[email protected] ENGR-43_Lec-08-1_AC-Steady-State.ppt 1 Bruce Mayer, PE Engineering-43:...

[email protected] • ENGR-43_Lec-08-1_AC-Steady-State.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Sinusoidal

AC SteadySt



Outline – AC Steady State

SINUSOIDS• Review basic facts about sinusoidal signals

SINUSOIDAL AND COMPLEX FORCING FUNCTIONS• Behavior of circuits with sinusoidal

independent current & voltage sources • Modeling of sinusoids in terms of

complex exponentials



Outline – AC SS cont.

PHASORS• Representation of complex

exponentials as vectors– Facilitates steady-state analysis of circuits

• Has NOTHING to do with StarTrek

IMPEDANCE AND ADMITANCE• Generalization of the familiar concepts of

RESISTANCE and CONDUCTANCE to describe AC steady-state circuit operation

phasEr



Outline – AC SS cont.2

PHASOR DIAGRAMS• Representation of AC voltages and

currents as COMPLEX VECTORS

BASIC AC ANALYSIS USING KIRCHHOFF’S LAWS

ANALYSIS TECHNIQUES• Extension of node, loop, SuperPosition,

Thevenin and other KVL/KCL Linear-Circuit Analysis techniques



Sinusoids Recall From Trig the

Sine Function Where

• XM “Amplitude” or Peak or Maximum Value– Typical Units = A or V

• ω Radian, or Angular, Frequency in rads/sec

• ωt Sinusoid argument in radians (a pure no.)

The function Repeats every 2π; mathematically

For the RADIAN Plot Above, The Functional Relationship )()2( txtx

tXtx M sin)(



Sinusoids cont. Now Define the

“Period”, T Such That

How often does the Cycle Repeat?• Define Next the CYCLIC

FREQUENCY

From Above Observe

)()()2( txTtxtx

) all(for ),()

and

2

2

ttTtxx(t

TT

Now Can Construct a DIMENSIONAL (time) Plot for the sine

TtXtx M 2sin)(



Sinusoids cont.2 Now Define the Cyclic

“Frequency”, f Quick Example

• USA Residential Electrical Power Delivered as a 115Vrms, 60Hz, AC sine wave

Describes the Signal Repetition-Rate in Units of Cycles-Per-Second, or HERTZ (Hz)• Hz is a Derived SI Unit

tVtv

tVtv

residence

residence

99.376sin6.162)(

602sin1152)( fT

f

2

2

1

Will Figure Out the 2 term Shortly• RMS “Root (of the)

Mean Square”



Sinusoids cont.3 Now Consider the

GENERAL Expression for a Sinusoid

Where• θ “Phase Angle” in

Radians

Graphically, for POSITIVE θ

tXtx M sin)(

"by leads"

"by lags"



Leading, Lagging, In-Phase Consider Two Sinusoids

with the SAME Angular Frequency

Now if > • x1 LEADS x2 by (−) rads

or Degrees• x2 LAGS x1 by ( −)

rads or Degrees

If = , Then The Signals are IN-PHASE

If , Then The Signals are OUT-of-PHASE

Phase Angle Typically Stated in DEGREES, but Radians are acceptable• Both These Forms OK

tXtx

tXtx

M

M

sin)(

sin)(

22

11

71.25sin

7sin)(

tX

tXtx

M

M



Sinusoid Phase Difference

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

Time (S)

xi (

V o

r A

)

x1 (V or A)

x2 (V or A)

file =Sinusoid_Lead-Lag_Plot_0311.xls

Out of Phase* 0.733 rads* 42° * 105 mS

PARAMETERS• T = 900 mS• f = 1.1111 Hz• = 6.981 rad/s• = 1.257 rads = 72°• = 0.5236 rads = 30°

x1 LEADs x2

x2 LAGs x1For Different Amplitudes, Measure Phase Difference• peak-to-peak• valley-to-valley• ZeroCross-to-ZeroCross

421

360

900

1105

PeriodmS

PeriodmS



Useful Trig Identities To Convert sin↔cos

To Make a Valid Phase-Angle Difference Measurement BOTH Sinusoids MUST have the SAME Frequency & Trig-Fcn (sin OR cos) • Useful Phase-Difference

