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

Knowing that C₁ = 12 µ F,   C₂ = 2 µ F and C₃ = 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 C₂ and C₃ 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 C₂ with C₃ we get C_eq = 6 µF. Now, we just compute the series association between C_eq = 6 µF and C₁ = 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:

So, doing the calculations we find the value of:

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:

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

We must remember that in a series of capacitors the charge on each capacitor in the circuit is the same. Therefore C₁ has a charge of 40 µC. With this, we can calculate what is the voltage on C₁, it is enough to divide the total charge by the value of C₁ and we find the value of 3.33 volts.

To compute the charges in C₂ and C₃ let's calculate the voltage on the parallel of C₂ and C₃. 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.

Note that the charge on the capacitor C₃ is double the load in C₂, as it should be, since C₃ = 2 C₂. Also, note that q₁ = q₂ + q₃, because we know that capacitors in series must have equal charges.