Question

If 3 capacitors each of
capacitance 1uF are connected
in such a way that the resultant
capacity is 1.5mF, then
1.All the 3 are connected in series
2.All the 3 are connected in parallel
3.
Two of them are in parallel and
connected in series to the third
Two of them are in series and then
4.
connected in parallel to the third​

Answer 1

Solution :-

As per the given data ,

• C 1 = C 2 = C 3 = C = 1 μF

Case I : .All the 3 are connected in series

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

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

Hence ,

⇒ 1 / Cs = 1 / C + 1 / C + 1 / C

⇒ 1 / Cs = 3 / C

⇒ Cs = 1 / 3

⇒ Cs = 0.33μF

Case II : All the 3 are connected in parallel

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

⇒ Cp = C 1 + C 2 ...+ Cn

Hence ,

⇒ Cp = 3C

⇒ Cp = 3μF

Case III : Two of them are in parallel and  connected in series to the third

C 1 and C 2 connected in series ,

⇒ 1 / C s = 1 / C 1 + 1 / C 2

⇒ 1 / C s =  1 / C + 1 /C

⇒ 1 / Cs = 2 /C

⇒ Cs = 1 / 2

⇒ Cs = 0.5 μF

Cs and C 3 connected in parallel

⇒ Cp = Cs + C3

⇒ Cp = 0.5 +1

⇒ Cp = 1.5 μF