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.

Step-by-step explanation:

The capacitors connected in series can be calculated by the formula

`(1)/(C)=(1)/(C1)+(1)/(C2)`

So, the equivalent capacitance of C1 and C2 will be

`\begin{gathered} (1)/(C)=(1)/(1.29)+(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