Post on 18-Dec-2021
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PHASORS
2019 Northwest Electric Meter School
Arlen Everist
Puget Sound Energy
Exchanging expertise since 1893
Objectives• What is a phasor?
• Why are phasors important in metering?
• Working with vectors or phasors
• Standard meter service phasor diagrams
• Troubleshooting
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Terms and Concepts
• Rotation – the direction around the center
Phasors rotate counter‐clockwise
• Sequence – the order of progressionPhase sequence can be ABC or ACB
A Phasor is a special type of Vector
A vector represents a quantity with magnitude and direction
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Handbook for Electricity Metering
• A phasor is a quantity which has magnitude, direction and time relationship. Phasors are used to represent sinusoidal voltages and currents by plotting on rectangular coordinates. If the phasors were allowed to rotate about the origin, and a plot made of ordinates against rotation time, the instantaneous sinusoidal wave form would be represented by the phasor.
Sine Wave
M7-7
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Three phase sine wave and corresponding phasors –
ideal condition
Three phase sine wave and corresponding phasors –unbalanced
load
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Quantities
• For our purposes the main uses include:
– Voltage
– Current
– Impedance
Lots of other uses
• Navigation: Seattle to Spokane 233 miles at 890
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Why Study Phasors?
• A simple visual representation of electrical phenomena
– Visual of what’s happening in the service and in the meter
– Understand necessary concepts for testing and billing
• A tool for troubleshooting
How do we work with phasors?
• Phasors are vectors in motion; treat them like vectors
• Vectors can be described in 2 ways: polar and rectangular (Cartesian)
• Vectors can be added, subtracted, multiplied and divided
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Vector DescriptionRectangular Coordinates: (1,2)
2
1
(1,2)
M7-1
Vector DescriptionPolar Coordinates: 2.24<63.40
1
2
1
63.40
M7-4,6
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Vector Exercise 1
• Draw vectors 2<00 and 3<300 (thin)
• Draw vectors (1,0) and (1.73, 1) (thick)
Vector Exercise 1
• Draw vectors 2<00 and 3<300 (thin)
1
1 200
900
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Vector Exercise 1• Draw vectors 2<00 and 3<300 (thin)
• They remain the same vectors regardless of position in space
1
1 200
900
Vector Exercise 1Reference changed – note position of 900
• Draw vectors 2<00 and 3<300 (thin)
1
1 200
900
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Vector Exercise 1
• Draw vectors 2<00 and 3<300 (thin)
• Draw vectors (1,0) and (1.73, 1) (thick)
1
1 200
900
Vector Exercise 1
• Draw vectors 2<00 and 3<300 thin
• Draw vectors (1,0) and (1.73, 1) thick
• Describe with both methods
1
1 200
900
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Vector Exercise 1• Describe with both methods – the vectors on the X axis are straightforward (1,0) = 1< 00
• 3 < 300 : 3 x cos 300 = 2.6 ; 3 x sin 300 = 1.5
• 3 < 300 = (2.6, 1.5)
1
1 200
900
1 < 00 (2,0)
(2.6, 1.5)
Vector Exercise 1• Describe with both methods
• (1.73, 1): 1.732 + 12 = 3.99; √ 3.99 = ~ 2
• 1 / 1.73 = 0.578; tan‐1 0.578 = 300
• (1.73, 1) = 2 < 300
1
1 200
900
1 < 00 (2,0)
2 < 300
(2.6, 1.5)
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Adding (or subtracting) Vectors
• Vectors can be added or subtracted
• Easiest in rectangular coordinates
M7‐3
M7‐17
Vector Exercise 1 extension• Add 1 < 00 + 2 < 00
• 1 < 00 = (1,0); 2 < 00 = (2,0)
• 1 + 2 = 3; 0 + 0 = 0
• 1 < 00 + 2 < 00 = (3,0) = 3 < 00
1
1 200
900
1 < 00 2 < 00
3
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• Add 2 < 00 + 3 < 300
• 2 < 00 = (2,0) ; 3 < 300 = (2.6, 1.5)
• Xs: 2 + 2.6 = 4.6; Ys: 0 + 1.5 = 1.5
• (4.6, 1.5) ; 4.62 + 1.52 = 23.41; √ 23.41 = ~4.8
• 1.5 / 4.6 = 0.326; tan‐1 0.326 = 180
• 2 < 00 + 3 < 300 = 4.8 < 180
1
1 200
900
(2,0)
(2.6, 1.5)
4.8 < 180
Vector Exercise 2
2 < 3100 + 10 < 350 = ?