ID’s

)sin(sin

)cos(cos

tt

tt

2cossin

2sincos

tt

tt

Additional Relations

sinsincoscos)cos(

sincoscossin)sin(

sinsincoscos)cos(

sincoscossin)sin(

(rads) rads

180(degrees)

360 radians 2



Example Phase Angles Given Signals

Find• Frequency in Standard

Units of Hz• Phase Difference

Frequency in radians per second is the PreFactor for the time variable• Thus

To find phase angle must express BOTH sinusoids using• The SAME trigonometric

function; – either sine or cosine

• A POSITIVE amplitude

)301000cos(6)(

)601000sin(12)(

2

1

ttv

ttv Hz 2.1592

)Hz(

S 1000 -1

f



Example – Phase Angles cont. Convert –6V Amplitude

to Positive Value Using

Then

It’s Poor Form to Express phase shifts in Angles >180° in Absolute Value

)180cos()cos(

)180301000cos(6

)301000cos(6

2

2

tv

tv

Next Convert cosine to sine using

)90sin()cos(

)902101000sin(6

)2101000cos(62

t

tv

601000sin6

3603001000sin6

)3001000sin(62

t

t

tv

So Finally

)601000sin(6)(

)601000sin(12)(

2

1

ttv

ttv

Thus v1 LEADS v2

by 120°



Sinusoid Phase Difference Example 7.2

-12

-8

-4

0

4

8

12

0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 0.020

Time (S)

Sig

nal

Lev

el (

V)

v1 (V)

v2 (V)

file =Sinusoid_Lead-Lag_Plot_0311.xls

120°



Effective or rms Values Consider Instantaneous

Power For a PurelyResistive Load

Since a Resistive Load Dissipates this Power as HEAT, the Effective Value is also called the HEATING Value for the Time-Variable Source• For example

– A Car Coffee Maker Runs off 12 Vdc, and Heats the Water in 223s.

– Connect a SawTooth Source to the coffee Maker and Adjust the Amplitude for the Same Time → Effective Voltage of 12V

Now Define The EFFECTIVE Value For a Time-Varying Signal as the EQUIVALENT DC value That Supplies The SAME AVERAGE POWER

)(ti

R Rtitp )()( 2



rms Values Cont. For The Resistive Case,

Define Ieff for the Avg Power Condition

If the Current is DC, then i(t) = Idc, so

The Pav Calc For a Periodic Signal by Integ

Now for the Time-Variable Current i(t) → Ieff, and, by Definition

)(ti

RRIP

Rtitp

effav2

2 )()(

Tt

t

Tt

t

av dttiT

RdttpT

P0

0

0

0

)(1

)(1 2

22

2

0

0

0

0

)(11

1

dc

Tt

t

dc

Tt

t

dcav

RIdttT

RI

dtIT

RP

avdceffav PRIRIP 22



rms Values cont.2 In the Pwr Eqn Examine the Eqn for Ieff

and notice it is Determined by• Taking the Square ROOT

of the time-averaged, or MEAN, SQUARE of the Current

In Engineering This Operation is given the Short-hand notation of “rms”

So

Equating the 1st & 3rd Expression for Pav find

2220

0

)(1

effdc

Tt

t

av RIRIdttiT

RP

Tt

t

eff dttiT

I0

0

)(1 2

This Expression Holds for ANY Periodic Signal rmseff II



RMS Value for Sinusoid

Find RMS Value for sinusoidal current• Note the Period

Sub Above into RMS Eqn:

Use Trig ID:

tIti M cos

2T

21

0

22 cos1

dttI

TI

T

Mrms

2cos2

1

2

1cos2




Sub cos2 Trig ID into RMS integral

Integrating Term by Term

21

0

22cos2

1

2

11

dtt

TII

T

Mrms

21

0 0

22cos2

11

2

11

dtt

Tdt

TII

T T

Mrms




Rearranging a bit

But the integral of a sinusoid over ONE PERIOD is ZERO, so the 2nd Term goes to Zero leaving

