Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-43_Lec-05c_Thevenin_AC_Power.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Thevenin/NortonAC

Power



Thevenin’s Equivalence Theorem Resistance to

Impedance Analogy

LINEAR C IRC U ITM ay contain

independent anddependent sources

with their co ntro ll ingvariablesPART A


independent anddependent so urces

with their contro ll ingvariablesPART B

a

b_Ov

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

THR

THv

PART AThevenin Equivalent Circuit for PART A

OOv V

IiTHTHV V

THTHR Z



AC Thevenin Analysis Same Circuit,

Find IO by Thevenin Take as “Part B” Load

the 1Ω Resistor Thru Which IO Flows

Now Use Src-Xform The Thevenin Ckt

VjjA

x

x

)22()1(02

VV



AC Thevenin Analysis cont. The Xformed

Circuit Another

SourceXform

Find ZTH by Zeroing the 8+j2V-Source

Now the Combined 8+j2 Source and all The Impedance are IN SERIES• Thus VOC Determined by

V-Divider

j28

2610)28(

)1()1(1 jj

jjj

OC

V



ReDraw the Voltage Divider

The V-Divider Formula:

OV1Z

2Z

212 ZZZVV SO

2

22882

128)1()1(

1282jjjjj

jjjjOC

V

jjjjjjjOC 35

2610

22828

22288 2

V



AC Thevenin Analysis Find ZTH by Zeroing the

8+j2 V-Source

IO is Now Simple

So The Thevenin Equivalent Circuit

122

111

1111

)1(||)1(

2j

jjjjjj

TH

TH

TH

Z

Z

Z

)(235

111

AjO

OC

TH

THO

I

VZ

VI



Norton’s Equivalence Theorem Resistance to

Impedance Analogy

OOv V

IiNNi I

NNR Z

LINEAR C IRC UITM ay contain

independent anddependent sources

with their co ntro ll ingvariablesPART A


independent anddependent so urces

with their contro ll ingvariablesPART B

a

b_Ov

i

LINEAR C IRC UIT

PART B

a

b_Ov

i

NRNi

PART ANorton Equivalent Circuit for PART A



AC Norton Analysis Same Circuit,

Find IO by Norton Take as the “Part B”

Load the 1Ω Resistor Thru Which IO Flows

Possible techniques to Find ISC:• Loops or Nodes• Source Transformation• SuperPosition