270
0
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Vector Exercise 22<310 = (1.29, ‐1.53) 10<35 = (8.19, 5.74)
Add X values Add Y values
1.29 + 8.19 = 9.48 ‐1.53 + 5.74 = 4.21
? = (9.48, 4.21)
4.21 / 9.48 = tan θ = 0.444
Θ = 23.95
Hyp = 9.48 / cos θ = 10.37
? = 10.37<23.95 = 10 < 24
Vector Exercise 2
2<3100 + 10<350 =10<24
270
0
10<24
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Proof of 1.73 relationship
• Side = 1; angle at bottom is 1200
• Perpendicular bisector of 120 also bisects red line
• Cos30 x 1 = half of red line = 0.866
• 0.866 x 2 = 1.73 = length of red line
3090
60
120
1
1.73
Proof of 1.73 relationship
• ‐0.5 – 1 = 1.5
• 0.866 – 0 = 0.866
• SQRT(1.52 + 0.8662) = 1.73
• 0.866 / 1.5 = Tan‐1 30
1<0
1<120
(1,0)
(-0.5, 0.866)
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Impedance
• To divide polar coordinates, divide magnitudes and subtract angles
V= 120 < 0; I = 10 < 30;
Z = V/I: 120/10 = 12; 0 – 30 = ‐30; Z = 12 < ‐30
What’s the reference?
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SCALE
What is the scale? Does the radius of the circle = 1v, 480v, 10a?Mfrs. handle this in different ways, showing the voltages relative to each other and the currents relative to each other, or scaling them to each other in some fashion.
Names
• Vector (and phasors) must be named or labeled properly to be useful
• Voltages are labeled with a V or E and the phase relationship of the potential difference: Van, ECB, etc.
• Currents are labeled with an I and the phase: Ia, IB, etc.
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Vector Review
• Magnitude and direction
• Rectangular coordinates
• Polar coordinates
• Add – head‐to‐toe
• Reference
• Scale
• Labels
Phasor rotation is counterclockwise
0
45
270
90
180
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Phasor rotation is counterclockwisePhase rotation or Phase sequencing
can be ABC or ACB
VA / VA
270
90
180
VB / VC
VC / VB
3 phase ACB rotation
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Phasor Review
• Phasors are vectors that rotate
• Represent cyclical phenomona
• In metering, rotation is counterclockwise
• In metering, degree notation is clockwise, with 0 at 3 o’clock
Real Life AdventuresHow does this relate to my meter work?
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Powermate Circuit Analyzer
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LED Billboard
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3 wire delta, 5S or 12S meter
Vab
Vcb
Ia
Ic
AB
C
W = E x I x √3 x cos θ
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3 wire delta
Or…W = E x I x cos (α+θ) per element α is the angle at unity between E and IΘ is the pf angle caused by the load.
