4-AC Circuit Analysis

8/14/2019 4-AC Circuit Analysis

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17 August 2005 Engineer M S Ayubi 1

ELECTRIC CIRCUITS

THEORY 1

These lecture slideshave beencompiled by

MohammedSalahUdDin Ayubi.

LECTURE 4 contd...AC Circuit Analysis


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8) AC Circuit AnalysisSinusoidal voltages and currents

8.1) Description of Sinusoidal Signal

3 parameters

Magnitude Vm

Phase

Frequency = 2 f

Ac circuit analysis is conducted at one frequency at a

time. Only two variables left, magnitude and phase.

If more than one frequency source is present. Use the

principle of superposition

)cos()( += tVtv m


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8.2 Advantages of Sinusoidal Signals

Sinusoidal functions yield further sinusoidalfunctions when integrated or differentiated

)cos()( += tVtv m

)sin()(

+= tV

dttv m

)sin()(

+= tVdt

tdvm


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Signal v(t) and integral (represented by iv(t)) and

differential (represented by dv(t))

0 0.5 1 1.5 2 2.5 3 3.5 460

48

36

24

12

0

12

24

36

48

6010V rms ac signal at 0.5 Hz

Time in seconds

Voltageinvolts

44.429

44.429

v t n( )dv t n( )

iv t n( )

40 t n


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8.3 Representation of Amplitude and Phase of a Signal on the

Complex Plane

+ j

+ real- real

- j

t

Vm

= 0

= 90 or /2

= -90 or - /2

= 180 or

imaginary


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+ j

+

real- real

- j

tV

m

= 0

= 90 or /2

= -90 or - /2

= 180 or

15 10 5 0 5 10 1515

13.5

12

10.5

9

7.5

64.5

3

1.5


voltage in volts

angularfrequencytimestim

einradians

0

12.566

tn

14.14214.142 v real t( ) n

0 5 10 1515

12

9

6

3

0

3

6

9

12


angular frequency times time in radians

Voltageinvolts

14.142

14.142

v imag t n( )

12.5660 t( )n

)sin(

isaxis)(imaginaryjon thephasor

rot

atingtheofprojectionThe

tVv mimag =

)cos(

isaxisrealon thephasor

rotatingtheofprojectionThe

tVv mreal =

( ) ( )

caseparticularIn this

5.02cos102cos ttfVv m ==


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8.4 Relationship between rms and peak

and i and j Electrical engineers generally use a slightly different

nomenclature to mathematicians for complexnotation

Firstly we use j, rather than i, for the square root of 1

(i gets confused with current)

Secondly, we normally use rms values rather than peakvalues to describe the amplitude. For sine waves the peakis always 1.414 times the rms. (rms values rather than peakare used to describe voltages and currents)


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8.5 Representation of d/dt by j We know that a signal represented by the differentiation

of a sinusoidal voltage or current is the equivalent of theoriginal signal Multiplied by

Advanced in phase by 90 degrees or /2

Now multiplying by j gives a phase shift of +90 degrees

Thus the operation of d/dt on a sinusoidal signal is likemultiplying by j

(in the complex plane)

)cos()( += tVtv m

)2

cos()sin()(

++=+= tVtVdt

tdvmm


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8.6 Impedance of resistors, inductors and capacitors

Impedance is a term used to describe therelationship between sinusoidal voltage and

current for

a single passive component

a group of passive components

(passive components are R, L and C. They

are passive because they are linear and

have no gain)


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Impedance is given the symbol Z

Note v and i are complex quantities here. (Some books use bold type to

distinguish between complex and absolute values of sinusoidal

voltages and currents or between complex and dc values)

Zv

i

v = i Z

v = i R

iLjdtdiLv == i

Cji

Cjv

thus

vCjdt

dvCi

==

==

1


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Thus

C

jZ

LjZ

RZ

C

L

R

=

=

=

We have two components to the impedance.One has no j term and is referred to as real

The other is all j terms and is referred to as imaginary

The same is true for voltage and current.

