Problem 3.1 Source:
Problem elaborated by the author of the site.

Knowing that C_{1} = 12 µ F,
C_{2} = 2 µ F and C_{3} = 4 µ F, compute:

a) the total capacitance of the circuit shown in Figure 03-01.1 between the points a and b.

b) if we connect a 10 volts battery between points a and b
calculate the accumulated load on each capacitor.

Solution of the Problem 3.1

Item a

As can be seen in the fFigure 03-01.1, capacitors
C_{2} and C_{3} are in parallel. To calculate the
total capacitance of a circuit
parallel, just add all the values of the capacitors that form the circuit.
It is identical to the association of resistors in series. Therefore, by adding C_{2} with
C_{3} we get C_{eq} = 6 µF.
Now, we just compute the series association between C_{eq} = 6 µF and
C_{1} = 12 µF, to obtain the total capacitance between points a
and b.

Remember the equation of the series of two capacitors (similar to two resistors
in parallel), we have:

C_{eq} = C_{1}. C_{2} / C_{1} + C_{2}

So, doing the calculations we find the value of:

C_{total} = 4 µF

Item b

In order to calculate the accumulated charge in each
capacitor, we first calculate the total circuit charge using
the value of the voltage of the source, which is 10 volts and the total capacitance we calculate in the item
previous. We know that:

q_{total} = C_{total} V

Thus, by multiplying the values we find that the total charge is:

q_{total} = 4 x 10^{-5} C = 40 µC

We must remember that in a series of capacitors the charge on
each capacitor in the circuit is the same. Therefore C_{1} has a charge of
40 µC. With this, we can calculate what is the voltage on C_{1},
it is enough to divide the total charge by the value of C_{1} and we find
the value of 3.33 volts.

To compute the charges in C_{2} and C_{3}
let's calculate the voltage on the parallel of
C_{2} and C_{3}. For this, it is enough to subtract the tension
of the source of the value of V_{C1}, that is, the voltage on the capacitor V_{C1}.
In this case, we find 10 - 3.33 = 6.67 volts. Now multiplying
this voltage by the capacitance value of the capacitors,
we will find the charges of each capacitor. Below is a summary of the calculated values
of the charges in each capacitor.

q_{1} = 4 10^{-5} C = 40 µC

q_{2} = 1.33 10^{-5} C = 13.33 µC

q_{3} = 2.67 10^{-5} C = 26.67 µC

Note that the charge on the capacitor C_{3} is double the load
in C_{2}, as it should be, since C_{3} = 2 C_{2}.
Also, note that q_{1} = q_{2} + q_{3}, because we know that
capacitors in series must have equal charges.