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