The differeence in the effective capacity of two similar capacitor when joined in series and then in paralllel is 6 `mu` F the capacity of each capacitor is
A
`2 mu F`
B
`4 mu F`
C
`8 mu F`
D
`16 mu F`
Text Solution
Verified by Experts
The correct Answer is:
a
When two similar capactors are joined in series then effective capacity is `(1)/(C_(s))=(1)/(c )+(1)/(C )+(1)/(C )=(2)/(C )` `therefore C_(s)=(C )/(2)` when capacitors are joined in parallel then `C_(p)=C+C=2C` since `C_(p)-C_(s)=6 mu f` `therefore 2c-(C )/(2)=6 rarr 4C-c =12` or C=4 muF`
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