Three Phase
Circuits
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Introduction:
• The generator , motor , transformer or rectifier have only one winding is
called a single phase system
• If the current or voltage follows a phase difference 900 in a two windings ,
called two phase systems
• If the phase difference is 1200 between voltages or currents in a three winding
, called as Three phase systems
• In poly-phase systems , there are more than three windings
Advantages of three phase system:
• More efficient than single phase system
• Cost is less
• Size is small . Compared to single phase system
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Adavantages of Three Phase Circuits
• The amount of conductor material is required less for transmitting same power,
over the same distance , under same power loss
• Three phase motors produce uniform torque , where as torque produced by
single motor is pulsating
• Three phase generators not produce the harmonics when they are connected in
parallel
• Three phase motors are self starting whereas single phase motors are not self
starting
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Generation of Three Phase Voltages
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• Vectorially r.m.s values of voltages
induced in three windings are represented
in the diagram
• ER = E ∟0o v,
• EY = E ∟-120o v
• EB=www.bookspar.com
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Balanced three phase supply:
• A three phase supply is said to be balanced, when all the three voltages have the
same magnitude but differ in phase by 120° with respect to one another.
• The three phase supply is said to be unbalanced, even if one of the above
conditions is not satisfied.
Balanced Load:
• A three phase load is said to be balanced, when the impedances of all the three
phases
are exactly| Website
the same.
Even if one of them is different from the other, then
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|phase
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| QUESTION
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theStudents
three
load
is
said
to
be
unbalanced
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• In a three phase balanced load, whether star connected or delta connected, the
magnitudes of the phase currents are the same but differ in phase by 120o with
respected to one another
• But in an unbalanced load, when a three phase balanced supply is given, the
magnitudes and phases of all the three phase currents will be different.
Three phase connections:
• There are two types of three phase connections
• Star connection (Y)
• Delta connection (Δ)
Star connection (Y):
• In this method of inter-connection, the similar ends, say, “start” ends of three coils (it
could be “finishing” ends also) are joined together at point ‘N’
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• The point ‘N’ is known as star point or
neutral point
• If this three-phase voltage is applied across
a balanced symmetrical load, the neutral wire
will be carrying three currents which are exactly
equal in magnitude but are 120o out of phase
with each other. Hence, their vector sum is zero
IR + IY + IB = 0
Voltages and Currents in Y-Connection:
• The voltage induced in each winding is called the
‘phase’ voltage and current in each winding is
known as ‘phase’ current.
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• The vector diagram for phase voltages and currents in a star connection shows
that
ER = EY = EB = Eph (phase e.m.f)
• Line voltage VRY between line 1 and line 2 is the vector difference of ER and EY.
• Line voltage VYB between line 2 and line 3 is the vector difference of EY and EB.
• Line voltage VBR between line 3 and line 1 is the vector difference of EB and ER.
•
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• The p.d. between lines 1 and 2 is
VRY = ER - EY (Vector difference)
• VRY is found by compounding ER and EY
reversed and its value is given by the
diagonal of the paral1elogram in figure.
• The angle between ER and EY reversed is 60°.
If ER = EY = EB = Eph the Phase e.m.f then,
•
VRY
cos30o 
C
O
2
E ph
VRY  2  E ph  cos30
o
3
 2  E ph 
 3E ph
2
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sim ilarly
VYB  EY  E B (Vector Difference)
VRY
 3E ph
 VYB  VBR
 line voltage, VL
and
VBR  E B  E R
(Vector Difference)
 3E ph
• Hence, in star connection
VL  3E ph
It will be noted from figure that
• (a) Line voltages are 120° apart.
• (b) Line voltages are 30° ahead of their respective phase voltages.
• (c) The angle between the line currents and the corresponding line voltages is (30
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+ ɸ)
with| VTU
current
Students
NOTES |lagging.
QUESTION
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Line Currents and Phase Currents:
• Current in line 1 = IR
• Current in line 2 = IY
• Current in line 3 = IB
Since IR = IY = IB = say,
Iph - the phase current
Line current IL = Iph
Power:
• The total power in the circuit is the sum of the three phase powers. Hence
• Total Power =3 x phase power=
•
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•
•
•
•
•
•
Delta (Δ) or Mesh Connection:
Phase sequence is R, Y, B
R leads Y by 120° and Y leads B by 120°.
The voltage between lines 1 and 2 as VRY
The voltage between lines 2 and 3 as VYB
VRY leads VYB by 120
VYB leads VBR by 120°.
VRY =VYB = VBR = line voltage VL
Then, it VL = Eph
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Line Currents and Phase Currents:
• Current in line 1 is I1 = IR – IB
• Current in line 2 is I2 = IY – IR
• Current in line 3 is I3 = IB - IY
• Current in line 1 is found by
compounding IR with IB reversed
and its value is given by the diagonal
of the parallelogram
• The angle between IR and IB reversed (-IB) is 60°.
• If IB = IR = IY = Iph phase current, then current in line 1 is
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Contd..
Since all line currents are equal in magnitude i.e., I1= I2 = I3 = IL
From Vector diagram, it should be noted that
• (a) Line currents are 120o apart.
• (b) Line currents are 30o behind the respective phase currents.
• (c) The angle between the line current and the corresponding line voltage is
(30 + ɸ) with the current lagging.
