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11
Alternating current
three-phasecircuits
Unit 3. AC THREE-PHASE CIRCUITS
22
Unit 3. AC THREE-PHASE CIRCUITS
Three-phase systems characteristics
Generation of three-phase voltages
Three-phase loads
-Y and Y- transformation Instantaneous power
Three-phase power: S, P and Q
Power measurement. Aaron connection
Power factor improvement
Electrical measurements
Exercises
CONTENTS:
33
Unit 3. AC THREE-PHASE CIRCUITS
THREE-PHASE SYSTEM CHARACTERISTICS
44
Unit 3. AC THREE-PHASE CIRCUITS
THREE-PHASE SYSTEM CHARACTERISTICS
The electricity grid is made up of four main components:
GENERATION: production of electricity from energy sources
such as coal, natural gas, hydropower, wind and solar.
TRANSMISSION: the transmission system carries the electricpower from power plants over long distances to a distribution
system.
DISTRIBUTION: the distribution system brings the power to the
customers.
COSTUMERS: these are the consumers of electric power
(industry, service sector and residential uses).
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55
Unit 3. AC THREE-PHASE CIRCUITS
THREE-PHASE SYSTEM CHARACTERISTICS
Instantaneous electric power has a sinusoidal shape with double
frequency than voltage or current.
SINGLE-PHASE AC CIRCUITS: instantaneous electric power is
negative twice a period (power flows from the load to thegenerator) and positive twice a period, falling to zero.
BALANCED THREE-PHASE AC CIRCUITS: instantaneous
electric power is constant. Three-phase power never fal ls to
zero.
Three-phase electric motors perform better than single-phase AC
motors.
Three-phase power systems allow two voltage levels (L-L, L-N).
When electric power is transmitted, three-phase AC systems
require 25% less Cu/Al than single-phase AC systems.
66
Unit 3. AC THREE-PHASE CIRCUITS
GENERATION OF THREE-PHASE VOLTAGES
Three-phase generators contain three sinusoidal voltage sources
with voltages of the same frequency but a 120-phase shift with
respect to each other.
This is achieved by positioning three coils separated by 120angles. There is only one rotor.
Amplitudes of the three phases are also equal.
The generator is then balanced.
77
Unit 3. AC THREE-PHASE CIRCUITS
INTRODUCTION
88
Unit 3. AC THREE-PHASE CIRCUITS
INTRODUCTION
R
S
T
N
N: neutral point
R S T (or A B C) direct sequence or sequence RST
VRS, VST, VTR: line voltages or line-to-line voltages
VRN, VSN, VTN: line-to-neutral voltages
Vline= Vline-to-neutral3
R
ST
N
V
V V
V
V
V
120
TN SN
RSTR
ST
RN
vRN
(t) = V0
cos(t + 90) VvSN(t) = V0cos(t - 30) VvTN(t) = V0cos(t +210) V
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99
Unit 3. AC THREE-PHASE CIRCUITS
INTRODUCTION
50 Hz Usual system
Vphase = Vline-to-neutral230 volt
Vline 400 volt
Frequency 50 Hz
phase
Ntolineline .V3V
R
ST
N
V
V V
V
V
V
120
TN SN
RSTR
ST
RN
1010
Unit 3. AC THREE-PHASE CIRCUITS
THREE-PHASE LOAD CLASSIFICATION
WYE(two voltages)
Balanced
3-wires
4-wires
Unbalanced
3-wires
4-wires
DELTA(one voltage)
Balanced
Unbalanced
Z Z
Z
R
N
S
T
O=N
IR
IN
IT
ISZ Z
Z
R
S
T
O=N
R
S
T
Z
Z
Z
RSTR
S
R
T
ST
I
I
I
I
I
I
R
S
T
R
S
T
Z
Z
Z
RSTR
S
R
T
ST
I
I
I
I
I
I
R
S
T
RS
ST
TR
1111
Unit 3. AC THREE-PHASE CIRCUITS
BALANCED WYE-CONNECTED LOAD
The wye or star connection is made by connecting one end of each of
the three-phase loads together.
