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The effective capacitances of two capacitors are 3 $\mu F$ and 16 $\mu F$, when they are connected in series and parallel respectively. The capacitance of two capacitors are:
Detailed Solution
Series: $\frac{C_1C_2}{C_1+C_2}=3$; parallel: $C_1+C_2=16$. So $C_1C_2=48$. $(C_1-C_2)^2=(C_1+C_2)^2-4C_1C_2=256-192=64 \Rightarrow C_1-C_2=8$. Hence $C_1=12\ \mu F$, $C_2=4\ \mu F$.