1. Three-Phase System
Three phase systems Faculty of IIITN Ratnakar N

2. 1.1: Review of Single-Phase System The Sinusoidal voltage
v1(t) = Vm sin wt i v1 v2
Load
AC generator

3. 1.1: Review of Single-Phase System The Sinusoidal voltage
v(t) = Vm sin wt
where
Vm = the amplitude of the sinusoid
w = the angular frequency in radian/s
t = time
Three phase systems Faculty of IIITN Ratnakar N

4. The angular frequency in radians per second

5. A more general expression for the sinusoid (as shown in the figure):
v2(t) = Vm sin (wt + q)
where q is the phase

6. A sinusoid can be expressed in either sine or cosine form
A sinusoid can be expressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes.
We can transform a sinusoid from sine to cosine form or vice versa using this relationship:
sin (ωt ± 180o) = - sin ωt
cos (ωt ± 180o) = - cos ωt
sin (ωt ± 90o) = ± cos ωt
cos (ωt ± 90o) = + sin ωt

7. Sinusoids are easily expressed in terms of phasors
Sinusoids are easily expressed in terms of phasors. A phasor is a complex number that represents the amplitude and phase of a sinusoid.
v(t) = Vm cos (ωt + θ)
Time domain
Phasor domain
Time domain
Phasor domain

8. Time domain Phasor domain or θ or v2(t) = Vm sin (wt + q)

9. 1.1.1: Instantaneous and Average Power
The instantaneous power is the power at any instant of time.
p(t) = v(t) i(t)
Where v(t) = Vm cos (wt + qv)
i(t) = Im cos (wt + qi)
Using the trigonometric identity, gives

10. The average power is the average of the instantaneous power over one period.

11. The effective value is the root mean square (rms) of the periodic signal.
The average power in terms of the rms values is
Where

12. 1.1.2: Apparent Power, Reactive Power and
Power Factor
The apparent power is the product of the rms values of voltage and current.
The reactive power is a measure of the energy exchange between the source and the load reactive part.

13. The power factor is the cosine of the phase difference between voltage and current.
The complex power:

14. 1.2: Three-Phase System
3-phase system will produce output of 1.5times of 1phase motor output.
The increase in efficiency from 1phase to 2phase is 41.4%
The increase in efficiency from 1phase to 3phase is 50%
A three-phase system is superior economically and advantage, and for an operating of view, to a single-phase system. In a balanced three phase system the power delivered to the load is constant at all times, whereas in a single-phase system the power pulsates with time.

15. eB = EmB sin (wt -240o) = EmBsin (wt +120o)
The instantaneous e.m.f. generated in phase R, Y and B:
eR = EmR sin wt
eY = EmY sin (wt -120o)
eB = EmB sin (wt -240o) = EmBsin (wt +120o) 15

17. IR VR ZR ER IN EB VB VY EY ZB ZY IY IB Three-phase AC generator
Three-phase Load

18. eR = EmR sin ωt Phase voltage
The instantaneous e.m.f. generated in phase R, Y and B:
eR = EmR sin ωt
eY = EmY sin (ωt -120o)
eB = EmB sin (ωt -240o) = EmBsin (ωt +120o)
In phasor domain:
120o
ER = ERrms 0o 0o
EY = EYrms
-120o
-120o
EB = EBrms
120o
Magnitude of phase voltage
ERrms = EYrms = EBrms = Ep

19. Line voltage IR ERY VR ZR ER IN EB VB VY EY ZB ZY IY IB ERY = ER - EY
Three-phase AC generator
Line voltage IR
ERY VR ZR ER IN EB VB VY EY ZB ZY IY IB
ERY = ER - EY
Three-phase Load

20. Line voltage
= Ep 0o
- Ep
-120o
= 1.732Ep
ERY
30o
= √3 Ep
= EL
ERY
-EY
120o 0o
-120o
ERY = ER - EY

21. Line voltage IR VR ZR ER IN EB VB VY EY ZB ZY IY IB EYB EYB = EY - EB
Three-phase AC generator
Line voltage IR VR ZR ER IN EB VB VY EY ZB ZY IY IB
EYB
EYB = EY - EB
Three-phase Load

22. Line voltage
= Ep
-120o
- Ep
120o
= 1.732Ep
EYB
-90o
= √3 Ep
= EL
120o 0o
-120o
-EB
EYB
EYB = EY - EB

23. Line voltage IR VR ZR ER EBR IN EB VB VY EY ZB ZY IY IB EBR = EB - ER
Three-phase AC generator
Line voltage IR VR ZR ER
EBR IN EB VB VY EY ZB ZY IY IB
EBR = EB - ER
Three-phase Load

