc-length times the radius, so: .t.X(t). This yields:dX(t)/dt = jX(t). This could also be done by means of determining the derivative of an


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With complex time functions U(t) = U.ejt:

Operation Complex modulus R Time function ejt

Adding vectorial adding unchanged

Subtracting vectorial subtracting unchanged

Multiplying product complex modulusvalues

Double frequency(sum-frequency)

Dividing quotient complex modulusvalues

Time independent

Differentiating jtimes thee complexmodulus value

unchanged

Integrating Complex modulus value dividedby j

unchanged


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The complex quantities can be written as the product of a modulus/angle determining part(the complex peak value ejuor the complex effective value U eju) and the time functiondetermining part (ejt).

For the complex voltage and current this gives:

u(t) = ejt met = ejui(t) = ejt met = eji

Instead of calculating with the complex peak value it is possible to calculate with thecomplex effective values (U = Uej_uen I = Iej_i). With energy applications, the effectivevalue is always used; whereas in electronics circuits the peak value is used.

From the complex voltage/current the real voltage current can be obtained via:

u(t) = Re{ u(t) } i(t) = Re{ i(t) }

5.2 Elementary laws for complex voltages and currents.

Related with the complex voltage and current, Kirchhoffs laws are observed more closely.

5.2.1 Kirchhoffs laws.

In a maze, for real voltages, applies: ut= u1+ u2+ u3.

The voltage law, identically applied on complex voltages gives: ut= u1+ u2+ u3

When using the real time function the real part of the complex time function is:

u(t) = Re{ u(t) } = Re{ u1(t) } + Re{ u2(t) } + Re{ u3(t) } = u1+ u2+ u3.

Conclusion: The Kirchhof voltage law is also applicable for complex voltages.

The same for the currents flowing in/to a node:

i(t) = Re{ i(t) } = Re{ i1(t) } + Re{ i2(t) } + Re{ i3(t) } = i1+ i2+ i3.

Conclusion: The Kirchhoff current law is also applicable for complex currents.


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5.2.2 The voltage current relation.

In network theory the relation between voltage and current is determined by impedance.For this understanding there is the following order:

Resistor R

Impedance Z Inductive XL(inductor)Reactance X

Capacitive XC(capacitor)

For every kind of these impedances, the results are considered for the complex quantities.For this, the understanding complex impedance Z is used.

Definition:

u(t) UZ = ------ = --- = --

i(t) I

In other words: the complex impedance is equal to the complex factor between thecomplex voltage and the complex current.

Resistor.When for the complex voltage applies: u(t) = R.i(t), the the result ejt= R. ejtcan befound. Via u(t) = Re{u} en i(t) = Re{i} applies thus:u(t)= cos(t+u) = Re{u(t)} = Re{R.i(t)} = R.Re{i(t)} = R.i(t) = R. cos(t+i)In other words: = R. en z= u- i.

With (stationary) sine wave signals the result of the calculation with complex voltage andcurrent for an ideal resistor is identical to the result of the conventional calculation.

Conclusion: For a resistor applies:

u(t) = ZR. i(t)

with ZR= R

Inductor.For an inductor applies: u(t)=L.di(t)/dt which results in: u(t) = L.di(t)/dt = jL.i(t) = jXL.i(t)with the characteristic = XL. en u= i+90This result is identical to the result of the conventional calculation.


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Conclusion: For an inductor applies:

u(t) = ZL. i(t)

with ZL = +j L = +j XL

Capacitor.For a capacitor applies: u(t) = 1/Ci(t) dt. This results in:

1u(t) = 1/Ci(t) dt = ---- . i(t) = - jXC. i(t)

jCIt can be seen that: = XC. en u= i 90. Also now the result is equal to that of theconventional calculation.

Conclusion: For a capacitor applies:

u(t) = ZC . i(t)

with ZC = 1/(jC) = -jXC

5.3 Impedance Z en admittance Y.

The complex impedance Z = U/I has the same function in alternating current as theresistive value in direct current. Comparable with direct current conductance admittance Y= 1/Z = I/U for alternating current with unity 'siemens' is known.In general, impedance is used in calculations on series circuits and admittance incalculations on parallel circuits.

