# Q: **Derive the relation between phase voltage and line voltage in case of balanced star connected load**.

# OR

# Q: **Derive an expression for total power input for n balance 3 phase load in terms of line quantities and power factors**.

**Star-connection**

As shown in the figure given below for the **star connection**, three equal impedances (load) are also connected in the star across the star-connected supply. If all the 3 impedances arc equal, then it is known as a balanced load.

All three voltages will have the same magnitude but 120^{0} **phase difference**. These voltages are measured between neutral and phase points (i.e. R, Y, B) and hence known as **phase voltage** (V ph ). Three more combinations are also possible. These arc voltages between (i) pts. R and Y V_{RY} (ii) pts. Y and B, V_{YB} (iii) pts. B and R, V_{BR}. These are known as **line voltage** ( V_{L }).

**The relation between** ** V**_{L} and V_{ph}

_{L}and V

_{ph}

**Phasor diagram for the phase voltage** will be as shown below for** the star connection**. We will calculate any one of the **line voltage**. V_{RY} is the voltage between R and Y. From the figure,

Hence, V_{RY} is obtained by adding ( -V_{Y} ) and The resultant V_{RY} will be 30^{0} ahead of V_{R}. Hence, V_{L} is 30^{0} ahead of V ph. The magnitude of V_{RY} can be calculated as,

Hence, magnitude of **line voltage **is magnitude under root 3 of **phase voltage**.

**Relation-between I**_{L} and I_{ph}

_{L}and I

_{ph}

The three currents I_{R} , I_{Y} and I_{B} can be calculated as,

Hence, all the three currents have the same magnitude but 120^{0} phase difference Now, the current through the neutral wire ( I_{N}) will be the sum of all the three currents in th**e star connection**.

Expression for power : For 3-phase load

Total power , P = 3 phase power

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