STEADY-STATE POWER ANALYSIS LEARNING GOALS Instantaneous Power For the special case of steady state sinusoidal signals Average Power Power absorbed or supplied during one cycle Maximum Average Power Transfer When the circuit is in sinusoidal steady state Effective or RMS Values For the case of sinusoidal signals Power Factor A measure of the angle between current and voltage phasors Complex Power Measure of power using phasors Power Factor Correction How to improve power transfer to a load by “aligning” phasors Single Phase Three-Wire Circuits Typical distribution method for households and small loads
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STEADY-STATE POWER ANALYSIS LEARNING GOALS Instantaneous Power For the special case of steady state sinusoidal signals Average Power Power absorbed or.
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STEADY-STATE POWER ANALYSISLEARNING GOALS
Instantaneous Power For the special case of steady state sinusoidal signals
Average Power Power absorbed or supplied during one cycle
Maximum Average Power Transfer When the circuit is in sinusoidal steady state
Effective or RMS Values For the case of sinusoidal signals
Power Factor A measure of the angle between current and voltage phasors
Complex Power Measure of power using phasors
Power Factor Correction How to improve power transfer to a load by “aligning” phasors
Single Phase Three-Wire Circuits Typical distribution method for households and small loads
INSTANTANEOUS POWER
)cos()(
)cos()(
iM
vM
tIti
tVtv
Statesteady In
)()()( titvtp Impedance to
SuppliedPower
ousInstantane
)cos()cos()( ivMM ttIVtp
)cos()cos(2
1coscos 212121
)2cos()cos(2
)( ivivMM t
IVtp
LEARNING EXAMPLE
)(),(
302
),60cos(4)(
tpti
Z
ttv
:Find
:Assume
)(302302
604A
Z
VI
))(30cos(2)( Atti
30,2
60,4
iM
vM
I
V
)902cos(430cos4)( ttp
constant Twice thefrequency
AVERAGE POWER
For sinusoidal (and other periodic signals)we compute averages over one period
2
T
)2cos()cos(2
)( ivivMM t
IVtp
)cos(2 iv
MM IVP
If voltage and current are in phase
MMiv IVP2
1
If voltage and current are in quadrature090 Piv
Purelyresistive
Purely inductive orcapacitive
It does not matterwho leads
LEARNING EXAMPLE Find the averagepower absorbedby impedance
)(1553.34522
6010
22
6010A
jI
15,60,53.3,10 ivMM IV
WP 5.12)45cos(3.35
Tt
t
dttpT
P0
0
)(1
RV
)(1506.7601022
2V
jVR
WP 53.306.72
1
Since inductor does not absorb powerone can use voltages and currents acrossthe resistive part
LEARNING EXAMPLEDetermine the average power absorbed by each resistor, the total average power absorbed and the average powersupplied by the source
)(4534
45121 AI
WP 183122
14
If voltage and current are in phase
MMiv IVP2
1 2
12
1MRI
R
VM2
2
1
)(57.7136.537.265
4512
12
45122 A
jI
)(36.522
1 22 WP W7.28
Inductors and capacitors do not absorbpower in the average
WPtotal 7.2818
absorbedsupplied PP WP 7.46supplied
Verification
57.7136.545321 III)(10.6215.8 AI
)cos(2 iv
MM IVP
)10.6245cos(15.8122
1 suppliedP
LEARNING EXTENSION Find average power absorbed by each resistor
I
iZI
6012
iZ )4||4(2 j
4524
6.713.25
44
1688
44
)4(42
j
jj
j
jZ i
6.2647.4iZ
)(6.8668.26.2647.4
6012AI
WRIP M 20.768.222
1
2
1 222
2I
6.8668.24524
904
44
42 I
j
jI
6.4190.12I
)(90.142
