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Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit
Theory - “ Polyphase Circuit”
Phase:- Phase of a particular value of an alternating quantity is the fractional part of time period or
cycle through which the quantity has advanced from the selected zero position of reference.
Single Phase:- The generator producing a single phase supply has only one armature winding. If
there is only one winding then only one voltage will be produced between the two ends of that
winding. This voltage is called as single phase voltage.
Polyphase: Poly means many
 Polyphase means many phases.
 Polyphase means It has many phases or windings.
 The generator producing two or more separate windings displaced from each other by
equal electrical angles called Polyphase.
 In 3phase system, they are produce three voltages of the same magnitude and frequency
but displaced 120 electrical from one another (Each Phase).
 In 2phase system, they are produce two voltages of the same magnitude and frequency but
displaced 90 electrical from second phase.
360
 The electrical displacement between different phases or windings is equal to
,
N
Where N = Number of phases.
 3phase is most popular method in electrical.
Advantages of 3-phase System:a) Constant Power:• In 1phase circuit, the instantaneous power varies sinusoidally from zero to peak value at
twice the supply frequency. This pulsating nature of power is objectionable for many
applications.
• In 3phase circuit, the power supplied at all instants of time is constant. This is very superior
in 3phase system than 1phase system.
b) Greater Output:• Output is high in 3phase machine than 1phase machine.
• 3phase machine is smaller than 1phase machines.
c) Cheaper:• 3phase motors are much smaller and less expensive than 1phase motors because less
material required.
• 3phase motors are self starting.
• 3phase motors not required any special provision to get them started as 1phase requires
internal starting device.
d) Power Transmission economics:• Transmission of electric power by 3phase system is cheaper than that of 1phase system.
• Example:- If we have to transmit the same amount of power over a fixed distance at a given
voltage, then 3phase system requires only 3/4th weight of copper than that required by
1phase system. This means a saving in the number and strength of transmission towers.
e) Three-phase Rectifier Service:-

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Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit
•
•
•

Rectification is very easy in 3phase system than 1phase system.
It is easier to filter out the ripple component of 3phase voltage.
This is useful where large a.c. power is to be converted into steady d.c.power example:Radio and Television transmitters.
f) Miscellaneous Advantages:• A 3phase system can set-up a rotating magnetic field in stationary windings. This is cannot
be done with a 1phase current.
• 3phase motors are more efficient and have a higher power factor than 1phase motors.
Concept of 1phase, 2phase and 3phase supply:Generation of 1phase Voltage

•
•
•

Consider, conductor A is placed between two pole say South and North as shown in figure.
It has only one winding or coil A rotating in anticlockwise direction with an angular velocity
( ) in the 2-pole field.
E.m.f. induced in the coil is,
e a1a2  E m . sin t

Generation of 2phase Voltage

•
•
•
•
•

Consider, conductor A & B (identical) are placed between two pole say South and North as
shown in figure.
They are displaced 90 electrical from each other. Coil is rotating in anticlockwise direction
with an angular velocity ( ) in the 2-pole field.
Since, the two coils are identical and have the same angular velocity, then e.m.f. induced in
them will be of the same magnitude and frequency.
E.m.f. in coil A leads that in coil B by 90
Emf equations are, e a1a2  E m . sin t and eb1b2  Em . sin(t  90)

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Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit
Generation of 3phase Voltage

•Consider, conductor A, B & C (identical) are placed between two pole say South and North as
shown in figure.
•They are displaced 120 electrical from each other. Coil is rotating in anticlockwise direction with
an angular velocity ( ) in the 2-pole field.
•Since, the three coils are identical and have the same angular velocity, then e.m.f. induced in
them will be of the same magnitude and frequency.
•E.m.f. in coil B will be 120 behind that of coil A and Coil C will be 240 behind that of coil A.
• E.m.f. Equations are, e a1a2  E m . sin t , eb1b2  Em . sin(t  90) and ec1cc  Em . sin(t  240)
Interconnection of three phases:-

•
•
•
•

In a 3phase alternator, there are three windings or phases.
Each has two terminals viz start and finish.
If separate load is connected then it requires 6 conductors for transmit power.
Due to this, system will be complicated and expensive.

Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit
•

To reduce the number of conductors, they are inter connected i.e. Star or Wye connection
and Delta or Mesh Connection.

Star Connection or Wye Connection:-

System is balanced and emf’s will be equal in magnitude and displaced 120 from one another.
• Junction of all three wires called as Neutral (N).
• Three wires are called Line wires say L1 , L2 , L3 i.e. R, Y, B supply.
• Voltage measured between two terminals are called as phase voltage ( V ph )say VR ,VY ,VB
• Voltage between any two lines are called line voltage ( VL ) say VRY ,VYB ,VBR
 Potential Difference between lines,
For R & Y
• VRY  VRN  VNY
………(Phasor Sum)
• VRY  VRN  VYN
……… (Phasor Difference)
For Y & B
• VYB  VYN  VNB
………(Phasor Sum)
• VYB  VYN  VBN
……… (Phasor Difference)
For B & R
• VBR  VBN  VNR
….……(Phasor Sum)
• VBR  VBN  VRN
……… (Phasor Difference)
From Phasor Diagram,
VR  VY  VB  VPh
We have,
VBR  VBN  VNR

VBR  E ph . cos 300  E ph . cos 300
VBR  2.E ph . cos 30 0
VBR  3.E ph
Similarly for other two phase voltage or line voltage,
VRY  3.E ph and VYB  3.E ph
OR

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Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit
By, the trigonometric,
2
2
2
VL  VR  VY  2.VR.VY . cos 60

VL  VPh  VPh  2.VPh .VPh .(0.5)
2

2

VL  3VPh
2

2

2

VL  3.VPh

………….Line Voltage for Star Connection

It shows, Line Voltage is 3 times the phase voltage.
From Phasor diagram,
Phase current is same as line current.
...……… Line Current for Star Connection
I L  I Ph
Power for one phase is,
Power = 3 X power in each phase
P  3.Vph .I Ph . cos 
We have,

VL  3.VPh Substitute in P, we will get,

……………Power of Star Connection
P  3.VL .I L . cos 
This is the relation between Phase current and Line current, Phase Voltage and Line Voltage in
Star Connection or Wye Connection..
Delta Connection or Mesh Connection:-

Fig. Balanced 3phase Delta Connection
Fig. Phasor Diagram
• System is balanced and current’s will be equal in magnitude and 120 displaced from one
another.
• There is no Junction for all three wires.
• Three wires are called Line wires say i.e. R, Y, B supply.
• Voltage measured between two terminals are called as phase voltage ( V ph )say VR ,VY ,VB
•
•

Voltage between any two lines are called line voltage ( VL ) say VRY ,VYB ,VBR
Currents between phases,
IR  IR  IB
IY  IY  I R

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Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit
I B  I B  IY
Now, for Current in R line,
IR  IR  IB
I R  I ph . cos 30 0  (  I ph . cos 30 0 )
I R  I ph . cos 30 0  I ph . cos 30 0
I R  2. I ph . cos 30 0
I R  3I ph
Similarly, I Y  3I ph & I B  3I ph
i.e. I L  3I ph

...……… Line Current for Delta Connection

This is required Line current in Delta Connection.
Line voltage is equal to phase voltage in Delta Connection.
...……… Line Voltage for Delta Connection
VL  V ph
Power for one phase is,
Power = 3 X power in each phase
P  3.Vph .I Ph . cos 
We have,

I L  3I ph

Substitute in P, we will get,

……………Power of Delta Connection
P  3.VL .I L . cos 
This is the relation between Phase current and Line current, Phase Voltage and Line
Voltage in Delta Connection or Mesh Connection..
Comparison of Star and Delta Connections:Sr. Star Connection
Delta Connection
No
1

VL  3.VPh

VL  VPh

2

I L  I ph

I L  3I ph

3

P  3.VL .I L . cos 

P  3.VL .I L . cos 

4

Alternator & distribution transformer
secondary is star connected.

