Programa

1. Introdução aos circuitos eléctricos2. Grafos e circuitos resistivos lineares3. Circuitos dinâmicos lineares4. Regime forçado sinusoidal

Amplitudes complexas (Phasors)Diagramas vectoriaisPotência instantânea, média, eficaz, complexaFactor de potência

5. Análise no domínio da frequência complexa6. Circuitos resistivos não-lineares

STEADY-STATE POWER ANALYSISLEARNING GOALS

1. Instantaneous PowerFor the special case of steady state sinusoidal signals

2. Average PowerPower absorbed or supplied during one cycle

3. Maximum Average Power TransferWhen the circuit is in sinusoidal steady state

4. Effective or RMS ValuesFor the case of sinusoidal signals

5. Power FactorA measure of the angle between current and voltage phasors

Complex PowerMeasure of power using phasors

6. Power Factor CorrectionHow to improve power transfer to a load by “aligning” phasors

7. Single Phase Three-Wire CircuitsTypical distribution method for households and small loads

INSTANTANEOUS POWER

)cos()()cos()(

iM

vM

tItitVtvθωθω

+=+=

Statesteady In

)()()( titvtp =Impedance to

SuppliedPower ousInstantane

)cos()cos()( ivMM ttIVtp θωθω ++=

[ ])cos()cos(21coscos 212121 φφφφφφ ++−=

[ ])2cos()cos(2

)( ivivMM tIVtp θθωθθ +++−=

LEARNING EXAMPLE

)(),(302

),60cos(4)(

tptiZ

ttv

:Find

:AssumeΩ°∠=

°+= ω

)(302302604 A

ZVI °∠=

°∠°∠

==

))(30cos(2)( Atti °+= ω

°==°==

30,260,4

iM

vM

IV

θθ

)902cos(430cos4)( °++°= ttp ω

constant Twice thefrequency

AVERAGE POWER

For sinusoidal (and other periodic signals)we compute averages over one period

ωπ2

=T

[ ])2cos()cos(2

)( ivivMM tIVtp θθωθθ +++−=

)cos(2 iv

MM IVP θθ −=

If voltage and current are in phase

MMiv IVP21

=⇒=θθ

If voltage and current are in quadrature090 =⇒°±=− Piv θθ

Purelyresistive

Purely inductive orcapacitive

It does not matterwho leads

LEARNING EXAMPLE Find the averagepower absorbedby impedance

)(1553.34522

601022

6010 Aj

I °∠=°∠°∠

=+

°∠=

°=°=== 15,60,53.3,10 ivMM IV θθ

WP 5.12)45cos(3.35 =°=

∫+

=Tt

tdttp

TP

0

0

)(1−

+

RV

)(1506.7601022

2 Vj

VR °∠=°∠+

=

WP 53.306.721

×=

Since inductor does not absorb powerone can use voltages and currents acrossthe resistive part

LEARNING EXAMPLE Determine the average power absorbed by each resistor, the total average power absorbed and the average powersupplied by the source

)(45344512

1 AI °∠=°∠

=

WP 1831221

4 =×=Ω

If voltage and current are in phase

MMiv IVP21

=⇒=θθ 212

1MRI=

RVM

2

21

=

)(57.7136.537.265

451212

45122 A

jI °∠=

°−∠°∠

=−

°∠=

)(36.5221 2

2 WP ××=Ω W7.28=

Inductors and capacitors do not absorbpower in the average

WPtotal 7.2818+=

⇒= absorbedsupplied PP WP 7.46=supplied

Verification

°∠+°∠=+= 57.7136.545321 III)(10.6215.8 AI °∠=

)cos(2 iv

MM IVP θθ −=

)10.6245cos(15.81221

°−°××=suppliedP

LEARNING EXTENSION Find average power absorbed by each resistor

I

iZI °∠=

6012

iZ )4||4(2 j−+=

°−∠°−∠

=−−−

=−−

+=4524

6.713.2544

168844)4(42

jjj

jjZi

Ω°−∠= 6.2647.4iZ

)(6.8668.26.2647.4

6012 AI °∠=°−∠

°∠=

WRIP M 20.768.2221

21 22

2 =××==Ω

2I

°∠×°−∠

°−∠=

−−

= 6.8668.24524

90444

42 I

jjI

°∠= 6.4190.12I

)(90.1421 2

4 WP ××=Ω

LEARNING EXTENSION Find the AVERAGE power absorbed by each PASSIVE component and the total power supplied by the source

