Delta-Wye Transformation of Capacitors

Question

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

Unfortunately this is the wrong answer. I'm not sure what i'm doing wrong, Please help.

Answer 1

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).

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

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