INTRODUCTION
This article treats phasors in a three phase wye circuit. It shows how to solve for current in the neutral line of a three phase wye circuit using polar coordinates. It also demonstrates the power savings in a three phase circuit.
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BRIEF REVIEW OF PHASORS
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CONVERTING PHASOR TO A COMPLEX NUMBER
magnitude/angle = (magnitude * cos angle) + j*(magnitude*sin angle)
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MULTIPLICATION OF PHASORS
mag1/angle a * mag2/angle b = mag1*mag2/angle a + angle b
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DIVISION OF PHASORS
(mag1/angle a) / (mag2/angle b) = mag1*mag2/angle a - angle b
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ADDITION OF PHASORS
mag1/angle a + mag2/angle b =
((mag1 * cos angle a) + (mag2*cos angle b)) +
j*((mag1*sin angle a) + (mag2*sin angle b))
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SUBTRACTION OF PHASORS
mag1/angle a - mag2/angle b =
((mag1 * cos angle a) - (mag2*cos angle b)) +
j*((mag1*sin angle a) - (mag2*sin angle b))
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WHAT IS THREE PHASE?
A three phase source is a source providing three voltages that are 120 degrees out of phase with each other. (See figure one) This article is focused on the three phase wye system with a neutral line.
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THREE PHASE SYSTEMS WITH BALANCED LOADS
The purpose of this example is to show that the current in the neutral line of a balanced three phase load is zero amperes and to show that the total power dissipated in a balanced three phase load is also zero watts.
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RESISTIVE LOADS BETWEEN EACH PHASE AND NEUTRAL
The voltage in each phase of the three phase wye source is 100 volts.. The resistive load 25/0 exists between each phase and neutral of a three phase circuit. (See Figure Two) Calculate the current in the neutral line and the total power dissipated.
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V1 = 100/0
V2 = 100/120
V3 = 100/240
R1 = R2 = R3 = 25/0 = (25 * cos 0) + j(0 * sin 0) = 25 * cos 0
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CALCULATE CURRENT IN PHASE ONE
I1 = (100/0) / (25/0)
I1 = 4/0 = (4 * cos 0) + j(0 * sin 0)
The current on the real axis is
4*cos 0 = 4
I1 = 4 + j0
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CALCULATE CURRENT IN PHASE TWO
I2 = (100/120) / (25/0)
I2 = 4/120 = (4*cos 120) + j(4*sin/120)
4 * cos 120 = 4*(-0.5) = -2
4*sin 120 = 4*(0.866) = 3.46
I2 = -2 + j3.46
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CALCULATE CURRENT IN PHASE THREE
I3 = (100/240) / (25/0)
I3 = 4/240 = (4 * cos 240) + j(4* sin/240)
4*cos 240 = 4*(-0.5)
4*sin 240 = 4*(-.866)
I3 = -2 -j3.46
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CALCULATE CURRENT IN NEUTRAL LINE
In = I1 + I2 + I3 (Algebraic Sum of currents on each axis.)
In = (4 + J0) + (-2 + j3.46) + (-2 -j3.46)
The algebraic sum of the currents is
In = 0 amperes
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POWER DISTRIBUTION
Calculate the total power Pt
P = V*I
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CALCULATE POWER IN PHASE ONE
P1/0 = V1*I1
V1 = 100/0
I1 = 4/0
P1/0 = 100/0 * 4/0 = 400/0
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CALACULATE IN PHASE TWO
P2 = V2* I2
V2 = 100/120
I2 = 4/120
P2 = 100/120 * 4/120 = 400/240
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CALCULATE POWER IN PHASE THREE
P3 = V3*I3
V3 = 100/240
I3 = 4/240
P3 = 100/240 * 4/240 = 400/480
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CALCULATE TOTAL POWER
Pt = P1 + P2 + P3
Pt = 400/0 + 400/240 + 400/480
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CONVERT PHASE ONE POWER TO COMPLEX NUMBER
P1/0 = 400/0 = (400 * cos 0) + (j*400 * sin 0)
P1/0 = 400 + j0
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CONVERT PHASE TWO POWER TO COMPLEX NUMBER
P2 = 400/240 = (400 * cos 240) + (j*400 * sin 240)
cos 240 = -0.5
sin 240 = -0.866
P2 = 400*(-0.5) + j400*(-0.866)
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CONVERT PHASE THREE POWER TO COMPLEX NUMBER
P3 = 400/480 = (400 * cos 480) + (j*400 * sin 480)
cos 480 = -0.5
sin 480 = 0.866
P3 = 400*(-0.5) + j400*(0.866)
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ADD PHASE ONE POWER, PHASE TWO POWER AND PHASE THREE POWER.
Pt = P1 + P2 + P3
P1/0 = 400 + j0
P2 = 400*(-0.5) + j400*(-0.866)
P3 = 400*(-0.5) + j400*(0.866)
Pt = (400 - 200 - 200) +j(0 + 400(-0.866) + 400(0.866)
Pt = 0
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Figure One: Three Phase Source
Figure Two: Three Phase Resister Load
Figure Three: Phasor Diagram