Choose Nodes

Shorting the Load Yields Isc= IN



AC Norton Analysis cont Short-Ckt Node On The

Norton Circuit

Use ISC=IN, and ZN to Find IO

• See Next Slide

As Before Deactivate Srcs to Find ZN=ZTH

)(1

281

0602 Ajj

jSC

I

1)1(||)1( jjZTH



AC Norton Analysis cont.2 The Norton

Equivalent Circuitwith LoadReattached

Same as all the Others

Find IO By I-Divider

2

351

44

11

14

14

1111

22

2 jj

jjj

jj

jj

jj

O

O

SC

NSCO

I

I

IZ

II



Outline – AC Steady State Power Instantaneous Power Concept

• For The Special Case Of Steady State Sinusoidal Signals

Average Power Concept• Power Absorbed Or Supplied During an

Integer Number of Complete Cycles Maximum Average Power Transfer

• When The Circuit is in Sinusoidal Steady State



Outline – AC SS Power cont. Power Factor

• A Measure Of The Angle Between the CURRENT and VOLTAGE Phasors

Power Factor Correction• Improve Power Transfer To a Load By

“Aligning” the I & V Phasors Single Phase Three-Wire Circuits

• Typical HouseHold Power Distribution



Instantaneous Power Consider in the Time

Domain a Voltage Source Supplying Current to an Impedance Load

Now Recall From Chp1 The Eqn for Power

In the General Case

vip

iM

vM

tItitVtv

cos)(cos)(

Then at any Instant for Time-Varying Sinusoidal Signals

ivMM ttIV

titvtp

coscos

Now Use Trig ID

)cos()cos(21coscos yxyxyx



Instantaneous Power cont. Rewriting the Power Relation for Sinusoids

Examine the TWO Terms of the Power Equation

)2cos(2

)cos(2

)(

)2cos()cos(2

)(

ivMM

ivMM

ivivMM

tIVIVtp

tIVtp

• The First Term is a CONSTANT, or DC value; i.e. There is No Time Dependence

• The Second Term is a Sinusoid of TWICE the Frequency of the Driving Source



Example For the Single Loop Ckt Use Phasors To find I

Assume

302302604 AV

ZVI

302604

60cos4)(

ZV

andVor

tVtv

To Obtain the Time Domain current Take the Real Part of the Phasor Current

30cos22Re302Re)( 30

tAmptieeti tj



Example cont Thus for This case In the Power Equation

Or

30602cos3060cos4

)(tW

titvtp

306024

iv

MM AIVV

The Amplitudes and Phase Angles

902cos446.3)( tWWtp

See Next Slide for Plots• v(t)• i(t)• p(t) = v(t)•i(t)

60cos4 tV j732.1



Sinusoidal Power Example 9.1

-4

-2

0

2

4

6

8

0.000 0.003 0.006 0.009 0.012 0.015 0.018 0.021 0.024 0.027 0.030

Time (S)

v(t)

or i

(t) o

r p(t)

v(t) (V) i(t) (V) P(t) (V)

file =Sinusoid_Lead-Lag_Plot_0311.xls

p(t) Calculated by p(t)=v(t)•i(t)

Avg-p = 3.46W

Max-p = 7.46W

NEGATIVE POWER – Inductive Load can Release stored Energy to the Circuit

p = 0 if either i or v are zero



Average Power For ANY Periodic Function, It’s Average Value Can be

calculated by Integrating Over at Least ONE COMPLETE PERIOD, T, and Then Dividing the Integrated Value by the Period

Then the Average POWER for Electrical Circuits With Sinusoidal Excitation

timeBaseLinearbitary an )(10

0

0

tdttxT

XTt

t

0

0

0

0

0

0

coscos1

11

tT

t ivMM

tT

t

tT

t

dtttIVT

dttitvT

dttpT

P

And Recall 2T



Average Power cont Also For ANY Periodic Function, the Average value May

be Calculated over any INTEGER number of periods. This is, in Fact, How most Electrical Power Values are MEASURED. Mathematically:

Now Sub Into the Average Power Integral The Simplified (n =1) Expression for Instantaneous Power

0

0

coscos1 tnT

t ivMM dtttIVnT

P

0

0

0

0

0

0

)2cos(2

1)cos(2

1

)2cos()cos(2

1

tT

t

tT

t ivMM

ivMM

tT

t ivivMM

dttIVT

dtIVT

P

dttIVT

P



Average Power cont.2 Examine the Two Terms from the Average Power Eqn

Thus the First Term is a CONSTANT that depends on the Relative Phase Angle

Also by Trig: cos(−) = cos()• Thus for the First Term It Does NOT Matter if the Current

LEADS or LAGs the Voltage

)cos(2

cos2

1

1)cos(2

1)cos(2

1

11

1

0

0

0

0

0

0

ivMMtT

tivMM

tT

tivMMtT

t ivMM

IVPtIVT

P

dtIVT

dtIVT

P

viiviviv coscoscoscos



Average Power cont.3 The Second Term from the Average Power Eqn

• As the The Integral of a sin or cos over an INTEGER number of periods is ZERO

Thus the Average Power is Described by the FIRST TERM ONLY

0)2cos(2

1 0

02

tT

t ivMM dttIV

TP

cos2

)cos(2

)cos(2

MM

viMM

ivMM

IVP

IVPIVP



Resistive & Reactive Power For a Purely

RESISTIVE Circuit There is NO Imaginary Component of the Impedance and thus no Phase Shift Between i & v. So for sinusoids

