Question

The difference in the effective capacitance of two equal capacitors when joined in parallel and series is 3uF. The value of each capacitor is * 1uF ЗuF 2uF 4uF​

Answer 1

Answer:

Ans should be C= 2 μF, for details see solution

Explanation:

Let the capacitance of each capacitor be C

For series combination :

1/Cs = 1/C + 1/ C

⟹Cs = C/2

For parallel combination : Cp = C+C

⟹Cp =2C

Given : Cp − Cs = 3 μF

⇒ 2C− C/2 = 3 μF

⟹C= 2 μF