Y-Delta conversion for Capacitors

  • #1
cupid.callin
1,137
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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
gneill
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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!

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
cupid.callin
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How do i find impedance of a capacitor?
 
  • #4
gneill
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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
cupid.callin
1,137
1
And to find Y form,

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

Thanks for the help gneill !
Thanks a lot !
 
  • #6
gneill
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And to find Y form,

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

Thanks for the help gneill !
Thanks a lot !

Right.

You're welcome.
 
  • #7
SammyS
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... 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
cupid.callin
1,137
1
Thanks for the help!
 
  • #9
PriyankB
12
0
Is there a simple result for delta to Y as well?
 
  • #10
gneill
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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
PriyankB
12
0
Thank you! :)
 

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