The voltage measured across a single load or phase is known as the
phase voltage.
The voltage measured between the lines is known as the line-to-line
voltage or the line voltage.
In a wye-connected system, the line voltage is higher than the load
phase voltage by a factor of the square root of 3.
In a wye-connected system, the phase current and line current are the
same.
1212
Unit 3. AC THREE-PHASE CIRCUITS
BALANCED DELTA-CONNECTED LOAD
This connection received its name from the fact that a schematic
diagram of it resembles the Greek letter delta ().
In the delta connection, the line voltage and phase voltage in the load
are the same.
The line current of a delta connection is higher than the phase currentby a factor of the square root of 3.
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1313
Unit 3. AC THREE-PHASE CIRCUITS
-Y TRANSFORMATION
R
S
T
Z
Z
Z
1
23
13
23
12
Z
Z Z
1
32
231312
1312
1
ZZZ
ZZZ
231312
2312
2
ZZZ
ZZZ
231312
2313
3
ZZZ
ZZZ
Balanced loads: ZY = Z/3
)(
)(
)1( 231312
231312
21ZZZ
ZZZ
ZZ
Z between nodes 1 and 2:
)(
)()2(
231312
23121331
ZZZ
ZZZZZ
)(
)()3(
231312
13122332
ZZZ
ZZZZZ
From expressions (1), (2) and (3) the result is:
)(
)(
231312
2313122,1
ZZZ
ZZZZ
212,1 ZZZY Y:
Z nodes 1-2:
Z nodes 1-3:
Z nodes 2-3:
(1) + (2) - (3) (1) - (2) + (3) -(1) + (2) + (3)
1414
Unit 3. AC THREE-PHASE CIRCUITS
BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD
Z Z
Z
R
N
S
T
O=N
IR
IN
IT
IS
Z Z
Z
R
S
T
O=N
0)III(I
Z
VI
Z
VI
Z
VI
TSRN
TNT
SNS
RNR
Q
lnialnia
P
lnialnia
lnialniaRNfasetotal
.sin.I.V3.cos.I.V3
.I.V3I.V3.S3.S *R
j
The three currents are balanced.
Thus the sum of them is always zero.
Since the neutral current of a balanced, Y-connected three-phase load is always zero,
the neutral conductor may be removed with
no change in the results.
1515
Unit 3. AC THREE-PHASE CIRCUITS
BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD
Z Z
Z
R
S
T
O=N
Example A three-phase, RST system (400 V, 50 Hz), has a
three-wire Y-connected load for which Z = 1030 Obtain theline currents and the complex power consumption
A3
40
10
3400/
Z
VI
60
30
90
RNR
A3
40
10
3400/
Z
VI
60
30
30SN
S
A3
40
10
3400/
Z
VI
180
30
210
TNT
(VAr)j8000(watt).4138561VA00061
)3)(40/33(400/IV3S3S
30
*6090
RNphasetotal*R
1watt4.1385630cos3
404003cosIV3P lltotal
VAr800030sin3
404003sinIV3Q lltotal
VA160003
40
4003IV3S lltotal
1616
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED FOUR-WIRE WYE-CONNECTED LOAD
Z Z
Z
R
N
S
T
O=N
R
S T
IR
IN
IS
IT
0)III(I
Z
VI
Z
VI
Z
VI
TSRN
T
TNT
S
SNS
R
RNR
*
T
*
S
*
R IVIVIVS TNSNRNtotal
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1717
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED THREE-WIRE WYE-CONNECTED LOAD
Z Z
Z
R
S
T
O=N
R
S T
IR
IS
IT
0III TSR
TSR
TTNSSNRRN
TSR
T
TN
S
SN
R
RN
ON
YYY
YVYVYV
Z
1
Z
1
Z
1
ZV
ZV
ZV
V
0Z
V
Z
V
Z
V
T
TO
S
SO
R
RO
0Z
VV
Z
VV
Z
VV
T
ONTN
S
ONSN
R
ONRN
1818
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED THREE-WIRE WYE-CONNECTED LOAD
Z Z
Z
R
S
T
O=N
R
S T
IR
IS
IT
T
ONTN
T
TO
T
S
ONSN
S
SOS
R
ONRN
R
ROR
Z
VV
Z
V
I
Z
VV
Z
VI
Z
VV
Z
VI2)
*
T
*
S
*
RIVIVIVS3) TOSOROtotal
TSR
T
TN
S
SN
R
RN
ON
Z
1
Z
1
Z
1
Z
V
Z
V
Z
V
V)1
1919
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED THREE-WIRE WYE-CONNECTED LOAD
Example A three-phase, RST system (400 V, 50 Hz), has a three-wire
unbalanced Y-connected load for which ZR= 100 , ZS= 100 and ZT= 1030
Obtain the line currents and the total complex power consumption.