Transmission transformer secondary is
delta connected.

5

Same power rating more current

Same power rating less current

6

For same line voltage, Star connected
alternator requires less insulation than
delta connection.

For same line voltage, Delta connected
alternator requires more insulation than star
connection.

7

We can get 3ph, 4 wire system. Its
flexible connection.

Suitable for rotary convertors.

Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit
8

Neutral point can be earthed and offers
measurement.

Most 3ph induction motors are delta
connected.

PROBLEMS – “Polyphase Circuit”
Formulae for polyphase Circuit: Z ph  R 2  X 2
 Power Factor = cos  


R
Z

Sr.
No

Star Connection

Delta Connection

1

VL  3.VPh

VL  VPh

2

I L  I ph

I L  3I ph

3

P  3.VL .I L . cos 

P  3.VL .I L . cos 

Q.1 A balanced 3phase star connected load consisting in each phase a resistance and inductive
reactance og 4ohm and 3ohm respectively, connected to 3phase 400V, 50Hz supply.
Find (a) Line Current, (b) Power Factor and (c) Total Power
Q.2 A balanced star connected load is connected across 230V, 3ph supply. If the resistance &
Inductive reactance per phase are 6ohm & 8ohm respectively.
Find (a) Line Current, (b) Power Factor and (c) Total Power
Q.3 A balanced 3ph star connected load is supplied from a 3ph, 400V, 50Hz supply. The
resistance of each coil is 6ohm and reactance 8ohm.
Find (a) Line Current, (b) Power Factor and (c) Total Power
Q.4 Three resistance each of 10ohm resistance are connected in (i) Star Connection and (ii) Delta
connected across 440V. 3ph a.c. supply.
Find the power consumed in each case and ratio of power in Delta to power in Star connection.
Q.5 A balanced 3ph, star connected load is supplied with 3ph, 400V, 50Hz Supply. The phase
current is 10A and p.f. is 0.6 lagging. Calculate total power.
---------------------- Read more for more knowledge ---------------------------------- Vijay Raskar ----

