# Delta-Wye Transformation of Capacitors

Homework: I am trying to wrap my head around converting a capacitor circuit into an equivalent capacitor. I have looked at resources saying that it is the same as delta-wye transformation of resistors, but in reverse. I thought I had the process right, but I keep getting the wrong answer.

      24       48
a-----)|---o---|(---o
|\       |
| \ 4    |
8 |(  |(   |( 24
|    \   |
|     \  |
|      \ |
b-----|(---o---)|---o
10       30


I combined the 48 and 24 into one, then took a Delta-wye transform:

      24        16
a-----)|---o----|(----o
\        /
\      /
C2 |(    )| C3
\  /
\/
o
|
|( C1
|
b-----|(--------o
10


The calculations I used were as follows:

4*8 + 4*30 + 8*30 = 392

C1 = 392/4 = 98

C2 = 392/30 = 13.1

C3 = 392/8 = 49

Plugging in those numbers the equivalent capacitance should be:

Ceq = (24^-1 + 98^-1 + 10^-1)^-1 + 13.1 + (16^-1 + 49^-1)^-1 = 31.7

• I don't see any need for D-Y here. – user_1818839 Jul 19 '15 at 22:01

Allow me to draw it in a different way: Now have another go at solving the problem.

A final hint: You will need to use the following two equations a total of 5 times (one equation is used twice, and the other three times).

$$\frac{1}{C_{series}} = \frac{1}{C_{1}} +\frac{1}{C_{2}} +...$$

$$C_{parallel} = C_1+C_2+...$$

• p.s. the answer starts with a 5. But that's all I'm saying ;) – Tom Carpenter Jul 19 '15 at 23:04
• Wow, I definitely made it more complicated than needed. Thank you @TomCarpenter. – Mark Walsh Jul 19 '15 at 23:35
• I also see where I made a mistake in my Ceq. It should have been: Ceq = 1/(1/24 + 1/10 + 1/98 + 1/(13.1+1/(1/16+1/49))) Which also equals the answer. – Mark Walsh Jul 19 '15 at 23:39