Question

Three capacitors of each capacity is 4uF How can they combine that resultant capacity will be 6uF.​

Answer 1

Solution :-

As per the given data ,

• C 1 = C 2 = C 3 = C = 4 μF

Two 4μF resistors are connected in series with each other .

Equivalent capacitance for n no of capacitor's connected in series is given by ,

⇒ 1 / Cs = 1 /C 1 + 1 / C 2 .. + 1 / Cn

Hence ,

⇒ 1 / Cs = 1 / C + 1 / C

⇒ 1 / Cs = 2 / 4

⇒ Cs = 2 μF

Now , Cs and C are connected in parallel to each other .

Equivalent capacitance for n no of capacitor's connected in parallel is given by ,

⇒  Cp = C 1 + C 2 .. + Cn

Hence ,

⇒ Cp = Cs + C

⇒ Cp = 2 + 4

⇒ Cp = 6  μF  

The net capacitance will be 6   μF   when two capacitors are connected in series and 1 in parallel .