5.3.1 RL circuit.

An RL circuit is composed of the ideal elements resistor and inductor. The ideal partialimpedances ZR= R en ZL= j.XL= j L are present.Note that the resistive part is frequency independent and that the inductive part isfrequency dependent.


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5.3.1.1 RL series circuit

For the total series impedance of a series RL circuit applies:Zs = ZR+ ZL

= Rs+jLs.When this impedance is considered as a function of the frequency, then it can be seen withcorner frequency in r/s that: =0 gives R


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5.3.1.2 RL parallel circuit

Here, for admittance Y applies:Y = 1/ZR+ 1/ZL= 1/Rp+ 1/jLp= 1/Rpj.1/Lp.When this admittance is considered as a function of thefrequency, then it can be seen:=0 gives


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Note: Series or parallel connection does not influence the resistive-inductive characterof Z.

5.3.1.3 Display as Z or Y

For an RL series circuit can be seen:

ZS = RS+ j XS= Z < Z

YS = 1/ZS< -S = A - j B

For an RL parallel circuit can be seen:

YP = 1/RP - j 1/XP = Y < Y

ZP = 1/Y < -Y

If ZS = ZP then the two circuits are equal for voltage and current at the consideredfrequency.This is the case when ZS=1/YPand Z= -P.

For convenience, a series impedance can be transformed into a parallel impedance.But in general, for every frequency, different values for the R and L are obtained. It is onlyat one frequency that a circuit can be transformed into another.

5.3.2 RC circuit.

An RCcircuit is composed of the ideal elements resistor and capacitor. So ZR=R en ZC=j.XC=-j/C.Again, the resistive part is frequency independent and the capacitive part is frequencydependent.


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5.3.2.1 RC series circuit

Now, for the total complex impedance applies: Zs=Rs-j1/Cs.When this impedance isconsidered as a function of thefrequency, it can be seen at thecorner frequency in r/s:If =0, then Z


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5.3.2.3 Display as Z or Y

Just as with the RL circuit, an RC circuit can be transformed from a series circuit into aparallel circuit and vice versa. The transformation is valid per specific frequency, since theimpedance is dependent on the frequency.

5.3.3 LC circuit

5.3.3.1 LC series circuit

The impedance of a series RC circuit is: j.(L - 1/C) = j.(Xl- XC). The voltages across theinductor and capacitor counteract each other. Due to this, the total reactive voltage islower than the individual voltage across the inductor and capacitor. At the resonancefrequency = 1/(Ls.Cs) this counteraction is complete. This is called series resonance.Here, the impedance of the inductor and capacitor together result in an impedance of 0 .Inductor and capacitor together seam to be a short circuit, but there is definitely a voltage

across each component.

5.3.3.2 LC-parallel circuit

For a parallel circuit of inductor and capacitor the admittance is j(C - 1/L) = 1/XC - 1/XL.In this case too, the capacitor and inductor counteract.The capacitor delivers a part of the inductor current, or vice versa: the inductor delivers apart of the capacitor current.

At parallel resonance (= 1/(Lp.Cp) this is complete. The currents through the inductorand capacitor are equal, but in counter phase. The total reactive current is zero.

At parallel resonance the admittance is 0 S and the impedance . It looks like an opencircuit, but in fact, current does flow through the components.

5.4 Summary.

The result of the calculation with the help of complex voltage and complex current, basedon . ejt, is correct forfor sine wave signals.

The characteristic values ( en ) of the complex voltage and current are identical to the

characteristic values of the real quantities.

The basis for the complex network theory is Kirchhoffs law and the relation betweencomplex voltage and current (Ohms law).


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This yields:

nComplex voltage law uj = 0in a maze j=1

nComplex current law ij = 0in a node j=1

Voltage/current relation u = Z . i

ZR = R (pure real)ZL = + j L = +j XL (pure positive imaginary)ZC = 1/jC = -j XC (pure negative imaginary)

ZL and ZCcan be in resonance


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6 Calculation on the basis of name plate data

6.1 Introduction.

The name plate on any specific machine displays electrical en mechanical data on themachine in question. In general, these are the nominal data.With the help of the name plate data, one should be able to determine currents andreplacing circuits for that machine. The electrical data are usually nominal power, nominalvoltage, operating frequency and cos(). Dependent on the application, nominal currentand/or efficiency can be given.The power on the name plate is the effective delivered power P at a given load pattern (i.e.continuous torque). Sometimes the complex power S is given.

