Chapter 7 Sinusoidal Steady-State Analysis

8/3/2019 Chapter 7 Sinusoidal Steady-State Analysis

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Sinusoidal SteadySinusoidal Steady--StateState

AnalysisAnalysis

Chapter 7

Artemio P. MagaboArtemio P. MagaboProfessor of Electrical EngineeringProfessor of Electrical Engineering

Department of Electrical and Electronics EngineeringUniversity of the Philippines - Diliman

Department of Electrical and Electronics Engineering

The Sinusoidal Function

T

)tcos(F)t(f m +=where

Fm = amplitude or peak value = angular frequency, rad/sec = phase angle at t=0, rad

The sinusoid is described by the expression

t, rad

-Fm

Fm

2


The sinusoid is generally plotted in terms oft,expressed either in radians or degrees. Considerthe plot of the sinusoidal function f(t)=Fmcos t.

-Fm

Fm

t, deg180 360

t, rad 2

T

When t=2, t=T. Thus we get T=2 or = .The function may also be written as

T2

tT

2cosFtcosF)t(f mm

==


The peak value of the voltage is Vm=311 volts. Theangular frequency is =377 rad/sec. The frequencyis f=60 Hz. The period is T=16.67 msec.

Define: The frequency of the sinusoid

T

1f= sec-1 or cycles/sec or Hertz (Hz)

Then, the sinusoid may also be expressed as

ft2cosFtcosF)t(f mm ==

Note:: The nominal voltage in the Philippines is asinusoid described by

V)t377cos(311)t(v +=


Leading and Lagging Sinusoids

Note: We say either f11(t) leads f2(t) by an angleof 30 or that f2(t) lags f1(t) by an angle of 30.

)30tcos(F)t(f m2 +=

Consider the plot of the sinusoidal functions

and )60tcos(F)t(f m1 +=

30

60

90 180-30

-60

-90t, deg

270 360

)t(f1

)t(f2


Note: The current is in phase with the voltage.

From Ohms law, we get tcosRIRiv mRR ==

Consider a resistor. Let thecurrent be described by

tcosIi mR =

R

+ -vR

iR

90 180-90t, deg

270 360

vRiR

The Resistor


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The Inductor

Consider an inductor. Let thecurrent be described by

tcosIi mL =

LiL

+ -vL

From vL= , we get tsinLIv mL =dt

diL L

90 180-90t, deg

270 360

iL

vL

Note: The current lags the voltage by 90o.


The Capacitor

iC

vC+ -

CConsider a capacitor. Let thecurrent be described by

tcosIi mC =

From vC= , we get tsinC

Iv mC

= dtiC1

C

90 180-90t, deg

270 360

iC

Note: The current leads the voltage by 90o.

vC


Note: We will show later that:

Summary:

In a resistor, iR and vR are in phase.1.

In an inductor, iL lags vL by 90o. In a capacitor,

iC leads vC by 90o (ELI the ICE man).

2.

For an RL network, the current lags thevoltage by an angle between 0 and 90.

1.

For an RC network, the current leads thevoltage by an angle between 0 and 90.

2.

For an RLC network, either 1 or 2 will hold.3.


Consider a DC (constant) current I and an AC(sinusoidal) current i(t)=Imcos t.

The sinusoidal current i(t) is said to be as effectiveas the constant current I if i(t) dissipates the sameaverage power in the same resistor R.

Since the current I is constant, then PAV,DC = I2R.

Consider R with the DC current I.

The power dissipated by R isRIP 2=

R

I

Effective Value of a Sinusoid


The average value of any sinusoidal function canbe shown to be equal to zero. Thus

RIP2

m21

AC,AV =

tcosRIRi)t(p 22

m2 ==

R

i(t)

Consider next R with the ACcurrent i(t). The instantaneouspower dissipated by R is

Simplifying, we get

+=

2

t2cos1RI)t(p

2

m

t2cosRIRI2

m212

m21 +=


Note: The same definition applies to a sinusoidalvoltage v(t)=Vm cos t.

Equating average power, we get

RIRI2

m212 =

orm

m I707.02

II =

Definition: The effective value of a sinusoidalcurrent with an amplitude Im is equal to

2

II mEFF =


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The effective value of a periodic function is also

called the Root-Mean-Square (RMS) value.That is, given a periodic function f(t), we get

==T

0

2RMSEFF dt)t(f

T

1FF

The effective or RMS value of the voltage is

V220)311(707.0V ==

Note:: The nominal voltage in the Philippines is asinusoid described by

V)t377cos(311)t(v +=


From KVL, we get

tcosVRidt

diL m =+

Network with Sinusoidal Source

Consider the network shown.Let v(t)=Vm cos t where Vmand are constant. Find thesteady-state current i(t).

R

Lv(t)

+

-

i

tcosBtsinAdt

di+=

tsinBtcosAi +=Let


Substitution gives

tsinLAtcosLB

tsinRBtcosRAtcosVm

+

+=

Comparing coefficients, we get

LARB0

LBRAVm

=

+=

Solving simultaneously, we get

m222V

LR

RA

+= m222 VLR

LB

+

=and


Thus, the steady-state current is

tsinVLR

LtcosV

LR

Ri m222m222 +

+

+=

Trigonometric Identity:

)tcos(KtsinBtcosA =+where

22 BAK += andA

Btan 1=

+=

R

Ltantcos

LR

Vi 1

222

m

or


( )= tcosK

= cosKA = sinKBProof: Let and

Substitution gives

)sintsincost(cosKtsinBtcosA +=+

)sin(cosKBA 22222 +=+

From the definitions of A and B, we get

22 BAK +=or

= tanA

B

A

Btan 1=

andor


8

0.6Hv(t)

+

-

i

Example: Find the currenti(t) and the average powerdissipated by the resistor.