1I2I

°∠++

+= 3010

24324

1 jjI

)(62.4014.6301095.1528.757.2647.4

1 AI °∠=°∠°∠°∠

=

)(14.6321

21 22

3 WRIP M ××==Ω

°∠−°∠= 62.4014.630102I

)(05.1412.495.1528.7

30303010243

32

Aj

I

°∠=°∠

°∠=°∠

++=

)(12.4421 2

4 WP ××=Ω

)(02 WPj =Ω

Power supplied by source

Method 1. absorbedsupplied PP =

WPPP 50.9043 =+= ΩΩsupplied

Method 2: )cos(21

ivMM IVP θθ −=

−

+

sV

°∠== 62.4042.183 1IVs

)3062.40cos(1042.1821

°−°×××=P

LEARNING EXAMPLE Determine average power absorbed or supplied by each element

)(30623012

2 AI °∠=°∠

=

)(366221

21 22

2 WRIP M =××==Ω

01 =ΩjP

To determine power absorbed/suppliedby sources we need the currents I1, I2

)(19.3643.7

39.466639.101

0630123

A

jj

jj

I

°−∠=

−=−+

=°∠−°∠

=

°−∠=−=−++=+=

07.728.11)(39.12.1139.46320.5321 AjjjIII

)(54

)07.730cos(28.111221

3012

W

P

−=

°+°×××=°∠

1836 +=

)cos(21

ivMM IVP θθ −=

Power Average

WP 18)19.360cos(43.7621

06 =°+××=°∠

Passive sign convention

RVRIP M

M

22

21

21

==

resistorsFor

LEARNING EXTENSION Determine average power absorbed/supplied by eachelement

1I2I

)012(42304012

22

1

°∠++−=°∠°∠=

IIjI

Equations Loop

°−∠−+

=−

°∠−°∠=

57.2647.448246.3

24048304

2j

jI

)(20497.957.2647.443.17758.44

2 AI °∠=°−∠°∠

=

)(5.5492.19))(055.4108.912(4))(20497.912(4)(4 214

VVjVIIV

°−∠=−−=°∠+=+=Ω

WR

VP M 6.49492.19

21

21 22

4 =×==Ω

)(02 WP j =Ω−

)(4.69)05.54cos(1292.1921

012 WP −=°−°−×××−=°∠Ω4V

)(8.19)20430cos()97.9(421

304 WP =°−°××−=°∠

Check: Power supplied =power absorbed

jVI

jVV

2304

02

3044

012

42

44

−−°∠

=

=−

°∠−++°∠−

Ω

ΩΩ

Equations NodeAlternative Procedure

)cos(21

ivMM IVP θθ −=

PowerAverage

RVRIP M

M

22

21

21

==

resistorsFor

LEARNING EXTENSION Determine average power absorbed/supplied by eachelement

V

042

0122

024=+

°∠−+

°∠− Vj

VVEquationNode

4j×

0)12(2)24(2 =+−+− jVVVj

)(85.177.14)(125.788.14

1324192

3232

324824

4824)32(

VjV

jjj

jjV

jVj

+=°∠=

+=

−−

×++

=

+=+

1I

925.062.42

85.177.14242024

1 jjVI −=−−

=−°∠

=

)(32.1171.41 AI °−∠=

2Ijj

jj

jVI

−−

×+−

=−°∠

=2

85.177.14122