For A Purely REACTIVE Circuit i&v are ±90° out of phase• So then Avg Power eqn

Thus the Resistors Absorb, or Dissipate, Power (as Heat) only

2MM

resIVP

!090cos2

MMreact

IVP Thus Purely reactive

Impedances absorb NO Power on Average• They STORE Energy

over one Half-Cycle, and then RELEASE it over the NEXT



Example Avg Power For This Circuit The Power Parameters

FIND: The Dissipated Power

Use Phasor Algebra to Find the Current

)(1554.3458284.2

601022

6010 Aj

V

I

1560

54.310

iv

MM AIVV

The Power Calculation

WWP

AVP

IVP ivMM

5.1245cos7.17

1560cos2

54.310

)cos(2



Example Avg Power cont Since Only the Resistor

Dissipates Power, Check the Previous Calc by using Pres=vres•i

Alternatively by OHM

Use Phasors for VR

WWP

AVP

IVP iv

MRM

5.12200.25

1515cos2

54.307.7

)cos(2,

RV

)(1507.7601022

2 VVVjR

V

For a Resistor the Current & Voltage WITHIN the Resistor are IN-PHASE; thus

1507.71554.32 VARR IV



Example Cap Circuit For This Circuit Find the Total

Impedance Across the V-Source

Find The Dissipated Power in Each Resistor

Start by Finding The Current In the Ckt Branches that Contain the Resistors

Then The total Current I

I

2I

565.264721.44524

6.713.2544

168844)4(42

jjj

jj

TZ

6.8668.26.2647.4

6012

A

V

TZVI

• i(t) LEADS vS(t)



Example Cap Circuit cont

So the 2Ω Resistor Power Dissipation

Then the Power Absorbed by the 4Ω R

I

2I

WRIP M 20.768.2221

21 22

2

6.4190.1

6.8668.24524

90444

42 II

jj

WP

WP

20.7

)(90.1421

4

24

Now Find I2 by Current Divider

2

,

21

0cos21

cos2

MR

MMR

ivMRM

R

RIP

IRIP

IVP



Maximum Avg Power Transfer Recall From the Study

of Resistive Ckts The Criteria for Max Power Xfer to a Load Resistor

Where RTH is the Thevenin Equivalent Resistance for the Driving Ckt

Now Try to Develop a Similar Relationship for Impedances

Consider This General Thevenin Equivalent Ckt

THL RR pwrmax,

For This Ckt the Avg Pwr Delivered to the Load

)cos(21

LL ivLLL IVP



Max Avg Power Transfer cont Now by Phasors

WhereLTH

LOCL

LTH

OCL

ZZZVV

ZZVI

Now By Euler Rln Recall

THTHTH

LLL

jXRjXR

ZZ

ZVVZ

VZZV

andMM

zVMM

ZV

ZV



Max Avg Power Transfer cont.2 Then the Load Voltage

& Current Magnitudes

Similarly Also by Euler

OCTHL

LL

THL

LL

VZZ

ZVor

OCV

ZZZV ||

L

LL Z

VI And

LIV

LVIVL

LLILLLL

Z

ZV

ZVI

LL

LLL

L

But

ZVI

LLLLLL RXZjXR tanZ Now a Useful Trig ID

LLLIV

IVIV

RXZLL

LLLL

tantanBut

tantantan11cos 2



Max Avg Power Transfer cont.3 ReArrange Trig ID to

Find

Again the Power Eqn Or

Substitute to Find And

22

2

)cos(

1

1)cos(

LL

LIV

LL

IV

XR

RRX

LL

LL

)cos(21

LL ivLLL IVP

2221

LL

L

L

OCTHL

L

OCTHL

LL

XR

RZ

VZZ

Z

VZZ

ZP

222

2

21

LL

L

THL

OCLL

XR

RZZVZP

222 )()(

)()(

THLTHLTHL

THLTHLTHL

XXRRZZ

XXjRR

ZZ



Max Avg Power Transfer cont.4 Finally The Power Eqn

Restated

Now To Maximize the Power Transfer, Set the Partial Derivatives to 0

at LAST The Optimized Load

• The Complex Conjugate And the Power

Transferred at Optimum

22

2

)()(21

THLTHL

LOCL XXRR

RVP

THL

THL

L

L

L

L

RRXX

RPXP

0

0

*TH

optL ZZ

TH

OCL R

VP42

1 2max



Max Avg Power Transfer cont.5 Check

At the Max Condition

22

2

)()(21

THLTHL

LOCL XXRR

RVP

THL

THL

L

L

L

L

RRXX

RPXP

0

0

TH

OCL

TH

OC

TH

THOC

TH

THOCL

THTHTHTH

THOCL

RVP

RV

RRV

RRVP

XXRRRVP

8

8421

0)2(21

)()(21

2max

2

2

2

22

2max

22

2max



Example Maximum Power Xfer Find ZL for the

Maximum Power Xfer And The Max Pwr Xfer

Recall The Max Power Criteria

Need to find • VOC = VTH

• RTH

*TH

optL ZZ

THOCL RVP 82max



Example Max Pwr Xfer cont. Remove Load and Find

VOC by Loop Current And by Ohm’s Law in

the Frequency Domain

Using KVL on the Loop

OCV

I

56.71974.1861812)1(92

0122

Vjjj

jOC IV

4573.12)1(98

)22(36022012024

jjj

I

II THZ

Find ZTH by Source Deactivation (Zeroing)



