Problem 3.5
Source: Adapted of Problem 6.24 - page 216 -
SADIKU, Matthew N. O. , ALEXANDER, Charles K. -
Book: Fundamentos de Circuitos Elétricos - McGraw Hill - 5ª edição - 2013.

a) calculate the total capacitance of the circuit shown in Figure 03-05.1 between the points a and b.

b) Calculate the voltage in each capacitor.

c) Calculate the acumulated energy in each capacitor.

Solution of the Problem 3.5

Item a

Looking at the circuit of the figure above we clearly notice that the capacitors
C_{4} and C_{5} are in series. To calculate the
equivalent capacitance of a series circuit of two capacitors we use the equation below.

C_{eq} = C_{4} C_{5} / (C_{4} + C_{5})

Substituting the numerical values and performing the calculations we find the value
of
C_{eq1}, or:

C_{eq1} = 20 80 / (20 + 80) = 16 µF

Look at the Figure 03-05.2 as the redesigned circuit was. It's also easy
to recognize that the capacitors of 14 and 16 µF are in parallel.
In order to calculate the equivalent capacitance of a parallel circuit,
sum the value of all the capacitors that make up the circuit. In our case, adding
14 and 16 µF we have the result of
C_{eq2} = 30 µF

In the Figure 03-05.3 we see how C_{eq2} is in series with the capacitor of
60 µF. Again we have the case of capacitors in series.
So just apply the equation mentioned above and we can calculate the value
equivalent of the association. After the calculation, we find the value of 20 µF.

Substituting these two capacitors (C_{2} e C_{eq2}) by a single
20 µF, we have to do the parallel
of this capacitor with that of 30 µF and finally adding its values
we find the value of the total capacitance
between the points a - b, or:

C_{total} = 50 µF

Item b

In this item we will calculate the voltages in each capacitor of the circuit. As C_{1} is in parallel with the voltage source of 90 volts,
of course the voltage on it is also 90 volts. Let's recall the eq. 03-02 linking
charge, voltage and capacitance. See below:

eq. 03-02

In this way, looking at the Figure 03-05.4 we know that C_{eq2} and
C_{2} are in series. To calculate the voltage across each capacitor
of a series circuit is enough to know that the voltage is inversely proportional
to the value of the capacitance, since the charge of the capacitors is the same.

So, we can write:

V_{C2} = V C_{eq2} / (C_{2} + C_{eq2})

V_{C2} = 90 30 / (30 + 60) = 30 volts

In the same way we can C_{eq2}, or:

V_{Ceq2} = V C_{2} / (C_{2} + C_{eq2})

V_{Ceq2} = 90 60 / (30 + 60) = 60 volts

As we can see in the Figure 03-05.5, the voltage V_{Ceq2}, voltage on
C_{eq2}, is the same about
C_{eq1} and C_{3}, since C_{eq2} is the
equivalent capacitance of the parallel association of C_{eq1} and C_{3}. Therefore:

V_{C3} = 60 volts

V_{Ceq1} = 60 volts

On the other hand, we know that C_{eq1} is the equivalent capacitance of the series
between the capacitors C_{4} and C_{5}. Therefore, applying the same technique
that we use to calculate the voltage on the capacitors C_{eq2} and
C_{2}, we find the values of:

V_{C4} = 60 80 / (20 + 80) = 48 volts

V_{C5} = 60 20 / (20 + 80) = 12 volts

With this, and following the notation of the initial figure we can write that:

V_{1} = 90 volts

V_{2} = 30 volts

V_{3} = 60 volts

V_{4} = 48 volts

V_{5} = 12 volts

Item c

To calculate the item c let's recall the eq. 03-05 which relates the energy of
a capacitor as a function of the voltage on the capacitor and its capacitance.

eq. 03-05

Then, to calculate the energy accumulated in each capacitor it is enough to apply the formula.
So:

W_{1} = (1/2) x 30 x 10^{-6} x 90^{2} = 0.1215 joules

W_{2} = (1/2) x 60 x 10^{-6} x 30^{2} = 0.027 joules

W_{3} = (1/2) x 14 x 10^{-6} x 60^{2} = 0.0252 joules

W_{4} = (1/2) x 20 x 10^{-6} x 48^{2} = 0.02304 joules

W_{5} = (1/2) x 80 x 10^{-6} x 12^{2} = 0.00576 joules

Adendo

Note that by adding the energy accumulated by each capacitor we must find the
total energy of the system. In addition, we find:

On the other hand, we calculate in the item a the total capacitance of the system, whose value is
50 µF, and we know that the voltage on this capacitance is 90 volts.
Let's calculate what the energy accumulated by the total capacitance.

W_{total} = (1/2)x 50 x 10^{-6} x 90^{2} = 0.2025 joules

As it could not be, the Energy Conservation Law prevailed.