In the given circuit the capacitance C1 and C2 are parallel with the capacitance C3 i.e
(C1 || C2) + C3
∴(20 × 30) ⁄ (20 + 30) + 20
CA= 12 + 20 = 32 μF
Now capacitance CA, C4, & C5 are in the series therefore
Ceqv = (1/30 + 1/20 + 1/20)
Ceqv = (60/8) = 7.5 μF