Two capacitors when connected in series the effective capacity is `2muF` and when in parallel is `9muF`. Calculate the value of each capacitors.

Explanation

Let `C_(1) and C_(2)` be the capacity of two capacitors.
Then in series `C_(1)=(C_(1)C_(2))/(C_(1)+C_(2))=2muF" and in parallel", C_(p)=C_(1)+C_(2)=9muF ......(1)`
`C_(1)C_(2)=2(C_(1)+C_(2))=2 xx 9 -18`
`C_(2)=(18)/(C_(1)) ......(2)`
Substituting eq. (2) in eq. (1) we get, `C_(1)+(18)/(C_(1))=9`
i.e, `C_(1)^(2)+18=9C_(1) " "C_(1)^(2)-9C_(1)+18=0" "i.e, (C_(1)-6) (C_(1)-3)=0`
`C_(1)=6 or 3`
Hence, `C_(2)=3 or 6`
If `C_(1)=6muF then C_(2)=3muF, or`
If `C_1=3muF, then C_(2)=6muF`