Question

You are given three capacitor of 4uF each. How they will be combined to obtain resultant
capacity of 6uF?​

Answer 1

Three capacitors of capacitance [C1 , C2 and C3 ] 4μF are given

We have to connect them in a manner so that the effective capacitance is

6μF.

It is a very simple question .

connect any two capacitors in series and the third one in parallel to each other , then

Total capacitance

C = $\frac{C1.C2}{C1 + C2}$ + C3

C = 6μF

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I hope you understand.

Answer 2

Answer:

We will have to combine two capacitors in series and one in parallel with the previous combination.

Let C1 and C2 be capacitors in series combination :

1/C1 + 1/C2 = 1/C

1/C = 1/4 + 1/4

1/C = 1/2

C = 2uF

C' is the capacitor which is to be connected in series with the other combination.

C + C'

= 2 + 4

= 6uF

Hope it helps!