ac power analysis

BENT 2113BENT 21131111

BENT 2113BENT 2113ELECTRIC CIRCUIT IIELECTRIC CIRCUIT II

AC Power AnalysisAC Power AnalysisAC Power AnalysisAC Power Analysis

1 Instantaneous and Average Power1 Instantaneous and Average Power2 Maximum Average Power Transfer3 Effective or RMS Value4 Apparent Power and Power Factor5 Complex Power 6 Conservation of AC Power6 Conservation of AC Power7 Power Factor Correction8 Power Measurement

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IntroductionIntroductionIntroductionIntroductionPrevious Chapters mainly focused on calculation ofvoltage and current.gThis chapter focus on Power analysisPower Analysis is of paramount importancePower is most important quantities in electricsutilities electronics and communications systemsutilities, electronics and communications systems –such sys involve transmission of power from onepoint to one point.every industries and household electrical device haspower rating that indicates how much power thepower rating that indicates how much power theequipment requires.Exceeding power rating can do permanent damageto an appliances.Most common form of electric power is 50 or 60Hzac power.The choice of ac over dc allowed high-voltagepower transmission from power generating plant to

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power transmission from power generating plant toconsumer.

4.1 Instantaneous and Average 4.1 Instantaneous and Average P (1)P (1)Power (1)Power (1)

• The instantaneously power, p(t) Is the power at any instant of time

)2(cos21)(cos

21

) ( cos ) ( cos )( )( )(

ivmmivmm

ivmm

tIVIV

ttIVtitvtp

θθωθθ

θωθω

+++−=

++==

Sinusoidal power at 2ωtConstant power

)(2

)(2 ivmmivmm

4p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.

4.1 Instantaneous and Average Power4.1 Instantaneous and Average PowerTh i t t l (t)• The instantaneously power, p(t) Is the power at any instant of time

)()()( titvtp =• Where v(t) = instantaneous voltage across the element• i(t)= instantaneous current through it

• Voltage and current at circuit terminals:• v(t)= Vm cos (ωt+θv)• i(t)= Im cos (ωt+θi)

• Therefore the instantaneous power absorbed by circuits elements:

• p(t)=v(t)i(t)= Vm Imcos (ωt+θv) cos (ωt+θi)• Applying the trigonometric identity:

•)]cos()[cos(

21coscos BABABA ++−=

5)2cos(Im21)cos(Im

21)( ivtVmivVmtp θθωθθ +++−=

4.1 Instantaneous and Average 4.1 Instantaneous and Average PPPowerPower

• The average power, P, is the average of the instantaneous

)( cos21 )(1

0 ivmm

TIVdttp

TP θθ −== ∫

power over one period.

)(2

)(0 ivmmp

T ∫1. P is not time dependent. 2 Wh θ θ it i l2. When θv = θi , it is a purely

resistive load case. 3. When θv– θi = ±90o, it is a

l ti l dpurely reactive load case. 4. P = 0 means that the circuit

absorbs no average power.

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4.1 Instantaneous and Average Power4.1 Instantaneous and Average Power

Example 1Calculate the instantaneous power and average power absorbed by a passive linear network if:power absorbed by a passive linear network if:

AttiVttv

)10377cos(10)()45377cos(120)(

°−=°+=

Solution:• The instantaneous power

Applying trigonometry identity:

Atti )10377cos(10)()10377cos()45377cos(1200 °−°+== ttvip

• Applying trigonometry identity:

Wtp )]35754cos()55[(cos600 °++°=

• The average power WP 2.34455cos600 =°=

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4.1 Instantaneous and Average Power4.1 Instantaneous and Average PowerExercise 1

Calculate the instantaneous power and average power absorbed by a passive linear network if:

AttiVttv

)6010sin(20)()2010cos(165)(

°+=°+=

Answer:

Atti )6010sin(20)( +

kWta )1020cos(65106061) °−+ kWta )1020cos(65.10606.1) −+

kWb 0606.1)

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Example 2


Example 2

For the circuit below, find the average power supplied by thesource and the average power absorbed by the resistor.

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• Solution: • Current I: °∠=

°∠= 57.56118.1305I

• Average power by

∠−

57.56118.124 j

I

WP 52)575630cos()1181)(5(1=°−°=g p y

voltage source :

• Current through resistor:

WP 5.2)57.5630cos()118.1)(5(2

AII °∠== 57561181• Current through resistor:• Voltage across it:

AIIR ∠== 57.56118.1

VIV RR °∠== 57.56472.44• Therefore average power absorbed by the resistor is :

WP 5.2)118.1)(472.4(21

==

10

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4.2 Maximum Average Power Transfer (1)4.2 Maximum Average Power Transfer (1)• Consider the circuit in figure, where an ac circuit is connected to a loadg ,ZL and is represented by its Thevenin equivalent.• The load is usually represented by an impedance, which may model anelectric motor, an antenna and etc.• In rectangular form the Thevenin impedance Z & load impedance Z

THTHTH X j R Z +=

• In rectangular form, the Thevenin impedance ZTH & load impedance ZL

LLL X j R Z +=

We come to the conclusion, for maximum average power transfer, ZL must be selected

XL = –XTH and RL = RTH

g p , Lso that

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4.2 Maximum Average Power Transfer4.2 Maximum Average Power Transfer

Therefore:

*THTHTHLLL ZjXRjXRZ =−=+=

For maximum average power transfer, the load impedance ZL must be equal to the complex conjugate of the Thevenin impedance ZTH.

