**
Tutorial 13 - Extension - Unbalanced 3-Phase systems**

So far we have kept the star system balanced, and those operating transmission
systems attempt to keep the system balanced as far as possible. However it is
possible to have different loads on each phase, which means that the system is
**unbalanced**. The diagram below shows how it is possible to have an
unbalanced star system.

__
Worked Example__

A 415 V, 3-phase star-connected system supplies three resistive loads as shown above. Determine:

a. The current in each line;

b. The current in the neutral line.

__
Answer (a)__

All the line voltages are 415 V. Calculate the phase voltage:

*V*_{P} = *V*_{L} ÷
Ö3 = 415 ÷
Ö3 = 240 V

Calculate the currents using *I = P/V*:

Red phase:

*I*_{R}* = *24000 ÷ 240 =
100 A

Yellow phase:

*I*_{Y} = 18000 ÷ 240 = 75 A

Blue phase:

*I*_{B} = 12000 ÷ 240 = 50 A

__
Answer (b)__

The answer is NOT 100 A + 75 A + 50 A = 225 A

By using Kirchhoff I, we can say that the neutral current is the **vector**
sum of all the currents.

*I*_{N} = *I*_{R}
+ *I*_{Y} + *I*_{B}

This is because all the vectors are 120^{o} apart. Our phasor diagram
would look like this:

So we can now put the vectors nose to tail:

The vector OC is the resultant, *I*_{N}. We can solve this by
accurate drawing, which gives a resultant current of about 43 A.

We can also solve this by adding the components of the vectors. This is shown below. Take upwards as positive and downwards as negative. Left to right is positive, right to left negative. The vectors are shown below:

The vector components are:

Vector | Vertical | Horizontal |

I_{R } |
+100 A | 0 |

I_{Y} |
-75 sin 30 | +75 cos 30 |

I_{B} |
-50 cos 60 | - 50 sin 60 |

The horizontal component of *I*_{N} is given by:

75 cos 30 + -50 sin 60 = (75 × 0.866) – (50 × 0.866) = 21.65 A

The vertical component of *I*_{N }is given by:

100 + -75 sin 30 + -50 cos 60 = 100 – (75 × 0.500) – (50 × 0.500) = 37.5 A

*
I*_{N} is the vector sum of
these two:

21.65^{2} + 37.5^{2} = 1875

*I*_{N }= 43.3 A

The neutral current in this case is 43.3 A.

Remember that the phase relationship is 120^{o}, so the vertical and
horizontal components can be found by simple geometry. Watch out for the signs
on the vertical and horizontal components. If in doubt, draw a diagram like the
one above.

A three phase industrial motor operates at 1500 V, 3-phase. Normally the loads of its coils are balanced, each coil taking a current of 200 A. a. What is the line voltage? b. What is the phase voltage? c. What is the neutral line current while in normal operation? The coils on the red phase develop a fault so that the line current of the red phase is now 250 A. The yellow and blue phases are not affected. Calculate the neutral current. The answer is not 650 A. |

Click HERE to go back to Tutorial 13 A

Click HERE to go back to Tutorial 13 B