Two condensers are once connected in parallel and then in series. If the equivalent capacitance in two cases are 16 F and 3F respectively, then capacity of each condenser is

16 F, 3F 12 F, 4 F 6F, 8F none of these

12 F, 4 F

Let C1 and C2 be the individual capacitance then C1 + C2 = 16              . . . .(i) $\frac{C_1 C_2}{C_1+C_2}$ = 3            . . . . (ii) From (i) and (ii) C1 = 12 F, C2 = 4F