21

0 0

22cos1

2

11

1

2

1

dtt

Tdt

TII

T T

Mrms

0

212

1

12

11

1

2

12121

0

21

0

MM

T

M

T

Mrms

IT

TI

t

TIdt

TII



Sinusoidal rms Alternative For a Sinusoidal Source

Driving a Complex (Z = R + jX) Load

Similarly for the rms Current If the Load is Purely

Resistive

Now, By the “effective” Definition for a R Load

2,

2,

2

1

2

1resM

resMav RI

R

VP

22

2

1

2

1M

Mav RI

R

VP

MMrms

rmsM

rmseffdcMav

VVV

VV

R

V

R

V

R

V

R

VP

707.02

2

2

1

22

2222

MMrms

rmsM

rmseffdcMav

III

II

RIRIRIRIP

707.02

2

2

1

22

2222



Sinusoidal rms Values cont In General for a

Sinusoidal Quantity Thus the Power to a

Reactive Load Can be Calculated using These Quantities as Measured at the SOURCE• Using a True-rms DMM

– The rms Voltage– The rms Current

• Using an Oscilloscope and “Current Shunt”– The Phase Angle

Difference

For the General, Complex-Load Case

By the rms Definitions

2

is valueeffective theand

)cos()(

Mrmseff

M

UUU

tUtu

)cos(22

)cos(2

1

ivMM

ivMMav

IV

IVP

)cos( ivrmsrmsav IVP



Example rms Voltage Given Voltage

Waveform Find the rms Voltage Value

During the 2s Rise Calc the slope• m = [4V/2s] = 2 V/s

Thus The Math Model for the First Complete Period

Find The Period• T = 4 s

Derive A Math Model for the Voltage WaveForm

Use the rms Integral

T

420

202

t

tttv

Tt

t

rms dttuT

U0

0

)(1 2



Example rms Voltage cont. Calc the rms Voltage

Numerically

T

dtdttdttvVrms 4

2

2

0

24

0

2 04

1)2(

4

1

4

1

VVtVrms 633.1)(3

8

3

12

0

3

0



Example Average Power Given Current

Waveform Thru a 10Ω Resistor, then Find the Average Power The “squared” Version

Find The Period• T = 8 s

Apply The rms Eqns

Then the Power

Tt

t

rms dttuT

U0

0

)(1 2 RIP rmsav

2

26

4

22

0

22 8448

1Adtdt

sI

s

s

s

rms

WARIP rmsav 80108 22



Sinusoidal Forcing Functions Consider the Arbitrary

LINEAR Ckt at Right. If the independent

source is a sinusoid of constant frequency then for ANY variable in the LINEAR circuit the STEADY-STATE Response will be SINUSOIDAL and of the SAME FREQUENCY

