Problem + Hard 3.1   Source:   Problem 6.24 - page 216 -    SADIKU, Matthew N. O. , ALEXANDER, Charles K. - Book: Fundamentals of Electric Circuits - McGraw Hill - 5th edition - 2013.

Knowing that C₁ = 30 µF ,  C₂ = 60 µF ,  C₃ = 14 µF ,  C₄ = 20 µF , C₅ = 80 µF , and V = 90 volts

a) Calculate the total capacitance of the circuit below, between points a and b;

b) Calculate the charge on each capacitor;

c) Calculate the voltage across each capacitor;

d) Calculate the accumulated energy in each capacitor.

Solution of the Problem + Hard 3.1

Item a

By the circuit of the figure above, it is noticed that the capacitors C₄ and C₅ are in series. To calculate the equivalent capacitance of a series circuit of two capacitors, use the equation below.

Substituting numerical values ​​and performing the calculation:

Note in the figure on the side how the redesigned circuit turned out. It's also easy recognize that the 14 and 16 µF capacitors are in parallel. To calculate the equivalent capacitance of a parallel circuit, just add the value of all the capacitors that make up the circuit.

In this case, adding 14 and 16 µF results in a value of C_eq2 = 30 µF

The figure on the side shows how C_eq2 is in series with the capacitor of 60 µF. So, just apply the equation mentioned above to calculate the value association equivalent. After the calculation, the value of 20 µF.

Replacing these two capacitors with a single one of 20 µF, lacks make the parallel of this capacitor with the one of 30 µF. Finally, when adding their values, find the value of the total capacitance between the  points a - b, or:

Item b

  Note that C₁ is in parallel with the voltage source. Soon, It's easy to calculate your load since we know that:

Then, substituting for numerical values:

To calculate the load on C₂, one should look at the last figure and note that C₂ is in series with the capacitor resulting from the association between C₃, C₄ and C₅, or C_eq2. On the other hand, it is known that in a series connection, the charge on each capacitor that make up the circuit is the same. Add that the result of this series is 20 µF, already calculated previously. Therefore, it is possible to calculate the load of this equivalent capacitance, and we will call it q₂₀, using the last equation above, or:

So this payload of 1.8 x 10^-3 C has to be present on the 60 µF capacitor and on C_eq2 = 30 µF, how could it not be.

However, notice that the 30 µF capacitor is the parallel of C₃ and C_eq1, where this is the series equivalent capacitance of C₄ and C₅, seen earlier. Therefore, based on the figure on the side, we can calculate the load on C₃ using a ratio, as the voltage across the two capacitors is the same since they are in parallel.

If at 30 µF we have a load of 1.8 x 10^-3 C, then in 14 µF we will have:

Substituting numerical values ​​and performing the calculation:

It is evident that in the 16 µF capacitor, that is, in C_eq1, the load will be the difference between the load on 1.8 x 10^-3C and the load on C₃ (q₃, value calculated above). Soon:

With this data in hand, and remembering that 16 µF is the value of the equivalent capacitance of the series association of C₄ and C₅, or C_eq1 , it is concluded that this load of 0.96 x 10^-3 C is the payload present in both C₄ and C₅.

Now, we can do a summary of the calculated values. See below.

Item c

In this item, we will calculate the voltages to which the capacitors that make up the the circuit. Since C₁ is in parallel with the 90 volts voltage source, of course the voltage across it is also 90 volts. Let's recall the equation that relates capacitance, voltage and charge. See below:

In this way, looking at the figure on the side, we know that C_eq2 and C₂ are in series. To calculate the voltage across each capacitor of a series circuit, it 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:

In the same way we can do for C_eq2, or:

As we can see in the figure on the side, the voltage V_Ceq2, voltage over C_eq2, is the same over C_eq1 and C₃, since C_eq2 is the equivalent capacitance of the parallel association of C_eq1 and C₃. Therefore:

On the other hand, we know that C_eq1 is the equivalent capacitance of the series between capacitors C₄ and C₅. Then, applying the same technique which we used to calculate the voltage across capacitors C_eq2 and C₂, we find the values ​​of:

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

Item d

To calculate the item d, let's remember the equation that relates the energy of a capacitor as a function of the voltage across the capacitor and its capacitance.

Therefore, to calculate the accumulated energy in each capacitor, just apply the formula. Then:

Adendo

Note that if we add the energy accumulated by each capacitor, we will find the total energy of the system. Adding, we find:

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

As it could not be otherwise, the law of Conservation of Energy prevailed.