_{P}= 230V) and line voltage (V

_{L}= 400V). ‘I

_{L}’ stands for line current and ‘I

_{P}’ stands for phase current. We can say that V

_{L}= √3*V

_{P}en I

_{P}= I

_{P}. If the generating set is connected in delta, the line voltage V

_{L}equals the phase voltage V

_{P}and I

_{L}= √3*I

_{P}.

Besides, the star configuration has 4 wires: 3 phase wires and 1 neutral wire starting from the star point. The delta configuration on the other hand, has no neutral wire.

__Star__:

V

_{Line}= √3*V

_{Phase}

I

_{Line}= I

_{Phase}

V

_{L1 }= V

_{F1 }- V

_{F2}

V

_{L2 }= V

_{F2 }- V

_{F3}

V

_{L3 }= V

_{F3 }- V

_{F1}

V

_{L1 }+ V

_{L2 }+ V

_{L3 }= 0

V

_{L1 }= 2*V

_{F1}*cos 30° = 2*V

_{F1}*√3/2 = √3*V

_{F1}

The line voltage leads the phase voltage by 30°.

The line currents equal the phase currents with symmetrical charges.

The active power when connected in Y: P = 3*(V

_{L}/√3)*I

_{L}*cos φ = √3*V

_{L}*I

_{L}*cos φ

__Delta__:

V

_{Line}= V

_{Phase}

I

_{Line}= √3*I

_{Phase}

Ī

_{L1 }= Ī

_{P1 }- Ī

_{P1}

Ī

_{L2 }= Ī

_{P2 }- Ī

_{P3}

Ī

_{L3 }= Ī

_{P3 }- Ī

_{P1}

Ī

_{L1 }+ Ī

_{L2 }+ Ī

_{L3 }= 0

I

_{L1 }= 2*I

_{P1}*cos 30° = 2*I

_{P1}*3/2 = √3*I

_{P1}

The line current leads the phase current by 30°.

The line voltages equal the phase voltages with symmetrical charges.

The active power when connected in Δ: P = 3*VL*(IL/√3)*cos φ = √3*VL*IL*cos φ