Post on 14-Mar-2020
EE 221 EE 221 Circuits IICircuits IIChapter 11Chapter 11AC Power AnalysisAC Power Analysis
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AC Power AnalysisAC Power AnalysisChapter 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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11.1 Instantaneous and Average Power (1)11.1 Instantaneous and Average Power (1)
Instantaneously power, p(t)
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Sinusoidal power at 2ωtConstant power
) 2( cos 21 ) ( cos
21
) ( cos ) ( cos )( )( )(
ivmmivmm
ivmm
tIVIV
ttIVtitvtp
θθωθθ
θωθω
+++−=
++==
p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.
11.1 Instantaneous and Average Power (2)11.1 Instantaneous and Average Power (2)
The average power, P, is the average of the instantaneous power over one period.
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) ( cos 21 )(1
0 ivmm
TIVdttp
TP θθ −== ∫
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.
11.1 Instantaneous and Average Power (3)11.1 Instantaneous and Average Power (3)
Example 1
Calculate the instantaneous power and average power absorbed by a passive linear network if:
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)60 10( sin 15 )()20 10( cos80 )(
°+=°+=
ttittv
Answer: 387.5W )W,10 600cos(20t385.7 °−+
11.1 Instantaneous and Average Power (4)11.1 Instantaneous and Average Power (4)
Example 2
A current flows through an impedance
. Find the average power
delivered to the impedance.
°∠= 30 10 I
Answer
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: 927.2W
Ω2220 °−∠=Z
11.2 Maximum Power Transfer (1)11.2 Maximum Power Transfer (1)
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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
2TH
max R 8V
P =
If the load is purely real, then TH2TH
2THL Z X R R =+=
11.2 Maximum Power Transfer (2)11.2 Maximum Power Transfer (2)
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Example 3For the circuit shown below, find the load impedance ZLthat absorbs the maximum average power. Calculate that maximum average power.
Answer: Zth =3.415 – j0.7317Ω, 1.429W
11.3 Effective or RMS Value (1)11.3 Effective or RMS Value (1)
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RIdtiTRRdti
TP rms
T T 2
0 0
221=== ∫ ∫
Hence, Ieff is equal to: rms
T
eff IdtiT
I 1
0
2 == ∫
The effective value of a periodic current is equal to the dc current that delivers the same average power to a resistor as the periodic current.
The rms value is a constant itself which depending on the shape of the function i(t).
The total power dissipated by R is given by:
11.3 Effective or RMS Value (2)11.3 Effective or RMS Value (2)
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2I I m
rms =
The average power can be written in terms of the rms values:
)θ (θ cos I V )θ (θ cos I V21 P ivrmsrmsivmm −=−=
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(ωt)is given by:
11.4 Apparent Power and Power 11.4 Apparent Power and Power Factor 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.
Power factor is defined as PF = P/S.In sinusoidal circuits, PF 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 −=−=
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Apparent Power, S Power Factor, pf
11.4 Apparent Power and Power 11.4 Apparent Power and Power Factor Factor
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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 or C)
θv– θi > 0θv– θi < 0
• Lagging - inductive load
• Leading - capacitive load
11.5 Complex Power (1)11.5 Complex Power (1)
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Complex power S is the product of the phasor voltage and the complex conjugate of the phasor current:
imvm θIθV ∠=∠= IV
)θ (θ I V I V21
ivrmsrms −∠=∗
11.5 Complex Power (2)11.5 Complex Power (2)
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ivrmsrms θ θ I V I V21 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 andthe 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
11.5 Complex Power (3)11.5 Complex Power (3)
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)θ (θsin I V j )θ (θ cosI VS ivrmsrmsivrmsrms −+−=⇒
Apparent Power, S = |S| = Vrms*Irms =
Real power, P = Re(S) = S cos(θv – θi)
Reactive Power, Q = Im(S) = S sin(θv – θi)
Power factor, pf = P/S = cos(θv – θi)
S = P + j Q
22 QP +
11.5 Complex Power (4)11.5 Complex Power (4)
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)θ (θsin I V j )θ (θ cos I V S ivrmsrmsivrmsrms −+−=⇒
S = P + j Q
Impedance Triangle Power FactorPower Triangle
11.6 Conservation of AC Power (1)11.6 Conservation of AC Power (1)
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The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads. The above statement does not apply to apparent power.
For parallel connection:
21*2
*1
*2
*1
* S S I V 21 I V
21 )I I( V
21 I V
21 S +=+=+==
The same results can be obtained for a series connection.
11.7 Power Factor Correction (1)11.7 Power Factor Correction (1)
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Power factor correction is the process ofincreasing the power factor without altering the voltage or current to the original load.
Power factor correction is necessary for economic reason.
11.7 Power Factor Correction (2)11.7 Power Factor Correction (2)
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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
ωVQ C −
==
11.8 Power Measurement (1)11.8 Power Measurement (1)
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The wattmeter is the instrument for measuring the average power.
If )cos()( vm tVtv θω += and )cos()( im tIti θω +=
)θ (θ cos I V )θ (θ cos I V P ivmm21
ivrmsrms −=−=
The basic structure Equivalent Circuit with load