Mathematically

Thus to Find iss(t), Need ONLY to Determine Parameters A &

)sin()(

)sin()(

tAti

tVtv

SS

M



Example RL Single Loop Given Simple Ckt Find

i(t) in Steady State Write KVL for Single

Loop

In Steady State Expect Sub Into ODE and

Rearrange

dt

tdiLtRitv )(

tAtAtdt

di

tAtAti

tAtAti

tAtAti

tAti

cossin)(

sincos)(

sinsincoscos)(

sinsincoscos)(

sinysinx-cosycosx y)cos(x using

)cos()(

21

21

tRAAL

tRAAL

tVM

cos)(

sin)(

cos

12

21



Example RL Single Loop cont Recall the Expanded

ODE

Equating the sin & cos PreFactors Yields Solving for

Constants A1 and A2

tRAALtRAAL

tVM

cos)(sin)(

cos

1221

MVRAAL

RAAL

12

21 0

Recognize as an ALGEBRAIC Relation for 2 Unkwns in 2 Eqns

222221 )(,

)( LR

LVA

LR

RVA MM

Found A1 & A2 using ONLY Algebra• This is Good



Example RL Loop cont.2

Using A1 & A2 State the i(t) Soln

Also the Source-V

If in the ID =x, and ωt = y, then

Comparing Soln to Desired form → use sum-formula Trig ID 222

221

)(sin

)(cos

LR

LVAA

LR

RVAA

M

M

tAtAti sincos)( 21

tVtv M cos)( Would Like Soln in

Form )cos()( tAti

yxAyxAyxA sinsincoscos )cos(

222 )( LR

LVA M

221 )( LR

RVA M



Example RL Loop cont.3 Dividing These Eqns

Find

Now

Elegant Final Result, But VERY Tedious Calc for a SIMPLE Ckt • Not Good

Subbing for A & in Solution Eqn

2222

22

sincos

sincos

AA

AA

Find A to be

R

L

A

A

cos

sintan

22 )( LR

VA M

R

Lt

LR

Vti M

1

22tancos

)()(



Complex Exponential Form Solving a Simple,

One-Loop Circuit Can Be Very Tedious for Sinusoidal Excitations

To make the analysis simpler relate sinusoidal signals to COMPLEX NUMBERS. • The Analysis Of the

Steady State Will Be Converted To Solving Systems Of Algebraic Equations ...

Start with Euler’s Identity (Appendix A)

sincos je j • Where 1j

Note:• The Euler Relation can

Be Proved Using Taylor’s Series (Power Series) Expansion of ej



Complex Exponential cont Now in the Euler

Identity, Let

So Notice That if

t

tjte tj sincos Next Multiply by a

Constant Amplitude, VM

tjVtVeV MMtj

M sincos Separate Function

into Real and Imaginary Parts

tVeV

tVeV

Mtj

M

Mtj

M

sinIm

cosRe

tVeVtv

tVtv

Mtj

M

M

cosRe

Then

cos

Now Recall that LINEAR Circuits Obey SUPERPOSITION



Complex Exponential cont.2 Consider at Right The

Linear Ckt with Two Driving Sources

By KVL The Total V-src Applied to the Circuit

tjM

MM

eI

tjItI

tititi

sincos21

Now by SuperPosition The Current Response to the Applied Sources

This Suggests That the…

tjVtv M sin2

tVtv M cos1 GeneralLinearCircuit

tjM

MM

eV

tjVtV

tvtvtv

sincos21



Complex Exponential cont.3 Application of the

Complex SOURCE will Result in a Complex RESPONSE From Which The REAL (desired) Response Can be RECOVERED; That is

tjMM eItI Recos

Thus the To find the Response a COMPLEX Source Can Be applied.

Then The Desired Response can Be RECOVERED By Taking the REAL Part of the COMPLEX Responseat the end of theanalysis

tjVtv M sin2




Realizability We can NOT Build

Physical (REAL) Sources that Include IMAGINARY Outputs

We CAN, However, BUILD These

We can also NOT invalidate Superposition if we multiply a REAL Source by ANY CONSTANT Including “j”

Thus Superposition Holds, mathematically, for

tjVtv M sin2


tjV

eV

M

tjM

sin

tV

tV

M

M

sin

costjVtVeV MM

tjM sincos



Example RL Single Loop This Time, Start with a

COMPLEX forcing Function, and Recover the REAL Response at The End of the Analysis• Let

In a Linear Ckt, No Circuit Element Can Change The Driving Frequency, but They May induce a Phase Shift Relative to the Driving Sinusoid