Angles α and θW = E x I x cos (α + θ)
Vab
Ia theoretical
Ia actual—add α and θ for watts
Angle α is the unity phase relationship = 300 for VAB - Ia
Angle θ caused by customer load
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Angles α and θW = E x I x cos (α + θ)
Vab
Angle α is the unity phase relationship 300 for IA in a 3W delta
If θ is leading the theoretical angle, consider it a negative angle when adding
If the theoretical angle is leading, consider angle αto be negative
3 wire delta with 300 lag
Wa = E x I x cos (30+30) (0.5) Wc = E x I x cos (-30+30) (1.0)Wt = E x I x 1.5Wt = E x I x {√3 x cos 30} ({}=1.5)
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3 Wire Delta
Several ways to approach this:
System PF = 0.816, so I lags V by 350
Or compare phases:- Ia lags Vab by 640, at unity this would be 300 so Ia lags by 340
- Ic lags Vcb by 50
but it should lead by 300 so Ic lags by 350
Is this a delta?
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2P-N service with CTs
Next service type
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Meter software example
4W Wye
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Ametek JemStar II
4W Wye
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Unbalanced 4W Wye
4 Wire Wye Power Calculations per Phase
V x I = VA x PF = W
• PhA: 123.9 x 3.3 = 408.9 x 0.999 = 408.3
• PhB: 124.0 x 0.1 = 12.4 x 0.438 = 5.4
• PhC: 123.5 x 0.5 = 61.8 x 0.906 = 56.0
• Totals 483.1 469.7
• PF = W / V = .972 or 97.2%
• OR by average: Iave =1.3, <ave = 30.60
(123.8 x 1.3) x 3 = 482.8 x 0.860 = 415.2
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Neutral Current
• Add current vectors
• Ia =3.31<2.83 = (3.31, ‐0.16)
• Ib = 0.11>304.40 = (0.06, 0.09)
• Ic = 0.51>145.36 = (‐0.49, ‐0.29)
• In = 2.88 > ‐180.04 = (2.88, ‐0.36)
• Angle of the neutral will be roughly opposite the largest current—current is flowing away from the meter in the neutral.
M-7 12
Phasors with Neutral (green)
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Example from phase angle lab tomorrow
4W Wye
• Ia‐EAN____ Ia‐EBN_____ Ia‐ECN_____
• Ib‐EBN____ Ib‐ECN_____ Ib‐EAN_____
• Ic‐ECN____ Ic‐EAN_____ Ic‐EBN_____
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What’s the reference?
LUNCH
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4 wire deltaPhasors
4 W Delta Xfrmrs
B N A C
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4 W Delta XfrmrsRearranged
N
AB
C
4 W Delta Service Representation
A
C
B
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4 W Delta Meter Phasors(3 phase balanced load)
C
AB
4 wire deltaPhasors
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4 wire deltaPhasors
4 wire delta, 3 element solid state auto‐detecting meter phasors
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4 wire delta, 3 element solid state auto‐detecting meter phasors
4 Wire Delta at UnityPhase to phase V to describe
service
Van and Vbn are ½ V Φ – Φ
Vcn = Van x √ 3
Example: 240v delta
Van & Vbn = 120v
Vcn = 208v
Angle α for A phase = 300
Angle α for B phase = ‐300
Angle α for C phase = 00
Angle θ is the pf caused by the customer load.
Van
IaIb
Vbn
Vcn
Ic
αα
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4 Wire Delta Power Calculations per Phase
• Wa = VAN * Ia * cos(α + θ) for A phase• Wb = VBN * Ib * cos(α + θ) for B phase• Wc = VCN * Ic * cos(α + θ) for C phase• Wt = Wa + Wb + Wc
• Angle α for A is 300, for B is ‐300, for C = 0• VAN and VBN = ½ Ø – Ø voltage, • VCN = VAN * √ 3• Like a wye, angle θ can be different for each phase
and results from the load connected to that phase
The VA Question
• There are at least 2 methods commonly used to calculate VA, and others as well.
• 1. The traditional method of adding watts from each phase and VARs from each phase, then creating a hypotenuse, VA. This is basically what the old wh meter / VARh meter installations did, and is called the vectorial method.