24. For star connected supply, EL= √3 Ep
Line voltage
= Ep
120o
- Ep 0o
= 1.732Ep
EBR
150o
EBR
120o 0o
150o
= √3 Ep
= EL
-ER
-120o
For star connected supply, EL= √3 Ep
EBR = EB - ER 24

25. Phase voltages
It can be seen that the phase voltage ER is reference.
ER = Ep 0o
EY = Ep
-120o
EB = Ep
120o
120o 0o
Line voltages
-120o
ERY = EL
30o
EYB = EL
-90o
EBR = EL
150o

26. Or we can take the line voltage ERY as reference.
Phase voltages
ER = Ep
-30o
EY = Ep
-150o
EB = Ep
90o
Line voltages
ERY = EL 0o
EYB = EL
-120o
EBR = EL
120o
Or we can take the line voltage ERY as reference. 26

27. Delta connected Three-Phase supply
Three-phase AC generator
Delta connected Three-Phase supply IR
ERY VR ZR ER EB VB VY EY ZB ZY IY IB
Three-phase Load
ERY
= ER
= Ep 0o

28. Delta connected Three-Phase supply
Three-phase AC generator
Delta connected Three-Phase supply IR VR ZR ER EB VB
EBR VY EY ZB ZY IY IB
EYB
Three-phase Load
For delta connected supply, EL= Ep

29. Connection in Three Phase System
4-wire system (neutral line with impedance)
3-wire system (no neutral line )
4-wire system (neutral line without impedance)
3-wire system (no neutral line ), delta connected load
Star-Connected Balanced Loads a) 4-wire system b) 3-wire system
Delta-Connected Balanced Loads a) 3-wire system

30. 4-wire system (neutral line with impedance)
Three-phase AC generator IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB
1.1
Three-phase Load
Voltage drop across neutral impedance:
VN = INZN

31. 4-wire system (neutral line with impedance)
Three-phase AC generator
Applying KCL at star point IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB
1.2
Three-phase Load
IR + IY + IB= IN

32. 4-wire system (neutral line with impedance)
Three-phase AC generator
Applying KVL on R-phase loop IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB
Three-phase Load

33. 4-wire system (neutral line with impedance)
Three-phase AC generator
Applying KVL on R-phase loop IR VR ZR ER ZN IN VN
ER – VR – VN = 0
ER – IRZR – VN = 0
Thus
ER – VN
1.3
IR =
Three-phase Load ZR

34. 4-wire system (neutral line with impedance)
Three-phase AC generator
Applying KVL on Y-phase loop IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB
Three-phase Load

35. 4-wire system (neutral line with impedance)
Three-phase AC generator
Applying KVL on Y-phase loop
EY – VY – VN = 0
Thus
EY – VN
1.4
IY =
EY – IYZY – VN = 0 ZY ZN IN VN VY EY ZY IY
Three-phase Load

36. 4-wire system (neutral line with impedance)
Three-phase AC generator
Applying KVL on B-phase loop
EB – VB – VN = 0
Thus
EB – VN
1.5
IB =
EB – IBZB – VN = 0 ZB ZN IN EB VB VN ZB IB
Three-phase Load

37. 4-wire system (neutral line with impedance)
Substitute Eq. 1.2, Eq.1.3, Eq. 1.4 and Eq. 1.5 into Eq. 1.1:
IR + IY + IB= IN
ER – VN
EY – VN
EB – VN VN = + + ZR ZY ZB ZN
ER – VN
EY – VN +
EB – VN = VN ZN ZR ZY ZB ER ZR + EY ZY EB ZB = 1 ZN VN

38. 4-wire system (neutral line with impedance)= ER ZR + EY ZY EB ZB 1 ZN
1.6 VN ER ZR + EY ZY EB ZB = 1 ZN VN

39. 4-wire system (neutral line with impedance)
VN is the voltage drop across neutral line impedance or the potential different between load star point and supply star point of three-phase system. = ER ZR + EY ZY EB ZB 1 ZN
1.6 VN
We have to determine the value of VN in order to find the values of currents and voltages of star connected loads of three-phase system.

