# Three Phase Circuits

```EE1302 Introduction to Electrical Engineering
EE1302 Introduction to Electrical Engineering
Unit 8: Three Phase Circuits
Dr.R.P.S. Chandrasena
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EE1302 Introduction to Electrical Engineering
Generation of three phase voltages
ππ = RMS phase voltage
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Balanced three phase voltages
In a balanced three phase supply, the phase voltages are equal in magnitude and are out of phase by 1200
with each other.
π£π΄π
π£π΅π
π£πΆπ
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abc or positive sequence
The phase sequence is the time order in which the voltages pass through their respective maximum values.
Hence, the phase sequence of the three phase voltages shown below is abc or positive sequence.
π£π΄π
π£π΅π
π£πΆπ
EE1302 Introduction to Electrical Engineering
acb or negative sequence
The phase sequence of the three phase voltages shown below is acb or negative sequence.
π£π΄π
π£πΆπ
π£π΅π
π£π΄π = 2ππ πππ ππ‘
π£πΆπ = 2ππ πππ  ππ‘ − 1200
π£π΅π = 2ππ πππ  ππ‘ + 1200
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Phase sequence: Example
Determine the phase sequence of the set of voltages given as
π£π΄π = 200 πππ  ππ‘ + 100
π£πΆπ = 200πππ  ππ‘ − 2300
π£π΅π = 200πππ  ππ‘ − 1100 .
EE1302 Introduction to Electrical Engineering
Star (Y) connected three phase source
Phase voltages:
Line voltages:
πππ is
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3 times πππ and leads by 300.
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Delta (β) connected three phase source
Line voltages = Phase voltages
EE1302 Introduction to Electrical Engineering
A balanced load is one in which the phase impedances are equal in magnitude and in phase. Star – Delta
transformation can be applied to transform a delta-connected load to an equivalent star-connected load
and vise versa.
For balanced three
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Star - Connected Three Phase Loads
For a star-connected load, phase currents and line currents are the same,
The line voltage is 3 times the phase voltage with 300 phase shift.
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Delta - Connected Three Phase Loads
For a delta-connected load, phase voltages and the line to line (line) voltages are the same. Line currents are 3
times the phase currents with -300 phase shift.
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Phase currents
πΌππ = πΌπ ∠π
Line currents
πΌπ = 3πΌπ ∠ π − 300
πΌππ = πΌπ ∠ π − 1200
πΌπ = 3πΌπ ∠ π − 1500
πΌππ = πΌπ ∠ π − 2400
πΌπ = 3πΌπ ∠ π + 900
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Delta - Connected Three Phase Loads: Example
EE1302 Introduction to Electrical Engineering
Delta - Connected Three Phase Loads
Taking IAB as the reference.
For a delta-connected load, phase voltages and the line to line (line) voltages are the same. Line currents are 3
times the phase currents with -300 phase shift.
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Phase currents
πΌππ = πΌπ ∠π
Line currents
πΌπ = 3πΌπ ∠ π − 300
πΌππ = πΌπ ∠ π − 1200
πΌπ = 3πΌπ ∠ π − 1500
πΌππ = πΌπ ∠ π − 2400
πΌπ = 3πΌπ ∠ π + 900
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Since both three phase source and three phase load can be either star or delta connected, there are four
• Y - Y connection
• Y - ο connection
• ο - ο connection and
• ο - Y connection.
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Balanced Y – Y connection
Circuit for phase A (Per phase equivalent circuit)
When solving balanced three phase circuits, it is convenient to analyze the per phase circuit and then
derive the parameters in other phases using the results obtained through per phase analysis.
For a balanced Y – Y connected system, the per phase equivalent circuit can be obtained simply by
considering one phase.
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Balanced Y – Y connection: Example
For the given figure, π¬π΄π = 120V∠00 .
a) Solve for line currents.
b) Solve for the phase voltages at the load.
c) Solve for the line voltages at the load.