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Power:
P  3  E ph I ph cos
but
E ph  VL
I ph
I
 L
3
P  3  VL 
IL
3
 cos
P  3 VL I L cos
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Measure of power in Three Phase Circuits:
• Wattmeter is the instrument which
is used to measure power in an electrical circuit.
• It consists of (i) a current coil ML’
through which the line current flows
• (ii) a potential coil PV, which is connected
across the circuit.
• The full voltage is applied across the potential coil and it carries a very small
current proportional to the applied voltage.
• Three single phase watt-meters may be connected in each phase
• The algebraic sum of their readings gives the total power consumed by the three
phase circuit.
• It can be proved that only two watt-meters are sufficient to measure power in a
three
phase
circuit.
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Two Wattmeter Method:
1. Balanced or Unbalanced Load:
• The current coils of the two watt meters are inserted in any two lines
• The potential coils joined to the third line.
• Sum of the instantaneous powers indicated by
W1 and W2 gives the instantaneous power
absorbed by the three loads L1, L2 and L3.
• Instantaneous current through W1 = IR
• Instantaneous P.D. Across
W1 = VRB = ER - EB
• Power read by W1 = IR (ER – EB)
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•
•
•
•
Contd..
Instantaneous current through W2 = IY
Instantaneous P.D. across W2 = VYB = EY - EB
Power read by
W2 = IY (EY – EB)
Therefore, W1 + W2= IR (ER – EB) + IY (EY – EB)
= IR ER + IY EY – EB (IR + IY)
I R + I Y + IB = 0
IR + IY = -IB
= IR ER + IY EY – EB (-IB)
= IR ER + IY EY + EB IB
= P1 + P2 + P3
Where P1 is the power absorbed by load L1, P2 that absorbed by L2 and P3 that
absorbed by L3.
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Balanced Load:
• The load is said to be balanced, when the impedances of the three phases are
equal
• The supply is said to be balanced,
if the three voltages are equal and
are displaced by 120o with respect
to one another.
• When a balance supply is given to
a balanced load, the currents flowing
through the three phases will be equal
in magnitude and are displaced by 120o
with respect to each other.
• Two wattmeter connections to measure power in a three phase balanced circuit
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as shown
above figure
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• Let ER, EY and EB be the r.m.s. values of the three-phase voltages and IR, IY and
IB be the r.m.s. values of the currents
• The currents lagging behind their phase voltages by ‘ɸ’.
• Current through wattmeter W1 = IR
• P.D. across voltage coil of W1 = VRB = ER – EB
-E
• Now reading W1 = VRB IR cos (30° - ɸ)
• Current through wattmeter W2 = IY
• P.D. across voltage coil of W2 = VYB = EY – EB
• W2 = VYB IY cos (30° + ɸ)
• Since the load is balanced, VRB = VYB = line voltage, VL
IR = IY = line current, IL
W1 = VL IL cos (30° - ɸ) --------(1)
W2 = VL IL cos (30° + ɸ)-------(2)
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•
B
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Contd..
W1 + W2 = VL IL cos (30° - ɸ) + VL IL cos (30° + ɸ)
= VL IL [cos 30° cos ɸ + sin 30° sin ɸ + cos 30° cos ɸ - sin 30° sin ɸ]
= VL IL (2 cos 30° cos ɸ)
W1  W2  P  3 VL I L cos
• Expression for Power Factor (p.f):
W1 - W2 = VL IL cos (30° - ɸ) - VL IL cos (30° + ɸ)
= VL IL [cos 30° cos ɸ + sin 30° sin ɸ - cos 30° cos ɸ + sin 30° sin ɸ]
= VL IL (2 sin 30° sin ɸ)
= VL IL sin ɸ
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Contd..
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• Effect of p.f. on W1 and W2:
W1 = VL IL cos (30° - ɸ) = VL IL cos 30°
= (√3/2) VL IL
W2 = VL IL cos (30° + ɸ) = VL IL cos 30°
= (√3/2) VL IL
(ɸ = 0)
(ɸ = 0)
The two wattmeter readings are positive and equal.
W1 = VL IL cos (30° - 60o) = VL IL cos (-30°) (ɸ = 60o)
= (√3/2) VL IL
W2 = VL IL cos (30° + 60o) = VL IL cos 90° (ɸ = 60o)
=0
One of the watt meters reads zero.
W1 = VL IL cos (30° - 90o) = VL IL cos (-60°)
= (1/2) VL IL
W2 = VL IL cos (30° + 90o) = VL IL cos (120°)
= - (1/2) VL IL
One of the watt-meters, reads negative (-ve).
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(ɸ = 90o)
(ɸ = 90o)
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Observations:
• One of the watt-meters reads -ve. The pointer of this watt-meter kicks back and
hence the reading can not be taken.
• Then, either the current coil connections or potential coil connections are
interchanged, pointer moves in the forward direction and the reading can be
taken.
• But this reading has to be considered as -ve.
Conclusion:
• For p.f. lying between 0 to 0.5, one of the watt-meters, reads negative (-ve).
• When p.f. = 0.5, one wattmeter reads zero (0).
• When p.f. lies between 0.5 to 1.0, both watt-meter readings are positive (+).
• When p.f. = 1, the readings of both watt-meters are equal.
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