1 24 WP
LEARNING EXTENSIONFind the AVERAGE power absorbed by each PASSIVE component and the total power supplied by the source
1I2I
3010243
241 j
jI
)(62.4014.6301095.1528.7
57.2647.41 AI
)(14.632
1
2
1 223 WRIP M
62.4014.630102I
)(05.1412.4
95.1528.7
30303010
243
32
A
jI
)(12.442
1 24 WP
)(02 WPj
Power supplied by source
Method 1. absorbedsupplied PP
WPPP 50.9043 supplied
Method 2: )cos(2
1ivMM IVP
sV
62.4042.183 1IVs
)3062.40cos(1042.182
1 P
LEARNING EXAMPLEDetermine average power absorbed or supplied by each element
)(3062
30122 AI
)(36622
1
2
1 222 WRIP M
01 jP
To determine power absorbed/suppliedby sources we need the currents I1, I2
)(19.3643.7
39.466639.10
1
0630123
A
jj
j
jI
07.728.11
)(39.12.1139.46320.5321 AjjjIII
)(54
)07.730cos(28.11122
13012
W
P
)1836(
)cos(2
1ivMM IVP
Power Average
WP 18)19.360cos(43.762
106
Passive sign convention
R
VRIP M
M
22
2
1
2
1
resistorsFor
LEARNING EXTENSIONDetermine average power absorbed/supplied by eachelement
1I
2I
)012(42304
012
22
1
IIj
I
Equations Loop
57.2647.4
48246.3
24
0483042
j
jI
)(20497.957.2647.4
43.17758.442 AI
)(5.5492.19))(055.4108.912(4
))(20497.912(4)(4 214
VVj
VIIV
WR
VP M 6.49
4
92.19
2
1
2
1 22
4
)(02 WP j
)(4.69)05.54cos(1292.192
1012 WP
4V
)(8.19)20430cos()97.9(42
1304 WP
Check: Power supplied =power absorbed
j
VI
j
VV
2
304
02
304
4012
42
44
Equations Node
Alternative Procedure
)cos(2
1ivMM IVP
Power Average
R
VRIP M
M
22
2
1
2
1
resistorsFor
LEARNING EXTENSIONDetermine average power absorbed/supplied by eachelement
V
042
012
2
024
V
j
VV
Equation Node
4j
0)12(2)24(2 jVVVj
)(85.177.14
)(125.788.14
13
24192
32
32
32
4824
4824)32(
Vj
V
j
j
j
j
jV
jVj
1I
925.062.42
85.177.1424
2
0241 j
jVI
)(32.1171.41 AI
2Ij
j
j
j
j
VI
2
85.177.1412
2
0122
)(73.12367.1)(385.1925.02
77.285.12 AAj
jI
)(18.2271.422
1 22 WP
)(67.274
88.14
2
1 2
4 WP
)(565.5)73.1230cos(67.1122
1012 WP
)(42.55)32.110cos(71.4242
1024 WP
)(42.55
)(565.567.2718.22
WP
WP
supplied
absorbed
:Check
)cos(2
1ivMM IVP
Power Average
R
VRIP M
M
22
2
1
2
1
resistorsFor
MAXIMUM AVERAGE POWER TRANSFER
)cos(||||2
1
)cos(2
1
LL
LL
IVLL
IVLMLML
IV
IVP
THTHTH jXRZ
LLL jXRZ
L
LL
OCTHL
LL
Z
VI
VZZ
ZV
|||| OC
THL
LL V
ZZ
ZV
LIV
LLL
Z
ZVI
LL
LLL jXRZ
22)cos(
)tan(
LL
LIV
L
LL
XR
R
R
XZ
LL
2tan1
1cos
||
||||
L
LL Z
VI
222
2
||
||||
2
1
LL
L
THL
OCLL
XR
R
ZZ
VZP
22
2
)()(
||
2
1
THLTHL
LOCL
XXRR
RVP
THL
THL
L
L
L
L
RR
XX
R
PX
P
0
0
*TH
optL ZZ
TH
OCL R
VP
4
||
2
1 2max
222 )()(||
)()(
THLTHLTHL
THLTHLTHL
XXRRZZ
XXjRRZZ
LEARNING EXAMPLEload the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
Remove the load and determine the Thevenin equivalent of remaining circuit
1I
64.926.5
64.908.6
03204
142
24
jVOC
93.1647.164.908.6
57.2694.8
16
48
37
1652
37
)16)(48(
16
48)12(||4
j
j
jjj
j
jjZTH
43.041.193.1647.1* jZL
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
)(45.241.14
26.5
2
1 2max WPL
We are asked for the value of thepower. We need the Thevenin voltage
LEARNING EXAMPLEload the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
Circuit with dependent sources!
SC
OCTH I
VZ
1'
1'
2
)42(04
IV
IjV
X
x
KVL
)(45707.04524
04
)4524()44(04
1
11
AI
IIj
KVL
5.1611013411042 1 jjIVOC
Next: the short circuit current ...
LEARNING EXAMPLE (continued)...
Original circuit
02)(204
04)(24"
SCSC
SCx
IjII
IIIjV
CURRENT CIRCUIT
SHORT FOR EQUATIONS LOOP
)(2" IIV SCx
VARIABLEGCONTROLLIN
4)22(2
44)44(
SC
SC
IjI
IIj
Substitute and rearrange
2)11( SCIjI
442)1()1(4 SCSC IIjj
57.1165)(21 AjISC
57.1611013411042 1 jjIVOC
11452 jZTH 11 jZ optL
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
)(25.14
)10(
2
1 2max WPL
LEARNING EXTENSION
load the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
OCV
I
4573.12)1(98
)22(36
)22(036
jj
I
Ij
186
)1(9212
2012
j
jj
IjVOC
)(57.71974.18 VVOC
22
42)2||2(2
j
jjjjZTH
)(18
88
22
4
jj
jZTH
)(1 jZ optL
)(454
360
2
1max WPL
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
360186|| 222 OCV
LEARNING EXTENSION
load the to suppliedpower average maximum the Compute
transfer.power average maximumfor Find LZ
TH
OCL R
VP
4
||
2
1 2max*
THoptL ZZ
OCV
2jV
024222
22 jj
jV j 9024
KVL
)(24129024012 VjVOC
7202412|| 222 OCV
THZ
222
)22(2)22(||2
jj
jjjjZTH
)(22 jZTH
)(22 jZ optL
)(4524
720
2
1max WPL
EFFECTIVE OR RMS VALUES
)(ti
R
Rtitp )()( 2
power ousInstantane
Tt
t
Tt
tav dtti
TRdttp
TP
T
0
0
0
0
)(1
)(1 2
period with periodic iscurrent If
2
)(
dcdc
dc
RIP
Iti
then )( DC iscurrent If dcaveff PPI :
Tt
teff dtti
TI
0
0
)(1 22
Tt
teff dtti
TI
0
0
)(1 2
The effective value is the equivalent DCvalue that supplies the same average power
Definition is valid for ANY periodicsignal with period T
2
)cos()(
Meff
M
XX
tXtx
is valueeffective the
signal sinusoidal aFor
If the current is sinusoidal the averagepower is known to be RIP Mav
2
2
1
22
2
1Meff II
)cos(2
1ivMMav IVP case sinusoidalFor
)cos( iveffeffav IVP
square) mean(root rmseffective
LEARNING EXAMPLECompute the rms value of the voltage waveform
One period
32)2(4
210
104
)(
tt
t
tt
tv
3T
3
2
21
0
2
0
2 ))2(4()4()( dttdttdttvT
The two integrals have the same value
3
32
3
162)(
1
0
33
0
2
tdttv
)(89.13
32
3
1VVrms
Tt
trms dttx
TX
0
0
)(1 2
)(ti
R
LEARNING EXAMPLECompute the rms value of the voltage waveform and use it todetermine the average power supplied to the resistor
2R
)(4 sT
40;16)(2 tti
Tt
trms dttx
TX
0
0
)(1 2
)(4 AIrms
)(322 WRIP rmsav
LEARNING EXTENSIONCompute rms value of the voltage waveform
tv 2
4T
Tt
trms dttx
TX
0
0
)(1 2
dttVrms 2
0
2)2(4
1VVt 633.1)(
3
8
3
12
0
3
)(ti
R
LEARNING EXTENSION
4R
Compute the rms value for the current waveforms and usethem to determine average power supplied to the resistor
6T
Tt
trms dttx
TX
0
0
)(1 2
RIP rmsav2
4
2
6
4
2
0
2 41646
1dtdtdtIrms 8
6
8328
)(3248 WP
8T
816168
1 6
4
2
0
2
dtdtIrms )(32 WP
THE POWER FACTOR
LZiMI
vMV
iv
VI
izv
IZVZIV
z
)cos()cos(2
1ivrmsrmsivMM IVIVP rmsrms IVP apparent
zivP
Ppf cos)cos(
apparent
inductive pure
inductiveor lagging
resistive
capacitiveor leading
capacitive pure
90
900
0
10
01
09010
900
z
z
z
pf
pf
pf
V
e)(capacitiv
leadscurrent 090 z
)(inductive
lagscurrent 900 z
pfIVP rmsrms
LEARNING EXAMPLEFind the power supplied by the power company.Determine how it changes if the power factor is changed to 0.9
Power company
pfIVP rmsrms
rmsAW
Irms )(3.259707.0480
)(1088 3
rmsV )(480
rmsA)(453.259
45707.0cos zz
Current lags the voltage
480453.25908.0
08.0
LrmsS VIVrms
)(7.14957.147.494
480)4.1834.183(08.0
Vj
jVrmsS
)(378.93)(000,88
378.508.03.259 32
kWWPP
kWRIP
lossesS
rmslosses
If pf=0.9
If pf=0.9
8.257.203rmsI
82.049448009.747.14 jVS
kWRIP
rmsAI
rmslosses
rms
32.3
)(7.2039.0480
000,88
2
Losses can be reduced by 2kW!
Examine also the generated voltage
LEARNING EXTENSION
1.0
707.0,)(480,100
line
LL
R
pfrmsVVkWP
Determine the power savings if the power factor can be increased to 0.94
pfIVP rmsrms
22
22 1
pfV
RPRIP
pfV
PI
rms
linelinermslosses
rmsrms
kW
WpfPlosses
34.42
)(707.0
1
480
1.010)707.0( 22
10
kW
WpfPlosses
34.413.1
)(94.0
1
480
1.010)94.0( 22
10
kWkWPsaved 77.334.487.0
COMPLEX POWER*rmsrms IVS
PowerComplex of Definition
ivrmsrms
irmsvrms
IVS
IVS
*
)sin()cos( ivrmsrmsivrmsrms IVjIVS
The units of apparent and reactive power are Volt-Ampere
2* ||)( rmsrmsrmsrmsrms IZIZISZIV Another useful form