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Polyphase circuit vbr

  • 1. Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit Theory - “ Polyphase Circuit” Phase:- Phase of a particular value of an alternating quantity is the fractional part of time period or cycle through which the quantity has advanced from the selected zero position of reference. Single Phase:- The generator producing a single phase supply has only one armature winding. If there is only one winding then only one voltage will be produced between the two ends of that winding. This voltage is called as single phase voltage. Polyphase: Poly means many  Polyphase means many phases.  Polyphase means It has many phases or windings.  The generator producing two or more separate windings displaced from each other by equal electrical angles called Polyphase.  In 3phase system, they are produce three voltages of the same magnitude and frequency but displaced 120 electrical from one another (Each Phase).  In 2phase system, they are produce two voltages of the same magnitude and frequency but displaced 90 electrical from second phase. 360  The electrical displacement between different phases or windings is equal to , N Where N = Number of phases.  3phase is most popular method in electrical. Advantages of 3-phase System:a) Constant Power:• In 1phase circuit, the instantaneous power varies sinusoidally from zero to peak value at twice the supply frequency. This pulsating nature of power is objectionable for many applications. • In 3phase circuit, the power supplied at all instants of time is constant. This is very superior in 3phase system than 1phase system. b) Greater Output:• Output is high in 3phase machine than 1phase machine. • 3phase machine is smaller than 1phase machines. c) Cheaper:• 3phase motors are much smaller and less expensive than 1phase motors because less material required. • 3phase motors are self starting. • 3phase motors not required any special provision to get them started as 1phase requires internal starting device. d) Power Transmission economics:• Transmission of electric power by 3phase system is cheaper than that of 1phase system. • Example:- If we have to transmit the same amount of power over a fixed distance at a given voltage, then 3phase system requires only 3/4th weight of copper than that required by 1phase system. This means a saving in the number and strength of transmission towers. e) Three-phase Rectifier Service:- Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
  • 2. Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit • • • Rectification is very easy in 3phase system than 1phase system. It is easier to filter out the ripple component of 3phase voltage. This is useful where large a.c. power is to be converted into steady d.c.power example:Radio and Television transmitters. f) Miscellaneous Advantages:• A 3phase system can set-up a rotating magnetic field in stationary windings. This is cannot be done with a 1phase current. • 3phase motors are more efficient and have a higher power factor than 1phase motors. Concept of 1phase, 2phase and 3phase supply:Generation of 1phase Voltage • • • Consider, conductor A is placed between two pole say South and North as shown in figure. It has only one winding or coil A rotating in anticlockwise direction with an angular velocity ( ) in the 2-pole field. E.m.f. induced in the coil is, e a1a2  E m . sin t Generation of 2phase Voltage • • • • • Consider, conductor A & B (identical) are placed between two pole say South and North as shown in figure. They are displaced 90 electrical from each other. Coil is rotating in anticlockwise direction with an angular velocity ( ) in the 2-pole field. Since, the two coils are identical and have the same angular velocity, then e.m.f. induced in them will be of the same magnitude and frequency. E.m.f. in coil A leads that in coil B by 90 Emf equations are, e a1a2  E m . sin t and eb1b2  Em . sin(t  90) Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
  • 3. Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit Generation of 3phase Voltage •Consider, conductor A, B & C (identical) are placed between two pole say South and North as shown in figure. •They are displaced 120 electrical from each other. Coil is rotating in anticlockwise direction with an angular velocity ( ) in the 2-pole field. •Since, the three coils are identical and have the same angular velocity, then e.m.f. induced in them will be of the same magnitude and frequency. •E.m.f. in coil B will be 120 behind that of coil A and Coil C will be 240 behind that of coil A. • E.m.f. Equations are, e a1a2  E m . sin t , eb1b2  Em . sin(t  90) and ec1cc  Em . sin(t  240) Interconnection of three phases:- • • • • In a 3phase alternator, there are three windings or phases. Each has two terminals viz start and finish. If separate load is connected then it requires 6 conductors for transmit power. Due to this, system will be complicated and expensive. Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