Assume v(t)=100cos 10t V.

+=

R

Ltantcos

LR

Vi 1

222

m

Earlier we got the steady-state current as

( )

( )

+=

8

6.010tant10cos

6.0108

100i 1

222

Substitution gives


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W400RIP2

RMSAV,R ==

The average power dissipated by R is

( ) A87.36t10cos10i o=

Simplifying, we get

A07.72

10IRMS ==

The RMS value of the current is


Definition: A complex number consists of a realpart and an imaginary part. For example, given

jbaA +=

A is a complex number with real part equal to aand an imaginary part equal to b. Note: j= .1

Example: The following complex numbers areexpressed in the rectangular-coordinate form.

3j5.0C = 25.4j6D +=

4j3A += 5.3j5.2B =

Algebra of Complex Numbers


The Complex Plane

Definition: The complex plane is a Cartesiancoordinate system where the abscissa is for realnumbers and the ordinate is for imaginary numbers.

ImaginaryAxis

Real Axis2 4 6 8-2-4-6-8

j4

j2

-j2

-j4

A=3+j4

D=0+j2

C=4+j0

B=2.5-j3.5F=-3-j3

E=-4+j3


Definition: In the polar-coordinate form, themagnitude and angle of the complex number isspecified.

Polar-Coordinate Form

Consider the complex number A=a+jb.

Imag

+ Reala

jb A

A

From the figure, we get

22 baA +=

abtan 1=

Thus,

=+= AjbaA


Trigonometric Form

Consider the complex number .=+= AjbaA

Imag

+ Reala

jb A

A

From the figure, we get

= cosAa

= sinAb

Thus, we can also write

)sinj(cosAA +=

For example, A=1036.87 can be expressed as

6j8)87.36sinj87.36(cos10A +=+=


Given A = a+jb and B = c+jd, then

( ) ( )dbjcaBA +=

Addition or Subtraction

For example, given A=8+j6 and B=4+j10

16j12)106(j)48(BA +=+++=+

4j4)106(j)48(BA =+=

Addition or subtraction of complex numbers canonly be done in the rectangular-coordinate form.

( ) ( )dbjcaBA +++=+


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Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, thenin the rectangular-coordinate form, we get

)jdc(jb)jdc(a +++=

Multiplication of complex numbers can be doneusing the rectangular-coordinate or polar form.

)jdc)(jba(AB ++=

)bcad(j)bdac(AB ++=

bdjjbcjadac 2+++=

Since j2=-1, the product is

Multiplication


Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, then

)B)(A(AB BA =

)(BAAB BA +=

The rule is multiply magnitude and add angles.We get

For example, given A=3+j4=553.13o andB=4+j3=536.87o

o2 902525j12j16j9j12

)3j4)(4j3(AB

==+++=

++=

orooo 9025)87.3613.53()5(5AB =+=


o13.5354j3*A ==

For example, given A=3+j4=553.13o andB=-4-j3=5-143.13o

o13.14353j4*B =+=

Definition: The conjugateof a complex numberA=a+jb=||||A||||A is definedas

AAjba*A ==a

Imag

Real

jb A

-jb A*

Conjugate of a Complex Number


Division

Division of complex numbers can be done using therectangular-coordinate or polar form.

Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, thenin the rectangular-coordinate form, we get

jdc

jba

B

A

++

=jdc

jdc

22 dc bdjbcjadac + ++=or

2222 dc

adbcj

dc

bdac

B

A

+

+++

=


Given A = a+jb =||||A||||A and B = c+jd =||||B||||B, then

B

A

B

A

B

A

=

)(B

A

B

ABA =

The rule is divide magnitude and subtract angles.We get

For example, given A=3+j4=553.13o andB=4-j3=5-36.87o

1j90187.365

13.535

B

A oo

o

==

=


Phasor Transformation

Define a transformation from the time domain tothe complex frequency domain such that

)tcos(F)t(f m +=

=2

F)j(F m

For example, given f1(t)=311 cos (377t+60o) volts

and F2(j)=1020o Amps

V60220)j(F o1 =

A)20tcos(14.14)t(f o2 +=


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From Ohms law, we get tcosRIRiv mRR ==


tcosIi mR =

R

+ -vR

iR

omR 0

2

I)j(I = omR 0

2

RI)j(V =

Transformation gives

and

=

R)j(I

)j(V

R

RDividing, we get

The Resistor



tcosIi mL =

LiL

+ -vL

From vL= , we get tsinLIv mL =dt

diL L


omL 0

2

I)j(I = omL 90

2

LI)j(V

=and

==

Lj90L

)j(I

)j(V o

L

LDividing, we get

The Inductor

=LImcos(t+90o)


iC

vC+ -


tcosIi mC =

From vC= , we get tsinC

Iv mC

= dtiC1

C

=

=

Cj

1

90C

1

)j(I

)j(Vo

C

CDividing, we get


and

om

C 02

I

)j(I =om

C 90C2

I

)j(V =

The Capacitor


Note:

(2) For an inductor, ZL = jL = jXL in

Impedance

)j(I

)j(VZ

=

Definition: The ratio of transformed voltage totransformed current is defined as impedance.

(1) For a resistor, Z(1) For a resistor, ZRR = R in

(3) For a capacitor, ZC = 1/jC = -jXC in

(4) XL and XC are the reactance of L and C,respectively.


Admittance

(3) For a capacitor, YC = jC = jBC in -1

)j(V

)j(I

Z

1Y

==

Definition: The ratio of transformed current totransformed voltage is defined as admittance.

(1) For a resistor, Y(1) For a resistor, YRR = 1/R in -1

(2) For an inductor, YL = 1/jL =-jBL in -1

Note:

(4) BL and BC are the susceptance of L andC, respectively.


2. All the methods of analysis developed forresistive networks (e.g. Mesh Analysis, NodalAnalysis, Superposition, Thevenins and NortonsTheorems) apply to the transformed network.

Summary

1. The equation describing any impedance isalgebraic; i.e. no integrals, no derivatives.

)j(IZ)j(V = (Ohms Law)

3. The phasor transformation was defined for acosine function. The magnitude is based on theRMS value. Other phasor transformations exist.


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Solution 1: Use network reduction to get the inputimpedance.

Transformed network

+= 12j6Z1 )j(I1

1Z

V(j)+

-2Z CZ

)j(I2

)j(I3 += 5j5Z2= 10jZC

10j5j5

)5j5(10j

ZZ

ZZZ

C2

C2eq +

+=

+=

=

= 105j5

50j50

+=+= 12j16ZZZ eq1in


Apply current division to get I2(j).

)87.36071.7(5j5

10j)j(I

ZZ

Z)j(I o1

C2

C2

=

+=

=

+==

87.3620042.141

12j16042.141

Z)j(V)j(I

oo

in

1

Solve for current I1(j).

A87.36071.7 =

45071.7

)87.36071.7)(9010(o

oo

=

A87.810.10 o=


)j(I)j(I)j(I 213 =

Use KCL to get I3(j).

oo 87.810.1087.3607.7 =

)9.9j41.1()24.4j66.5( =

A13.5307.766.5j24.4 o=+=

Inverse transform I1(j), I2(j), and I3(j).

A)36.87-(10tcos10)t(i o1 =A)81.87-(10tcos14.14)t(i o2 =

A)53.13(10tcos10)t(i o3 +=


Solution 2: Use meshanalysis.

)j(I1

1Z

V(j)+

-2Z CZ)j(I3

+= 12j6Z1+= 5j5Z2= 10jZC

)]j(I)j(I[Z)j(IZ)j(V 31211 +=mesh 1:

mesh 2: )](jI-)(j[IZ)j(IZ0 1323C +=

Substitution gives

)]j(I)j(I)[5j5()j(I)12j6(2.141 311 +++=

)]j(I)j(I)[5j5()j(I10j0 133 ++=


)j(I)5j5()j(I)17j11(2.141 31 ++=

Simplifying the equations, we get

(1)

)j(I)5j5()j(I)5j5(0 31 ++= (2)

)j(I1j)j(I901 11o ==

From (2), we get

)j(I45071.7

45071.7)j(I

5j5

5j5)j(I 1o

o

13

=+

=

Substitute in (1)

)j(jI)5j5()j(I)17j11(2.141 11 ++=


Solve for I1(j). We get

)j(I)12j16(2.141 1 +=

o1 87.3620

2.141

12j16

2.141)j(I

=

+=

or

A87.36071.7 o=

A1.53071.7 o=


)(jI)901()j(jI)j(I 1o

13 ==

)87.36071.7)(901( oo =


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Finally, I2(j) can be found using KCL.

)j(I)j(I)j(I 312 =

A87.810.1090.9j41.1 o==

oo 53.137.071-87.36071.7 =

)66.5j24.4()24.4j66.5( +=

Inverse transform I1(j), I2(j), and I3(j).

A)36.87-(10tcos10)t(i o1 =

A)81.87-(10tcos14.14)t(i o2 =

A)53.13(10tcos10)t(i o3 +=


Xi10

5 H5.0

vs

+

-

.01F is

Example: Given

vs=100cos10t voltsis=10cos(10t+30

o)amps. Find iX.

Transform the network

V071.70)j(V oS =

A30071.7)j(I oS =

==

= 10j)01.0)(10(j

1

Cj

1ZC

=== 5j)5.0)(10(jLZL


Transformed network 1Z

Vs(j)+

-2Z 3Z Is(j)

Ix(j)Z1=5+j5

Z2=10

Z3=-j10

)j(VX +

Solution 1: Nodal AnalysisREF

3

X

2

X

1

SXS

Z

)j(V

Z

)j(V

Z

)j(V)j(V)j(I

+

+

=

Substitution gives

5j5

71.70)j(V

10j

1

10

1

5j5

130071.7 X

o

+

++

+=


Evaluate the coefficient of VX(j)

2.0

1.0j1.050

5j51.0j1.0

5j5

5j5

5j5

1

=

++

=++

+

o

o

oo

451045071.7

071.70

5j5

071.70=

=+

Evaluate the constant term

Substitution giveso

Xo 4510)j(V2.030071.7 =

]451030071.7[)j(V oo2.0

1X +=

or


Solve for Ix(j).

A1583.610

)j(V)j(I oXX =

=

V1533.687.17j0.66 o==

Simplifying, we get

]07.7j07.754.3j12.6[5)(jVx ++=

Thus, using inverse transformation, we get

A)15-(10tcos66.9)t(i oX =


Solution 2: Superposition

=++

= 5j5

1j1

1j1

1j1

10j

Get the input impedance.

10j10

)10j(10

ZZ

ZZZ

32

32eq

=

+=

=+= 105j5ZZ 1in

Thus,

Consider the voltagesource alone.

1Z

Vs(j)

+

- 2Z 3Z

Ix1(j)Is1(j)


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A0071.710

071.70Z

)j(V)j(I oo

in

s1s ===

The source current is

A54.3j54.3455 o ==

Using current division, we get

)j(IZZ

Z)j(I 1s

32

31X +

=

)0071.7(4514.14

9010 oo

o

=


)j(V10j

1

10

1

5j5

1)j(I XS

++

+=

From KCL, we get

)(VX +

REF

Consider the current

source alone.

Is(j)

1Z

2Z 3Z

Ix2(j)

Substitution gives

)j(V2.030071.7 Xo =

V3036.35)j(V oX =or


A77.1j06.33054.3 o +==

Applying, superposition, we get

)j(I)j(I)j(I 2X1Xx +=

A1583.677.1j6.6 o==

Solving for the current, we get

10

3036.35

Z

)j(V)j(I

o

2

X2x

=

=

A77.1j06.354.3j54.3 ++=

A)15-(10tcos66.9)t(i oX =Thus,


1Z

Vs(j)+

-3Z Is(j)

I1(j)Vth(j)

+

-

Solution 3:Thevenins Theorem

For mesh 1,we get

)](jI)(j[IZ)j(IZ)j(V s1311s ++=

)](jI)(j[I10j)j(I)5j5()j(Vs11s

++=

Substitution gives

5j5

)j(I10j)j(V)j(I ss1

+=



5j5

24.61j36.35

+

=

Simplifying, we get

5j5

)3007.7(10j071.70)j(I

oo

1

+=

o

o

o

1051045071.7

6071.70=

=

The Thevenin voltage is

)]j(I)j(I[10j)j(V s1th +=

]3007.710510[10j oo +=


V156.13636.35j94.131 o==

=

+=

+= 10

5j5

)10j)(5j5(

ZZ

ZZZ

31

31th

Simplifying, we get

)54.3j12.666.9j59.2(10j)j(Vth +++=

Find the Theveninimpedance 1

Z

3Z

a

babth ZZ =


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A1583.620

156.136 oo

=

=

V156.136)j(V oth =

= 10Zth

The Thevenin equivalent networkZ

th

Vth(j)+

-10)j(IX

Finally, we put back the 10resistor and solve for thecurrent.

10Z

)j(V)j(I

th

thX +

=


Power Equations

Consider a voltage source, a current source or anetwork of passive elements (R, L and/or C). Let

i(t)=Im cos (t+ I) and v(t)= Vm cos (t+V).

VoltageSource

i(t)v(t)

+

-

CurrentSource

i(t) v(t)

+

-

PassiveNetwork

i(t) v(t)

+

-

Note: The current flows from positive to negativeterminal for the passive network.


)tcos()tcos(IVv(t)i(t)p IVmm ++==

The instantaneous power supplied by the voltageor current source or delivered to the passivenetwork is

Trigonometric Identities:

(1) =+ sinsincoscos)cos(

(3) +=+ sincoscossin)sin(

(4) )2cos1(cos212 +=

(2) += sinsincoscos)cos(


From (1) and (2), we get

)]cos()[cos(coscos21 ++=

The instantaneous power can be expressed as

)]cos()t2[cos(IVp IVIVmm21 +++=

)]()2t2cos[(IV IVImm21 ++=

)cos(IV IVmm21 +

Simplify using trigonometric identity (1). We get

)cos()t2[cos(IVp IVImm21 ++++====

)]cos()sin()t2sin( IVIVI ++++++++


Collecting common terms, we get

)]t(2cos1)[cos(IVp IIVmm21 ++=

)t(2sin)sin(IV IIVmm21 +

Using the RMS values of the voltage and current,we get

)]t(2cos1)[cos(VIp IIV ++=

)t(2sin)sin(VI IIV +

Note: The instantaneous power consists of aconstant term plus two sinusoidal components.


From Ohms law, we get

)tcos(RIRiv ImRR +==


)tcos(Ii ImR +=

R

+ -v

R

iR

The instantaneous power delivered to (dissipatedby) the resistor is

)]t(2cos1[RI)t(cosRIp I2m2

1I

22mR ++=+=

The Resistor

or )]t(2cos1[RIp I2

R ++=


12/24


The Inductor


)tcos(Ii ImL +=

LiL

+ -vL

From vL= , we get )tsin(LIv ImL +=dt

diL L

The instantaneous power delivered to the inductoris

)tcos()tsin(LIp II2mL ++=

)t(2sinLI I2m2

1 +=or )t(2sinXIp IL

2L +=


The Capacitor

iC

vC+ -


)tcos(Ii ImC +=

From vC= , we get )tsin(C

Iv I

mC += dtiC

1C

The instantaneous power delivered to the capacitoris

)tcos()tsin(C

Ip II

2m

C ++=

or )t(2sinXIp IC2

C +=


Real or Active Power

Definition: Real or active power is defined as theaverage value of the instantaneous power. It is thepower that is converted to useful work or heat.

Recall the instantaneous power supplied by asource or delivered to a passive network.

)]t(2cos1)[cos(VIp IIV ++=

)t(2sin)sin(VIIIV

+

)cos(VIP IV = in Watts

Since the average of any sinusoid is zero, the realor active power is


Recall the instantaneous power delivered to aresistor, inductor or capacitor.

)]t(2cos1[RIp I2

R ++=

)t(2sinXIp IL2

L +=

)t(2sinXIp IC2

C +=

Since the average of any sinusoid is zero, the realor active power delivered to R, L and C are

RIP 2R = in Watts

0PL =

0PC =


Consider a series LC circuit. Letthe current be described by

i=Im cos (t+I). The voltagesvL and vC can be shown to be

vC

+ -

CLi

+ -vL

ivL

vC

Reactive Power

)tsin(C

Iv I

mC +=

)tsin(LIv ImL +=

90 180-90t, deg

270 360

I


The energy stored in the magnetic and electricfields are

)t(cosLiW I22

L21

L +=

Plots of the energy are shown below.

i

WLWC

90 180-90t, deg

270 360

I

)t(sinCvW I22

C21

C +=


13/24


14/24


Complex Power

)(VI)I)(V(*IVS IVIV ===vv

)sin(jVI)cos(VI IVIV +=

Note: The complex power S is a complex numberwhose real and imaginary components are the realand reactive power, respectively.

Definition: The product of the phasor voltage andthe conjugate of the phasor current is defined asthe complex power S. Let and .III =

v

VVV =v

jQPS +=or


Recall the transformation from the time domainto the complex frequency domain defined by

)tcos(F)t(f m +=

= F)j(F

Note: Unlike complex numbers, a phasor quantityis a complex representation of a sinusoidal function.

The quantity F(j) is referred to as a phasor. Forsimplicity, we make a change in notation:

== F)j(FFv

Phasor Notation


Transform the network.

volts0220V o=v

== 154.1j)00306.0(377jZ 1L== 846.4j)012854.0(377jZ 2L

Example: In the circuit shown, v(t) = 311 cos377tvolts. Find the power and reactive power deliveredto the load.

v(t)

+

-

i(t)

0.5 3.06 mH

7.5

12.854 mHLoad


+

-

0.5 j1.154

7.5

j4.846Vv I

v

+

-

xVv

TransformedNetwork

Get the totalimpedance.

846.4j5.7154.1j5.0Zeq +++=

=+= 87.360.106j8 o

Find the current

A87.362287.3610

0220

Z

VI o

o

o

eq

=

==

v

v


Find the voltage across the load

)846.4j5.7(IVX +=vv

oo

32.87(8.929)87.3622( = volts0.445.196 o=

Find the complex power delivered to the load.

)36.87(22)445.196(IVjQP oo*XLL ==+vv

o87.328.4321 =

346,2j630,3 +=

Thus, PL=3,630 Watts and QL=2,346 Vars.


Example: Givenv(t)=100 cos 10t volts,find all Ps and Qs. v(t)

+

-

i(t)

8

0.6H

Transform the network

V071.70V o=v

=+= o87.36106j8Z

+

-

8

j6Vv I

v

A87.36071.787.3610

071.70

Z

VI o

o

o

=

==

v

v

Solve for the current


15/24


Power and Reactive Power delivered to R and L

watts400)8(071.7RIP 22R ===

arsv300)6(071.7XIQ 2L2

L ===

Power and reactive power supplied by the source

oIVs 87.36cos)071.7(71.70)cos(VIP ==

watts400=

Vars30087.36sin)071.7(71.70Q os ==

300j40087.36500 o +==

)87.36071.7)(071.70(*IVS oo ==vvor


Example: Given

v=200 cos 10tVolts. Find allreal power andreactive power.

i1+

-

R1=6 L1=1.2H

R2=5

L2=0.5H0.01F

i2 i3

v

The transformed network

V042.141V o=v

= 12jZ1L

= 5jZ2L

= 10jZC

Vv 1I

v

+

-

6 j12

5

j5-j10

2Iv

3Iv


In a previous example, we found

A87.36071.7I o1 =v

A87.8110I o2 =v

A13.53071.7I o3 =v

watts500)5(10RIP 22222R ===

Average power dissipated by the resistors

watts300)6(071.7RIP 212

11R ===

Reactive Power delivered to the capacitor

vars500)10(071.7XIQ 2C2

3C ===


Reactive Power delivered to the inductors

vars600)12(071.7XIQ 21L2

11L ===

arsv500)5(10XIQ 22L2

22L ===


ooIV 87.36)87.36(0 ===

o1S 87.36cos)071.7(42.141cosVIP ==

watts800=

vars600sinVIQ 1S ==


We can also use the complex power formula

*

1SS IVjQPvv

=+

600j80087.361000 o +==)87.3607.7)(042.141( oo =

Thus, PS=800 watts and QS=600 vars.

Note: Real and reactive power must always bebalanced. That is,

watts800PPP 2R1RS =+=

vars600QQQQ C2L1LS =++=


From a previous example, we found

V153.68V oX =v

A1583.6I oX =v

and

Example: Given

V071.70V oS =v

A30071.7I oS =v

Find all P and Q.

+= 5j5Z1= 10Z2= 10jZ3

XVv

+

SIv

SVv

REF

1Iv

XIv

CIv

1Z

+

-

2Z 3Z


16/24


We can also find and .1Iv

CIv

( )