Example Max Pwr Xfer cont.2 Then ZTH

Taking The Conjugate

THZ

22

42)2||2(2jjjjjTHZ

)(18

8822

4

jj

jTHZ or

)(1 joptLZ

Then The Power Transferred to this Load

W

WV

RVP THOCL

458

36018

974.18

82

2max



Power Factor Consider a Complex

Current Thru a Complex Impedance Load

The Current and Load-Voltage Phasors (Vectors) Can Be Plotted on the Complex Plane

By Ohm & Euler

in the Electrical Power Industry θZ is the Power Factor Angle, or Simply the Phase Angle

LZiMI

vMV

iv

VIz

ivz

izv

or

IZV

ZIV



Power Factor cont The Phase Angle Can

Be Positive or Negative Depending on the Nature of the Load

Typical Industrial Case is the INDUCTIVE Load• Large Electric Motors are

Essentially Inductors Now Recall The

General Power Eqn

Measuring the Load with an AC DMM yields• Vrms

• Irms

Zrmsrmsivrmsrms

ZMM

ivMM

IVIVP

IVIVP

cos)cos(

cos22

)cos(21

0MVV

e)(capacitivleadscurrent

090 z

)(inductivelagscurrent

900 z

ZZZVI

0

V is the BaseLine



Power Factor cont.2 The Product of the

DMM Measurements is the APPARENT Power

The Apparent Power is NOT the Actual Power, and is thus NOT stated in Watts.• Apparent Power

Units = VA or kVA

Now Define the Power Factor for the Load

Some Load Types

rmsrmsapparent IVP pfIVP

PPpf

rmsrmsactual

zivapparent

actual

and

cos)cos(

inductive pureinductiveor lagging

resistivecapacitiveor leading

capacitive pure

90900

010

0109010

900

z

z

z

pf

pf

pf



pf – Why do We Care? Consider this

case • Vrms = 460 V• Irms = 200A• pf = 1.5%

Then• Papparent = 92kVA• Pactual =1.4 kW

This Load requires The Same Power as a Hair Dryer or Toaster

However, Despite the low power levels, The WIRES and CIRCUIT BREAKERS that feed this small Load must be Sized for 200A!• The Wires would be nearly an INCH in

Diameter



Example Power Factor The Local Power

Company Services this Large Industrial Load

Find Irms by Pwr Factor

Then the I2R Loses in the 100 mΩ line

Improving the pf to 94%

1.0

kW100010V5

Power company

rmsrms

rmsrms

rmsrms

VkWIVpfPIpfIVP

480707.0100

22

22 1

pfVRPRIPrms

linelinermslosses

234.4

)(707.01

4801.010)707.0( 22

10

kW

WpfPlosses

13.134.4

)(94.01

4801.010)94.0( 22

10

kW

WpfPlosses

kWkWPsaved 77.334.487.0

I lags V



Example - Power Factor cont For This Ckt The

Effect of the Power Factor on Line Losses

1.0

kW100010V5



Complex Power Consider a general Ckt

with an Impedance Ld Mathematically

For this Situation Define the Complex Power for the Load:

Converting to Rectangular Notation

*rmsrmsIVS

Ziv

ivrmsrms

irmsvrms

irmsvrms

IVIVIV

:recall

*

SSS

)sin()cos( ivrmsrmsivrmsrms IVjIVS

PActive Power

QReactive Power



Complex Power cont Thus S in Shorthand Alternatively,

Reconsider the General Sinusoidal Circuit

S & Q are NOT Actual Power, and Thus all Terms are given Non-Watt Units• S→ Volt-Amps (VA)• Q → Volt-Amps,

