Question

Three capacitors are connected as follows: 1.29 F capacitor and 3.17 F capacitor are connected in series, then that combination is connected in parallel with a capacitor of 8.36 F. What is the capacitance of the total combination?

Answer 1

Given:

The capacitor C1 = 1.29 F

The capacitor C2 = 3.17 F

The capacitor C3 =8.36 F

Capacitors C1 and C2 are connected in series while C3 is connected in parallel.

To find the capacitance of the total combination.

Explanation:

The capacitors connected in series can be calculated by the formula

$\frac{1}{C}=\frac{1}{C1}+\frac{1}{C2}$

So, the equivalent capacitance of C1 and C2 will be

$\begin{gathered} \frac{1}{C}=\frac{1}{1.29}+\frac{1}{3.17} \\ C=0.92\text{ F} \end{gathered}$

The capacitors connected in parallel can be calculated by the formula

$C=C1+C2$

On substituting the values, the capacitance of the total combination will be

$\begin{gathered} C_T=C+C3 \\ =0.92+8.36 \\ =9.28\text{ F} \end{gathered}$

Thus, the total capacitance of the combination is 9.28 F