Thus Assume Current Response of the Form

tjMeVtv )(

tjMeVtv )(

tjjM

tjM eeIeIti )(

Then The KVL Eqn

)()()( tdt

diLtRitv



Example – RL Single Loop cont. Taking the 1st Time

Derivative for the Assumed Solution

Then the Right-Hand-Side (RHS) of the KVL

Then the KVL Eqn

tjMeVtv )(

Canceling ejt and Solving for IMej

tjM

tjM eIjeI

dt

d

dt

tdi

tjjM

tjM

tjM

tjM

eeIRLj

eIRLj

eIReIjL

tRitdt

diL

)(

)(

)()(

tjM

tjjM eVeeIRLj )(

RLj

VeI Mj

M



Example – RL Single Loop cont.2 Clear Denominator of

The Imaginary Component By Multiplying by the Complex Conjugate

Then the Response in Rectangular Form: a+jb

tjMeVtv )(

Next A & θ

LjR

LjR

LjR

VeI Mj

M

bLR

LV

aLR

RV

M

M

22

22

)(

)(

22 )(

)(

LR

LjRVeI Mj

M



Example – RL Single Loop cont.3 First A

And θ

tjMeVtv )(

222

2

222

2

22

)(

LR

V

LR

LVRV

baA

M

MM

RL

RL

ab

1

1

1

tan

tan

tan

A & Correspondence with Assumed Soln• A → IM

• θ → Cast Solution into

Assumed Form

aLR

RVM 22 )(

bLR

LVM

22 )(



Example – RL Single Loop cont.4 The Complex

Exponential Soln

Where

tjMeVtv )(

R

Lj

MjM e

LR

VeI

1tan

22

Recall Assumed Soln

R

L

LR

VI MM

1

22tan,

Finally RECOVER the DESIRED Soln By Taking the REAL Part of the Response

tjtI

eIeeIti

M

tjM

tjjM

sincos

)(



Example – RL Single Loop cont.5 By Superposition

Explicitly

tjMeVtv )(

R

Lt

LR

Vti M

1

22tancos

)(

SAME as Before

)cos()(

}Re{)(

}Re{cos)(

tIti

eIti

eVtVtv

M

tjM

tjMM



Phasor Notation If ALL dependent

Quantities In a Circuit (ALL i’s & v’s) Have The SAME FREQUENCY, Then They differ only by Magnitude and Phase• That is, With Reference to

the Complex-Plane Diagram at Right, The dependent Variable Takes the form

Borrowing Notation from Vector Mechanics The Frequency PreFactor Can Be Written in the Shorthand “Phasor” Form

tjjtj eAeAetx )(

A

b

a Real

Imaginary

AAe j



Phasor Characteristics Since in the Euler Reln

The REAL part of the expression is a COSINE, Need to express any SINE Function as an Equivalent CoSine• Turn into Cos Phasor

Examples

90)sin(

So

)90cos()sin(

ID Trig theand

)cos(

AtA

AA

AtA

Phasors Combine As the Complex Polar Exponentials that they are

42512)425377cos(12)( ttv

8.8518)902.42513cos(18

)2.42513sin(18)(

t

tty

)())(( 21212211 VVVV

)( 212

1

22

11

V

V

V

V



Example RL Single Loop As Before Sub a

Complex Source for the Real Source

The Form of the Responding Current

For the Complex Quantities

Then The KVL Eqn in the Phasor Domain

tjM

M

eVv

V

0V

tjjM

M

eeIi

I

I

Recall The KVL

vtRitdt

diL )()(

tjM

tjjM eVeeIRLj )(

LjR

eIRLj jM

V

I

IVII as

• Phasor Variables Denoted as BOLDFACE CAPITAL Letters



Example RL Single Loop cont. To recover the desired

Time Domain Solution Substitute

Then by Superposition Take

This is A LOT Easier Than Previous Methods

The Solution Process in the Frequency Domain Entailed Only Simple Algebraic Operations on the Phasors

tjM VevV 0V

tjjMM eeIiI I

tI

eI

eeIti

M

tjM

tjjM

cos

Re

Re



Resistors in Frequency Domain The v-i Reln for R

Thus the Frequency Domain Relationship for Resistors

MM

tjM

tjM

M

RIV

eVeRI

tIti

tRitv

Then

cos if

)()(

)()(

IV R



Resistors in -Land cont. Phasors are complex

numbers. The Resistor Model Has A Geometric Interpretation

In the Complex-Plane The Current & Voltage Are CoLineal• i.e., Resistors induce

NO Phase Shift Between the Source and the Response

• Thus resistor voltage and current sinusoids are said to be “IN PHASE”

R → IN Phase



Inductors in Frequency Domain The v-i Reln for L

Thus the Frequency Domain Relationship for Inductors

iv

i

iv

jM

jM

tjjM

tjM

tjM

eLIjeV

eLIj

eIdt

dLeV

or

)(

IV Lj



Inductors in -Land cont. The relationship

between phasors is algebraic.• To Examine This Reln

Note That

Thus

909011 jej

Therefore the Current and Voltage are OUT of PHASE by 90°• Plotting the Current and

