Solving Delta-to-Wye Circuit: Find Current Across Inductor

In summary: I was calculating the total impedance, I got 82-64.07i but here you wrote 82-64.7i, a difference of 0.07i. Just wanted to make sure if it's a typo or not.In summary, the conversation involves a question on solving a current across an inductor in a Delta circuit using the Delta-to-Wye method. The person asking the question provides details of their method and calculations, seeking help and feedback from the other person. They discuss the steps involved in finding the current and potential difference across the inductor, including using complex numbers and the voltage divider rule. The other person offers guidance and suggests checking calculations and using different methods such as node analysis. The conversation ends
  • #1
MohdAziz
11
0
Hello, I have this question that I've been working on. The question requires me to find the current across an inductor in a Delta circuit, so I want to solve this question using Delta-to-Wye method. so I transformed it using the method and I found the total impedance of the whole circuit, then I divided the phasor voltage by the impedance and I found the current, but I still have no idea how I can find the current across the inductor. I attached a picture of the question so you guys can see my problem.
The result of the total impedance was around ( 82 - 64.07i)
and the phasor current after dividing by the phasor voltage is 0.6731 < 38

Thanks a lot, your help is really appreciated.
 

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  • #2
Say that you transform R1 R2 and ZL1 from delta to star. As you've found, you now have a series - parallel circuit that is amenable to "ladder style" collapsing. You do lose track of L1 in the process (it gets tangled up with R1 and R2 in the transform), more importantly however you do still have access to nodes A and B.

So you can calculate the voltages at A and B from the transformed circuit, then using this knowledge of [itex]V_{AB}[/itex] you can then go back to the original circuit to solve for Ix
 
  • #3
Okay. I got your idea, by that Va should the voltage across the capacitor and Vb should the voltage across R3 but the thing how could I find the voltage or apply the voltage divider rule in case of a sinosoidal voltage source.
just to tell you
Z1= 16 - 21.33i
Z2= 51.2 + 38.4i
Z3 = 64 - 48i

when I apply the voltage divider rule to Z1, VZ1 = Vin * Z1/ Ztot
all of Z1, Z total and Vin should be in phasor form?

Thanks a lot man.
 
  • #4
MohdAziz said:
Okay. I got your idea, by that Va should the voltage across the capacitor and Vb should the voltage across R3 but the thing how could I find the voltage or apply the voltage divider rule in case of a sinosoidal voltage source.

Yeah that's right. Just use complex numbers for all the impedances and use the voltage divider rule with complex numbers. For the source voltage use 70 + j0 (70 angle 0).
 
  • #5


If you apply a Δ-Y transform to the circuit you will "transform away" the component through which you want to find the current (L1). However, nodes A and B still exist in the "new" layout, so you could calculate their node voltages. With the two node voltages in hand you can go back to the original layout to determine the potential difference across L1 and hence its current...

A more straightforward approach might be to write KVL mesh or KCL node equations for the original circuit.
 
  • #6
Okay that's great, I think I got the picture. I'll solve the whole thing and I'll scan and attach it to show it to you so I can know your remarks, of course if you don't mind?
 
  • #7


Okay. The thing I'm supposed to do it using delta to wye, however; If I ever have time, I'll try the node analysis. And if you don't mind, can I solve the whole thing, and attach it so I can show and you could tell me what are my mistake, if I have any?

And thanks a lot for your help;.
 
  • #8


MohdAziz said:
Okay. The thing I'm supposed to do it using delta to wye, however; If I ever have time, I'll try the node analysis. And if you don't mind, can I solve the whole thing, and attach it so I can show and you could tell me what are my mistake, if I have any?

Yes, you are encouraged to show your work :smile:
 
  • #9
There is this one weird thing that I came across when solving the question, as you know to find the voltage of node A, you have to find the voltage of the capacitor, therefore; firsr you need to find the voltage of the whole parallel part by applying voltage divider rule like this (Zparallel/ Zparallel + Z1)* 70<0 ... The answer would be 52.92<5.076. Then you use this in voltage divider rule again to find the capacitor voltage Like this, (Xc/Xc + Z2)* 52.92<076. However the weird thing that the voltage of node A (capacitor) is actually higher than the voltage of whole parallel part, and this should be impossible. I tried to check all of my results, but I'm quite sure I haven't made any calculation mistakes?
 
  • #10
Here you go, this is my work. BTW I tried also to find the current through the capacitor and then find the voltage but I got a different result as well :S
 

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  • #11
MohdAziz said:
Here you go, this is my work. BTW I tried also to find the current through the capacitor and then find the voltage but I got a different result as well :S

You'll want to recheck your calculation for Z3. Otherwise, your method of work looks fine.
 
