Y-Delta conversion for Capacitors

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In summary, the conversation discusses solving a problem using Y-Delta conversion for resistors and capacitors. The process involves converting the capacitances to their equivalent impedances and using specific formulas. The conversation also mentions a simple result for Delta to Y conversion.
  • #1
cupid.callin
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This was a ques in my book ... (Pic)

After thinking a while, i realized that it can be solved using Y-Delta (by converting Y to delta form) conversion. But i couldn't get the answer ... i don't know why ...

I used this http://en.wikipedia.org/wiki/Y-%CE%94_transform" [Broken]

It is given for resistors ... i guess its same for resistors and capacitors ... Right?

Please tell me if the eqn used is wrong or something else!
 

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  • #2
cupid.callin said:
This was a ques in my book ... (Pic)

After thinking a while, i realized that it can be solved using Y-Delta (by converting Y to delta form) conversion. But i couldn't get the answer ... i don't know why ...

I used this http://en.wikipedia.org/wiki/Y-%CE%94_transform" [Broken]

It is given for resistors ... i guess its same for resistors and capacitors ... Right?

Please tell me if the eqn used is wrong or something else!

Remember how capacitors combined differently in series and parallel than do resistors?

What you can do is convert all your capacitances to their equivalent impedance, then use those formulae. Impedances mix and match like resistances.
 
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  • #3
How do i find impedance of a capacitor?
 
  • #4
cupid.callin said:
How do i find impedance of a capacitor?

[tex]Zc = 1/(j\omega C[/tex])

[tex]\omega[/tex] is the operating frequency. The result is in Ohms, and will be an imaginary value.

Don't panic! You don't need to know the frequency for the math to work out; it's a constant for the given Y to Delta situation. If you do the algebra, a pretty simple result obtains. If the resistor version is:

Ra = (R1*R2 + R2*R3 + R3*R1)/R2
Rb = (R1*R2 + R2*R3 + R3*R1)/R3
Rc = (R1*R2 + R2*R3 + R3*R1)/R1

Then the capacitor version looks like:

Ca = C1*C3/(C1 + C2 + C3)
Cb = C1*C2/(C1 + C2 + C3)
Cc = C2*C3/(C1 + C2 + C3)
 
  • #5
And to find Y form,

C1 = (CaCb + CbCc + CcCa) / Cc
Right?

Thanks for the help gneill !
Thanks a lot !
 
  • #6
cupid.callin said:
And to find Y form,

C1 = (CaCb + CbCc + CcCa) / Cc
Right?

Thanks for the help gneill !
Thanks a lot !

Right.

You're welcome.
 
  • #7
gneill said:
... for the given Y to Delta situation. If you do the algebra, a pretty simple result obtains. If the resistor version is:

Ra = (R1*R2 + R2*R3 + R3*R1)/R2
Rb = (R1*R2 + R2*R3 + R3*R1)/R3
Rc = (R1*R2 + R2*R3 + R3*R1)/R1

Then the capacitor version looks like:

Ca = C1*C3/(C1 + C2 + C3)
Cb = C1*C2/(C1 + C2 + C3)
Cc = C2*C3/(C1 + C2 + C3)
attachment.php?attachmentid=31194&d=1294597370.jpg


The three capacitors enclosed in red form a Y. So does the other set of 1, 3, and 4 μF capacitors. Converting each of these sets to Δ configuration, as shown by gneill above, will allow you analyze the circuit as a combination of parallel and series capacitors.
 
  • #8
Thanks for the help!
 
  • #9
Is there a simple result for delta to Y as well?
 
  • #10
PriyankB said:
Is there a simple result for delta to Y as well?

Sure. Just substitute the appropriate capacitor impedances into the formulas for resistance, stir and serve. Note that a capacitor impedance is of the form [itex] 1/(j \omega C) [/itex].

attachment.php?attachmentid=38864&stc=1&d=1316032546.gif


So for example, given that for resistors
[tex] R_1 = \frac{R_a R_b}{R_a + R_b + R_c} [/tex]
then
[tex] \frac{1}{C_1} = \frac{C_c}{C_b C_c + C_a C_c + C_a C_b} [/tex]

and so on.

EDIT: Fixed up the expression. Should have been 1/C1 on the LHS.
 

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  • #11
Thank you! :)
 

What is Y-Delta conversion for Capacitors?

Y-Delta conversion for capacitors is a method used to simplify complex capacitor networks by reconfiguring them into either a 'Y' or 'Delta' configuration. This allows for easier analysis and calculations of the circuit.

Why is Y-Delta conversion used for capacitors?

Y-Delta conversion is used for capacitors because it reduces the complexity of the circuit, making it easier to analyze and calculate the values of the components. It also helps to reduce the overall cost and size of the circuit.

How is Y-Delta conversion performed for capacitors?

Y-Delta conversion is performed by first identifying the three capacitors in the circuit. Then, the capacitors are reconfigured into either a 'Y' or 'Delta' configuration, depending on the circuit's requirements. The values of the new capacitors are calculated using specific formulas.

What are the advantages of using Y-Delta conversion for capacitors?

The main advantage of using Y-Delta conversion for capacitors is that it simplifies the circuit, making it easier to analyze and calculate values. It also reduces the overall cost and size of the circuit, making it more efficient and practical to use in various applications.

Are there any limitations to using Y-Delta conversion for capacitors?

One limitation of using Y-Delta conversion for capacitors is that it can only be used for circuits with three capacitors. It also requires a certain level of understanding of circuit analysis and calculations. Additionally, the conversion may introduce some errors in the values of the capacitors, depending on the complexity of the circuit.

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