0122

)(73.12367.1)(385.1925.02

77.285.12 AAjjI °∠=+−=

+−=

)(18.2271.4221 2

2 WP =××=Ω

)(67.27488.14

21 2

4 WP =×=Ω

)(565.5)73.1230cos(67.11221

012 WP =°−°××−=°∠

)(42.55)32.110cos(71.42421

024 WP −=°+°×××−=°∠

)(42.55)(565.567.2718.22

WPWP

=++=

supplied

absorbed

:Check

)cos(21

ivMM IVP θθ −=

PowerAverage

RVRIP M

M

22

21

21

==

resistorsFor

MAXIMUM AVERAGE POWER TRANSFER

)cos(||||21

)cos(21

LL

LL

IVLL

IVLMLML

IV

IVP

θθ

θθ

−=

−=

THTHTH jXRZ +=

LLL jXRZ +=

L

LL

OCTHL

LL

ZVI

VZZ

ZV

=

+= |||| OC

THL

LL V

ZZZV+

=⇒

LIV

LLL

ZZVI

LL∠=−⇒∠−∠=∠⇒

θθ

LLL jXRZ +=

22)cos(

)tan(

LL

LIV

L

LL

XRR

RXZ

LL +=−∴

=∠⇒

θθθ

θ2tan1

1cos+

=

||||||

L

LL Z

VI =⇒

222

2

||||||

21

LL

L

THL

OCLL

XRR

ZZVZP

++=

22

2

)()(||

21

THLTHL

LOCL XXRR

RVP+++

=

⎩⎨⎧

=−=

⇒

⎪⎪⎭

⎪⎪⎬

⎫

=∂∂

=∂∂

THL

THL

L

L

L

L

RRXX

RPXP

0

0

*TH

optL ZZ =∴

⎟⎟⎠

⎞⎜⎜⎝

⎛=

TH

OCL R

VP4

||21 2

max

222 )()(||

)()(

THLTHLTHL

THLTHLTHL

XXRRZZ

XXjRRZZ

+++=+

+++=+

LEARNING EXAMPLEload the to suppliedpower average maximum the Compute

transfer.power averagemaximumfor Find LZ

Remove the load and determine the Thevenin equivalent of remaining circuit

1I °−∠=°∠

°∠=°∠

+×= 64.926.5

64.908.603204

1624

jVOC

Ω°∠=°∠°∠

=++

=

Ω+

=−+

=++

=+=

93.1647.164.908.657.2694.8

1648

371652

37)16)(48(

1648)12(||4

jj

jjjjjjZTH

Ω−=°−∠= 43.041.193.1647.1* jZL

⎟⎟⎠

⎞⎜⎜⎝

⎛=

TH

OCL R

VP4

||21 2

max*TH

optL ZZ =∴

)(45.241.14

26.521 2

max WPL =×

×=We are asked for the value of thepower. We need the Thevenin voltage

LEARNING EXAMPLEload the to suppliedpower average maximum the Compute

transfer.power averagemaximumfor Find LZ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

TH

OCL R

VP4

||21 2

max*TH

optL ZZ =∴

Circuit with dependent sources!

SC

OCTH I

VZ =

1'

1'

2

)42(04

IV

IjV

X

x

−=

++−=°∠

KVL

)(45707.04524

04)4524()44(04

1

11

AI

IIj

°−∠=°∠

°∠=

°∠=+=°∠KVL

°−∠=−−=−−=°∠−= 5.1611013411042 1 jjIVOC

Next: the short circuit current ...

LEARNING EXAMPLE (continued)...