11

XS1

Z

68.17j97.6571.70

Z

VVI

=

=

vv

v

o

o

45071.7

753.18

5j5

68.17j737.4

=++

=

A3059.2 o=

o

o

C

XC

9010

153.68

Z

VI

==

v

v

A7583.6 o=


Power and reactive power delivered to Z1

watts5.33)5(59.2RIP 21Z2

11Z ===

vars5.33XIQ 1Z2

11Z ==

watts5.466)10(83.6RIP 222

X2R ===

Power dissipated by the resistor R2

vars5.466XIQ C2

CC ==

Reactive power delivered to the capacitor


SVv

Power and reactive power supplied by

)3059.2)(71.70(IVjQP o*

1SVV ==+vv

watts5.158PV = vars5.91QV =

5.91j5.15830183 o ==

)30071.7)(153.68(IVjQP oo*

SXII ==+vv

Power and reactive power supplied by SIv

o45-95.824 = 5.341j5.341 =

watts5.341PI = vars5.341QI =


vars100,3)pf(costanPQ 11

11 ==

V0220V o1 =v

Example: Findthe power andpower factorof generator 2.Assume

Gen. 1

1Vv

1Iv

LIv

+

-2V

v

+

-

0.3+j0.4 0.2+j0.2

LVv

+

-

2Iv

Gen. 2Load

P1=5 kW PL=10 kW

pf1=0.85 lag pfL=0.8 lag

For generator 1,watts000,5P1 =


From the complex power formula, we get

1

11*

1V

jQPI vv +

=

08.14j73.22220

100,3j000,5+=

+=

08.14j73.22I1 =v

A79.3174.26 o=

Thus,

From KVL, we get the voltage at the load

11L I)4.0j3.0(VVvvv

+=


Substitution gives

)31.79-)(26.7413.535.0(VV oo1L =vv

o1 34.2137.13V =v

)86.4j45.12(220 +=

86.4j55.207 =

V34.16.207V oL =v

vars7,5000.8)tan(cos000,10Q -1L ==

At the load,

watts000,10PL =


17/24


o

o

34.16.207

87.36500,12

=

A24.37j31.47 =

o34.16.207

500,7j000,10

+

=

L

LL*

LVjQPI v

v

+=


A21.3821.60I oL =v

Thus,


1L2 IIIvvv

=

From KCL, we get the current supplied by Gen 2.

)08.14j73.22()24.37j31.47( =

16.23j58.24 =

A3.4377.33I o2 =v

Thus,

From KVL, we get the voltage of Generator 2

L22 VI)2.0j2.0(Vvvv

++=

LV)16.23j58.24)(2.0j2.0(v

++=


Simplifying, we get

86.4j55.20728.0j55.9V2 ++=v

V21.115.21758.4j1.217 o==

Applying the complex power formula,

*

2222 IVjQPvv

=+

)3.4377.33)(21.115.217( oo =

915,4j443,51.42334,7o

+==

lag74.0)P

Qcos(tanpf

2

212 ==

The power is 5,443 watts while the power factor is


Example: A small industrial shop has the followingconnected load:

Load L1: Induction motor 2 kW, 0.85 pf lagLoad L2: Electric Heater 3 kW, 1.0 pfLoad L3: Lighting Load 500 W, 0.9 pf lagLoad L4: Outlets 1 kW, 0.95 pf lag

The voltage across the load is 220 V RMS. Find thecurrent through each load and the total currentsupplied to the shop.

+

220VtI

v

1Iv

2Iv

3Iv

4Iv

L4L3L2L1

-


For load L1, P1=2,000 watts, pf1=0.85 lag

vars1,2400.85)(costan000,2Q -11 ==

A79.317.1063.5j09.9 o==

For load L2, P2=3,000 watts. Since pf2=1, thenQ2=0. Thus

A064.13220

0j000,3I o2 =

=

v

220

240,1j000,2

V

jQPI

*

1

111

=

= v

v



For load L3, P3=500 watts, pf3=0.90 lag

vars2420.9)(costan500Q -13 ==

A84.2552.21.1j27.2 o==220

242j500I3 =v

For load L4, P4=1,000 watts, pf4=0.95 lag

vars3290.95)tan(cos000,1Q -14 ==

220

329j1000I4

=

v

A19.1878.449.1j54.4 o==


18/24


From KCL, the total current is

4321t IIIIIvvvvv

+++=

A55.1566.3022.8j54.29 o==

)0j64.13()63.5j09.9( +=

)49.1j54.4()1.1j27.2( ++

or

watts500,6PPPPP 4321t =+++=

vars811,1QQQQQ 4321t =+++=

A22.8j54.29220

811,1j500,6It =

=

v


Phasor Diagrams

v(t)

+

-

i(t)

8

0.6H

vR +

-

+ -

vL

Consider the circuit shown.Let v=100 cos(10t+) V.

Phasor diagrams show graphically how KVL andKCL equations are satisfied in a given circuit.

The transformed network

V71.70V =v

= 8ZR= 6jZL

+

-

+

-

+ -

Vv

RVv

LVv

ZL

ZR

Iv


)8)](87.36(071.7[ZIV oRR ==vv

V)87.36(57.56 o=

V)13.53(43.42 o+=

)906)](87.36(071.7[ZIV ooLL ==vv

oT 87.3610

71.70

6j8

71.70

Z

VI

=+

==

v

v

Solve for the phasor current and voltages. We get

A)87.36(07.7 o=


Assume =0o. We get

V071.70V o=v

V36.87-56.57V oR =v

V53.1342.43V oL =v

A87.3607.7I o=v

The phasor diagramis shown.

Note:LR

VVVvvv

+=

Iv

is in phase with RVv

LVv

Iv

lags by 90o

RVv

V

v

LVv

Iv 36.87o

53.13o


RVv

Vv

LVv

Iv

36

.87o

53

.13o

Assume =60o. We get

V6071.70V o=v

V3.13256.57V oR =v

V3.131142.43V oL =v

A13.2307.7I o=v

The phasor diagramis shown.

Note: The entire phasordiagram was rotated byan angle of 60o.


Assume =120o. We get

V12071.70V o=v

V3.13856.57V oR =v

V3.131742.43V oL =v

A13.8307.7I o=v

The phasor diagram is shown.

Note: The entire phasordiagram was rotated byanother 60o.

RVv

Vv

LVv

Iv3

6.8

7o

53.1

3o

Note: The magnitude and phase displacementbetween the phasors is unchanged. The phasorsare rotating in the counterclockwise direction.