EE1302 Introduction to Electrical Engineering
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Balanced Y – Y connection: Example
For the given figure, π¬π΄π = 120V∠00 .
a) Solve for line currents.
b) Solve for the phase voltages at the load.
c) Solve for the line voltages at the load.
EE1302 Introduction to Electrical Engineering
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Balanced Y – ο or ο - Y connection
Transform the ο connected source/load to equivalent Y connected source/load to transform the circuit to
Y – Y connected system. Then apply the per phase analysis, considering one phase and used the voltage
and current relationships for ο connected system to obtain the required parameters of the original circuit.
EE1302 Introduction to Electrical Engineering
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Balanced Y – ο or ο - Y connection
A balanced abc – sequence Y-connected source with πππ = 100∠100 V is connected to a ο-connected
balanced load 8 + π4 β¦ per phase. Calculate the phase and line currents.
EE1302 Introduction to Electrical Engineering
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Balanced Y – ο or ο - Y connection
A balanced ο-connected load having an impedance 20 – j15 β¦ is connected to a ο-connected, positivesequence generator having πππ = 330∠00 V. Calculate the phase currents of the load and the line
currents.
EE1302 Introduction to Electrical Engineering
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Balanced Y – ο or ο - Y connection
A balanced Y-connected load with a phase impedance of 40 + j25 β¦ is supplied by a balanced, positive
sequence ο-connected source with a line voltage of 210 V. Calculate the phase currents. Use πππ as a
reference.
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Balanced Y – ο or ο - Y connection
The line voltage of the generator is 207.8 V.
I.
Find the line voltage πππ at the load.
II.
Find the ο currents.
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Power in a Balanced Three Phase System
For a Y-connected three phase load with a per phase impedance π∠π ,
ππ and πΌπ are rms voltage and rms current.
Thus the total instantaneous power,
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Power in a Balanced Three Phase System
Thus the total instantaneous power,
Using the trigonometric identity
EE1302 Introduction to Electrical Engineering
Power in a Balanced Three Phase System
Thus the total instantaneous power,
π = 3ππ πΌπ πππ π
Thus the total instantaneous power in a balanced three-phase system is constant.
Hence, average total active and reactive power
ππ = 3ππ πΌπ πππ π
ππ = 3ππ πΌπ π πππ
where ο± is the power factor angle.
The average per phase active and reactive power become
ππ = ππ πΌπ πππ π
ππ = ππ πΌπ π πππ
The total real and reactive power,
These expressions are valid for both Y and ο connected loads.
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Power in a Balanced Three Phase System
The total complex power
πΊπ» = 3ππ πΌπ πππ π + π3ππ πΌπ π πππ
πΊ π = ππ + πππ
πΊ π = 3π½π π°π ∗
πΊ π = 3 ππΏ πΌπΏ ∠π
ππ = 3 ππΏ πΌπΏ
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EE1302 Introduction to Electrical Engineering
Power in a Balanced Three Phase System
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EE1302 Introduction to Electrical Engineering
Power in a Balanced Three Phase System
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EE1302 Introduction to Electrical Engineering
Power in a Balanced Three Phase System
EE1302 Introduction to Electrical Engineering
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Power Factor Correction in Three Phase System
In three phase systems, three capacitors are needed and hence Qcap represent the reactive power
supplied by all three capacitors. Also note that the voltage applied across the capacitor vary based on
the type of connection of the capacitors.
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Power Factor Correction in Three Phase System
Two balanced loads are connected to a 240 kV rms, 60 Hz line, as shown in the figure below. Load 1
draws 30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8
lagging. Assuming the abc sequence, determine
(a) The complex, real, and reactive power absorbed by the combined load
(b) The line currents
(c) The kVAR rating of the three capacitors β connected in parallel with the load that will raise the
power factor to 0.9 lagging and
(d) The capacitance of each capacitor.
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