All complex values can be expressed as a magnitude and phase angle

Or as a complex quantity.

Complex voltages and currents are called phasors


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8.7 Phasors, Reference Phasor and CIVIL

A phasor is a term given to a voltage or current which has a real and complex part. Represented as one of the following forms

( ) ( )

partimaginarytheofsignthebetoof

signtherequiresfunctioncostheUsingNote

coscosarccos

sincos

1

22

=

=

=

+=

=

+=

=

+=

m

real

m

real

m

real

imagrealm

m

mm

j

m

imagreal

v

v

v

va

v

v

vvv

wherevv

vjvv

evv

vjvv

+real

-real

- j

vm

= 0

= 90 or /2

= -90 or - /2

= 180 or

In circuits with several voltages and currents

We need to define we need to take

a current or voltage as a reference (i.e. let =0 for that variable)


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Taking the same voltage applied across a parallel combination of aresistor, a capacitor and an inductor

v ZR

iRZC

iCZL

iL

Plotting the voltage and three current phasors on the complex plane

produces a phasor diagram. Voltage is common to all so we take thatas the reference phasor in this case

+ j

+ real- real

- j

= 0

= 90 or /2

= -90 or - /2

= 180 or

C

jZ

LjZ

RZ

C

L

R

=

=

=

Z

vi =

v


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v ZR

iRZC

iCZL

iL



+ j

+ real- real

- j

= 0

= 90 or /2

= -90 or - /2

= 180 or

C

jZ

LjZ

RZ

C

L

R

=

=

=

Z

vi =

viR


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v ZR

iRZC

iCZL

iL



+ j

+ real- real

- j

= 0

= 90 or /2

= -90 or - /2

= 180 or

C

jZ

LjZ

RZ

C

L

R

=

=

=

Z

vi =

viR

iC


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v ZR

iRZC

iCZL

iL



+ j

+ real- real

- j

= 0

= 90 or /2

= -90 or - /2

= 180 or

C

jZ

LjZ

RZ

C

L

R

=

=

=

Z

vi =

viR

iC

iL


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8.8 Phasor Addition and Complex Arithmetic to Find the

Combined Current for the Previous Circuit

Find i

v ZR

iRZC

iCZL

iL

i

+=

++=

++=

++=

LCj

Rvi

LjCjR

v

ZZZ

vi

iiii

LCR

LCR

11

1

1

11111

- real

j

By complex algebra and arithmetic

+ real

- j

v

iL

By phasor

addition


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Find i

v ZR

iRZC

iCZL

iL

i

+=

++=

++=

++=

LCj

Rvi

LjCjR

v

ZZZ

vi

iiii

LCR

LCR

11

1

1

11111

j


+ real- real

- j

v

iRiL

By phasor

addition


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Find i

v ZR

iRZC

iCZL

iL

i

+=

++=

++=

++=

LCj

Rvi

LjCjR

v

ZZZ

vi

iiii

LCR

LCR

11

1

1

11111

j


+ real- real

- j

v

iR

iC

iL

By phasor

addition


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Find i

v ZR

iRZC

iCZL

iL

i

+=

++=

++=

++=

LCj

Rvi

LjCjR

v

ZZZ

vi

iiii

LCR

LCR

11

1

1

11111

+ real- real

- j

v

iR

iC

iL

j

i i


By phasor

addition


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8.9 General Circuit Solution

We can solve ac networks with the same tools andmethods used for dc networks

Must use complex representation of voltages, currentsand impedances

Must choose one current or voltage as reference phasorand relate all others to it in terms of the phase angle

Evaluation of power needs care (rms values help)

Evaluation of stored energy needs care

8 10 I l d i


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8.10 Instantaneous power, real power and reactive power

0 1 2 3 42

1

0

1

22

2

v t n( )

i t n( )

P t n( )

