Question

Three capacitors are connected as follows: 2.15 F capacitor and 4.14 F capacitor are connected in series, then that combination is connected in parallel with a capacitor of 8.46 F. What is the capacitance of the total combination?

Answer 1

Step-by-step explanation:

Capacitance,
`C_1=2.15\ F`

Capacitance,
`C_2=4.14\ F`

Capacitance,
`C_3=8.46\ F`

Capacitors C₁ and C₂ are connected in series. The equivalent capacitance is given by :

`(1)/(C_s)=(1)/(C_1)+(1)/(C_2)`

`(1)/(C_s)=(1)/(2.15)+(1)/(4.14)`

`C_s=1.41\ F`

The combination is now connected with C₃ in parallel. Now the equivalent capacitance is given by :

`C_p=C_s+C_3`

`C_p=1.41+8.46`

`C_p=9.87\ F`

So, the equivalent capacitance of the total combination is 9.87 F. Hence, this is the required solution.