When ZL=ZTH, we say that the load is matched to the source.

Maximum average power transferTH

2TH

max R 8V

P =

(means maximum average power transfer to a purely resistive load, theIf the load is purely real, then TH

2TH

2THL Z X R R =+=

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load impedance (or resistance) is equal to the magnitude of the TheveninImpedance.

4.2 Maximum Average Power Transfer4.2 Maximum Average Power Transfer

E l 3Example 3

Determine the load impedance ZL that maximizes the

average power drawn from the circuit What is the maximumaverage power drawn from the circuit. What is the maximum

average power?

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Maximum Average Power TransferMaximum Average Power Transfer

S l ti• Solution:• First, obtain the Thevenin equivalent at the load

terminals because *ZjXRjXRZ +terminals because *THTHTHLLL ZjXRjXRZ =−=+=

Ω+=−+= 467.4933.2)68(45 jjlljZTH

• Therefore, the load impedance draws the maximum power from the circuit when

Ω−== 467.4933.2* jZZ THL

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• Next, to find the maximum average power.• Formula: 2

THV

• Have to find VTH, using the voltage divider:TH

THmax R 8

V P =

Vj

jVTH °−∠=−+

−= 3.10454.7)10(

68468j

• Therefore, the maximum average power is

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15W368.2

)933.2(8454.7

R 8V

P2

TH

2TH

max ===

4.2 Maximum Average Power 4.2 Maximum Average Power T fT fTransferTransfer

Exercise 2Exercise 2

For the circuit shown below, find the load impedance ZL that absorbs the maximum average power Calculate that maximum absorbs the maximum average power. Calculate that maximum average power.

Answer: Z =3 415 j0 7317Ω

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Answer: ZL=3.415 – j0.7317Ω

: Max average power=1.4278W

4.3 Effective or RMS Value (1)4.3 Effective or RMS Value (1)

The effective value is the measure of ff ti f lt teffectiveness of a voltage or current source

in delivering power to a resistive load.

Hence, Ieff is equal to:rms

T

eff IdtiT

I 1

0

2 == ∫

The effective value of a periodic signal is itsroot mean square (rms) value.

The rms value is a constant itself whichdepending on the shape of the function i(t).

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4.3 Effective or RMS Value4.3 Effective or RMS Value

The rms value of a constant is the constant itself

Th l f i idThe rms value for a sinusoid :

cosI)( m= tti ω cos)( m tVtv = ω

2I

)(m

=rmsI2

)(Vm

rmsV =

• The average power absorbed by a resistor R

RVRIP rms

rms

22 ==

18


Example 4

Determine the rms value of the current waveform in figure. If

the current is passed through a 2Ω resistor, find the average

power absorbed by the resistor.

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• Solution

The period of the waveform is T=4 Therefore the current The period of the waveform is T=4. Therefore, the current waveform:

4220

.........105

)(<<<<

−=

ttt

ti

• The rms value is

0 t

AdtdttdtiT

IT

eff 165.8])10()5([41 1

4

2

22

0

2

0

2 =−+== ∫∫∫

• The power absorbed by 2Ω:

WRIP 3133)2()1658( 22

20

WRIP rms 3.133)2()165.8( 22 ===

4.4 Apparent Power and Power Factor4.4 Apparent Power and Power Factor

• Apparent Power, S, is the product of the r.m.s. values of voltage and current.

• It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in watts.

)θ (θ cos S )θ (θ cos I VP ivivrmsrms −=−=

Apparent Power, S Power Factor, pf

• Power factor is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle f th l d i d

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of the load impedance.

4.4 Apparent Power and Power Factor4.4 Apparent Power and Power FactorPower

)cos( ivSPpf θθ −==

Power factor angle

Purely resistive load (R)

θv– θi = 0, Pf = 1 P/S = 1, all power are consumedload (R) consumed

Purely reactive load (L or C)

θv– θi = ±90o, pf = 0

P = 0, no real power consumptionp

Resistive and reactive load

θv– θi > 0θv– θi < 0

• Lagging - inductive load

(R and L/C) v i • Leading - capacitive load

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Leading power factors-the current leads voltageLagging power factors- the current lags voltage

4.4 Apparent Power and Power Factor4.4 Apparent Power and Power FactorExample 5Example 5

• A series connected load draws a current i(t)=4 cos

(100∏t+100)A when the applied voltage is v(t)=120 cos(100∏t+10 )A when the applied voltage is v(t) 120 cos

(100∏t-200)V. Find the apparent power and the power

factor of the load.