40 t n

0 1 2 3 42

1

0

1

22

2

v t n( )

i t n( )

P t n( )

40 t n

0 1 2 3 42

0

2

43

2

v t n( )

i t n( )

P t n( )

40 t n

Instantaneous power waveforms for a voltage of 2V peak and a current of 1.5A peak

Flowing separately in a resistor, a capacitor and an inductor

Resistor case

Average power

Pav =0.5Vm*ImPav =vrms *irms

Inductor case

Pav = 0

Capacitor case Pav = 0


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Zvi

v = i Z Consider a series connected resistor and inductor

( ) ( ) ( )

( )( ) ( )

( ) ( ) ( )( )

( )

( )

sin

Q.symbol

given theisThis.componentsreactiveinstorageenergyforeresponsiblproductcurrent

voltagetheofcomponentthedefineshpower whicreactivetheispartimaginarytheAnd

cos

powertheaspartrealonly thetakeWe

sincos

Thus

referenceasvoltagethetakeweifand

cosandiwhere

1

0

0

2222

22

=

=

===

=

+=+==

+

=

+

=

+==

+=

ivQ

ivP

jiveiveeiviv

evv

LR

Ra

LR

veii

LR

LjR

LjRLjR

LjR

LjRv

Zvi

LjRZ

jjj

j

j

8 11 Solutions to Problems with Power Evaluation When


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8.11 Solutions to Problems with Power Evaluation When

Neither Voltage or Current Phasor is the Reference Phasor

( ) ( )

( ) ( )( )

( )

( )

( ) ( )

( )

0)(tryQforsigncorrectthegivesthissince

sinhavemustBut we

functioncostheofsignthechangenotdoesangletheofsignthesince

cosorcos

beshouldanswercorrectThe

sin

cos

powertheaspartrealonly thetakeWesincos

now

isphasorscurrentandvoltageebetween thangleThe

referenceasvoltagethenot takedoweifand

=

=

=

+=

+=

++=

===

=

=

=

+

v

vi

ivvi

iv

iv

iviv

jj

vi

j

j

ivQ

ivivP

ivQ

ivP

jiv

eiveeivivP

evv

eii

ivij

v

v

i


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Solutions to Problems with Power Evaluation When

Neither Voltage or Current Phasor is the Reference Phasor

( ) ( )

( ) ( )( )

( )

( )vi

iv

iviv

jjj

ivQ

ivQ

ivP

jiv

eiveeiviv

iv

iviv

=

=

=

==

sin

ofpartimaginarytheispowerreactiveThus

onlyvoltagetheofconjugatethemust takewe

Q,powerreactivefor thesigncorrectwant theweIf

cos

powertheaspartrealonly thetakeWe

sincos

product.in thecurrenttheofconjugateor thevoltagethe

ofconjugatethetakecircuit wein thepowertheevaluatetoThus

*

*


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8.12 Power Dissipated in Resistors With Complex

Currents and Voltages

In dc circuits the power dissipated in resistors was

R

vP

RiP

2

2

=

=

In ac circuits with complex phasor notation the

power dissipated in resistors is

R

vP

RiP

2

2

=

=

Rv i v = i R

8 13 P F i P S


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8.13 Power Factor in Power Systems The angle between the voltage and current ( in our previous

calculation) in a power network is called the power factor angle and isusually given the symbol .

The term cos( ) is called the power factor

Most power networks in factories and shops etc. have a lagging powerfactor. That is the load is inductive and resistive and is negative.

The charge for industrial electricity is based on the power consumed

and the maximum VA product (Maximum demand).

This is because

The cables and transformers required to supply a load are rated accordingto their maximum current (power lost in device conductors is dependenton current squared).

The bigger the rated VA the greater the cost of the equipment required tosupply it

Thus it is necessary to charge for maximum demand because otherwiseyou could have a high current demand with virtually no power and norevenue to cover your investment in expensive power supply equipment.

4-AC Circuit Analysis

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