Original circuit

02)(20404)(24"

=−−+°∠=∠°−−++−

SCSC

SCx

IjIIIIIjV

CURRENT CIRCUITSHORTFOREQUATIONSLOOP

)(2" IIV SCx −=

VARIABLEGCONTROLLIN

4)22(244)44(−=−+−

=−+

SC

SC

IjIIIj

Substitute and rearrange

2)11( +−=⇒ SCIjI

[ ] 442)1()1(4 =−+−+ SCSC IIjj

°−∠=−−= 57.1165)(21 AjISC

°−∠=−−=−−=°∠−= 57.1611013411042 1 jjIVOC

Ω−=°−∠= 11452 jZTH Ω+=⇒ 11 jZ optL

⎟⎟⎠

⎞⎜⎜⎝

⎛=

TH

OCL R

VP4

||21 2

max*TH

optL ZZ =∴

)(25.14

)10(21 2

max WPL =×=

LEARNING EXTENSION

load the to suppliedpower average maximum the Computetransfer.power averagemaximumfor Find LZ

−

+

OCV

I

°−∠=−=−

=

+=°∠

4573.12)1(98

)22(36)22(036

jjI

Ij

186)1(9212

2012

jjjIjVOC

+=−×+−=

+°∠−=

)(57.71974.18 VVOC °∠=

Ω+

+−=+−=22

42)2||2(2jjjjjZTH

)(18

8822

4Ω−=

−=

+= jj

jZTH

)(1 Ω+= jZ optL

)(454

36021max WPL =×=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

TH

OCL R

VP4

||21 2

max*TH

optL ZZ =∴

360186|| 222 =+=OCV

LEARNING EXTENSION

load the to suppliedpower average maximum the Computetransfer.power averagemaximumfor Find LZ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

TH

OCL R

VP4

||21 2

max*TH

optL ZZ =∴

−

+

OCV

2jV

=°∠+−

= 024222

22 jj

jV j °∠9024

KVL

)(24129024012 VjVOC +=°∠+°∠=

7202412|| 222 =+=OCV

THZ

222)22(2)22(||2

jjjjjjZTH −+

−=−=

)(22 Ω+= jZTH

)(22 Ω−= jZ optL

)(4524

72021max WPL =

××=

EFFECTIVE OR RMS VALUES

)(ti

R

Rtitp )()( 2=

powerousInstantane

⎟⎟⎠

⎞⎜⎜⎝

⎛== ∫∫

++ Tt

t

Tt

tav dtti

TRdttp

TP

T0

0

0

0

)(1)(1 2

periodwith periodiciscurrent If

2

)(

dcdc

dc

RIP

Iti

=

= then)( DCiscurrent If dcaveff PPI =:

∫+

=Tt

teff dtti

TI

0

0

)(1 22 ∫+

=Tt

teff dtti

TI

0

0

)(1 2

The effective value is the equivalent DCvalue that supplies the same average power

Definition is valid for ANY periodicsignal with period T

2

)cos()(

Meff

M

XX

tXtx

=

+=is valueeffective the

signalsinusoidalaFor θω

If the current is sinusoidal the averagepower is known to be RIP Mav

2

21

=

22

21

Meff II =∴

)cos(21

ivMMav IVP θθ −= case sinusoidalFor

)cos( iveffeffav IVP θθ −=

square)mean(root rmseffective ≈

Nota: rede 230 V (AC) => Vm aprox 325 V

LEARNING EXAMPLE Compute the rms value of the voltage waveform

One period

⎪⎩

⎪⎨

⎧

≤<−−≤<≤<

=32)2(4

210104

)(tt

ttt

tv

3=T

∫∫∫ −+=3

2

21

0

2

0

2 ))2(4()4()( dttdttdttvT

The two integrals have the same value

332

3162)(

1

0

33

0

2 =⎥⎦⎤

⎢⎣⎡×=∫ tdttv

)(89.1332

31 VVrms =×=

∫+

=Tt

trms dttx

TX

0

0

)(1 2

)(ti

R

LEARNING EXAMPLE Compute the rms value of the voltage waveform and use it todetermine the average power supplied to the resistor

Ω= 2R

)(4 sT =

40;16)(2 <≤= tti ∫+

=Tt

trms dttx

TX

0

0

)(1 2

)(4 AIrms =

)(322 WRIP rmsav ==

LEARNING EXTENSION Compute rms value of the voltage waveform

tv 2=

4=T∫+

=Tt

trms dttx

TX

0

0

)(1 2

dttVrms ∫=2

0

2)2(41 )(

38

31 2

0

3 Vt =⎥⎦⎤

⎢⎣⎡=

)(ti

R

LEARNING EXTENSION

Ω= 4R

Compute the rms value for the current waveforms and usethem to determine average power supplied to the resistor

6=T

∫+

=Tt

trms dttx

TX

0

0

)(1 2

RIP rmsav2=

=⎥⎦

⎤⎢⎣

⎡++= ∫ ∫∫

4

2

6

4

2

0

2 416461 dtdtdtIrms 8

68328=

++ )(3248 WP =×=

8=T

8161681 6

4

2

0

2 =⎥⎦

⎤⎢⎣

⎡+= ∫∫ dtdtIrms )(32 WP =

THE POWER FACTOR

LZiMI θ∠

−∠

+

vMV θ

iθvθ

VI

izv

IZVZIVθθθ +=∠+∠=∠⇒=

zθ

)cos()cos(21

ivrmsrmsivMM IVIVP θθθθ −=−= rmsrms IVP =apparent

zivPPpf θθθ cos)cos( =−==

apparent

inductive pureinductiveor lagging

resistive

capacitiveor leadingcapacitive pure

°°<<°<<

°

°<<°−<<°−

90900

010

0109010

900

z

z

z

pf

pf

pf

θ

θ

θ

V

e)(capacitivleadscurrent

°<≤°− 090 zθ

)(inductivelagscurrent

°≤< 900 zθ

pfIVP rmsrms ××=

LEARNING EXAMPLE Find the power supplied by the power company.Determine how it changes if the power factor is changed to 0.9

Power company

pfIVP rmsrms ××=

rmsAWIrms )(3.259707.0480

)(1088 3=

××

=

rmsV )(480

rmsA)(453.259 °−∠

°−=⇒=⇒ 45707.0cos zz θθ

Current lags the voltage

480453.25908.0

08.0

+°−∠×=

+= LrmsS VIVrms

)(7.14957.147.494

480)4.1834.183(08.0

Vj

jVrmsS

°−∠=−=

+−×=

)(378.93)(000,88378.508.03.259 32

kWWPPkWRIP

lossesS

rmslosses

=+==×==

If pf=0.9

If pf=0.9

°−∠= 8.257.203rmsI

°−∠=+−= 82.049448009.747.14 jVSkWRIP

rmsAI

rmslosses

rms

32.3

)(7.2039.0480

000,88

2 ==

=×

=

Losses can be reduced by 2kW!

Examine also the generated voltage

LEARNING EXTENSION

Ω====

1.0707.0,)(480,100

line

LL

RpfrmsVVkWP

Determine the power savings if the power factor can be increased to 0.94

pfIVP rmsrms ××=

22

22 1

pfVRPRIP

pfVPI

rms

linelinermslosses

rmsrms

×==

×=

kW

WpfPlosses

34.42

)(707.01

4801.010)707.0( 22

10

×=

××

==

kW

WpfPlosses

34.413.1

)(94.01

4801.010)94.0( 22

10

×=

××

==

kWkWPsaved 77.334.487.0 =×=

COMPLEX POWER*rmsrms IVS =

PowerComplex ofDefinition

[ ]ivrmsrms

irmsvrms

IVSIVSθθ

θθ−∠=

°∠×°∠= *

)sin()cos( ivrmsrmsivrmsrms IVjIVS θθθθ −+−=

The units of apparent and reactive power are

Volt-Ampere

2* ||)( rmsrmsrmsrmsrms IZIZISZIV ==⇒=

Another useful form

⎩⎨⎧

==

⇒+= 2

2

||||

rms

rms

IXQIRPjXRZ

PActive Power

QReactive Power

inductive

capacitive

| || |

rms msS V IS P pf=

= ×

ANALYSIS OF BASIC COMPONENTS

RESISTORS

∴ 0Q⇒ =

INDUCTORS

CAPACITORS

Supplies reactive power!!

