Resolved Problems about Capacitors

Problem 3.7 Source: Adapted of Problem 6.26 - page 244 - IRWIN, J. David - Book: Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson - 2013.

Determine the value of C to produce the desired total capacitance of C_total = 4.5 µF in the circuit shown in Figure 03-07.1. All values are in µF.

Solution of the do Problem 3.7

Initially, let's look closely at the circuit highlighted in yellow. We realized that the capacitor C is in parallel with that of 9 µF. And this association 9 + C is in series with the capacitor of 4 µF. To calculate the equivalent capacitance of this circuit, which we will call C_eq₁, we can write:

C_eq₁ = (9 + C) 4 / (9 + C) + 4 = (36 + 4 C) / (C + 13)

Let's turn our attention to the circuit highlighted in green. Here we have the parallel of two capacitors: one of 1 µF and another of 2 µF, totalizing a capacitance equivalent of 3 µF. This association is in series with the capacitor C. Soon, we can write the value of this association, which we will call C_eq₂.

C_eq₂ = 3 C / (3 + C)

Looking at the circuit it is easy to see that the circuits highlighted in green and yellow are in parallel. And this association must be equal to the total value of the desired capacitance, that is, 4.5µF. So:

C_eq₁ + C_eq₂ = 4.5 µF

Replacing C_eq₁ and C_eq₂ by respective equations, we have:

After some algebraic manipulations we arrive at a equation of the second degree, or:

2,5. C² + 15 C - 67.5 = 0

Solving this equation we find two numerical values: one of the values is 3 and the other is -9. Since we can not physically have negative values for capacitance, we can only the positive value. Then the value of C is equal to:

C = 3 µF

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