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The VA Question
• 2. With microprocessor meters, we can now calculate VA per phase, then add the 3 VA values. This is called the arithmetic method. The Handbook considers this to be more “accurate.” (See VA Metering)
• 3. One meter mfr. multiplies V x I x 0.93 on all three phases to arrive at VA.
The VA Question
• Your utility has a method built into its rate structure. It is probably the vectorial method, based on a wh reading and a varh reading. Your meter may be calculating by another method. This will only matter if you use the meter’s power or PF calculations for billing.
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4 Wire Wye Power Calculations per Phase
V x I = VA x PF = W var• PhA: 123.9 x 3.3 = 408.9 x 0.999 = 408.3 22.1• PhB: 124.0 x 0.1 = 12.4 x 0.438 = 5.4 11.6• PhC: 123.5 x 0.5 = 61.8 x 0.906 = 56.0 26.1• Arithmetic Totals 483.1 469.7 59.8• Vectoral: 408.3 + 5.4 + 56 = 469.7w
»22.1 + 11.2 + 26.1 = 59.4 var»469.72 + 59.42 = 473.42 va
• Is PF 97.2% or 99.2% ?
Vectorial vs Arithmetic • How is your meter calculating power factor?
• 97.2% or 99.2%
• What if the difference is at the PF cutoff for var adjustment?
• Which is more accurate?
483.1 va469.7 w
473.4 va
469.7 w
59.9 var
59.9var
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4 wire delta, 2 element meterphasor analysis
VAB,
IA-B
VC, IC
4 wire delta, 2 element meter, phasor analysis
VAB,
IA-B
VC, IC
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What if IB is not reversed?
VAB,
VC, IC
IA+B
Full torque on C0 torque on AB
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4 wire delta with roughly equal 1 phase and 3 phase loads
Phase C is a good indicator of 3p load
Bp 2.34-1.37=0.97Ap 2.95-1.37=1.58
Service has roughly 1.4a of 3p load and 1-1.5a single phase load
Check
• P‐P voltage = 230 x 1.37 x 1.73 = 545w
• A = 115 x 2.95 = 339w
• B = 115 x 2.34 = 269w
• C = 199 x 1.37 = 273w
» Total 881w
»3p ‐545w
» 336w / 2 = 163 / 115=1.4
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3 phase component1 phase component
Varied Load
• This is just a way to understand what’s happening behind the phasors
• Don’t count on reverse analyzing load with any confidence
• There may be phase to phase load buried in there
• Understanding the relationship between the load and the phasors is important
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Typical 4 Wire Delta
More load on A and B phases, but significant load on C phase indicating 3 phase loadNotice 3rd harmonic distortion on B phase
4 Wire Delta
Small 3 phase load
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Note mfr’s use of phasors to indicate angle but not magnitude
4 Wire Delta
Notice lack of C phase current – A and B phase roughly balanced
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4 Wire Delta
Notice lack of C phase current –most of load is on B phase
4 Wire Delta Service
Note distortion of current waveform and smooth voltage waveform
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Harmonic Analysis
Lots of 11th and 13th harmonic in current but relatively little voltage distortion
Reactive MeteringElectromechanical
W =Van x Ia x cos θ
VAR = Van x Ia x sin θ or Vra x cos(90‐ θ)
Van
Ia
Vra
θ
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Phase shift schematic from Handbook for Electricity Metering
Shifting a Wye: Wiring and Phasors from HEM
Phasor analysis can help verify expected voltage angles and troubleshoot mis-wiring
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Shifting a Delta: Wiring and Phasors from HEM
Reactive Metering
VAR measurements are made by integrating the voltage waveform to obtain a 90°phase shift, then each voltage sample is multiplied by the coincident current sample and the product is accumulated to obtain a VARs.