Reactive (VAR) P is Actual Power and

hence has Units of W

First: U vs. Urms

jQP S

rmsrms

Mrms

M

UUU

U

U

U

2and



Complex Power cont.2 Now in the General Ckt

By Ohm’s Law

In the Last Expression Equate the REAL and Imaginary Parts

And Again by Ohm

iviv

rmsrms

ivrms

rms

irms

vrms

rmsrms

jZZj

soZIV

IV

IV

sincosImRe

and

ZZZ

Z

IVZ

Z

Z

iv

iv

Z

Z

Imsin

Recos

ZVI

ZV

ZV

rmsrms

irms

iv

vrmsrms

rmsrms

So

I

ZVI



Complex Power cont.3 And by Complex Power

Definition

Using the Previous Results for P

Similarly for Q

ivrmsrms

ivrmsrms

IVQIVP

jQP

sinImcosRe

then

SS

S

RIIP

IZV

ZIVP

rmsrms

rmsrms

rmsrms

22 Re

Re

ReRe

Z

Z

ZS

XIIQ rmsrms22 Im Z

So Finally the Alternative Expression for S

ZS 2rmsI



Complex Power Triangle The Expressions for S

Plotting S in the Complex Plane

From The Complex Power “Triangle” Observe

ZSS

2rmsI

jQP

PQ

iv tan Note also That Complex

Power is CONSERVED

kkrmsktot I ZSS 2,



Example - Complex Power For the Circuit At Right

• Zline =0.09 Ω + j0.3 Ω• Pload = 20 kW• Vload = 2200°• pf = 80%, lagging• f = 60 Hz → ω = 377s-1

Lagging pf → Inductive

From the Actual Power

Thusinductive

capacitive

pfSP iv )cos(||Re SS

kVAkWpfPSL 25

8.020

And Q from Pwr Triangle

kVAR15222 QPSQ L



Example - Complex Power cont Then SL

Recall the S Mathematical Definition

Alternatively

87.36251520 kVAkVAjLS

*LLL IVS

Note also that [U*]* = U In the S Definition,

Isolating the Load Current and then Conjugating Both Sides

)(86.3664.1130220

87.3625 **

AV

kVA

L

L

LL

IVSI

)(18.6891.90220

000,15000,20 *

Aj

j

L

L

I

I

Lagging



Example - Complex Pwr cont.2 Now Determine VS

Then VS Then The Phase Angle

To find the Src Power Factor, Draw the I & V Phasor Diagram

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

Vjjj

S

LS

LlineS

VIV

VVV

86.453.24914.2163.248

22014.2163.28

rmsS

S

S

Vjj

VVV

86.4

86.36

SV

LI

7464.0

72.41coscosalso and

Load Inductive V Lags I 72.4186.3686.4

pfpf iv

iv



Example - Complex Power kVAR For the Circuit At

Right, Determine• Real & Reactive Power

losses in the Line• Real & Reactive Power

at the Source Lagging pf → Inductive

From the Actual Power

Thusinductive

capacitive

kVAkWpfPSL 62.47

84.040

And by S Definition

laggingpfkW

84.040

1.0 25.0j

pfSP iv )cos(||Re SS

rmsL

LL A

VSI )(45.216* VIS



Example - Complex kVAR cont. Also from the S

Relation & Pythagorus

Now the PowerFactor Angle Then for Line Loses

Quantitatively

laggingpfkW

84.040

1.0 25.0j

)(839,25|||| 22 VARPLL SQ

86.3284.0acosiv

pf = cos(θv − θi); hence

2

** )(

Lline

LLlinelinelineline

IZ

IIZIVS

VAjj

line

line

117134685)45.216)(25.01.0( 2

SS

I Lagging V



Example - Complex kVAR cont.2 Find Power Supplied

by Conservation of Complex Power

Then to Summarize the Answer• Pline = 4.685 kW• Qline = 11.713 kVAR• PS = 44.685 kW• QS = 37.552 kVAR

laggingpfkW

84.040

1.0 25.0j

kVAkVAj

j

jjSup

04.4037.58 552.37685.44

839.25713.1140685.4

839.2540713.11685.4S

LoadlineSupplied SSS

In this Case



Power Factor Correction As Noted Earlier, Most

Industrial Electrical Power Loads are Inductive• The Inductive

Component is Typically Associated with Motors

The Motor-Related Lagging Power Factor Can Result in Large Line Losses

The Line-Losses can Be Reduced by Power Factor Correction

To Arrive at the Power Factor Correction Strategy Consider A Schematic of a typical Industrial Load

