Kumar developed a group of capacitors as shown in figure. Calculate its equivalent capacitance between the points Mand K.

Explanation

Equivalent circuit diagram can be drawn as
Let `C_(1) = C_(2) = 8 muF`
`C(3) = C_(4) = 10 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) = 1/8 + 1/8`
`1/C_(5) = 2/8 = 1/4`
`C_(5) = 4 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) = 1/10 +1/10 =2/10 = 1/5`
`rArr C_(6) = 5 muF`
Now `C_5` and `C_6` are in parallel with each other and let their net capacitance be C.
`rArr C= C_(5) + C_(6)`
`rArr C = 4+5 = 9 muF`