From Landis + Gyr manual
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Reactive MeteringElectromechanical
VAR = Van x Ia x sin θ = 0
or VAR = Vra x Ia x cos(90‐ θ) = 0
Ia
Vra
Dk. Blue lags by 90 degrees
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Power Factor = 0
Current (orange) lags voltage (blue) by 900 so positive and negative power (light green) cancel out – no real power, all reactive power
Reactive MeteringElectromechanical
W =Van x Ia x cos 10
VAR = Van x Ia x sin 10 or Vra x Ia x cos(80)
Van
Ia
Vra
Θ = 10
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100 lagging current, WH meter
VARH meter sees 800 leading current
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TROUBLESHOOTING
How Phasors Can Help Solve Your Problems
Examples of troubleshooting:What’s wrong here?
4W Y with CTs
Van
Vbn
VcnIc
Ia
Ib
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One polarity reversed
What’s wrong here?
4W Y with CTs
Van
Vcn
VbnIc
IaIb
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What’s wrong here?
4W Y with PTs and CTs
Van
Ic
Ib Ia
Vbn
Vcn
Miswired 4 wire delta with 2 element EM 15S Self‐Contained Meter
What will the meter
register if C phase
(high leg) is in the
center at the
socket?
A C B
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Miswired 4 wire delta 2 element EM
C
B A
B
AC
Miswired 4 wire delta 2 element EM
• Expected ActualVcn
IcVab
Ia-IbVac
Vbn
Ib
Ia-Ic
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Power calculations for 15S
• E x I x SQRT3 = 240v x 10a x 1.73 = 4157
• Correct meter
• “2S” element sees (240v x ½ Ia x 0.866) + (240v x ‐ ½ Ib x 0.866) = 2078va
• “1S” element sees 208v x 10a = 2080va
• 2078 + 2080 = 4158VAt
Miswired 15S meter socket
• “2S” element sees the same as before, just A and C instead of A and B = 2078va
• “1S” element is now B phase with a 300
lag and 120v: 120v x 10a x 0.866 = 1039va
• 2078va + 1039va = 3117
• 3117 / 4157 = 75% registration on balanced 3 Φ load! Lighting load is ok.
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What’s This?
Solar System with 2 inverters to create a 3 phase feed
Here B phase is common – an inverter is connected A-B and another C-B. There is no A-C load / generation, but B current is combined from both sources.
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Schematic
BA C
Inverter Inverter
PanelsPanels
• 12S16S
Ib is unmetered in a 12S or 5S meter
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Phase B is vector sum of –Ia and ‐Ic
Use of Phasors or Vectors in Related Equipment
Transformer connections
Instrument transformer performance
Relay applications
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Power Transformers
Power Transformer Connections
Depending on the winding configuration, the phase angle relationship changes from high side to low. Normal shift in a delta—wye is +/‐ 300.
This is important for transformer relays.
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Instrument Transformers
Vector Analysis of a PT from the Handbook for Electricity Metering
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Electromechanical relays
Testing microprocessor relays
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Sequence Phasors in Relaying
In relaying, phasors are used to analyze fault current into positive, negative and zero sequence currents; and there are many other situations where phasors make things clearer.
C B
C
A A
B
A
B C
Arlen Everistarlen.everist@pse.com
206‐550‐6706
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TI‐36X Scientific Calculator
Polar – Rectangular Conversion
You can do the rectangular-polar conversion on some calculators. Look for the expressions R>P and P>R as second and/or third functions. There should also be an X<>Y key. To convert 1,2 to polar coordinates:1(X<>Y) 2(3rd) (R>P)Display should show 2.2360… which is the magnitude(X<>Y)Display should show 63.4349… which is the angle
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Rectangular to Polar
Rectangular to Polar
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Rectangular to Polar
Rectangular to Polar
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Polar to Rectangular
• Reverse the process:
• Magnitude (2.236) X/Y angle (63.434)
• 2nd P>R
• Display = 1 X/Y 1.99999
L+G terminology
• VAtd: Time delay (lagging) measurement of Volt‐Amperes. At unity power factor VAtd is equal to watts.
• VARtd: Time delay (lagging) measurement of Volt‐Amperes reactive.