The name plate data are valid under normalized ambient conditions.For ambient temperature, temperatures between -25C en +40C apply with a yearaverage lower than 30C and a day average of 35C.Occurring temperatures under operation can reach more than 100C. In many cases one

calculates with an operational average of 75.

6.2 Impedance display.

For the resistance and reactance in 3.6.1 it is already given how to extract these valuesfrom the name plate data U, Pconsumeden cos.

For series display applies:RS = P/I = S.cos/I = U.cos/I = U.cos/SXS = Q/I = S.sin/I = U.sin/I = U.sin/S

For parallel display applies:RP = U/P = U/(S.cos) = U/(I.cos)XS = U/Q = U/(S.sin) = U/(I.sin)

The choice between series and parallel display is dependent on the desired circuit to makecalculations with. Most of the times, one will choose the series display, but in situationswith cos() compensation one will prefer the parallel display.Since most electric constructions are using electromagnetic fields, the replacing ReactanceX will be inductive.

6.3 Currents.

The connection with the public grid gives the energy at an almost constant voltage ( 5%)and at an almost exact frequency (Europe: 50 Hz).The currents adapt to the need. The grid, however, is dimensioned on a maximumapplicable current.A distinction must be made between:a) The grid at the electricity company


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b) The grid inside a building.

The component sizing in the electricity company is based on the power that is written in acontract. When expanding, attention has to be paid to this power, or a new maximum hasto be written in a contract. The responsibility is of the electricity company.The dimensioning inside a building is the responsibility for the user. In many cases thecircuits are constructed for a maximum current of 16A. Group connections of 25A, 35A ,50A and 63A are also possible.When making changes in a grid, one always has to be mindful that the total current insome point does not exceed the maximum allowed value. If necessary, one has to look atspreading power demands over more groups or spread power demand of differentmachines in time.

6.4 Energy costs.

The energy bill of an electricity company consists of more components, dependent on the

type of user

The components can be:

* Maximum active power (peak value of the 15 minute averages)* Maximum reactive power (peak value of the 15 minute averages)* Consumed kWhs* Consumed kVArhs (dependent on the normal cos())

The extra costs for the (borrowed) reactive power are justified, since reactive powercauses extra losses in the grid. Also the operation at generation is made more difficult. Fora company it can be an economical consideration whether to take own action for cos()compensation (reducing reactive power) or to pay the higher electricity bill for the use ofreactive power.

6.5 Cos() compensation.

A bad cosine() can be compensated towards the electricity company. This meanscompensation of low cos() (relatively high inductive power) by connecting a capacitorbank in parallel with the consumers.Since capacitors can deliver the reactive power directly to the consumers, less reactivepower has to be drawn from the grid.

Qgrid= Qconsumption - Qcompensation

Cos() compensation is theoretically identical to bringing better in resonance of the RCLcircuit.

Chapter 4 showed that a capacitor and inductor can compensate each other. In a seriescircuit the voltage across an inductor is in counter phase with the voltage across a


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capacitor. In a parallel circuit the inductor current is in counter phase with the capacitorcurrent.In both cases the effect on current/voltage of the resulting reactance is smaller. When aninductor and capacitor compensate each other, full resonance is obtained.Parallel compensation is when a replacing parallel circuit is brought to parallel resonance bymeans of adding parallel components to that circuit.Series compensation is when a replacing series circuit is brought to series resonance bymeans of adding series components to that circuit. The disadvantage of seriescompensation is that the voltage across the inductor or capacitor can exceed the nominalvoltage. Due to this, series resonance is rarely used, except in tubular lamp circuits.

Distinction can be made between:Series resonance: L and C in series with ULC=0 V en ILCnot 0 A.Parallel resonance: L and C parallel with ILC=0 A en U not 0 V.

The final choice is an economical question: investment in compensation capacitor bankversus kVArh bill.

Example exercise:* The energy company applies cos() measurements when the cos() is lower than

0,85. The grid delivers a 230 V, 50 Hz alternating current.The number of operation hours is around 2200.

* A machine workshop uses every year around 175.000 kWh active energy and 98.500kVArh reactive energy.