  • #12
You're right Z3 here is wrong, but when I calculated the total Z I used the right calculation which is 64 + 48i , and the answer of the Z of Parallel would still be 66 - 42.74. And thus the total Z is 82 - 64.07. And when we calculate Va we just need the Z parallel and Z tot so the mistake in calculation Z3 won't have any effect on the result of Va.
 
  • #13
MohdAziz said:
You're right Z3 here is wrong, but when I calculated the total Z I used the right calculation which is 64 + 48i , and the answer of the Z of Parallel would still be 66 - 42.74. And thus the total Z is 82 - 64.07. And when we calculate Va we just need the Z parallel and Z tot so the mistake in calculation Z3 won't have any effect on the result of Va.

It does, however, impact Vb which in turn effects Va-Vb for the calculation of Ix.
 
  • #14
Anyway I was concerned bcuz when we find the magnitude of Va in polar form it's greater than the magnitude of the parallel voltage in polar form. However; I decided to check the total voltage of Vz2 + Va in rectangular form and it was equal to the parallel voltage which I guess my answer is correct. Anyway I resolved everything and I corrected my mistakes and here is my final work.
Thanks a lot for your help, could you please check it for the last time. I really appreciate your help.
 

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  • #15
MohdAziz said:
Anyway I was concerned bcuz when we find the magnitude of Va in polar form it's greater than the magnitude of the parallel voltage in polar form. However; I decided to check the total voltage of Vz2 + Va in rectangular form and it was equal to the parallel voltage which I guess my answer is correct. Anyway I resolved everything and I corrected my mistakes and here is my final work.
Thanks a lot for your help, could you please check it for the last time. I really appreciate your help.

It all looks fine except for your final calculation of Ix; the impedance of the inductor should be an imaginary value (240j Ω), but it seems that you've treated it as a real value.
 
  • #16
oh missed that. SO final answer should be (28.93 - 22.27j)/240j = -0.093 +0.120i which equals to 0.152 < -127.59 in phasor form. Thanks a lot gneil, and sorry for that coin in my last solution sheet :D
 
  • #17
MohdAziz said:
oh missed that. SO final answer should be (28.93 - 22.27j)/240j = -0.093 +0.120i which equals to 0.152 < -127.59 in phasor form. Thanks a lot gneil, and sorry for that coin in my last solution sheet :D

Yes, that looks better :approve:

Next time post a cheque; paper coins are hard to cash :smile:
 
  • #18
Hahaha, you got it :biggrin: take care
 

1. How do you convert a delta circuit to a wye circuit?

To convert a delta circuit to a wye circuit, you can use the delta-to-wye transformation method. This involves replacing each delta connection with an equivalent wye connection by using the following equations:

Rab = Ra + Rb + (Ra*Rb)/Rc

Rac = Ra + Rc + (Ra*Rc)/Rb

Rbc = Rb + Rc + (Rb*Rc)/Ra

Once you have the new resistor values, you can redraw the circuit as a wye circuit and proceed with solving it.

2. Why is it important to solve for the current across an inductor in a delta-to-wye circuit?

The current across an inductor in a circuit is important because it determines the magnetic field strength and the amount of energy stored in the inductor. In a delta-to-wye circuit, the inductor is part of the load, and its current can affect the overall performance of the circuit. Therefore, solving for the current across the inductor is crucial for understanding the behavior of the circuit.

3. How do you calculate the current across an inductor in a delta-to-wye circuit?

The current across an inductor in a delta-to-wye circuit can be calculated using Ohm's Law (I = V/R) and the voltage and resistance values of the inductor. You can also use Kirchhoff's Current Law to determine the current at each node in the circuit and then apply the delta-to-wye transformation to find the current across the inductor.

4. What factors can affect the current across an inductor in a delta-to-wye circuit?

The current across an inductor in a delta-to-wye circuit can be affected by the voltage and resistance values of the inductor, as well as the voltage and resistance values of other components in the circuit. The frequency of the power source and the presence of any capacitors or resistors in parallel with the inductor can also impact the current across the inductor.

5. What are some common mistakes when solving for the current across an inductor in a delta-to-wye circuit?

One common mistake when solving for the current across an inductor in a delta-to-wye circuit is forgetting to convert the resistor values using the delta-to-wye transformation, resulting in incorrect current calculations. Another mistake is not considering the full circuit and ignoring the impact of other components on the inductor's current. It is also important to pay attention to the direction of the current flow and ensure that all calculations are consistent with the chosen direction.

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