  • 4. Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit • To reduce the number of conductors, they are inter connected i.e. Star or Wye connection and Delta or Mesh Connection. Star Connection or Wye Connection:- System is balanced and emf’s will be equal in magnitude and displaced 120 from one another. • Junction of all three wires called as Neutral (N). • Three wires are called Line wires say L1 , L2 , L3 i.e. R, Y, B supply. • Voltage measured between two terminals are called as phase voltage ( V ph )say VR ,VY ,VB • Voltage between any two lines are called line voltage ( VL ) say VRY ,VYB ,VBR  Potential Difference between lines, For R & Y • VRY  VRN  VNY ………(Phasor Sum) • VRY  VRN  VYN ……… (Phasor Difference) For Y & B • VYB  VYN  VNB ………(Phasor Sum) • VYB  VYN  VBN ……… (Phasor Difference) For B & R • VBR  VBN  VNR ….……(Phasor Sum) • VBR  VBN  VRN ……… (Phasor Difference) From Phasor Diagram, VR  VY  VB  VPh We have, VBR  VBN  VNR VBR  E ph . cos 300  E ph . cos 300 VBR  2.E ph . cos 30 0 VBR  3.E ph Similarly for other two phase voltage or line voltage, VRY  3.E ph and VYB  3.E ph OR Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
  • 5. Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit By, the trigonometric, 2 2 2 VL  VR  VY  2.VR.VY . cos 60 VL  VPh  VPh  2.VPh .VPh .(0.5) 2 2 VL  3VPh 2 2 2 VL  3.VPh ………….Line Voltage for Star Connection It shows, Line Voltage is 3 times the phase voltage. From Phasor diagram, Phase current is same as line current. ...……… Line Current for Star Connection I L  I Ph Power for one phase is, Power = 3 X power in each phase P  3.Vph .I Ph . cos  We have, VL  3.VPh Substitute in P, we will get, ……………Power of Star Connection P  3.VL .I L . cos  This is the relation between Phase current and Line current, Phase Voltage and Line Voltage in Star Connection or Wye Connection.. Delta Connection or Mesh Connection:- Fig. Balanced 3phase Delta Connection Fig. Phasor Diagram • System is balanced and current’s will be equal in magnitude and 120 displaced from one another. • There is no Junction for all three wires. • Three wires are called Line wires say i.e. R, Y, B supply. • Voltage measured between two terminals are called as phase voltage ( V ph )say VR ,VY ,VB • • Voltage between any two lines are called line voltage ( VL ) say VRY ,VYB ,VBR Currents between phases, IR  IR  IB IY  IY  I R Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
  • 6. Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit I B  I B  IY Now, for Current in R line, IR  IR  IB I R  I ph . cos 30 0  (  I ph . cos 30 0 ) I R  I ph . cos 30 0  I ph . cos 30 0 I R  2. I ph . cos 30 0 I R  3I ph Similarly, I Y  3I ph & I B  3I ph i.e. I L  3I ph ...……… Line Current for Delta Connection This is required Line current in Delta Connection. Line voltage is equal to phase voltage in Delta Connection. ...……… Line Voltage for Delta Connection VL  V ph Power for one phase is, Power = 3 X power in each phase P  3.Vph .I Ph . cos  We have, I L  3I ph Substitute in P, we will get, ……………Power of Delta Connection P  3.VL .I L . cos  This is the relation between Phase current and Line current, Phase Voltage and Line Voltage in Delta Connection or Mesh Connection.. Comparison of Star and Delta Connections:Sr. Star Connection Delta Connection No 1 VL  3.VPh VL  VPh 2 I L  I ph I L  3I ph 3 P  3.VL .I L . cos  P  3.VL .I L . cos  4 Alternator & distribution transformer secondary is star connected. Transmission transformer secondary is delta connected. 5 Same power rating more current Same power rating less current 6 For same line voltage, Star connected alternator requires less insulation than delta connection. For same line voltage, Delta connected alternator requires more insulation than star connection. 7 We can get 3ph, 4 wire system. Its flexible connection. Suitable for rotary convertors. Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in
  • 7. Vijay Balu Raskar – Electrical Maintenance – Polyphase Circuit 8 Neutral point can be earthed and offers measurement. Most 3ph induction motors are delta connected. PROBLEMS – “Polyphase Circuit” Formulae for polyphase Circuit: Z ph  R 2  X 2  Power Factor = cos    R Z Sr. No Star Connection Delta Connection 1 VL  3.VPh VL  VPh 2 I L  I ph I L  3I ph 3 P  3.VL .I L . cos  P  3.VL .I L . cos  Q.1 A balanced 3phase star connected load consisting in each phase a resistance and inductive reactance og 4ohm and 3ohm respectively, connected to 3phase 400V, 50Hz supply. Find (a) Line Current, (b) Power Factor and (c) Total Power Q.2 A balanced star connected load is connected across 230V, 3ph supply. If the resistance & Inductive reactance per phase are 6ohm & 8ohm respectively. Find (a) Line Current, (b) Power Factor and (c) Total Power Q.3 A balanced 3ph star connected load is supplied from a 3ph, 400V, 50Hz supply. The resistance of each coil is 6ohm and reactance 8ohm. Find (a) Line Current, (b) Power Factor and (c) Total Power Q.4 Three resistance each of 10ohm resistance are connected in (i) Star Connection and (ii) Delta connected across 440V. 3ph a.c. supply. Find the power consumed in each case and ratio of power in Delta to power in Star connection. Q.5 A balanced 3ph, star connected load is supplied with 3ph, 400V, 50Hz Supply. The phase current is 10A and p.f. is 0.6 lagging. Calculate total power. ---------------------- Read more for more knowledge ---------------------------------- Vijay Raskar ---- Suggestions or difficulties are most welcome at vijayraskar2003@yahoo.co.in