40. Example IR EL = 415 volt VR ZR = 5 Ω ER ZN =10 Ω IN EB VB VN ZY= 2 ΩEY
ZB = 10 Ω IY IB
Find the line currents IR ,IY and IB. Also find the neutral current IN.
Three-phase Load

41. 3-wire system (no neutral line )
Three-phase AC generator IR VR ZR ER ZN IN EB VB VN VY EY ZB ZY IY IB
Three-phase Load

42. 3-wire system (no neutral line )
Three-phase AC generator IR VR ZR ER VN EB VB VY EY ZB ZY IY IB
Three-phase Load
No neutral line = open circuit , ZN = ∞

43. 3-wire system (no neutral line )ER ZR EY ZY EB ZB + +
1.6 VN = 1 1 ZR 1 ZY 1 ZB + + + ZN ZN = ∞ ∞ 1 ∞ =

44. 3-wire system (no neutral line )ER ZR EY ZY EB ZB + +
1.7 VN = 1 ZR + ZY ZB

45. Example IR EL = 415 volt VR ZR = 5 Ω ER EB VB VN ZY= 2 Ω EY ZB = 10 ΩIY IB
Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB.
Three-phase Load

46. 3-wire system (no neutral line ),delta connected load
Three-phase AC generator IR VR ZR ER EB VB VY EY ZB ZY IY IB
Three-phase Load

47. 3-wire system (no neutral line ),delta connected load
Three-phase AC generator IR Ir ER
VRY
VBR
ZRY
ZBR EB Ib
ZYB EY Iy IY
VYB IB
Three-phase Load

48. 3-wire system (no neutral line ),delta connected load
Three-phase AC generator IR Ir
ERY
=VRY ER
VRY
VBR
ZRY
EBR
ZBR
=VBR EB EY Ib
ZYB Iy IY
VYB IB
EYB
=VYB
Three-phase Load

49. 3-wire system (no neutral line ),delta connected load
Phase currents
VRY
ZRY
ERY
ZRY EL
ZRY
30o
Ir = = =
VYB
ZYB
EYB
ZYB EL
ZYB
-90o
Iy = = =
VBR
ZBR
EBR
ZBR EL
ZBR
150o
Ib = = =

50. 3-wire system (no neutral line ),delta connected load
Three-phase AC generator IR Ir
Line currents
ERY
=VRY ER
IR = Ir - Ib
VRY
VBR
ZRY EL
ZRY
30o
EBR EL
150o
ZBR
=VBR - EB = EY
ZBR Ib
ZYB Iy IY
IY = Iy - Ir
VYB EL EL
ZYB
-90o
30o
EYB
=VYB IB - =
ZRY
Three-phase Load

51. 3-wire system (no neutral line ),delta connected load
Three-phase AC generator IR Ir
Line currents
ERY
=VRY ER
IB = Ib - Iy
VRY
VBR
ZRY EL
ZBR
150o
EBR EL
-90o
ZBR
=VBR - EB = EY
ZYB Ib
ZYB Iy IY
VYB
EYB
=VYB IB
Three-phase Load

52. Star to delta conversion
ZRY =
ZRZY +
ZYZB
ZBZR ZB
ZYB ZR
ZBR ZY ZR
ZBR
ZRY ZY ZB
ZYB

53. Use star-delta conversion.
Example
Use star-delta conversion. IR
EL = 415 volt VR
ZR = 5 Ω ER EB VB VN
ZY= 2 Ω EY
ZB = 10 Ω IY IB
Find the line currents IR ,IY and IB .
Three-phase Load

54. 4-wire system (neutral line without impedance)
Three-phase AC generator IR VR ZR ER ZN
0 Ω IN = EB VB VN VY EY ZB ZY IR IB
Three-phase Load
VN = INZN = IN(0) = 0 volt

55. 4-wire system (neutral line without impedance)
For 4-wire three-phase system, VN is equal to 0, therefore Eq. 1.3, Eq. 1.4, and Eq. 1.5 become, ER
ER – VN
1.3
IR = ZR
EY – VN EY
1.4
IY = ZY EB
EB – VN
1.5
IB = ZB

56. Example IR EL = 415 volt VR ZR = 5 Ω ER IN EB VB VN ZY= 2 Ω EY
ZB = 10 Ω IY IB
Find the line currents IR ,IY and IB . Also find the neutral current IN.
Three-phase Load

57. eB = EmB sin (wt -240o) = EmBsin (wt +120o)
The instantaneous e.m.f. generated in phase R, Y and B:
eR = EmR sin wt
eY = EmY sin (wt -120o)
eB = EmB sin (wt -240o) = EmBsin (wt +120o)

58. 1.4: Phase sequences RYB and RBY
(a) RYB or positive sequence
VR leads VY, which in turn leads VB.
This sequence is produced when the rotor rotates in
the counterclockwise direction.

59. (b) RBY or negative sequence
VR leads VB, which in turn leads VY.
This sequence is produced when the rotor rotates in
the clockwise direction.

