Hello, I am looking for some help with this Delta-Wye source/load problem.
I attached a picture for better visualization.
Here's my thought process:
1) Convert the delta source to a single phase equivalent Wye circuit.
This means V_y would be 1/√3 * V_delta.
So V_y = 714/√(3)
It also means the source impedance in my delta source would be 1/3 the impedance in my equivalent Wye source.
So Z_src = (1/3)(4.5+j3) = 1.5 + j1
Now I construct my single phase equivalent Wye circuit, with:
V_y=714/√3
Z_src=1.5 + j1
Z_line=6.5 + j15
Z_load=1192 + j1584
So, for part 1, it asks for the line to line voltage at the terminals of the load.
First, I find the magnitude of the phase current going through my equivalent circuit.
I_a = V_y / Z_total = (714/√3) / (1200 + j1600) = .206A
Now, I find the magnitude of the line to neutral voltage, V_an, across my Z_load.
V_an = I_a * Z_load = .206(1192+j1584) = 408V
I know that for a Wye-load, the line to neutral voltage and line to line voltage of a single phase is related by this:
V_ab = √3 * V_an
Where V_ab is my line-line voltage.
So, my final answer for the line-to-line voltage, V_ab, should be:
V_ab = √3 * V_an = √3 * 408 = 708V
I tried this answer, and also several other answers that I got with different thinking but none of them were correct. Any help would be much appreciated.
I attached a picture for better visualization.
Here's my thought process:
1) Convert the delta source to a single phase equivalent Wye circuit.
This means V_y would be 1/√3 * V_delta.
So V_y = 714/√(3)
It also means the source impedance in my delta source would be 1/3 the impedance in my equivalent Wye source.
So Z_src = (1/3)(4.5+j3) = 1.5 + j1
Now I construct my single phase equivalent Wye circuit, with:
V_y=714/√3
Z_src=1.5 + j1
Z_line=6.5 + j15
Z_load=1192 + j1584
So, for part 1, it asks for the line to line voltage at the terminals of the load.
First, I find the magnitude of the phase current going through my equivalent circuit.
I_a = V_y / Z_total = (714/√3) / (1200 + j1600) = .206A
Now, I find the magnitude of the line to neutral voltage, V_an, across my Z_load.
V_an = I_a * Z_load = .206(1192+j1584) = 408V
I know that for a Wye-load, the line to neutral voltage and line to line voltage of a single phase is related by this:
V_ab = √3 * V_an
Where V_ab is my line-line voltage.
So, my final answer for the line-to-line voltage, V_ab, should be:
V_ab = √3 * V_an = √3 * 408 = 708V
I tried this answer, and also several other answers that I got with different thinking but none of them were correct. Any help would be much appreciated.
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