AC Circuits [Compatibility Mode]

8/3/2019 AC Circuits [Compatibility Mode]

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AC Circuits

Electrical power is generated, transmitted and

distributed in the form of alternating current (AC).Contents

• Sinusoidal AC waveform parameters

– Peak, rms, period, frequency, phase

• Phasors

– Complex numbers, impedance, leading/lagging.

• Stead state AC circuit anal sis.• Power in AC circuits.

•





It is the equivalent DCi I

generates the same mean

amount of power loss P inv V R

R

a resistor. AC circuit Equiv. DC circuit

circuitDCinPower

But v=iRV v

T 221==

V R I VI

22P === thatshowcanYou

coset pv −= π

Instantaneous power

in AC circuit R RT

0isvoltagermsthe

pV

T T

circuitACinpowerMean=

T

dt vV 21

2

currentrmsSimilarily

∫ ∫ == vidt T pdt T 00

PT

0

2 p I =



Analysis of an RL circuit R L

i

(1) cos2cosLet t V t V v p ω ω ==

v

0

di LiRv =++−

or t ss iii +=let

(2) viRdt

L =+ We are normally interested in the

steady state solution i ss only.qua on s a rs or er

ordinary differential equation illsolution wstatesteadyThe

substitution.(3) )cos(2)cos( φ ω φ ω +=+= t I t I i p ss



Analysis of RL circuit continued

Substitute (1) and (3) in (2)

=−

The objective is to find I and φ.

BUT this is not easy!

V

iswork)of lota(afteranswerThe

+

− L

L R

ω

ω 22 )(

Phasors.

⎠⎝

−= R

an



Phasors

φ t

functionComplex functionReal

⇔ re )sin(ω +t r

ACsinusoidalhere rsent ttoPhasorUse jφ re

analysishesimplify ttocurrentsandvoltages





com lex numbers

y

Imaginary axisCartesian form

x+ 1

x,formPolar

r x

y x

θ r

3 22 xr +=Real axis

(4) tan

1

x

y−

=θ

(5) cosθ r x =(6)and(5)(2),(1),From

s nr y =θ θ θ sin jcos1 +=∠



Euler’s identity

?1isWhat θ ∠ω =+=

∴

Recos2

can writenowWe

t i i

θ θ θ sin jcos1

thatknowWe

+=∠{ }φ ω = +Re2 )t j(e I

t

seriesTaylorusecanWe { }φ ω ∠×∠=

=

11Re2 t I

thatproveto

exponential function instead of

sin and cos function!θ θ sin jcos1 +==∠ eθ

v

Note the use the bold symbols to

represent complex function

Euler’s identityi.e. i

ss

is the complex function

equivalent of i ss.



Circuit

R Li

t V v ω cos2=

equationaldifferentiorderfirsta

yescr esc rcue

:functionsRealusing

viRdi

L =+dt



continued

RL

i

t Ve ω j2=v

equationaldifferentiorderfirsta

yescr esc rcue

:functionsComplexusing

vii

=+d

Ldt



differential e uation

(4) ) j( j0 j Ve R L Ie =+ω φ

(1)2 ) j( Ve Rdt

L ωt φ +==+ vi I φ andunknowsThe

(3) 2Let ) j( φ ω += t Iess

tss

i Ve Ie φ

0 j j =

t t

t

Ied ω φ ω

φ ω

j)( j

)( j

)2(

(1)into(3)Substitute

=+

+

ω

Which is easier thandt

t t t Ve RIe LIe φ φ ω j) j() j( j =+ ++

using the real functions

sin and cos.

t t t

Vee RIee LIe

φ ω φ

ω

j j j j j

j =+



’Phasors

:slidepreviousfromRecall Z is known as impedance.

(4) ) j( j0 j Ve R L Ie =+ω φ The complex variables V,

I are called pahsors.

