Three capacitors of value 8µF , 16µF and 32µF are connected in series, the total capacitance will be

Three capacitors of value 8µF , 16µF and 32µF are connected in series, the total capacitance will be 

(A) 32/7 µF . 
(B) 7.32 µF . 
(C) 56 µF . 
(D) 32 µF . 

Explanatory answer:

For series connection of capacitors

C = C1C2C3 / ( C1C2+C2C3+C3C1)

C = 32/7 µF .