WARNING ( )0IF X ≠

LEARNING EXAMPLE

HzfjZrmsVlaggingpfkWP LLL 60,3.009.0,0220,8.0,20 =Ω+=°∠===:Given

Determine the voltage and power factor at the input to the line

}Re{SP = pfSS iv ×=−= ||)cos(|| θθ

inductive

capacitive

inductive

kVAQPSQ L 15222 =⇒−= °∠=+= 87.3625)(1520 kVAjSL

*LLL IVS =

)(86.3664.1130220

87.36000,25 **

AVSI

L

LL °−∠=⎥⎦

⎤⎢⎣⎡

°∠°∠

=⎥⎦

⎤⎢⎣

⎡=⇒

)(220)18.6891.90)(3.009.0(0220)3.009.0(

VjjVIjV

S

LS

+−+=°∠++=

)(18.6891.90220

000,15000,20 *

AjjIL −=⎥⎦⎤

⎢⎣⎡ +

=

°∠=+= 86.453.24914.2163.248 jVS

°86.4

°− 86.36

SV

LI

746.0)72.41cos( =°=sourcepflagging

LEARNING EXAMPLE Compute the average power flow between networksDetermine which is the source

1012030120

jZVVI BA °∠−°∠

=−

=

rmsVVA )(30120 °∠= rmsVVB )(0120 °∠=

Ω1j

rmsAjjjI )(08.1660120)6092.103(

+=−+

=

rmsAI )(1512.62 °∠=

rmsBB VAIVS °−∠=°−∠×°∠== 15454,71512.620120)( *

)(200,7)165cos(454,7 WPA −=°=

)(200,7)15cos(454,7 WPB =°−=

A supplies 7.2kW average power to B

rmsAA VAIVS °−∠=°−∠×°∠=−= 165454,719512.6230120)( * Passive sign convention.Power received by A

LEARNING EXTENSION

laggingpfkW

84.040

=

Ω1.0 Ω25.0j

Determine real and reactive power losses andreal and reactive power supplied

inductive

capacitive}Re{SP = pfSS iv ×=−= ||)cos(|| θθ

kVApfPSL 62.47

84.40|| ===

rmsL

LL A

VSIVIS )(45.216

||||||* ==⇒=

°=−⇒−= 86.32)cos( ivivpf θθθθ

)(839,25|||| 22 VAPSQ LL =−=

rmsL AI )(86.3245.216 °−∠=

2* ||)( LlineLLline IZIIZS ==losses

VAj 713,11685,4 +=

Balance of power

kVAjjj

SSS

552.37685.44839.2540713.11685.4 +=+++=

+= loadlossessupplied

2)45.216)(25.01.0( jS +=losses

laggingpfkW

85.060

=

Ω12.0 Ω18.0j

LEARNING EXTENSION Determine line voltage and power factor at the supply end

pfIVS LLiv ××=−= ||||)cos(|| θθ

rmsL

L ApfV

PI )(86.320||

|| =×

=

}Re{SP =*LLL IVS =

⇒=− − )(cos 1 pfiv θθ °=− 79.31iv θθrmsL AI )(79.3186.320 °−∠= rmsAj )(03.16972.272 −=

