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Chapter 11Chapter 11

AC Power AnalysisAC Power Analysis

電路學電路學 (( 二二 ))

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AC Power AnalysisAC Power Analysis Chapter 11Chapter 11

11.1 Instantaneous and Average Power11.2 Maximum Average Power Transfer11.3 Effective or RMS Value11.4 Apparent Power and Power Factor11.5 Complex Power 11.6 Conservation of AC Power11.7 Power Factor Correction11.8 Power Measurement

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Sinusoidal power at 2t Constant power

) 2( cos 2

1 ) ( cos

2

1

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

ivmmivmm

ivmm

tIVIV

ttIVtitvtp

11.1 Instantaneous and 11.1 Instantaneous and Average Power (1)Average Power (1)

• The instantaneously power, p(t)

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

1cos cos [cos( ) cos( )]

2A B A B A B

4

) ( cos 2

1 )(

1

0 ivmm

TIVdttp

TP

11.1 Instantaneous and 11.1 Instantaneous and Average Power (2)Average Power (2)

• The average power, P, is the average of the instantaneous power over one period.

1. P is not time dependent.

2. When θv = θi , it is a purely

resistive load case.

3. When θv– θi = ±90o, it is a

purely reactive load case.

4. P = 0 means that the circuit

absorbs no average power.

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11.1 Instantaneous and 11.1 Instantaneous and Average Power (3)Average Power (3)

*1 1

2 21

[cos( ) sin( )]2

m m v i

m m v i v i

V I

V I j

VI

*1 1Re[ ] cos( )

2 2 m m v iP V I VI

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Example 1

Given that

find the instantaneous power and average power absorbed by a passive linear network .

)10 377( sin 01 )(

)45 377( cos201 )(

tti

ttv

11.1 Instantaneous and 11.1 Instantaneous and Average Power (4)Average Power (4)

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)60 10( sin 15 )(

)20 10( cos80 )(

tti

ttv

11.1 Instantaneous and 11.1 Instantaneous and Average Power (5)Average Power (5)

Example 2

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

Answer: 387.5W )W,10600cos(20t385.7

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Example 3

Calculate the average power absorbed by an impedance Z = 30 j70 when a voltage V = 1200 is applied across it.

11.1 Instantaneous and 11.1 Instantaneous and Average Power (6)Average Power (6)

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Example 4

A current flows through an impedance . Find the average power delivered to the impedance.

11.1 Instantaneous and 11.1 Instantaneous and Average Power (6)Average Power (6)

Answer: 927.2W

30 10 IΩ2220 Z

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11.2 Maximum Average Power 11.2 Maximum Average Power Transfer (1)Transfer (1)

LLL X j R Z

THTHTH X j R Z

The maximum average power can be transferred to the load if

XL = –XTH and RL = RTH

TH

2

THmax R 8

V P

If the load is purely real, then TH2TH

2THL Z X R R

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11.2 Maximum Average Power 11.2 Maximum Average Power Transfer (2)Transfer (2)

Example 5For the circuit shown below, find the load impedance ZL that absorbs the maximum average power. Calculate that maximum average power.

Answer: 3.415 – j0.7317, 1.429W

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11.3 Effective or RMS Value (1)11.3 Effective or RMS Value (1)

RIdtiT

RRdti

TP rms

T T 2

0 0

221

Hence, Ieff is equal to:rms

T

eff IdtiT

I 1

0

2

The effective of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.

The effect value of a periodic signal is its root-mean-square (rms) value.

The total power dissipated by R is given by:

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11.3 Effective or RMS Value (2)11.3 Effective or RMS Value (2)

2I

I mrms

The average power can be written in terms of the rms values:

2222

coscos21

XRR

VRI

) θ (θ IV) θ (θ IVP

rmsrms

ivrmsrmsivmmeff

Note: If you express amplitude of a phasor source(s) in rms, then all the answer as a result of this phasor source(s) must also be in rms value.

The rms value of a sinusoid i(t) = Imcos(t) is given by:

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

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11.3 Effective or RMS Value (3)11.3 Effective or RMS Value (3)

Example 6Determine the rms value of the current waveform. If the current is passed through a 2- resistor, find the average power absorbed by the resistor.

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11.3 Effective or RMS Value (4)11.3 Effective or RMS Value (4)

Example 7The waveform is a half-wave rectified sine wave. Find the rms value and the average power dissipated in a 10- resistor.