O N
R
S
T
IR
IS
IT
ZR
ZS
ZTV40.93
1.01.01.0
0.12300.12300.1230
YYY
YVYVYVV
114.89
3000
30-210030-090
TSR
TTNSSNRRNON
VRO= VRNVON= 23090 - 40.93114.89 = 193.6484.90 V
VSO= VSNVON= 230-30 - 40.93114.89 = 264,54-35.10 V
VTO= VTNVON= 230210 - 40.93114.89 = 237,18-140.10 V
IR= VRO/ZR= 193.6484.90/100 = 19,3684.90A
IS= VSO/ZS= 264.54-35.10/100 = 26,45-35.10A
IT= VTO/ZT= 237.18-140.10/1030 = 23.72-170.10A
Stot= VROIR* + VSOIS* + VTOIT* = 15619.56 W + j2812.72 VAr
1)
2)
3)
2020
Unit 3. AC THREE-PHASE CIRCUITS
BALANCED DELTA-CONNECTED LOAD
R
S
T
Z
Z
Z
RSTR
S
R
T
ST
I
I
I
I
I
I
R
S
T
Z
VI
Z
VI
Z
VI
TRTR
STST
RSRS
Q
lnialnia
P
lnialnia
lnialniaRSfasetotal
sinIV3cosIV3
IV3IV3S3S *RS
j
STTRT
RSSTS
TRRSR
III
III
III
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2121
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED DELTA-CONNECTED LOAD
R
S
T
Z
Z
Z
RSTR
S
R
T
ST
I
I
I
I
I
I
R
S
T
RS
ST
TR
TR
TRTR
ST
STST
RS
RSRS
Z
VI
Z
VI
Z
VI
STTRT
RSSTS
TRRSR
III
III
III
*TR
*ST
*RS
IVIVIVS TRSTRStotal
2222
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED THREE-WIRE -CONNECTED LOAD
ExampleA three-phase, RST system (400 V, 50 Hz), has an unbalanced -connected load for which ZRS= 10
0 , ZST= 1030 i ZTR= 10-30 Obtainthe line currents and the complex power consumption
R
S
T
Z
Z
Z
RSTR
S
R
T
ST
I
I
I
I
I
I
R
S
T
RS
ST
TRA4010
400
Z
VI
A4010
400
Z
VI
A40
10
400
Z
VI
90
30
120
TR
TRTR
30
30
0
ST
STST
120
0
120
RS
RSRS
A40.00IIIA77.29III
A77.29III
150STTRT
45RSSTS
105TRRSR
43712.81VA
(VAr)j0)43712.81(WIVIVIVS *TR
*
ST
*
RS TRSTRStotal
R
ST
N
V
V V
V
V
V
120
TN SN
RSTR
ST
RN
2323
Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT. Four-wire load
Unbalanced wye-connected, four-wire load
Z Z
Z
R
S
T
O=N
N
W
R
S TW
W
R
S
T
WR = VRNIRcos( VRN - IR)
WS= VSNIScos( VSN - IS)
WT= VTNITcos( VTN - IT)
Ptotal = WR + WS + WT
Z Z
Z
R
S
T
O=N
N
W
Balanced wye-connected, four-wire load
Ptotal = 3W
W = VRNIRcos( VRN - IR)
2424
Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT. ARON CONNECTION
General 3-wire load. Two-wattmeter method (ARON connectio n)
R
ST
N
V
V V
V
V
V
120
TN SN
RS
TR
ST
RNVRT
VST
VSR
)30cos(VI)cos(IVW
)30cos(VI)cos(IVW
SST
RRT
IVSST2
IVRRT1
cosVI3)]30cos()30cos(VI[WWP 21TOTAL
,
LOAD
inVI)]cos()cos(VI[]W[WQ 21TOTAL s3303033
Demonstration done for a balanced 3-wire load
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2525
Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT. BALANCED LOAD
,
LOAD
)WW
WW3arctg(
21
21
)W(W3Q
WWP
21total
21total
Balanced load, g eneral (Y/D, 3/4 wires). Two-wattmeter method
(ARON connection)
,
LOAD
Unbalanced wye/delta-connected, three-wire load
Ptotal = W1 + W2
BALANCED
UNBALANCED
2626
Aron cycl ic permutat ions
Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT: THE TWO-WATTMETER METHOD
W3QTOT
W1 W2
V I V I
RT R ST S
SR S TR T
TS T RS R
WV I
ST R
TR S
RS T
Q measurement: cyclic permutations
3inIV)os(90-IV)cos(IVW TOTALlinelinelinelineRST
QscIRVST
2727
Unit 3. AC THREE-PHASE CIRCUITS
INSTANTANEOUS THREE-PHASE POWER
Single-phase load: cosAcosB = 05[cos(A+B) + cos(A-B)]
p(t) = v(t)i(t) = V0cos(wt + V)I0cos(wt + )
p(t) =1/2V0I0cos(V-I) + 1/2V0I0cos(2wt + V+ I) watt
Constant Oscillates at twice the mains frequency!
Three-phase wye balanced load:p(t)= vRN(t)iR(t) + vSN(t)iS(t) + vTN(t)iT(t) =
=2Vpcos(wt +V)2Ipcos(wt +)
+2Vpcos(wt -120+V)2Ipcos(wt -120+)
+2Vpcos(wt +120+V)2Ipcos(wt +120+)
=
VpIpcos(V-I) + VpIpcos(2wt +V+I)
+ VpIpcos(V-I) + VpIpcos(2wt -240+V+I)+ VpIpcos(V-I) + VpIpcos(2wt +240+V+I)
= 3/2VpIpcos(V-I) = 3/2VpIpcos = constant! 2828
Unit 3. AC THREE-PHASE CIRCUITS
INSTANTANEOUS POWER: SINGLE-PHASE LOAD
v(t)
i(t)
p(t)Average p ower = P
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2929
Unit 3. AC THREE-PHASE CIRCUITS
INSTANTANEOUS THREE-PHASE POWER
v(t)
i(t)
p(t)
pTOTAL = pR(t) + pS(t) + pT(t)
3030
Unit 3. AC THREE-PHASE CIRCUITS
POWER LOSSES: THREE-PHASE/SINGLE PHASE
Single-phase lineR1
R
N
LOAD
R1
Three-phase line
Supposing same losses1p3p
3p1p
21 S2
1S
SS2R2R ll
Single-phase line: 2 conductors of length land section S1p
Three-phase line: 3 conductors of length l and section S3p= 1/2S1p
As a result: weight3p-cables= 3/4weight1p-cables
22
2
load1
2
1losses
cosV
P2RI2RP V
V
22
2
load2222
2
load2
2
2losses
cosV
PR
cos.V)3(
P3RI3RP
cosV
PI load
cosV3
PI load
3131
Unit 3. AC THREE-PHASE CIRCUITS
Example 1
A
Balanced load 1capacitive
C
K
CC
UU1
ZL
ZL
ZL
R
S
T A1
W1
Balanced load 2inductive
A2
W2
Three-phase balanced RST system for which A1= 1.633 A, A2= 5.773 A, W1= 6928.2
W, W2= 12000 W, U = 6000 V and Z line=4+j3 a) Obtain the complex power in theloads, as well as the ammeter A and the voltmeter U1 readings. b) Obtain the value of
Cto improve the loads PF to 1, assuming U = 6000 V.
