**Difference Between Star Connection and Delta Connection **

Star & Delta connections are two different configuration methods of the interconnection of the three-terminal of a three-phase system. Both the above said types of connection are discussed below one by one.

**Star Connection or Way Connection(Y)**

In the star connection method, the similar terminals of the winding say star ends are connected together at one point N as shown in the figure. The point N is called a neutral point or star point.

In the given figure A, B and C are the windings of a three-phase system, and I_{A}, I_{B,} and I_{C} are the electric current flowing in the three phases called the line current, therefore for any balanced load value of line current ‘I’_{L}= I_{A}= I_{B}= I_{C}.

It means the current in each line is in series with its individual phase winding hence the line current in each line is the same as the current in the phase winding to which the line is connected.

i.e I_{L} = I_{P}

Where I_{P }= Phase current

Thus in the case of star connection line current is equal to phase current.

Star connection is also called Way (Y) connection.

**Line voltage and Phase voltage in Star Connection**

Let the emf produced in each phase of the star-connected winding be E_{A}, E_{B,} and E_{C} respectively. EMFs produced here are at a phase difference of 120^{0} as shown in the phasor diagram.

Here the line voltage is the vector difference between E_{A} and E_{B} and to find the line voltage, E_{B} is extended from the point O till E’_{B} which is of the same length as E_{B}.

From the phasor diagram, it is clear that points O, E_{A },E_{AB,} and E’_{B} form a parallelogram and the resultant of the parallelogram is the vector E_{AB} which the resultant emf.

Then by using parallelogram law we can write,

(E_{AB})^{2} = (E_{A})^{2 }+ (E^{,}_{B})^{2} +2 E_{A }E^{,}_{B }cos 60^{0}

Or, (E_{AB})^{2} =E^{2}_{A }+ E^{,}_{B}^{2}+ E_{A} E^{’}_{B} (As, cos60^{0}= ½)

Let E_{A}= E^{’}_{B}= E (phase voltage)

Then, (E_{AB})^{2} = E^{2 }+ E^{2} + E.E

Or, (E_{AB})^{2} = 3 E^{2}

E_{AB} = √3 E

Therefore the line voltage E_{L} = √3 E_{P}

Thus from the above equation, we finally concluded that the line voltage in the case of star connection is √3 times the phase current.

i.e E_{L}= √3 E_{P }.

**Power in case of star connection**

In the case of an AC circuit, the average power consumption is given by the product of supplied voltage and the component of electric current which is in phase with the voltage.

i.e Power in AC circuit is given by,

P_{AC} = V I cos φ

In the case of 1- phase circuit, Power (P) = V_{P} I_{P} cosφ

And in case of three-phase circuit Power is,

P = 3 × power of 1- phase

P = 3 × V_{P}× I_{P}× cosφ

We know that in star connection phase current (I_{P}) = Line current (I_{L}) and phase voltage (V_{P}) = V_{L}/√3

Then power in star connection is

P = 3 × V_{L}/√3 × I_{L}× cosφ

Therefore power ‘P’ = √3 × V_{L} × I_{L} × cosφ.

**Delta (∆) Connection or Mesh connection**

In this method of interconnection, the dissimilar ends of 3-phase winding are connected together. It means the starting end of one phase is connected to the finishing end of the other phase as shown in the figure.

In the given figure the terminal of the winding A’ – B, B’- C, and C’- A is connected with the **phase**– R, Phase – Y, Phase-B respectively.

It is also said that the three windings are connected in series to form a closed mesh as shown in the figure. So this delta connection is also called a Mesh connection.

In the case of ∆ connection, the line voltage is equal to phase voltage (for balanced load) because here in the case of delta connection there is no neutral point.

i.e E_{L} = E_{P}

Where E_{L}= line voltage and E_{P} = Phase voltage

**Line current and Phase current in Delta connection**

Let the emf produced in each phase be E_{A },E_{B,} and E_{C} respectively as shown in the figure and these EMFs are at a phase difference of 120^{0} with each other.

Consider on each phase there is equal and opposite load, then the current I_{A} is in phase with the phase emf E_{A}. Similarly, current I_{B} is in phase with the phase emf E_{B} and I_{C} is in phase with the phase emf E_{C}.

Here in this case line current is calculated as by the vector difference of the current I_{A} and I_{C} for this calculation current I_{C} is extended in opposite direction till I_{C}’ as shown in the phasor diagram. The length of I_{C}’ is the same as that of I_{C}.

By using parallelogram law expression for line current can be written as,

(I_{AC})^{2} = I_{A}^{2} + I^{2}_{C}’ + 2 I_{A}× I_{C}’× cos60^{0}

(I_{AC})^{2} = I_{A}^{2} + I^{2}_{C}’ + I_{A}× I_{C}’ [cos60^{0}= ½]

Let I_{A }= I_{C}’= I

Put these values in the above equation we get

(I_{AC})^{2} = I^{2}+ I^{2}+ I^{2}

(I_{AC})^{2} = 3I^{2}

I_{AC }= √3 I

As I = I_{Phase}(I_{p})

Then Line current I_{L} = √3 I_{P}

Here from the above result, we can say that in the case of delta connection line current is equal to √3 times the phase current.

**Power in case of Delta connection**

we know that in the case of delta connection line current I_{L} = √3 I_{P }

and V_{P} =V_{L}

Then power in the case of Delta connection is given by

‘P’ = √3 × V_{L} × I_{L} × cosφ.

Overall it is concluded that there are various parameters on which Star & Delta connections are differentiated. So points of comparison between star and delta connection are discussed below in the table.

**Comparison between Star & Delta Connection**

S.NO | Star Connection |
Delta connection |

1. |
Similar ends are joined together. | Dissimilar ends are joined together. |

2. |
Line voltage is √3 times the phase voltage. | Line voltage is equal to the phase voltage. |

3. |
Line Current is equal to the phase current. | Line current is √3 times the phase current. |

4. |
Neutral point is present in this connection. | Neutral point is absent in this connection. |

5. |
Power (P) =√3 × V_{L} × I_{L} × cosφ. |
Power (P) =√3 × V_{L} × I_{L} × cosφ. |

6. |
In this connection 4-wire 3-Φ system is possible | In this connection, 4-wire 3-Φ system is not possible. |

7. |
Chances of breakdown are high in this connection. | Chances of breakdown are low in this connection. |

8. |
Both industrial load and domestic load can be handled. | Only industrial loads can be handled. |

9. |
This connection looks like Way or Y. | This connection looks like a triangle or ∆. |

10. |
By earthing the neutral wire relay and protective devices can be provided with the alternator for safety. | Due to the absence of neutral wire, it is not possible in this type of connection. |