Intentionally Added



Power Factor Correction cont. Prior to The Addition

of the Capacitor

For The Capacitive Load

)cos(||

oldold

oldoldoldoldold

pfSjQPS

22

2

C

Im

90

LL

CCC

LC

LLLC

L

CVC

CVIQ

CVI

CVCj

Z

VZVI

After Addition of the Capacitor

)cos(||

newnew

newnew

Coldold

Coldnew

pf

jQjQPS

SSS



Power Factor Correction cont.2 Find θnew

• Cap is a Purely REACTIVE Load

The Vector Plot Below Shows Power Factor Correction Strategy

old

new

old

Coldnewtan

PQ

PQQ

Use Trig ID to find QC to give desired θnew

2tan1

1cos

1cos

1

new2

old

Cold

PQQ

QL

QC

QL-QC

P

Qnew



Trig ID Digression Start with the ID

2tan11cos

Solve for tan

old

Coldnewtan

PQQ

Recall tanθnew

1cos

1

new2

old

Cold

PQQ

Or

new

new2

new2

new2

new2

new2

new2

old

Cold

coscos1

coscos1

coscos

cos1

P

QQ

But: cosθnew = pfnew

new

new

pfpf 2

new

1tan

1cos

1tan 22

Substituting



Example pf Correction Kayak Centrifugal

Injection-Molding Power Analysis• Improve Power Factor

to 95%

Find Sold

Now Qold

Adding A Cap DoesNOT Change P• Use Trig ID to Find Tan(new)

And by S Relation

laggingpfVkW rmsL

8.00220,50

Roto-molding

process

pfSSP iv )cos(Re S

kVAkWpfPSold 5.62

80.050

)(5.37|||| 22 kVARPoldold SQ

329.01

tan95.0cos2

new

newnewnewnew pf

pfP

Q

kVARPQP

Qnew

new 43.16329.0



Example pf Correction cont Then the Needed QC

Recall The Expression for QC

laggingpfVkW rmsL

8.00220,50

Roto-molding

processkVAQQQQ

C

newoldC

07.2143.165.37

CVQ LL 2CC |||| IV

Then C from QC

FFCVQC

L

C

1155)(001155.0)220()602(

1007.212

3

2



All Done for Today

PowerFactor

Correction



WhiteBoard Work

Let’s Work Problem similar to P5.78• Determine at the input

SOURCE – Voltage & Current– Complex Power– Δθ & pf

19V220 rms

60 kVA





























WhiteBoard Work

Z L1 20 V

2 j1

- j2

2 I x

I x

F ig u re P 9 .3 2

Let’s Work This Nice Problem

Find ZL for Pav,max, & the value of Pav,max



Example Use Nodal Analysis

to Find VO

The KCL at Node-1

Subbing for V1 Yields

Now KCL At VO

1V

0212

3012 1111

jjOVVVVV

O

OO

jj

VV

VVV

1

01

1

1

Multiply Node-1 KCL by 2j to Obtain

0)(22)3012( 1111 Ojj VVVVV

Factoring out VO Yields 3012901))1)(31(2( Ojj V

3012)1)(221(2 jjjj OO VV

75121.245657.5

1201244

12012

Vj

O

O

V

V

Isolating VO



Example Alternative This Time Use Thevenin

to Find VO

Take Cap & Res in Series as the Load Find VOC by V-Divider

Deactivate V-Src to Find ZTH

2.06.062

62462

4232

342||1||2

22

j

jjj

jj

jj

TH

TH

TH

Z

Z

Z

3012)2||1(2

2||1j

jOCV

30122)21(2

2jj

jOCV

435.48795.33112012

6212024

Vjj

OC

OC

V

V



Example Alternative cont. Now Apply the Thevenin

Equivalent Circuit to the Load

Calculate VO By V-Divider

+-OCV

THZ 1j

1

0V

00.75121.2435.48795.3565.26559.0

8.06.11

12.06.01

11

VV

j

jj

jZ

O

O

OCO

OCO

OCTH

O

VV

VV

VV

VV

2.06.0 j

Same as Before



P5.57 Graphics



P5.81 Graphics

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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