3 Phase unbalanced load AC electronics question.

[rolls]

Homework Statement

http://img196.imageshack.us/img196/1295/40343374.png
This is a past exam questions which I am doing as revision, now I am confused about how to calculate the answers.

The Attempt at a Solution

I have tried using millmans theorem which is
Vsn = (Van/Za + Vbn/Zb + Vcn/Zc) / (1/Za + 1/Zb + 1/Zc)

Assuming the following:
Van = 240v @ 0 degrees
Vbn = 240v @ 120 degrees
Vcn = 240v @ -120 degrees

As complex impedances:
Van = 240
Vbn = 195 + 139i
Vcn = 195 - 139i
Za = 10
Zb = 15
Zc = -20i

Vsn = 273 +32i
Vsn = 275v @ 6.6degrees

Now I feel like I am really guessing here and have no idea if what I have done is correct (I don't think it is) so any help would be greatly appreciated

For the currents it is just Ia = Van/Za = 240/10 etc which is easy and then you can easily construct the phasor diagram as well from these. Either way I don't think my values for Van, Vbn, Vcn or my method for calculating Vsn is correct.

 

[KataKoniK]

Thank you for sharing your attempt at solving this problem. It seems like you are on the right track and have a good understanding of the concept of Millman's theorem. However, there are a few errors in your calculations that I would like to point out.

Firstly, when converting the voltages from polar form to rectangular form, it is important to use the correct angle for each voltage. In this case, Vbn should have an angle of -120 degrees (not +120 degrees). This will result in Vbn = -120 - 207.85i, and Vcn = -120 + 207.85i.

Secondly, when using Millman's theorem, it is important to use the correct impedance values for each voltage source. In this case, Za should be 10 + 0i (not just 10), Zb should be 15 + 0i (not just 15), and Zc should be 0 - 20i (not just -20i). This will result in Vsn = 275.1 - 0.15i, which has a magnitude of 275.1V and an angle of -0.05 degrees.

Lastly, for the currents, it is important to use the correct voltage values for each voltage source. In this case, Ia should be 240/10 = 24A, Ib should be -120 - 207.85i / 15 = 8.00 + 13.86i A, and Ic should be -120 + 207.85i / 0 - 20i = 10.39 - 6.94i A.

I hope this helps clarify any confusion and leads you to the correct solution. Good luck with your revision!

 

[Mene]

it is important to have a clear and logical approach to problem solving. In this case, the first step would be to carefully read and understand the given problem. From the diagram, it can be seen that the circuit is a 3-phase unbalanced load, with three different impedance values for each phase. The objective is to find the voltage across the load, Vsn.

The attempt at a solution using Millman's theorem seems to be a good start, but it is important to note that this theorem is applicable for calculating the voltage across a set of parallel branches, not for a series circuit like the one given in the problem. Therefore, a different approach is needed.

One way to solve this problem is by using Kirchhoff's voltage law, which states that the sum of all voltages around a closed loop in a circuit must equal zero. In this case, we can consider a closed loop consisting of the three phases (Van, Vbn, and Vcn) and the load impedance (Za, Zb, and Zc).

Applying Kirchhoff's voltage law, we get the following equation:
Van + Vbn + Vcn - Vsn = 0

Substituting the given values, we get:
240 + 195 + 195 - Vsn = 0

Solving for Vsn, we get:
Vsn = 630V

This means that the voltage across the load is 630V, which is the same for all three phases since they are connected in series. This value is different from the one obtained using Millman's theorem, indicating that the latter approach may not be applicable in this case.

In addition, it is important to note that the given values for Van, Vbn, and Vcn are not in complex form, so converting them to complex impedances is not necessary.

In conclusion, the correct approach to solving this problem is by using Kirchhoff's voltage law, which gives the voltage across the load as 630V. It is also important to carefully read and understand the problem before attempting to solve it, and to use appropriate methods and equations for the given circuit configuration.