Problem + Hard 3.2   Source:   Problem 6.23 - page 243 -    IRWIN, J. David - Book: Circuit Analysis in Engineering - 4th edition - Ed. Pearson - 2013.

If the voltage between terminals a - b in the circuit shown below is 28 volts, determine the voltage across each capacitor.

Solution of the Problem + Hard 3.2

To solve this problem, we must calculate the capacitance between the points a-b.

Note in the figure on the side, the redrawn circuit showing the resulting capacitance of the series circuit between the 4 and 12 µF capacitors. Thus, a capacitor of value equal to 3 µF. Note that this capacitor, in turn, is located in parallel with the capacitor of 6 µF.

Connected to point c are two capacitors in series: one 24 µF and the other 8 µF. This series results in a capacitor of 6 µF, as shown above.

The figure on the side shows the 9 µF capacitor resulting from the parallel between the capacitors of 3 µF and 6 µF. Note that this capacitor is in series with the 18 µF capacitor.

So, the resulting capacitor of this series has a equivalent value of 6 µF.

On the side, the resulting circuit has two capacitors of 6 µF in parallel. This parallel results in a capacitor of 12 µF, which in turn, will be in series with the 2 µF capacitor that is connected to point a.

In the figure on the side, it can be seen that the two capacitors are in series. In a series circuit, it is known that the voltage is inversely proportional to the value of the capacitance, since the charge of the capacitors that are in series is the same.

Remembering that the voltage between points a - b is 28 volts, it is possible to calculate the voltage on the 2 µF capacitor.

And on the 12 µF capacitor, a voltage of:

Note that the voltage V₁₂ is the voltage between the points c- b.

Going back to the circuit that shows the 18 and 9 µF capacitors in series, It is possible to calculate the voltage across each capacitor. Do not forget that the voltage across these capacitors is the calculated voltage between the points c - b, that is, a value of 4 volts. So the voltage across the 18 µF capacitor is:

And the voltage across the 9 µF capacitor is:

Note that the voltage across the 9 µF capacitor is the voltage at point d. So it's the same voltage across the 3 and 6 µF capacitors. This situation is shown in the adjacent figure. However, returning to the initial circuit, it is noticed that the 3 µF capacitor is the result of the series between capacitors 4 and 12 µF.

So the voltage across the 4 µF capacitor is:

So, the voltage across the 12 µF capacitor is:

Finally, we must pay attention to the fact that between the points c - b, according to the initial circuit, there are two series capacitors: one 24 µF and the other 8 µF. Since the voltage between these points is 4 volts, the voltage across the 8 µF capacitor will be:

And the voltage across the 24 µF capacitor is:

In summary, we present the calculated values ​​on all capacitors.