Set the kWh costs on 15 ct/kWh en 10 ct/kVArh

a) Should reactive power taken into account when making calculations on the energyconsumption? ( tg() = Q/P = 0,563 dus cos() = 0,872)

b) If yes,over how many kVArh should then be paid? .How much capacitance should be installed in order to not receive a bill for reactivepower? (cos() gives Qmax=108.500 kVAr: this is more than the actual consumption,so compensation is not necessary).

If no, motivate the answer.

How much capacitance should be installed in order to achieve a cos() of 1 ? (Q.t =98.500 --> Q = 44.773 VAr --> C=2,94 mF so in practice 3 mF).


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modeling, since it displays process descriptions with proportional, integrating anddifferentiating actions (Laplace equation for processes).

7.2 Some understandings that describe the frequency.

Transfer function.

This is the complex function that generates an outputsignal when an input signal is offered. For a voltagedivider this is: Uuit/Uin

Remark:For the transfer the input and output can be defined.

In general, for input the control signal or supplying voltage is taken.Most of the times the output signal is the voltage across a specified impedance or the

current through that impedance.

Bode diagram.This diagram gives the modulusvalue of the transfer function as afunction of the frequency.Here the crossover frequencies,band width and possible resonancefrequencies are characteristicvalues.

Remark:Bode-diagrams are often used incontrol engineering to display thebehavior of a control system.

Crossover frequencies (at Rs = Xs or Rp = Xp).At a crossover frequency the derivative of a transfer function is maximal (positive or

negative). In this case this equals the frequency where the complex description of thetransfer exists of a real part and an imaginary part of equal size.At crossover frequencies the modulus values are thus 2 times the maximum valueat a phase angle of 45.


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Resonance frequencies (at Xs=0 or Xp=).At a resonance frequency the imaginary part of the impedance is 0 or . The circuit

behaves like a pure resistor at this frequency. The possible reactances counteract eachother in their effect on current and voltage.

Band widthThe band width is the frequency band between two crossover frequencies thatneutralize each other.

7.3 RL circuits.

Many practical circuits can be seen first and foremost as an RL circuit. Examples are electricmotors and filter inductors.

7.3.1 RL series circuit.

The construction of a filter inductor in general is a copper conductor wound on aferromagnetic core. When alternating current is flowing through the conductor, a voltagedrop across the conductor will occur as a result of the winding resistance and a voltagedrop as a result of the induced magnetic field will occur.

A practical inductor can thus be seen as an RL circuit.The voltage will divide over both components accordingto:UZ= UR+ UL= (ZR+ZL).I = (RS+jXL).I = ZRL.ISo U = Z.I en Z= U-Iwith |Z| = (Modulus Z) = Z = (RS+XL)Z = arctg (XL/RS)

The impedance of a non ideal inductor can be seen as areal part (resistance Rs) in series with a frequencyproportional reactive imaginary part XL.

The display of these impedance components gives a right angled triangle in the impedance

triangle. This is the triangle that fits between the voltage and current.

Example:Given: Across a filter inductor, there exists a sine wave voltage (50 Hz) with effective

value 220 V. Via a current meter it is found that the current is 1,3A while bymeans of phase shift measurement is found that the current lags the voltagewith 64.


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Question: Determine the coefficient of self induction and the resistance of the filterinductor.

Solution:Choose time t=0 s on the moment that the phase angle of the current is equal to 0. Forthe complex (effective) current then applies: I = (1,3 < 0) A.For the voltage follows automatically: U = (220 < 64) volt.

The impedance has to be then: Z = U/I = (230/1,3 < 64-0) So Z = (169,23 < 64) = 74,19 + 152,1 j = R + j XL

From the reactance XL= 2..50.L follows the coefficient of self induction L = 484 mH.The resitance is 74,19 . P=I.R=125,4 watt is converted into heat.

Frequency influence on ZRLS.

When the impedance triangle is considered as a function of the frequency it can be seenthat the imaginary part XLchanges linearly with corner frequency (since XL= .L).The modulus of the impedance changes on the basis of the hypotenuse in the diagram.

At the crossover frequency the resistance and reactance value are equal. Thus applies: k= R/LIn this case the modulus of the impedance is 2*R. At agiven voltage at crossover frequency the current is thus r2 times smaller than the current at =0 r/s.