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11.4 Apparent Power and 11.4 Apparent Power and Power Factor (1)Power Factor (1)

• Apparent Power 視功率 , 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.

• Power factor is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle of the load impedance.

)θ (θ cos S )θ (θ cos I VP ivivrmsrms

Apparent Power, S Power Factor, pf

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11.4 Apparent Power and 11.4 Apparent Power and Power Factor (2)Power Factor (2)

Purely resistive load (R)

θv– θi = 0, pf = 1 P/S = 1, all power are consumed

Purely reactive load (L or C)

θv– θi = ±90o,

pf = 0

P = 0, no real power consumption

Resistive and reactive load (R and L/C)

θv– θi > 0

θv– θi < 0

• Lagging - inductive load

• Leading - capacitive load

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11.4 Apparent Power and 11.4 Apparent Power and Power Factor (3)Power Factor (3)

Example 8A series-connected load draws a current i(t)=4cos(100t + 10) A when the applied voltage is v(t)=120cos(100t 20) V. Find the apparent power and the power factor of the load. Determine the element values that from the series-connected load.

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11.4 Apparent Power and 11.4 Apparent Power and Power Factor (4)Power Factor (4)

Example 9Determine the power factor of the entire circuit of the figure as seen by the source. Calculate the average power delivered by the source.

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11.5 Complex Power (1)11.5 Complex Power (1)Complex power S is the product of the voltage and the complex conjugate of the current:

imvm θIθV IV

ivrmsrms θ θ I V I V2

1

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11.5 Complex Power (2)11.5 Complex Power (2)

ivrmsrms θ θ I V I V2

1 S

)θ (θsin I V j )θ (θ cos I VS ivrmsrmsivrmsrms

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 and the reactive part of the load. It is measured in VAR.

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

S = P + j Q

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11.5 Complex Power (3)11.5 Complex Power (3)

)θ (θsin I V j )θ (θ cosI VS ivrmsrmsivrmsrms

Apparent Power, S = |S| = Vrms*Irms =

Real power 實功率 , P = Re(S) = S cos(θv – θi)

Reactive Power 虛功率 , Q = Im(S) = S sin(θv – θi)

Power factor 功率因子 , pf = P/S = cos(θv – θi)

S = P + j Q

22 QP

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11.5 Complex Power (4)11.5 Complex Power (4)

)θ (θsin I V j )θ (θ cos I V S ivrmsrmsivrmsrms

S = P + j Q

Impedance Triangle Power FactorPower Triangle

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11.5 Complex Power (5)11.5 Complex Power (5)Example 10The voltage across a load is v(t)=60cos(t 10) V and the current through the element in the direction of the voltage drop is i(t)=1.5cos(t + 50) A. Find: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance.

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11.5 Complex Power (6)11.5 Complex Power (6)Example 11A load Z draws 12 kVA at a power factor of 0.856 lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance.

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11.6 Conservation of AC Power (1)11.6 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

* 21

21

)(21

21

SSVIVIIIVVIS

The same results can be obtained for a series connection.

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Example 12The figure shows a load being fed by a voltage source through a transmission line. The impedance of the line is represented by the (4 + j2) impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load.

11.6 Conservation of AC Power (2)11.6 Conservation of AC Power (2)

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Example 13In the circuit, Z1 = 6030 and Z2 = 4045 . Calculate the total: (a) apparent power, (b) real power, (c) reactive power, and (d) pf, supplied by the source and seen by the source.

11.6 Conservation of AC Power (3)11.6 Conservation of AC Power (3)

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11.7 Power Factor Correction (1)11.7 Power Factor Correction (1)Power factor correction is the process of increasing the power factor without altering the voltage or current to the original load.

Power factor correction is necessary for economic reason.

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11.7 Power Factor Correction (2)11.7 Power Factor Correction (2)

Q1 = S1 sin θ1 = P tan θ1

Q2 = P tan θ2

Qc = Q1 – Q2

= P (tan θ1 - tan θ2)

= ωCV2rms

P = S1 cos θ1

2rms

212rms

c

V ω

)θtan θ(tan P

ωV

Q C

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Example 14When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

11.7 Power Factor Correction (3)11.7 Power Factor Correction (3)

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11.8 Power Measurement (1)11.8 Power Measurement (1)The wattmeter is the instrument for measuring the average power.

If )cos()( vm tVtv and )cos()( im tIti

)θ (θ cos I V )θ (θ cos I V P ivmm21

ivrmsrms

The basic structure Equivalent Circuit with load