Voltage Vectors in the Complex Plane

90

90

90

IV

IV

L

eLIeV

eeLI

ejLIeV

Lj

iv

i

iv

jM

jM

jjM

jM

jM



Inductors in -Land cont.2 In the Time Domain

Phase Relationship Descriptions• The VOLTAGE LEADS

the current by 90°• The CURRENT LAGS

the voltage by 90°

Short Example

)( Find

)20377cos(12)(,20

Given

ti

ttvmHL

90

2012

901 with so

AVL

V

j

I

)(701020377

123

A

I

In the Time Domain

)70377cos(593.1)( tAti

Lj

VI

V

2012

377

L → current LAGS



Capacitors in Frequency Domain The v-i Reln for C

Thus the Frequency Domain Relationship for Capacitors

vi

v

vi

jM

jM

tjM

tjM

tjM

eCVjeI

eCVj

eVdt

dCeI

or

)(

VI Cj



Capacitors in -Land cont. The relationship

between phasors is algebraic.• Recall

Thus

909011 jej

90

90

90

VI C

eCV

eCVe

eCVjeI

v

v

vi

jM

jM

j

jM

jM

Therefore the Voltage and Current are OUT of PHASE by 90°• Plotting the Current and

Voltage Vectors in the Complex Plane



Capacitors in -Land cont.2 In the Time Domain

Phase Relationship Descriptions• The CURRENT LEADS

the voltage by 90°• The VOLTAGE LAGS

the current by 90°

Short Example

In the Time Domain

)( Find

)15314cos(100)(,100

Given

ti

ttvFC

15100901

901with

C

j

I

VI

V

Cj

15100

314

)(10510010100314 6 A I

)105314cos(14.3)( tAti

C → current LEADS



All Done for Today

Root ofthe MeanSquare

rms voltage = 0.707 peak voltage rms voltage = 1.11 average voltagepeak voltage = 1.414 rms voltage peak voltage = 1.57 average voltageaverage voltage = 0.637 peak voltage average voltage = 0.9 rms voltage



WhiteBoard Work

2

+

_

v (t)

i( t)

Let’s Work Text Problem 8.5

45377

sin12

180377

cos10

2 1

ts

Vt

v

ts

Vt

v



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

AppendixComplex

No.s



Complex Numbers Reviewed Consider a General

Complex Number

This Can Be thought of as a VECTOR in the Complex Plane

This Vector Can be Expressed in Polar (exponential) Form Thru the Euler Identity

Where

)sin(cos

jA

Aejba j

jban

Ab

a Real

Imaginary

11 jjj Then from the Vector Plot

a

b

baA

1

22

tan



Complex Number Arithmetic Consider Two Complex

Numbers

The SUM, Σ, and DIFFERENCE, , for these numbers

The PRODUCT n•m

j

j

Dejdcm

Aejban

Complex DIVISION is Painfully Tedious• See Next Slide

dbjcamn

dbjcamn

jjj ADeDeAemn

adbcjbdac

bdjadbcjac

jdcjbamn2



Complex Number Division For the Quotient n/m in

Rectangular Form

The Generally accepted Form of a Complex Quotient Does NOT contain Complex or Imaginary DENOMINATORS

Use the Complex CONJUGATE to Clear the Complex Denominator

jdc

jba

m

n

The Exponential Form is Cleaner• See Next Slide

22

22

2

dc

adbcjbdac

m

n

dcddcjc

bdjadbcjac

m

n

jdc

jdc

jdc

jba

m

n



Complex Number Division cont. For the Quotient n/m in Exponential Form

However Must Still Calculate the Magnitudes A & D...

j

j

j

eD

A

De

Ae

m

n



Phasor Notation cont. Because of source

superposition one can consider as a SINGLE source, a System That contains REAL and IMAGINARY Components

The Real Steady State Response Of Any Circuit Variable Will Be Of The Form

Or by SuperPosition

Since ejt is COMMON To all Terms we can work with ONLY the PreFactor that contains Magnitude and Phase info; so

tjjM

M

eeU

tUtu

Re

)cos()(

)cos()( tYty M

}Re{}Re{ tjjM

tjjM eeYeeU

)cos(}Re{)(

)cos()(

tYty

YU

tUtu

M

MM

M

Y

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[email protected] ENGR-43_Lec-08-1_AC-Steady-State.ppt 1 Bruce Mayer, PE Engineering-43:...

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Transcript of [email protected] ENGR-43_Lec-08-1_AC-Steady-State.ppt 1 Bruce Mayer, PE Engineering-43:...