Y-Delta conversion for Capacitors


by cupid.callin
Tags: capacitors, sta delta conversion
cupid.callin
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#1
Jan9-11, 12:22 PM
P: 1,135
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 dont know why ...

I used this equation

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!!!
Attached Thumbnails
Y-Delta.jpg  
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gneill
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#2
Jan9-11, 01:13 PM
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Quote Quote by cupid.callin View Post
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 dont know why ...

I used this equation

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.
cupid.callin
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#3
Jan10-11, 10:43 AM
P: 1,135
How do i find impedance of a capacitor?

gneill
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#4
Jan10-11, 11:49 AM
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Y-Delta conversion for Capacitors


Quote Quote by cupid.callin View Post
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)
cupid.callin
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#5
Jan10-11, 02:16 PM
P: 1,135
And to find Y form,

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

Thanks for the help gneill !!!!!!!!!!!!!!!!!!!!!
Thanks a lot !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
gneill
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#6
Jan10-11, 02:25 PM
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Quote Quote by cupid.callin View Post
And to find Y form,

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

Thanks for the help gneill !!!!!!!!!!!!!!!!!!!!!
Thanks a lot !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Right.

You're welcome.
SammyS
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#7
Jan10-11, 05:57 PM
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Quote Quote by gneill View Post

... 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)


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.
cupid.callin
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#8
Jan11-11, 04:27 AM
P: 1,135
Thanks for the help!!!!
PriyankB
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#9
Sep14-11, 02:18 PM
P: 12
Is there a simple result for delta to Y as well?
gneill
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#10
Sep14-11, 03:42 PM
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Quote Quote by PriyankB View Post
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].



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

and so on.
Attached Thumbnails
Fig1.gif  
PriyankB
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#11
Sep18-11, 02:28 PM
P: 12
Thank you! :)


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