L j ,Let

0 j j

ω

φ

+===

RVe Ie

Z

VI Phasors are not functions of time!Im

aswritten

thenbecan(4)Equation I Stationary

arrow I

Im ω

(5) ZIV = Re

Phasor IRe

φ+ jω t arrow

generalized Ohm’s Law. φ

φ

∠== 1

j

IeI

timeof functionComplex) j( ω φ ω +∠== + t Ie t

I



1. Draw the hasor e uivalent circuit

i- replace v and i with their

equivalent phasors.v ω cos=- replace and L with their

equivalent impedances.

ea c rcu t. na yse e a res st ve c rcu t

but using complex numbers. R

jω Lφ ∠=I+=∠=

L R ω φ

jI

0∠=V V⎟ ⎠

⎜⎝

⎟ ⎠

⎜⎝

−∠+

= −

R L R ω

1

22tan

)(

⎟ ⎠⎜⎝ −=+=−

R L R I

ω

φ ω

1

22 tan,)(



Impedance of R

Time domain Phasor Domain

i

R

t j2Re

)cos(2

ω

φ ω

e

t I i

I=

+=I

z Rφ ∠== IeI

φ == j

v

)cos(2 φ ω +== t IRiRv R R ==V

z

)cos(2 ω += t V

j t jφ ω

{ }t j2Re ω eV=



hase relationshi

v V

m

it Re

v and i are in phase




)cos(2 φ ω t I i += φ

φ

∠== I Ie

jI L z L

{ }t j

j j

2Re

2Re

ω

φ

e

e Ie t

I=

=

⎞⎛ ==⎟ ⎠ ⎞

⎜⎝ ⎛ +

2 j π

π φ

v V

+−== )sin(2 φ ω ω t LI di

Lv ⎠⎝ 2

== 2 j

ω π

Le L

Vz

⎟

⎞

⎜

⎛

++= cos2

π

φ ω ω t LI ⎟

⎞

⎜

⎛ +=2

sin j2

cos π π

ω L

==

X L is

calledt j2Re ev V= L L X L j j === ω V

z

reactance



hase relationshi

v 2

π

V

m

i t Re

i lags v




)cos(2 φ ω t I i += φ

φ

∠== I Ie

jI

C zC

{ }t j

j j

2Re

2Re

ω

φ

e

e Ie t

I=

= ⎞⎛ −∠==

⎟ ⎠ ⎞

⎜⎝ ⎛ − 11 2

j π π

φ

I IeV

v V

= dt dvC i 2ω ω == 1 2

j

ω

π

eC

-

LI

Vz

=-1=( )+== ∫ sin

21φ ω

ω t

C idt

C v

⎟ ⎞

⎜⎛

⎟ ⎞

⎜⎛ −+⎟

⎞⎜⎛ =

2sin j

2cos

1

π π

ω C ⎟⎜

⎝ −+=

2cos

C

π φ ω

ω t

C C X j1 j

−==−

==V

z

2Re ev V=



hase relationshi

v 2

π VI

m

t Re

i leads v



,=

I

VV and I are in hase

Inductance z L= jω L jω L

I lags V V

Capacitance 11=

C

z

II leads V



Capacitive loads

• Their equivalentExample

mpe ance as a negat ve

imaginary part of the form2Ω j1ΩI

- , w ere >• They can be made of a

- j6ΩV

combination of R, L and

C !

Equivalent impedance

= - = -• Current leads the voltage.

V10Let =V

)2.68(86.1 j52

0

+∠=−== ZI



Inductive loads

• Their equivalentExample

mpe ance as a pos t ve

imaginary part of the form1Ω j5ΩI

+ , w ere >• They can be made of a

- j2ΩV

combination of R, L and

C !

Equivalent impedance

Z=1+j5-j2=1+j3• Current lags the voltage.

V10Let =V

)6.71(16.3 j31

0

−∠=+== ZI



Complex number arithmetic

• Before we can analyse AC circuits usingphasors we need to review complex number

arithmetic.

– Conversion from polar to Cartesian forms and.

– Addition, subtraction, multiplication and

.