19/24



SSSS cosIVP =

W40087.36cos)071.7(71.70 o ==

vars30087.36sin)071.7(71.70Q oS ==

Power dissipated by the resistor

watts400)8()071.7(RIP 22R ===

vars300)6()071.7(XIQ 22L ===

Reactive power delivered to the inductor


The total impedance seen by the source

C21T Z//ZZZ +=

10j5j5

)5j5(10j12j6

++

++=

=+= 87.362012j16 o

Example:

V042.141V o=v

+= 5j5Z2

+= 12j6Z1 1Vv

2Vv

Vv

1Iv

+

-

6 j12

5

j5-j10

2Iv

3I

v

+ -

+

-

Note: Refer to a previous problem.


Source current

A87.3607.787.3620

042.141

Z

VI o

o

o

T

1 =

==

v

v

Apply current division to get

A87.810.10I o2 =v

A13.5307.7I o3 =v

)12j6)(87.3607.7(ZIV o111 +==vv

V56.2687.94 o=

Solve for the voltage1

Vv


)5j5)(87.8110(ZIV o222 +==vv

Solve for the voltage2V

v

V87.3671.70 o=

Phasor Diagram

Vv

1Vv

2Vv

21 VVVvvv

+=

Note:

1Iv

2Iv

3Iv

321 IIIvvv

+=


)87.36071.7)(042.141(oo

=600j80087.361000 o +==

Power supplied by the source

*

1SS IVjQPvv

=+

or

watts80087.36cos)071.7(42.141P oS ==

vars60087.36sin)071.7(42.141Q oS ==

vars500)10()07.7(XIQ 2C2

CC ===

Power delivered to ZC


We can also use the complex-power formula

*

1111 IVjQPvv

=+

)87.36071.7)(56.2687.94( oo =

600j30043.6382.670 o +==

Power delivered to Z1

watts300)6()071.7(RIP 212

11 ===

vars600)12()071.7(XIQ 212

11 ===


20/24


Power delivered to Z2

watts500)5()10(RIP 222

22 ===

vars500)5()10(XIQ 222

22 ===

or

*

2222 IVjQPvv

=+

)87.8110)(87.3671.70( oo =

500j500451.707 o +==

Note: It can be shown that power balance issatisfied.


Let volts, the reference phasoro

R 0120V =v

A0620

0120

20

VI o

oR

S =

==

v

v

Then

Example: A voltmeter reads these voltages for the

network shown below:RMSV220VS =

RMSV120VR =

RMSV150VC =CV

v

SIv

+

-

+

-SV

v

RVv

+ -

20

R C

a) Find R and C.

b) Find P and Q supplied by the source.


Note: (1) Since the network is capacitive, iS mustlead vS.

CRS VVVvvv

+=(2) From KVL,

Phasor DiagramRV

v

SIv

SVv

CV

v

Apply the cosine law

+= cos(220)(120)2-120220150 222

o25.40=763.0cos = orWe get


o12.31=86.0cos =We get or

Check: 14.142j92.47120VV CR +=+vv

14.142j92.167 =

So V25.40220

v

==

Apply the cosine law again

volts0120V oR =v

volts25.40220V oS =v

volts37.71150V oC =v

Thus


038.0j013.0 += CjR

1+=

o

o

o

C

S 37.7104.0

37.71150

06

V

I=

=v

v

The admittance for the parallel RC branch

o25.401320 =

853j1008 =

We get R=78.26 and C=0.04-1.

*

SSSS IVjQPvv

=+ )06)(25.40220( oo =

Power supplied by the source


Balanced Three Phase Voltages

Three sinusoidal voltages whose amplitudes areequal and whose phase angles are displaced by120o are three-phase balanced.

)tcos(V(t)v ma +=

The RMS value of the voltages is

mm V71.02

VV =

120)tcos(V(t)v mc ++=

120)tcos(V(t)v mb +=


21/24


Balanced Three Phase Voltages

Note: The synchronousgenerator is a three phasemachine that is designed togenerate balanced three-phase voltages.

Transforming to phasors, we get

Va = V Vb = V -120Vc = V +120

aVr

bVr

cVr


Balanced Three-Phase Currents

are three-phase balanced.

The currents )t(cosIi ma +=

)120tcos(Ii mb +=

)120tcos(Ii mc ++=

In phasor form, we get

= IIar

ob 120II =

r

oc 120II +=

r

aIr

bIr

cIr


Balanced Three-Phase SystemA balanced three-phase system consists of :

1. Balanced three-phase sinusoidal sources;

3. The connecting wires have equal impedances.

2. Balanced three-phase loads; and

a) Equal impedances per phase or

A balanced three-phase load has:

Note: The load may be connected in wye or delta.

b) Equal P and Q per phase


ExampleExample: Given Va=200/0

o volts, Vb=200/-120o

volts and Vc=200/120o volts. Find the phasor

currents Ia, Ib and Ic. Also, find P and Q supplied toZa, Zb and Zc.

n n'

aVr

bVr

cV

r

cIr

aIr

bIr

Za

Za=Zb=Zc=7+j5

Zb Zc

2Ir

1Ir

ZF

ZF

ZF=1+j1


Example

Mesh equations using loop currents I1 and I2.Va Vb = 2(Zf+ ZL)I1 (Zf+ ZL)I2Vb Vc = -(Zf+ ZL)I1 + 2(Zf+ ZL)I2

Substitution gives300 + j173.2 = (16 + j12)I1 (8 + j6)I2

- j346.4 = -(8 + j6)I1 + (16 + j12)I2

Solving simultaneously we getI1 = 20-36.87AI2 = 20-96.87 A


Example

Solving for currents Ia, Ib and Ic, we getIa = I1 = 20-36.87AIb = I2 I1 = 20-156.87AIc = -I2 = 2083.13A

Note: Currents Ia, Ib and Ic are balanced.

