Solving for Line Currents in a Wye-Delta Circuit

[scothoward]

Hi,

In the attached file, the question asks to find the three line currents. I understand the solution using Mesh Analysis that was used. However, I am unsure as to why when I convert the Delta Load to a Wye load, I get the wrong answer.

Using Zwye = (1/3)Zdelta, adding the series line impedance with the load impedance, and then finding the line current using a single phase equivalent.


Essentially the Balanced Delta load becomes a Balance Wye load of 4 + j0.666. Then the line impedances will be in series with each individual load. So each impedance should be (4 + j0.666) + (1 + j2). Finally, the line currents will be the phase voltages (100 angle0, 100 angle-120, 100 angle120) divided by the series line and load impedance, 5 + j2.666.

Am I doing something wrong?

 

[delta123]

I can't help you here, because there is no question to what need to be solve.

 

[Mene]

I can understand your confusion. It is important to note that when converting a Delta load to a Wye load, the line currents will remain the same but the load currents will change. This is because the Delta load is connected in a series, while the Wye load is connected in parallel. Therefore, the line currents in both cases will be the same, but the load currents will be different.

In your calculation, you have correctly converted the Delta load to a Wye load and found the equivalent impedance. However, when finding the line currents, you have used the phase voltages divided by the series line and load impedance, which is incorrect. The correct way to find the line currents in a Wye-Delta circuit is to use the current divider rule, where the line current is equal to the load current divided by the impedance of the branch that it is connected to. In this case, the line current will be the load current of 100 angle0 divided by the impedance of (4 + j0.666) + (1 + j2). This will give you a line current of approximately 16.67 angle-18.5 degrees.

In conclusion, your approach was correct but the calculation for the line current was incorrect. By using the current divider rule, you will be able to find the correct line currents in a Wye-Delta circuit. I hope this clarifies your doubt and helps you in your future calculations. Keep up the good work!