from Cartesian to Polar forms

y

Imaginary axisθ j

x,PolartoCartesian

r x

y x

θ 22 y xr +=

Real axis tan 1

x

y−=θ

Complex number z cos

θ r x =

s nr y =



Complex Number Summation

Let ⎟ ⎞⎜⎛ ⎟ ⎞⎜⎛ ∠+= − 1122 tan y xz

22

j

222

11111

2

1

j

j

θ

θ

θ ∠==+=

∠==+=

r er y x

r er y x

2

1

z

z

⎞⎛ ⎞⎛

⎝

− 2122

1

y

x

Summation⎝ ⎠⎝ 2

22

x2

)( j)(

) j() j(

2121

2211

y y x x

y x y x

+++=

+++=+21

zz

formPolar

⎟⎟ ⎠⎜

⎜⎝

⎟ ⎠

⎜⎝ +

∠+++=+ −

21

2112

21

2

21

tan)()( x x

y y x x21

zz



Complex Number Subtraction

Let

⎟ ⎞⎜⎛ ⎟ ⎞⎜⎛ ∠+= − 1122 tan y xz

22

j

222

11111

2

1

j

j

θ

θ

θ ∠==+=

∠==+=

r er y x

r er y x

2

1

z

z

⎞⎛ ⎞⎛

⎝

− 2122

1

y

x

nSubtractio⎝ ⎠⎝ 2

22

x2

)( j)(

) j() j(

2121

2211

y y x x

y x y x

−+−=

+−+=−21

zz

formPolar

⎟⎟

⎜⎜⎝

⎟ ⎠

⎜⎝ −

−∠−+−=− −

21

2112

21

2

21

tan)()( x x

y y x x21

zz



Complex Number Multiplication

Let

⎟ ⎞⎜⎛ ⎟ ⎞⎜⎛ ∠+= − 1122 tan y xz

22

j

222

11111

2

1

j

j

θ

θ

θ ∠==+=

∠==+=

r er y x

r er y x

2

1

z

z

⎞⎛ ⎞⎛

⎝

− 2122

1

y

x

tionMultiplica⎝ ⎠⎝ 2

22

x2

)( j)(

) j() j(

12212121

2211

x y x y y y x x

y x y x

++−=

+×+=×21

zz

formPolar

θ θ

)( 2121

)( j

21

221121

21 θ θ θ θ +∠==

×=×=×+ r r er r

r r er er 21

zz



Complex Number Division

11 j

ormartes an

y x +=1

z

22

rdenominatotheof conjugateby theMultiply

j y x +2

z

2

2

2

2

12212

2

2

2

2121

22

22

22

11 j j j

j j

y x x y x y

y x y y x x

y x y x

y x y x

+−+++=−−×++=2

1

z

z

formPolar

j 1θ

22

j

22 θ

θ ∠==

r er 2

z

)( 21

2

1

2

1 21 θ θ −∠== −

r

e

r



Example – AC Circuit Analysis

10Ω 1mH 3mH

( )t v 502cos141 π = 2μF 5Ω



–Phasor dia rams

Find V

( ) A302 −∠

?

Source

V

2Ω j1.5ΩloadInductive



Summary

• Most electric power is AC, which is cheaper and.

• The transformer is a device that increases or.

• Stepping-up AC voltage steps-down the AC

losses.

cos functions. Phasors offer an easier alternative.



Power in AC Circuits

• Instantaneous power in R, L and C.• Active and Reactive power.

.

• Power factor correction.



Power in a urel resistive load

t V v ω cos2

Let

= v R

0 t

ii

vt V v

i == ω cos2

IRV

t I

=

= ω cos2

0 t

p

t VI iv ω 2cos2powerousInstantane

==)2cos1( t VI p ω +=

oweractiveorAvera e P is called active power measured in1

P0

pdt T

T

= ∫ Watts [W] because it corresponds to

heat/mechanical energy output from a

[W] 2

2

R I V VI ===

motor or ue use y a generator.



Lt V v = ω cos2

Letv

φ

ivt

L

V vdt

Li ==

∫ ω

ω sin

21 i

0 t

I V

t I

==

=

ω

ω sin2 p

t PowerousInstantaneElectrical reactive ower measured in

t VI ω 2sin =

==

[VAR] used to magnetise the inductor is

recovered when it is demagnetised. On

V

t L ω 2sin

2

=

=

orAverage powerReactive

average no energy is lost.

t

X

L

L

ω 2sinQ = 0Ppoweractive

=[VAR] Q

2

2

L

L L X

V X I ==





Average or Active Power P in an



Average or Active Power P in an

n uc ve oa2P =

+

∠=−∠=

L

V I

ω

φ

j

0I

R

jω L−∠=I

⎟ ⎞

⎜⎛

⎟ ⎞

⎜⎛ −∠

+= − LV ω

ω

1

22tan 0∠=V V

⎟ ⎞⎜⎛ == − LV I ω φ 1

22tan, Phasor circuit

P IV = ω L

( )2 L R ω +

Lω + R

φ

[W] cosP φ VI =

( )22cos L R ω φ +=

R i Q i i d i



Reactive power Q in an inductive

load22

∠

=−∠=

V

I φ

0

I

L ==

jω L−∠=I

⎟ ⎞

⎜⎛

⎟ ⎞

⎜⎛ −∠= − LV ω 1tan 0∠=V V

⎞⎛ ==

+

− LV ω

ω

1tan,Phasor circuit

+ Lω )(

Lω ω L

( )2 L R ω +

( )22 L R ω += R

φ

[VAR] sinQ φ VI = ( )22sin L R

L

ω

ω φ +=

Active and Reactive Power in a



Active and Reactive Power in a

capacitive load

R C ω jCan show:

φ ∠=Ipower(average)Active

2 == =

Phasor circuit

powerReactive

Q2

2

ω C

I X I C −=−=

[VAR] sin φ VI −=



Trian le

[VA] jQP

aspowercomplexDefine

+=S

RecallQ

S

ee o

remember that

[VAR] sinQ

[W] cosP

φ

φ

VI

V I

=

=P

for a capacitiveload

22

PS

powerApparant

+=== VI S

Power triangle

anglefactorPowerfactorPower

Ptan 1-=φ

φ cosSPF ==

A P d P F



A arent Power and Power Factor

PowerActive powerApparant factorPower

[W] cosP φ VI = [VA] S VI =φ cos

S

PPF ==

PPF=cosφ PF angle φ Element

(Unity)

es s ance

R

cos<

(lagging)

n uct ve oa

R+ j X 0

2≤≤− φ

π

VI cosφ <1

(leading)

Capacitive load

R- j X 20

π φ ≤≤

Power factors near 1 are good.Small power factors are poor.





–factor: ca acitive load case.

Find V

I

?

Source

V

0.2Ω j1ΩV010

loadCapacitive

∠

50 W

PF=0.866

ea ng



–factor: inductive load case.

Find V

I

?

Source

V

0.2Ω j1ΩV010

loadInductive

∠50 W

PF=0.866

agg ng

Example summary: Effect of type



Example summary: Effect of type

of load on transmission lossesan an vo tage regu at on

load

factor

line loss

W

Voltage V

V

regulation

100xΔV/V

%Resistive 1 5W 21%o4.241.12 ∠

o.lagging

. ..

o.

leading

. -.



Effect of Power Factor Summary

• “Poor” power factor increases transmissionne osses.

• Extra line losses are caused by flow of

reactive power.•

higher generator voltage

lower generator voltage.



Generator voltage control

flux φ / 2

co

N

dt

dN E

φ

−= ω i

Can control generator S

i to control the flux φ.

Increasing i increase

field losses.



roblem?

• Utility charges for active power P, and customer

pay per , ecause s proport ona to ue

used to generate the electricity.

• But poor power factor increases transmissionlosses and field losses.

• Customers with poor power factors (less that 0.85)

have to a a enalt or correct the ower factor.





Loads and generatorsi

i

vv

load Generator

powerabsorbsloadA powersuppliesgeneratorA

positiveispower

0>= vi p

negativeispower

<= vi p



Reactive Power

ii

VI

V

X sin

sgeneratescapac tor

22

−=−=−=V

VARsabsorbsinductorAn

2

2C

1= L

X L L L

ω =C ω



Ca acitor and Inductor i

iC

i LWhen the inductor power p L

is positive, the capacitorpower pC is negative, i.e. the

capacitor provides part of thev

iC

magnet s ng current o t e

inductor.

pC

i L

e power rom t e supp y s

less than the power consumed

p L.

In fact when the currents

= p

C L

the supply p=0.



Power factor correction example

An inductive load with a power factor of 0.5 draws 80 kVA at

400 V 50 Hz. Determine the ower and reactive ower and draw the power triangle. If a power factor correction capacitor

of 1mF is connected across the load, determined the new

power factor.



Summary

• Instantaneous AC power varies sinusoidally atou e t e requency.

• The average power in a resistance P=VI=I 2 R.

• The average power in a capacitance of inductanceis zero.

• An inductors absorbs reactive power Q= I 2ω L.

• A Capacitor generates reactive power Q=-I 2(ω C )-1

• Average power in a load is in general given byP=VI cosφ [Watt].



Summary continued• Reactive power in a load is in general given by

Q=VI sinφ [VAR].• Q is positive for an inductor and negative for a

capacitor.

• Apparent power S=VI , complex powerS

=P+jQ.• Poor ower factor cos increases transmission

losses and voltage regulation.

• Power factor can be corrected b connectin acapacitor across a load to generate VARs locally.

Date post:	06-Apr-2018
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AC Circuits [Compatibility Mode]

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