Power and Reactive Power supplied to load impedancesZa, Zb and Zc.

Pa = Pb = Pc = (20)2(7) = 2800 Watts

Qa = Qb = Qc = (20)2(5) = 2000 Var


22/24


Comments1. The sum of 3 balanced phasors is zero.2. If a neutral wire is connected between n and n, no

currents will flow through the neutral wire.

3. The nodes n and n are at the same potential.4. We can analyze the circuit using single-phase

analysis.

n n'neutral line

aVr

Za=7+j5

ZF=1+j1

aIr


CommentsUsing KVL, we get

Va = (Zf+ Za)Ia + Vnn

Since Vnn = 0, we get forphase a

Ia =Zf + Za

Va= 20-36.87A

8 + j6

2000=

Pa = (20)2(7) = 2800 Watts

Qa = (20)2(5) = 2000 Watts

Forphases b and c, we getIb = Ia-120= 20-156.87 AIc = Ia120 = 2083.13 A


Line-to-line and Phase Voltages

Consider a 3-phase wye-connected generator or a3-phase wye-connected load.

Van, Vbn and Vcn are line-to-neutral (or phase) voltagesVab, Vbc and Vca are line-to-line (or line) voltages

nnb

a

c


Line-to-line and Phase Voltages

From KVL, Vab = Van + Vnb = Van VbnVbc = Vbn + Vnc = Vbn VcnVca = Vcn + Vna = Vcn Van

If Van, Vbn and Vcnare balanced3-phase voltages,we get the phasordiagram shown.

Vca

VcnVab

Van

Vbn

Vbc


Comments1. The line-to-line voltages Vab, Vbc and Vca are also

balanced 3-phase voltages;

2. The magnitude of the line-to-line voltage issquare root of three times the magnitude of theline-to-neutral voltage; and

3. Vab leads Van by 30, Vbc leads Vbn by 30and Vcaleads Vcn by 30.

oanab 30V3V =

rr

obnbc 30V3V =

rr

o

cnca 30V3V =rr


-Y Conversion for GeneratorsGiven a balanced 3-phase delta connectedgenerator, what is its equivalent wye ?

a a

Note: The line-to-line voltages must be the same.

c

b

c

b

Vca Vab

Vbc

Vcn

Van

Vbn

If Vab = VL, then Van = (1/3) VL-30


23/24


-Y Conversion for Loads

Note: For equivalence, Zab, Zbc and Zca must bethe same for both networks.

If the impedance of the delta load is specified,convert the impedance to wye.

Y Z3

1Z =

a

c b

YZ

YZYZ

c

a

b

Z Z

Z


Power EquationsFor a single-phase systemPp = VpIp cos Qp = VpIp sin

VAp = VpIp

For a three-phase systemP3 = 3Pp = 3VpIp cos = 3VLIp cos Q3 = 3Qp = 3VpIp sin = 3VLIp sin

VA3 = 3VAp = 3VpIp = 3VLIpwhere

Vp = magnitude of voltage per phaseIp = magnitude of current per phase = angle of Vp minus angle of Ip


ExampleA balanced 3-phase load draws a total power of 75KW at 0.85 pf lag from a 440-volt line-to-linesupply. Find the current drawn by the load?

A115.8IP =

cosIV3P PL3 =

cos V3

PI

L

3

P =

From

we get

)(440)(0.853

75,000=

or Department of Electrical and Electronics Engineering

Example

For load 1, P1=75 kw

For load 2, P2=60 kw

kvars48.46)85.0tan(cosPQ 111 ==

kvars06.29)9.0tan(cosPQ 122 == Get P and Q drawn by combined load

kw135PPP 21T =+=kvars54.75QQQ 21T =+=

Another 3-phase load rated 60 KW at 0.9 pf lag isconnected in parallel with the load in the previousexample. Find the total current drawn.


Example

A203(440)3

154,700IP ==

The total volt-ampere is

kva7.154QPVA2

T

2

TT =+=

Since

PL3 IV3VA =

we get


Single-Phase Analysis of a BalancedThree-Phase System

A balanced three-phase system can bereplaced by a single-phase equivalent circuitprovided:

1. All generators are connected in wye; and2. All loads are connected in wye.

Note: To get the single-phase equivalent,draw the neutral line and isolate one phase.


24/24


ExampleFind the phasor current Is, I1 and I2 and the total power andreactive power supplied by the three-phase voltage source.

Source Load 1 Load 2

Van = 120 /0oV PT = 4.5 kW Z = 8 + j6

at 0.92 pf lag

c

b

a

IsI1 Z

ZZ

I2


ExampleSingle-Phase Equivalent Circuit

Van = 1200 V P1=1,500 W Z=8+j60.92 pf lag

For load 1, P1 = 1,500 wattsQ1 = P1tan(cos

-1 0.92) = 639 vars

A1.236.13 o=

Is I1 I2Van

I1 =1500 j369

120= 12.5 j5.3 A= 13.6-23.1 A

Z


ExampleFor load 2,

For the source,Is = I1 + I2 = 22.1 j12.5 A = 25.4 -29.5Ps + jQs = VanIs* = 120 (25.4 -29.5)

= 2652 + j1503For the three phase source

arsv509,4)503,1(3Q3 ==

watts956,7)652,2(3P3 ==

I2 =VanZ

120 0120 0=

= 12 -36.87 = 9.6 j7.2 A

Chapter 7 Sinusoidal Steady-State Analysis

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