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Three capacitors are connected as follows: 3.64 F capacitor and 4.67 F capacitor are connected in series, then that combination is connected in parallel with a capacitor of 7.16 F. What is the capacitance of the total combination?

Respuesta :

First, let's find the equivalent capacitance of the capacitors connected in series.

We can find it using the following formula:

[tex]\begin{gathered} \frac{1}{Ceq1}=\frac{1}{C1}+\frac{1}{C2} \\ so: \\ \frac{1}{Ceq1}=\frac{1}{3.64}+\frac{1}{4.67} \\ Ceq1=2.05F \end{gathered}[/tex]

Now we can find the equivalent capacitance of the capacitors connected in parallel using the following formula:

[tex]\begin{gathered} C_{eq}=Ceq1+C3 \\ C_{eq}=2.05F+7.16F \\ C_{eq}\approx9.21F \end{gathered}[/tex]

Answer:

9.21 F

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