48000VArsinUI3Q
12.53
cosUI3W36000W3P
W12000cosUI3P
45
sinUI3VAr12000Q
VAr12000W3Q
222
2
2222
111
1
111
11
VA6000060003j48000)Qj(Q)P(PS 36.87212121
A774.5773.5633.1III 85.369012.5390459021
TOTAL
6050VU3 UV97.34923
6000)34(774.5UZ.IU phase1,1
90
90
85.3690
phaseL,1
jphase
F1.06C).50.C1/(2
(6000)-r-36000/3VAQ/3Q
2
,1
C
R
ST
3232
Unit 3. AC THREE-PHASE CIRCUITS
Example 2
Three-phase 50 Hz system for which V = 400 V, W1= -8569.24 W, W2= -5286.36 W, AS= 21.56 A. Obtain a) the value of R. b) the reading of A R.c) the value of the inductance
L
22
11
W'W
and
W'W
23.095R:inresultingR
4002W13855.6'W'WPa)
2
21otalt
A32.17095.23
400
R
VIA32.17
095.23
400
R
VIb) 60
60
RTRT
6060
SRSR
A30I-II:resultsIt90
SRRTR
L
90
L
180
60
L
TS60TSSRS
X400jj1566.8
X40032.17
jXV32.17IIIc)
H0.2684L:isresultThe
50L284.308X)X
40015(8.66A21.56I L
2
L
2
S
ARR
S
T
W1
R L
W2
R
AS
AT
V
R
S
T
R
ST
W1 W2V I V I
RT R ST S
SR S TR T
TS T RS R
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3333
Unit 3. AC THREE-PHASE CIRCUITS
Example 3
Varley phase-sequence indicator. Calculate the voltage in each element and deduce
the practical consequences Three-phase 400 V/50 Hz system
V171.55
5290/15290/13183/1
/5290230/5290230/3183230
YYY
YVYVYVV
171.31
0090
0210030-90-90
TSR
TTNSSNRRNON
VRO= VRNVON= 23090 - 171.55171.31 = 265.3450.28 V
VSO= VSNVON= 230-30 - 171.55171.31 = 394.78-20.91 V
VTO= VTNVON= 230210 - 171.55171.31 = 144.00-101.86 V
Conductor R is s ituated where the capac itor is p laced, conductor S
is si tuated where the brigh ter bu lb is p laced and T is the remaining
conductor.
C = 1F XC= 3183
R2bulbs= 2V2/P = 22302/20 = 5290
R
ST
0V0N
VR0
VT0 VS0
3434
Unit 3. AC THREE-PHASE CIRCUITS
Example 4
A 400 V and 50 Hz three-phase line feeds two balanced loads through a line which has
an internal impedance of ZL=0.5 + j1 The-connected load has phase impedanceswhose values are 45+j30 , whereas the Y-connected load has phase impedances of
15j30 Determine: a) the reading of the ammeter A, b) the reading of the voltmeter
V and c) the readings of watt-meters W1and W2.
R
ST
A30.13365.17
3/400
Z
VIa) 875.82
125.7
90
TOT
RNR
386.33V.7706)(13.30163)(IZ3Vb) //
1.154313.30)W(W3Q
16.731313.30WWPc)
2
21LOAD
2
21LOAD
The Aron connection results in:
W1 = 4616,1 W, W2 = 4262,5 W
17.365j1.154)(16.731j1)(0.5Z)Z//Z(ZZ7.125
//LYLTOT ZY
W1 W2
V I V I
RT R ST SSR S TR T
TS T RS R
W1
W2ZL
ZL
ZL
AV
R
S
T
3535
Unit 3. AC THREE-PHASE CIRCUITS
Example 5
Three-phase 400 V-50 Hz line. When switch K2is closed , WA= 4000 W. When K1and
K3are closed, WA= 28352.6 W and WB= -11647.4 W. Determine: a) R2,b) R1and c) AT.