LLS VIZV += line 220)03.16972.272)(18.012.0( +−+= jj

rmsS VjV )(81.2815.283 += rmsV )(81.561.284 °∠=

792.0)6.37cos( =°=sourcepflagging

°81.5

°− 79.31

SV

LILVThe phasor diagram helps in visualizing

the relationship between voltage and current

Low power factors increase losses and are penalized by energy companies

Typical industrial loadsare inductive

Simple approach to power factor correction

)cos(||

oldold

oldoldoldoldold

:capacitorWithout

θθ

=∠=+=

pfSjQPS

)cos(||

newnew

newnew

capacitoroldold

capacitoroldnew

capacitorWith

θθ

=∠=

−+=

+=

pfS

jQjQP

SS S CV

IVQ

L

L

ω2||

||||

=

= capacitorcapacitor

capacitorICj

VL ω1

=

old

capacitoroldnew P

QQ −=θtan

θθ

2tan11cos

+=

POWER FACTOR CORRECTION

LEARNING EXAMPLE Economic Impact of Power Factor Correction

Operating Conditions

Current Monthly Utility Bill

Correcting to pf=0.9

New demand charge

Additional energy charge due to capacitor bank is negligible

Monthly savings are approx $1853 per month! A reasonable capacitor bank should Itself in about a year

LEARNING EXAMPLE

laggingpfVkW rmsL

8.00220,50

=°∠=Roto-molding

process

lagging 0.95 tofactorpower the increase to requiredcapacitor the Determine

.60Hzf =

}Re{SP = pfSS iv ×=−= ||)cos(|| θθ

kVApfPSold 5.62

80.50|| === ).(5.37|||| 22 kVAPSQ oldold =−=

329.01

tan95.0cos2

=−

=⇒=new

newnewnew pf

pfθθ kVAPQ

PQ

newnew 43.16329.0 =×=⇒=

kVAQQQ newoldcapacitor 07.2143.165.37 =−=−=∴

CV

IVQ

L

L

ω2||

||||

=

= capacitorcapacitorFF

VQ

CL

capacitor μπω

1156)(001156.0)602()220(

1007.21|| 2

3

2 ==××

×==∴

LEARNING EXTENSION

HzfRpfrmsVVkWP

line

LL

60,1.0707.0,)(480,100

=Ω====

Determine the capacitor necessary to increase the power factor to 0.94

}Re{SP = pfSS iv ×=−= ||)cos(|| θθ

kVApfPSold 44.141

707.100|| === ).(02.100|||| 22 kVAPSQ oldold =−=

363.01

tan94.0cos2

=−

=⇒=new

newnewnew pf

pfθθ kVAPQ

PQ

newnew 3.36363.0 =×=⇒=

kVAQQQ newoldcapacitor 72.633.3602.100 =−=−=∴

CV

IVQ

L

L

ω2||

||||

=

= capacitorcapacitorFF

VQ

CL

capacitor μπω

733)(000733.0)602()480(

1072.63|| 2

3

2 ==××

×==∴

Power circuit normallyused for residencialsupply

Line-to-line usedto supply majorappliances (AC, dryer). Line-to-neutral for lightsand small appliances

Basic circuit.

Generalbalancedcase

General case by sourcesuperposition

Neutral current is zero

Neutral currentis zero

An exercise insymmetry

SINGLE-PHASE THREE-WIRE CIRCUITS

LEARNING EXAMPLE Determine energy use over a 24-hour period and the costif the rate is $0.08/kWh

rmsA1= Lights on

rmsA2.0=

Stereo on

rmsA30=

LSnN

RSbB

RLaA

IIIIII

III

−=−−=

+=KCL

Assume allresistive

rmsrms IVP =

∫ ×== TimePdttpEnergy average)(

kWhHrkWHrkWElights 8.1712.0812.0 =×+×=

kWhHrkWErange 8.28)112(2.7 =++×=

kWhHrkWEstereo 192.0)35(024.0 =+×=

kWhEdaily 792.30= dayCost /46.2$= ∫∫ == dtIVpE rmsrmssuppliedsupplied

Outline of verification

aAIA31

A1

SAFETY CONSIDERATIONS

Average effect of 60Hz current from hand tohand and passing the heart

Required voltage depends on contact, personand other factors

Typical residential circuit with ground andneutral

Ground conductor is not needed fornormal operation

Increased safety due to grounding

When switched on the tool case is energized

without the ground connector theuser can be exposed to the fullsupply voltage!

