Question

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?

Answer 1

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

We can find it using the following formula:

`\begin{gathered} (1)/(Ceq1)=(1)/(C1)+(1)/(C2) \\ so: \\ (1)/(Ceq1)=(1)/(3.64)+(1)/(4.67) \\ Ceq1=2.05F \end{gathered}`

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

`\begin{gathered} C_(eq)=Ceq1+C3 \\ C_(eq)=2.05F+7.16F \\ C_(eq)\approx9.21F \end{gathered}`

Answer:

9.21 F