Solution

• The apparent power is VAsIVS rmrms 2404120===The apparent power is

• The power factor is

VsVS mms 022

• The power factor is

866.0)1020cos()cos( =°−°−=−= ivpf θθ

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Since θv– θi < 0, it is leading

4 5 C l P (1)4 5 C l P (1)4.5 Complex Power (1)4.5 Complex Power (1)

Complex power S is the product of the voltage and theComplex power S is the product of the voltage and thecomplex conjugate of the current:It is use to find the total effect of parallel loads.

imvm θIθV ∠=∠= IV

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4 5 C l P4 5 C l P4.5 Complex Power4.5 Complex Power

θθIVIV1S ∠∗ivrmsrms θθI VIV

21S −∠== ∗

)θ(θsin IVj )θ (θcosIVS ivrmsrmsivrmsrms −+−=⇒ )(j)( ivrmsrmsivrmsrms

S = P + j Q

P: is the average power in watts delivered to a load and it is the only useful power.

Q: is the reactive power exchange between the source andthe reactive part of the load. It is measured in VAR.(volt-

ampere- reactive).

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• Q = 0 for resistive loads (unity pf).• Q < 0 for capacitive loads (leading pf).• Q > 0 for inductive loads (lagging pf).

4.5 Complex Power4.5 Complex Powerpp

1

Spowerapparent21power complex

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ivrmsrms

QPIV

IVjQP θθ

+====

−∠==+== ∗

S

VIS

)(iS)I (Qti)cos( S)Re(P power real

Spowerapparent

v i

rmsrms QPIV

θθθθ −===

+====

SS

S

)cos(SP factor power

)(sin S)Im(Qpower reactive v

iv

i

θθ

θθ

−==

−=== S

S

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4.5 Complex Power4.5 Complex Power

Example 6

• The voltage across a load is v(t)=60 cos (ωt 100)V • The voltage across a load is v(t)=60 cos (ωt-100)V

and the current through the element i(t) =1.5 cos

( 0) d(ωt+500)A. Find:

(a)Complex and apparent power

(b)Real and reactive powers( ) p

(c)Power factor and the load impedance

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((Solution)Solution) 4.5 Complex Power4.5 Complex Power

• The rms values of the voltage and current:• The rms values of the voltage and current:

°−∠= 102

60Vrms °+∠= 5025.1Irms

(a) the complex power

2 2

ivrmsrms IVjQP θθ −∠==+== ∗VIS21power complex 2

)5010()25.1)(

260( °−°−∠=S VA)60(45 °−∠=S

• Apparent power

22

S== Spowerapparent

therefore VA45S == S

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(b) Real and reactive powers

• Real power )Re(PpowerReal S==

VA°−∠= 6045S• Real power

• Reactive power

)Re(Ppower Real S==WP 5.22=

)Im(Qpower reactive S== )(Qp

)(97.38 reactiveamperevoltVARQ −−−=

(c) Power factor & load impedance • Power factor

)cos(SP factor power iv θθ −==S

leadingpf ),60cos()5010cos( °−=°−°−=

• Load impedance

Ω°∠°−∠ 60401060VZ

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Ω°−∠=°+∠

== 6040505.1I

Z

4.5 Complex Power4.5 Complex Power• Exercise 3For a load, .154.0,85110 AIrmsVVrms °∠=°∠=

Determine : (a) the complex and apparent powers

(b) the real and reactive powers(b) the real and reactive powers

(c) the power factor and the load

impedance impedance.

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4 6 C ti f AC P (1)4 6 C ti f AC P (1)4.6 Conservation of AC Power (1)4.6 Conservation of AC Power (1)The complex real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads.

For parallel connection:

21*2

*1

*2

*1

* SS IV21 I V

21 )II(V

21 I V

21 S +=+=+==

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212121 22)(

22The same results can be obtained for a series connection.

4 7 P F t C ti (1)4 7 P F t C ti (1)4.7 Power Factor Correction (1)4.7 Power Factor Correction (1)Power factor correction is the process of increasing the gpower factor without altering the voltage or current to the original load.

32Power factor correction is necessary for economic reason.

4 7 Power Factor Correction (2)4 7 Power Factor Correction (2)4.7 Power Factor Correction (2)4.7 Power Factor Correction (2)

Qc = Q1 – Q2

P (tan θ tan θ )= P (tan θ1 - tan θ2) = ωCV2rms

Q1 = S1 sin θ1= P tan θ1

2rms

212rms

c

V ω)θtan θ(tan P

ωVQ C −

==

33Q2 = P tan θ2P = S1 cos θ1

4 8 P M t (1)4 8 P M t (1)4.8 Power Measurement (1)4.8 Power Measurement (1)The wattmeter is the instrument for measuring the averageThe wattmeter is the instrument for measuring the average power.

If )cos()( vm tVtv θω += and )cos()( im tIti θω +=

The basic structure Equivalent Circuit with load

34)θ (θ cos I V )θ (θ cos I V P ivmm2

1ivrmsrms −=−=

ac power analysis

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