Basic Electrical Engineering (3110005)

Question-3c

BE | Semester-1 Winter-2019 | 11-01-2020

Q3) (c)

Let

Line Voltage,

V_{RY} = V_{BY} = V_{BR} = V_{L}

Phase Voltage,

V_{R (ph)} = V_{Y (ph)} = V_{B (ph)} = V_{ph}

Line Current,

I_{R (line)} = I_{Y (line)} = I_{B (line)} = I_{line}

Phase Current,

I_{R (ph)} = I_{Y (ph)} = I_{B (ph)} = I_{ph}

For star connection, line current and phase current both are same.

I_{R (line)} = I_{R (ph)}

I_{Y (line)} = I_{Y (ph)}

I_{B (line)} = I_{B (ph)}

I_{L} = I_{ph}

So, Line Current = Phase Current.

V_{RY} = V_{R (ph)} - V_{Y (ph)}

V_{YB} = V_{Y (ph)} - V_{B (ph)}

V_{BR} = V_{B (ph)} - V_{R (ph)}

From parallelogram,

V_{RY} = \sqrt{{V_{R (ph)}}^{2} + {V_{Y (ph)}}^{2} + V_{R (ph)} V_{Y (ph)} \cos θ}

\Rightarrow V_{L} = \sqrt{{V_{ph}}^{2} + {V_{ph}}^{2} + 2 V_{ph} V_{ph} \cos 60 °}

\Rightarrow V_{L} = \sqrt{{V_{ph}}^{2} + {V_{ph}}^{2} + 2 {V_{ph}}^{2} (\frac{1}{2})}

\Rightarrow V_{L} = \sqrt{{V_{ph}}^{2} + {V_{ph}}^{2} + {V_{ph}}^{2}}

\Rightarrow V_{L} = \sqrt{3 {V_{ph}}^{2}}

\Rightarrow V_{L} = \sqrt{3} V_{ph}

\Rightarrow V_{YB} = V_{BR} = \sqrt{3} V_{ph}