To derive the **relations between line and phase currents and voltages of a star connected system**, we have first to draw a balanced star connected system.

Suppose due to load impedance the current lags the applied voltage in each phase of the system by an angle ϕ. As we have considered that the system is perfectly balanced, the magnitude of current and voltage of each phase is the same. Let us say, the magnitude of the voltage across the red phase i.e. magnitude of the voltage between neutral point (N) and red phase terminal (R) is V_{R}.

Similarly, the magnitude of the voltage across yellow phase is V_{Y} and the magnitude of the voltage across blue phase is V_{B}.

In the balanced star system, magnitude of phase voltage in each phase is V_{ph}.

∴ V_{R} = V_{Y} = V_{B} = V_{ph}

We know in the star connection, line current is same as phase current. The magnitude of this current is same in all three phases and say it is I_{L}.

∴ I_{R} = I_{Y} = I_{B} = I_{L}, Where, I_{R} is line current of R phase, I_{Y} is line current of Y phase and I_{B} is line current of B phase. Again, phase current, I_{ph} of each phase is same as line current I_{L} in star connected system.

∴ I_{R} = I_{Y} = I_{B} = I_{L} = I_{ph}.

Now, let us say, the voltage across R and Y terminal of the star connected circuit is V_{RY}.

The voltage across Y and B terminal of the star connected circuit is V_{YBBR}.

From the diagram, it is found that

V_{RY} = V_{R} + (− V_{Y})

Similarly, V_{YB} = V_{Y} + (− V_{B})

And, V_{BR} = V_{B} + (− V_{R})

Now, as angle between V_{R} and V_{Y } is 120^{o}(electrical), the angle between V_{R} and – V_{Y} is 180^{o} – 120^{o} = 60^{o}(electrical).

Thus, for the star-connected system line voltage = √3 × phase voltage.

Line current = Phase current

As, the angle between voltage and current per phase is φ, the electric power per phase is

So the total power of three phase system is