60. 1.5: Connection in Three Phase System
1.5.1: Star Connection a) Three wire system

61. Star Connection b) Four wire system

62. Wye connection of Load

63. 1.5.2: Delta Connection

64. Delta connection of load

65. 1.6: Balanced Load Connection in 3-Phase System

66. Wye-Connected Balanced Loads b) Three wire system
Example IR
EL = 415 volt VR
ZR = 20 Ω ER EB VB
ZY= 20 Ω VN EY
ZB = 20 Ω IY IB
Find the line currents IR ,IY and IB . Also find the voltages VR, VY and VB.
Three-phase Load

67. Wye-Connected Balanced Loads b) Three wire system
VN = = 0 volt
VR = ER
VY = EY
VB = EB

68. 1.6.1: Wye-Connected Balanced Loads a) Four wire system
Example IR
EL = 415 volt VR
ZR = 20 Ω ER IN EB
ZY= 20 Ω VB VN EY
ZB = 20 Ω IY IB
Find the line currents IR ,IY and IB . Also find the neutral current IN.
Three-phase Load

69. 1.6.1: Wye-Connected Balanced Loads a) Four wire system
For balanced load system,
IN = 0 and Z1 = Z2 = Z3

70. Wye-Connected Balanced Loads b) Three wire system

71. 1.6.2: Delta-Connected Balanced Loads
Phase currents:
Line currents:

72. 1.7: Unbalanced Loads

73. 1.7.1: Wye-Connected Unbalanced Loads Four wire system
For unbalanced load system,
IN  0 and Z1  Z2  Z3

74. 1.7.2: Delta-Connected Unbalanced Loads
Phase currents:
Line currents:

75. Power in a Three Phase System

76. The three phase power is equal the sum of the phase powers
Power Calculation
The three phase power is equal the sum of the phase powers
P = PR + PY + PB
If the load is balanced:
P = 3 Pphase = 3 Vphase Iphase cos θ

77. 1.8.1: Wye connection system:
I phase = I L and
Real Power, P = 3 Vphase Iphase cos θ
Reactive power,
Q = 3 Vphase Iphase sin θ
Apparent power,
S = 3 Vphase Iphase
or S = P + jQ

78. 1.8.2: Delta connection system
VLL= Vphase
P = 3 Vphase Iphase cos θ

79. The Single-Phase Equivalent
By now it should be apparent that if you know the solution for one phase of a balanced system, you effectively know the solution for all three phases.

80. Basic Three-Phase Relationships

81. Tutorial Problems

82. Practice problem 1

83. Three phase systems Faculty of IIITN Ratnakar N

84. Practice problem 1
Three phase systems Faculty of IIITN Ratnakar N

93. Three phase power measurement

94. Power measurement
In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters.

95. Three Phase Circuit Four wire system, Each phase measured separately

96. watt-meter connection
Current coil (low impedance)
voltage coil (high impedance) 96

97. a) Four wire system Example WR IR VR ER EL = 415 volt IN EB VB EY VN
ZR = 5
30o Ω ER
EL = 415 volt IN EB VB
ZY = 10
90o EY VN Ω WY
ZB = 20
45o Ω IY IB WB
Three-phase Load
Find the three-phase total power, PT.

98. b) Three wire system Example WR IR VR ER EL = 415 volt EB EY VN WY IY
ZR = 5
30o Ω ER
EL = 415 volt EB
ZY = 10
90o EY VN Ω WY
ZB = 20
45o Ω IY IB WB
Three-phase Load
Find the three-phase total power, PT.

99. b) Three wire system Example WR IR VR ER EL = 415 volt EB VB EY VN WY
ZR = 5
30o Ω ER
EL = 415 volt EB VB
ZY = 10
90o EY VN Ω WY
ZB = 20
45o Ω IY IB WB
Three-phase Load
Find the three-phase total power, PT.

100. Three Phase Circuit Three wire system,
The three phase power is the sum of the two watt-meters reading

101. Proving:
The three phase power (3-wire system) is the sum of the two watt-meters reading
Instantaneous power:
pA = vA iA
pB = vB iB
pC = vC iC
pT = pA + pB + pC = vA iA + vB iB +vC iC
= vA iA + vB iB +vC iC = vA iA + vB (-iA -iC) +vCiC
101

102. Proving:
The three phase power (3-wire system) is the sum of the two watt-meters reading
Instantaneous power:
pT = vA iA + vB (-iA –iC) +vCiC
= (vA – vB )iA + (vC – vB )iC
= vAB iA + vCBiC
pT = pAB + pCB
102

103. Phasor diagram for a three-phase balanced star-connected circuit

105. Comments on Two Wattmeter Readings(V.V.IMP)

107. Power measurement
In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters.
In a three-wire system (three phases without neutral) the power is measured using only two single phase watt-meters.
The watt-meters are supplied by the line current and the line-to-line voltage.
107

108. Q1 Q2

109. Practice Problems P2 P1

110. P3 P4

111. Thank you