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}
_{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 IL 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_{YB
The voltage across B and R terminal of the star connected circuit is VBR.
From the diagram, it is found that
VRY = VR + (− VY)
Similarly, VYB = VY + (− VB)
And, VBR = VB + (− VR)
Now, as angle between VR and VY is 120o(electrical), the angle between VR and – VY is 180o – 120o = 60o(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
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