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YDelta conversion for Capacitors 
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#1
Jan911, 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 YDelta (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!!! 


#2
Jan911, 01:13 PM

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P: 11,837

What you can do is convert all your capacitances to their equivalent impedance, then use those formulae. Impedances mix and match like resistances. 


#3
Jan1011, 10:43 AM

P: 1,135

How do i find impedance of a capacitor?



#4
Jan1011, 11:49 AM

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P: 11,837

YDelta conversion for Capacitors
[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
Jan1011, 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 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 


#6
Jan1011, 02:25 PM

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P: 11,837

You're welcome. 


#7
Jan1011, 05:57 PM

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P: 7,819

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
Jan1111, 04:27 AM

P: 1,135

Thanks for the help!!!!



#9
Sep1411, 02:18 PM

P: 12

Is there a simple result for delta to Y as well?



#10
Sep1411, 03:42 PM

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P: 11,837

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. 


#11
Sep1811, 02:28 PM

P: 12

Thank you! :)



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