Mangal developed a group of capacitors as shown in figure. Calculate its equivalent capacitance between the points J and D.

Explanation

Equivalent circuit can be drawn as
Let `C_(1) = C_(2) = 2muF`
`C_(3) = C_(4) = 16 muF`
Now `C_(1)` and `C_2` are in series combination and let their net capacitance in series be `C_(5)`

`1/C_(5) = 1/C_(1) + 1/C_(2) = 1/2 + 1/2`
`1/C_(5) = 2/2=1`
`C_(5) = 1 muF`
Also `C_3` and `C_4` are in series combination and let their net capacitance be `C_6`.
`1/C_(6) = 1/C_(3) + 1/c_(4) = 1/16 + 1/16`
`1/C_(6) = 2/16 = 1/8`
`C_(6) = 8 muF`
Now `C_5` and `C_6` are in parallel with each other and let their net capacitance be C.
`C = C_(5) + C_(6)`
`rArr C= 1+ 8 = 9 muF`