__Problem 3.6__
Source: Problem 6.28 - page 216
- SADIKU, Matthew N. O. , ALEXANDER, Charles K. -
Book: Fundamentos de Circuitos Elétricos - McGraw Hill - 5ª edição - 2013.

Given the values of the capacitors in the circuit shown in Figure 03-06.1
calculate the total capacitance of the circuit between points a and b.

__Solution of the Problem 3.6__

This problem presents a circuit that, to solve it, we need to do
a star-delta transformation. If you are not familiar with this kind of
circuit, you can review the theory Clicking Here.
We can use the same principles and equations used for circuits with resistors.
However, we must point out that in the equation, whenever R, we should
replace by 1 / C. In the Figure 03-06.2, we see the circuit transformed. So, we can
calculate the capacitor values. To remember, we also show,
one of the equations, the eq. 05-10, for the calculation of
C_{a}. A eq. 05-11 and the eq. 05-12 are similar.

By doing the numerical substitution, we find the value for C_{a}, or:

To calculate C_{b} we must substitute, in the numerator of the above equation, the value of
C_{2} by C_{3} and for the calculation of C_{c},
we substitute in the numerator of the above equation the value of C_{1} by C_{3}.
And so, after the calculations we find the values of:

Now that we know the capacitor values we will calculate the total capacitance. For this,
let's start by calculating the parallel of the capacitors C_{b} = 15 µF with
C_{4} = 50 µF and the value of
parallel to the capacitors C_{c} = 3.75 µF
with C_{5} = 20 µF. For the first parallel, we find 65 µF and
for the second, 23.75 µF.

Look at the Figure 03-06.3. Notice that we now have two
capacitors in series in the right side of the circuit. Solving, we found a capacitor
equivalent of 17.4 µF. Thus, to find
the total capacitance value, we must add this value found with the value of
C_{a}, resulting in C_{total}, or: