__Problem 3.8__ Source:
Problem 6.29 - page 244
- IRWIN, J. David -
Book: Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson - 2013.

Determine the value of the total capacitance that the circuit shown in Figure 03-08.1 presents between the terminals a-b
knowing that: C_{1} = 4.0 µF, C_{2} = 6.0 µF,
C_{3} = 1.0 µF, C_{4} = 8.0 µF and
C_{5} = 3.0 µF.

__Solution of the Problem 3.8__

To calculate the total capacitance that appears between the points a e b,
we must pay attention to the fact that the three capacitors, C_{3},
C_{4} and C_{5}, are in parallel. Therefore, we can
calculate the equivalent capacitance of these capacitors by adding
their values, or:

See in the Figure 03-08.2, how was the circuit when we replaced the capacitors that
were in parallel by their equivalent C_{eq1}. Note that this capacitor
is in series with C_{1}. Solving the series of these two capacitors,
we will check that it is in parallel with C_{2}. After this
we are able to calculate the total capacitance between points a and b.
First, let's calculate the series between C_{1} and C_{eq1}.

Now adding the found value of C_{eq2}
with C_{2}, since they are in parallel,
we find the value of the total capacitance of the circuit.