# 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
𝑉𝑝 = 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.
𝑣𝐴𝑁
𝑣𝐵𝑁
𝑣𝐶𝑁
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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.
𝑣𝐴𝑁
𝑣𝐵𝑁
𝑣𝐶𝑁
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acb or negative sequence
The phase sequence of the three phase voltages shown below is acb or negative sequence.
𝑣𝐴𝑁
𝑣𝐶𝑁
𝑣𝐵𝑁
𝑣𝐴𝑁 = 2𝑉𝑝 𝑐𝑜𝑠𝜔𝑡
𝑣𝐶𝑁 = 2𝑉𝑝 𝑐𝑜𝑠 𝜔𝑡 − 1200
𝑣𝐵𝑁 = 2𝑉𝑝 𝑐𝑜𝑠 𝜔𝑡 + 1200
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Phase sequence: Example
Determine the phase sequence of the set of voltages given as
𝑣𝐴𝑁 = 200 𝑐𝑜𝑠 𝜔𝑡 + 100
𝑣𝐶𝑁 = 200𝑐𝑜𝑠 𝜔𝑡 − 2300
𝑣𝐵𝑁 = 200𝑐𝑜𝑠 𝜔𝑡 − 1100 .
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Star (Y) connected three phase source
Phase voltages:
Line voltages:
𝑉𝑎𝑏 is
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3 times 𝑉𝑎𝑛 and leads by 300.
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Delta (∆) connected three phase source
Line voltages = Phase voltages
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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
𝐼𝑎𝑏 = 𝐼𝑝 ∠𝜃
Line currents
𝐼𝑎 = 3𝐼𝑝 ∠ 𝜃 − 300
𝐼𝑏𝑐 = 𝐼𝑝 ∠ 𝜃 − 1200
𝐼𝑏 = 3𝐼𝑝 ∠ 𝜃 − 1500
𝐼𝑐𝑎 = 𝐼𝑝 ∠ 𝜃 − 2400
𝐼𝑐 = 3𝐼𝑝 ∠ 𝜃 + 900
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Delta - Connected Three Phase Loads: Example
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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
𝐼𝑎𝑏 = 𝐼𝑝 ∠𝜃
Line currents
𝐼𝑎 = 3𝐼𝑝 ∠ 𝜃 − 300
𝐼𝑏𝑐 = 𝐼𝑝 ∠ 𝜃 − 1200
𝐼𝑏 = 3𝐼𝑝 ∠ 𝜃 − 1500
𝐼𝑐𝑎 = 𝐼𝑝 ∠ 𝜃 − 2400
𝐼𝑐 = 3𝐼𝑝 ∠ 𝜃 + 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 -  connection
•  -  connection and
•  - 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, 𝑬𝐴𝑁 = 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.
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Balanced Y – Y connection: Example
For the given figure, 𝑬𝐴𝑁 = 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.
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Balanced Y –  or  - Y connection
Transform the  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  connected system to obtain the required parameters of the original circuit.
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Balanced Y –  or  - Y connection
A balanced abc – sequence Y-connected source with 𝐕𝑎𝑛 = 100∠100 V is connected to a -connected
balanced load 8 + 𝑗4 Ω per phase. Calculate the phase and line currents.
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Balanced Y –  or  - Y connection
A balanced -connected load having an impedance 20 – j15 Ω is connected to a -connected, positivesequence generator having 𝐕𝑎𝑏 = 330∠00 V. Calculate the phase currents of the load and the line
currents.
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Balanced Y –  or  - Y connection
A balanced Y-connected load with a phase impedance of 40 + j25 Ω is supplied by a balanced, positive
sequence -connected source with a line voltage of 210 V. Calculate the phase currents. Use 𝐕𝑎𝑏 as a
reference.
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Balanced Y –  or  - Y connection
The line voltage of the generator is 207.8 V.
I.
Find the line voltage 𝑉𝑎𝑏 at the load.
II.
Find the  currents.
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Power in a Balanced Three Phase System
For a Y-connected three phase load with a per phase impedance 𝑍∠𝜃 ,
𝑉𝑝 and 𝐼𝑝 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
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Power in a Balanced Three Phase System
Thus the total instantaneous power,
𝑝 = 3𝑉𝑝 𝐼𝑝 𝑐𝑜𝑠𝜃
Thus the total instantaneous power in a balanced three-phase system is constant.
Hence, average total active and reactive power
𝑃𝑇 = 3𝑉𝑝 𝐼𝑝 𝑐𝑜𝑠𝜃
𝑄𝑇 = 3𝑉𝑝 𝐼𝑝 𝑠𝑖𝑛𝜃
where  is the power factor angle.
The average per phase active and reactive power become
𝑃𝑝 = 𝑉𝑝 𝐼𝑝 𝑐𝑜𝑠𝜃
𝑄𝑝 = 𝑉𝑝 𝐼𝑝 𝑠𝑖𝑛𝜃
The total real and reactive power,
These expressions are valid for both Y and  connected loads.
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Power in a Balanced Three Phase System
The total complex power
𝑺𝑻 = 3𝑉𝑝 𝐼𝑝 𝑐𝑜𝑠𝜃 + 𝑗3𝑉𝑝 𝐼𝑝 𝑠𝑖𝑛𝜃
𝑺 𝑇 = 𝑃𝑇 + 𝑗𝑄𝑇
𝑺 𝑇 = 3𝑽𝑝 𝑰𝑝 ∗
𝑺 𝑇 = 3 𝑉𝐿 𝐼𝐿 ∠𝜃
𝑆𝑇 = 3 𝑉𝐿 𝐼𝐿
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Power in a Balanced Three Phase System
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Power in a Balanced Three Phase System
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Power in a Balanced Three Phase System
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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 ∆ 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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