R
ST
j34.64A6010
400
10
400
R
V
R
VIII
60180
1
RT
1
TSRTTST1
R
S
T
N
WA
AT
WB
K1
C1 R1
R1 R2
K2 K3
AS
a) K2 closed: WA = VSTIScos(VST -IS)
4000 = 400IScos(0+30) IS = 11.55 A
R2 = VSN/IS = (400/3)/11.55 = 20
b) K1 and K3 closed:
PTOT = W1 + W2 = -WB + WA = 40000 W = 24002/R1 + 400
2/R2 R1 = 10
c) K1 and K3 closed:
j0A2020
400
R
VI180
2
TST2
A87.18j34.64-80j0)(-20j34.64)60(III -156.6T2T1Ttotal
This results in AT = 87.18 A
W1 W2
V I V I
RT R ST S
SR S TR T
TS T RS R
36
Unit 3. AC THREE-PHASE CIRCUITS
Question 1
An electrical lineman is connecting three single-phase transformers in a Y(primary)-
Y(secondary) configuration, for a businesss power service. Draw the connecting
wires necessary between the transformer windings, and those required between the
transformer terminals and the lines. Note: fuses have been omitted from this
illustration for simplicity.
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37
Unit 3. AC THREE-PHASE CIRCUITS
Question 2
Identify the primary-secondary connection configuration of these pole-mounted
power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).
HV
LV
R ST
These transformers are connected in a Yyconfiguration.
38
Unit 3. AC THREE-PHASE CIRCUITS
Question 3
Identify the primary-secondary connection configuration of these pole-mounted
power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).
These transformers are connected in a Ydconfiguration.
39
Unit 3. AC THREE-PHASE CIRCUITS
Question 3
Identify the primary-secondary connection configuration of these pole-mounted
power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).
These transformers areconnected in open-delta
configuration.
Three single-phase transformers are not normally used because this is more
expensive than using one three-phase transformer.
However, there is an advantageous method called the open-Delta or V-
connection
It functions as follows: a defective single-phase transformer in a Dd three-phase
bank can be removed for repair. Partial service can be restored using the open-Delta configuration until a replacement transformer is obtained.
Three-phase is still obtained with two transformers, but at 57.7% of the original
power.
This is a very practical transformer application for emergency conditions.
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Unit 3. AC THREE-PHASE CIRCUITS
Question 4
One of the conductors connecting the secondary of a three-phase power distribution
transformer to a large office building fails when open. Upon inspection, the source of
the failure is obvious: the wire overheated at a point of contact with a terminal block,
until it physically separated from the terminal. What is strange is that the overheated
wire is the neutral conductor, not any one of the line conductors. Based on this
observation, what do you think caused the failure?
After repairing the wire, what would you do to verify the cause of the failure?
Heres a hint (pista): if you were to repair the neutral wire and take current
measurements with a digital instrument (using a clamp-on current probe, for safety), you
would find that the predominant frequency of the current is 150 Hz, rather than 50 Hz.
This scenario is all too common in modern power systems, as non-linear loads such asswitching power supplies and electronic power controls have become more prevalent.
Special instruments exist to measure harmonics in power systems, but a simple DMM
(digital multimeter) may be used as well to make crude assessments.
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Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT. ARON CONNECTION
General 3-wire load. Two-wattmeter method (ARON connection)
p(t) = vRN(t)iR(t) + vSN(t)iS(t) + vTN(t)iT(t)
p(t) = vRN(t)iR(t) + vSN(t)iS(t) + vTN(t)[-iR(t) - iS(t)]
p(t) = iR(t)[vRN(t) - vTN(t)] + iS(t)[vSN(t) - vTN(t)] = vRT(t)iR(t) + vST(t)iS(t)
Mean value
Ptotal= W1+ W2= VRTIRcos(VRT-IR) + VSTIScos(VST-IS)
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