Example:Given: a low pass filter. The filter is based on an RL circuit.The resistor has a value of 1 k. For this filter a crossoverfrequency k= 400 r/s is desired; in order to pass 50 Hzsignals well and 100Hz signals badly.

Question: How should this filter be constructed?Answer: With an inductor the reactance and voltage are proportional to the frequency.With an RL series circuit the voltage across the resistor is higher at lower frequencies. Thehighest low frequent voltage is thus across the resistor; the inductor is between the inputand output.


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Question: Determine the needed L and determine the transfer uout/uinat 50 Hz,100 Hz and 200 Hz.

Solution: At the crossover frequency R=XAnd thus =R/L.

This gives L=R/=2,5 H.Since uout = i.R and uin = i.(R+jL) applies,follows via elimination of the i as transfer:A=R/(R+jL), where the modulus value is equalto R/((R+X) with X=L,

This gives:

Remarks:1) An inductor of L = 2,5 H is veryhard to construct, since the value isvery high (large inductor).

2) The phase shift between inputvoltage and output voltage increasesfrom 0, via 45 at the crossoverfrequency, with increasing frequency to90.

frequency XL Transfer

50 Hz 785,4 0,7864

100 Hz 1570 0,5372

150 Hz 2356 0,3907


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7.3.2 RL parallel circuit.

In electronics, one uses power supplies with constant (effective) voltage. The differentconsumers are connected in parallel to the power supply. An RL circuit is obtained when anideal resistor is connected in parallel with an ideal inductor.

With his parallel circuit the voltage is equal across both components.The supplying current splits through the separatecomponents.

IZ = IR+ IL= (YR+YL).U = (1/RP+1/jXL).U= (1/RP-j/XL).U = (GP+ j BP).U= YRL. U

So I =Y.U en Y= I-Uwith |Y| =(Modulus Y)=Y= ((1/RP)+(1/XL))and Y = arctg (RS/XL)

The admittance of a parallel circuit exists of a real part1/RPand a negative frequency proportional, reactive part1/XL.

The display of the admittance components gives in thecomplex plane a right angled triangle; the admittancetriangle that first between the current and voltage.

With the admittance Y belongs the impedance Z = 1/Y.It can be written that: Z = 1/Y en Z= - Y

Example:

Given: Across a resistor (100 ) parallel to an ideal inductor (L=250 mH) a sinewave voltage (50 Hz) with effective value of 24V is put.

Question: Determine the partial currents and the total current.Calculate the replacing impedance.

Solution: Set the complex voltage to (24 < 0). ZR=R= 100 , ZL=78,54 .Via I=U/Z per element can be seen: IR=(0,240


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7.3.3 Conclusions RL circuits.

For an RL series circuit the impedance is:

ZRLS= RS+ jL = RS+ jXL= Z


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7.4 RC circuits.

7.4.1 RC series circuit.

With a series circuit the voltage divides across thecomponents according to:UZ= UR+UC=(ZR+ZC).I=(RS-jXC).I= ZRC.IDus weer U=Z.I en Z= U-I met|Z| = (Modulus Z) = Z = (RS+XC)Z= - arctg (XC/RS)

The impedance of an RC circuit exists of resistanceRSand a frequency proportional negative reactivepart XC.

Again, the display of the impedance components inthe complex plane results in an impedance triangle

that fits between voltage and current.

Example:Given: The voltage 100 cos(1000t+31) stands across a series circuit of 200

resistance and a capacitance of 10 F.Question: Calculate the current through the circuit and the voltage across the capacitor.

Solution: ZC= -jXC= 1/C = -100j ==> Z = 200-100j = 223,6


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At very high frequencies, the capacitor acts as a short circuit. In this case the impedance ismainly determined by the value of the resistor.

Example:

Given: A low pass filter on the basis of an RC series circuit. The resistor is 1 k.For this filter a crossover frequency k = 400 r/s is desired, in order totransmit 50 Hz signals well and 100 Hz badly.

Question: How should this filter be constructed? Determine the desired capacitance Cand the transfer uout/uinas a function of .

Answer: With and RC series circuit the voltage is mainly across the capacitor at lowfrequencies; at high frequencies the voltage across the resistor will becomehigher. The capacitor is the output; the resistor is between input and output.

