Calculate the equivalent capacitance of the combination between the points P and Q as shown in fig. given below :

Explanation

Here, `C_(1)/C_(2) =C_(3)/C_(4)` or `1/2 = 1/2`
This forms the balanced Wheatstone bridge, hence, the capacitor in arm CD, i.e., `3muF` becomes ineffective. So equivalent circuit can be drawn as :
`rArr 1/C_(6) = 1/C_(1) + 1/C_(2) =1/1 + 1/2`
`1/C_(6) =3/2`
`C_(6) = 2/3 muF`
So `C_(1)` and `C_2` are in series and let their net resistance in series be `C_(7)` and is given by :
`1/C_(7) =1/C_(3) + 1/C_(4) =1/1 + 1/2`
`1/C_(7) = 3/2`
`C_(7) = 2/3 muF`
Now `C_6` and `C_7` are in parallel combination and let their net resistance be C.
`rArr C = C_(6) + C_(7)`
`C = 2/3 + 2/3 = 4/3 = 1.33 muF`