Conducting due to wet floor

If case is grounded then the supply is shorted and the fuse acts to openthe circuit

LEARNING EXAMPLE

More detailed numbers in a related case study

LEARNING EXAMPLE

Ground prong removed

R(dry skin) 15kOhmR(wet skin) 150OhmR(limb) 100OhmR(trunk) 200Ohm

Suggested resistancesfor human body

Ω150Ω150

Wetskin

Ω400

Limbstrunk

Ω≈1

mAIbody 171701120

==

Can cause ventricular fibrillation

LEARNING EXAMPLE Ground Fault Interrupter (GFI)

In normal operating mode the two currents induce canceling magnetic fluxes

No voltage is induced in the sensing coil

coil sensing the in induced voltagea is there thenfault)atodue(e.g.,different becomeandIf 21 ii

LEARNING EXAMPLE

xGroundfault

Vinyl lining (insulator)

Circuit formed when boy inwater touches boy holding grounded rail

A ground fault scenario

While boy is alone in the poolthere is no ground connection

LEARNING EXAMPLE Accidental groundingOnly return path in normaloperation

New path created by the grounding

Using suggested values of resistance thesecondary path causes a dangerous currentto flow through the body

mA150

LEARNING EXAMPLE A grounding accident

After the boom touches the live line the operator jumps down and starts walkingtowards the pole

7200 V

Ground is not a perfectconductor

10m

720V/m

One step applies 720 Volts to theoperator

LEARNING EXAMPLE A 7200V power line falls on the car and makes contact with it

Wet Road

Option 1.Driver opens door and steps down

Option 2: Driver stays inside the car

7200V

Car body is good conductor

Tires are insulators

R(dry skin) 15kOhmR(wet skin) 150OhmR(limb) 100OhmR(trunk) 200Ohm

Suggested resistancesfor human body

trunklimbskindry body RRR

I++

=27200

0=bodyI

dangerous! Very mAI 460≈

LEARNING EXAMPLE Find the maximum cord length

CASE 1: 16-gauge wire ft

mΩ4 CASE 2: 14-gauge wireft

mΩ5.2

Ω==×

382.051.13

cord

cord

RVAR

ftRLft

mcord 5.95

4==

Ω ftRLft

mcord 8.152

5.2==

Ω

Minimum voltage for properoperation

Working with RMS values the problem is formally the same as a DC problem

LEARNING EXAMPLE Light dimming when AC starts

Typical single-phase 3-wire installation

rmslight AI 5.0≈

Circuit at start of AC unit. Current demand is very high

A40≈

rmsAN VV 100≈

AC off

rmsAN VV 5.119120241240

=×=

AC in normal operation

rmsAN VV 115≈

LEARNING EXAMPLE DRYER HEATING AND TEMPERATURE CONTROL

Temperature is controlled by disconnecting the heatingelement and letting it cool off.

Vary power to thermostatheater

If temperature from dryer is highthan the one from thermostat thopen switch and allow heater tocool off

Safety switch in case controlfails.

LEARNING BY DESIGN Analysis of single phase 3-wire circuit installations

Option 1

Option 2

||1205000|| HL II ==

67.412×=nI

°−∠= 9.3667.41mI°−∠+°∠+°∠= 9.3667.41067.41067.41aI )(1.124.119 A°−∠=

mb II =

[ ]kVAjkVA

SA

3141.1238.141.124.1190120 *

+=°∠=°−∠×°∠=

[ ]kVAjkVA

SB

349.3659.3667.410120 *

+=°∠=°−∠×°∠=

°−∠+°∠= 9.3667.41067.41aI )(4.1807.79 A°−∠=

ab II =

HL II =0=nI

[ ]kVAjkVA

SS BA

394.185.94.1807.790120 *

+=°∠=°∠×°∠==

( ) kWIIIP

R

nbalosses

lines

147.1||||||05.0

05.0222 =++×=

Ω=

( ) kWIIP

R

balosses

lines

625.0||||05.0

05.022 =+×=

Ω=

)/$08.0(@366/$522.0

kWhyearkWPsaved

==

Steady-statePower Analysis