At the crossover frequency is R=X, this means that C = 1/R = 2,5 F.Via uout= -jX.i en uin= i.(R-jX) follows by means of elimination of i for the complex transferfunction A = -jX/(R-jX) with modulus A = 1/((1+(RC)).

7.4.2 RC parallel circuit.

Calculations done on a parallel RC circuit is left to the reader. This should not present anytrouble at this point. Here applies: at low frequencies the capacitor acts as an open circuit;where at high frequencies the ressitance is bridged by the lower impedance of thecapacitor.

7.4.3 Conclusions RC circuits.

For an RC series circuit the impedance is equal to the sum ofthe partial impedances;

ZRCS= RS+1/(jC) = RS-jXC= Z


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7.5 The general series and parallel circuits.

7.5.1 The RCL series circuit.

The impedance Z exists of an R, L en C in series.For this circuit applies:UZ = UR+ UL+ UC = (ZR+ ZL+ ZC).I

= (RS+ jXL- jXC).I= (RS+ j(XL-XC) ).I= ( RS+ j XS) . I = ZS. I

So U = Z.I and U= I- Z

with modulus Z = Z = (R2S+ X2S)

and hoek Z= arctg (XS/RS)

The impedance of an RCL series circuit consists of a real

part RSand a frequency dependent partl XS.The display of these impedance components gives aright angled triangle in the complex plane.

Frequency influence on ZS .

When the triangle is considered to be a function of thefrequency, it can be seen that the reactive part XS= L -1/C changes in value. when XSis positive, the inductivecharacter dominates.

This replacing reactance XSis 0 at the frequency S=1/(LS.CS).This special frequency is called the (series) resonancefrequency.At this resonance frequency the circuit behaves like apure resistance RS .Across the inductor and capacitor the voltages are: UL en UC , but these voltages are incounter phase, which results in a total voltage of 0 V. The influence of the inductor and theinfluence of the capacitor counteract their effect in voltage completely.

At frequencies lower than the series resonance frequency the capacitive reactance is higherthan the inductive reactance. The inductor compensates a part of the influence of the

capacitor, but the whole displays the behavior of an RC series cruit

At frequencies higher than the series resonance frequency the capacitive reactance is lowerthan the inductive reactance. Now the circuit acts as an RL series circuit.


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Example:

Given: u = 100 cos(t-60) V; = 1000 r/sR=50 ; L=100 mH; C= 20 F

Questions:a) Determine the impedance triangle.b) Calculate the current through the circuit.c) Determine the most simple replacing circuit at

a frequency of =1000 r/s.d) Determine the voltage across the resistor.e) Draw the phasor diagram for voltage and current.f) Determine the resonance frequency.g) Determine the cross over frequencies.h) Draw the voltage across the LC part as a function of the corner frequency.

Solution:

a) XS = 1000*0,1 - 1/(1000*20/1000000) = 50 . The resulting reactance isthus 50 inductive. Combined with the 50 this gives an impedance of(50+50) = 70,71 with angle Z= 45.

b) From u=(100


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7.5.2 The RCL parallel circuit.

The impedance Z exists of an R, L en C in parallel.

For this circuit can be written:

IZ = IR+ IL+ IC = (YR+ YL+ YC).U= (GP+ jBL-jBC).U = (GP+ j(BL-BC) ).U= (GP+ j BP) . I = YP. U

with G = conductance en B = susceptance

So I = Y.U and I= U+ Y and U = Z.I and U= I- Y

Here the modulus Y = Y = (G2P+ B2P)

and argument Y = Y= arctg(BP/GP) = arctg(RP/XP)

By definition a positive reactance means an inductivereactance. A positive B indicates a capacitive reactance.

Thus, the impedance of an RCL parallel circuit exists of a realpart with conductance GP parallel to a reactive part withsusceptance BP.

Frequency influence on ZP .

When the triangle is considered to be a function of thefrequency it can be seen that the imaginary part, which isdescribed with the susceptance BP= C - 1/L,changes

This susceptance B is equal to 0 siemens at the specialfrequency P = 1/(LP.CP). The reactance X thatresembles this B is now ..

This frequency is called the parallel resonance frequency.

At the resonance frequency the impedance behaves likean ideal resistor RP. There are currents flowing through

the capacitor and inductor, but these are equal and incounter phase. The total reactive current is therefore 0A.


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