Question

Net capacitance of three identical capacitors in series is `1 muF`. What will be their net capacitance in parallel ? Find the ratio of energy stored in two configurations if they are connected to the same source.

Answer 1

Net capacitance=`1muF=10^(-6)F`
`"if "C_(1)=C_(2)=C_(3)=C`
Let be the capacitance of each of three capacitors and `C_(3) and C_(R)` be the capacitance of series and parallel combination respectively.
`"then", (1)/(C_(s))=(1)/(C)+(1)/(C)+(1)/(C)=(3)/(C) Rightarrow C_(1)=(C)/(3)" "[C_(1)=1muF]`
`" "1muF=(C)/(3)`
`" "C=3muF`
`"Also in"" "C_(P)=C+C+C+=3+3+3 Rightarrow C_(P)=9muF`
Energy stored in capacitor
`E=(1)/(2)CV^(2) Rightarrow (E_(S))/(E_(P))=(1/(2)C_(S) V^(2))/((1)/(2)C_(P)V^(2))=(C_(S))/(C_(P))=(1)/(9)=1:9`
i.e. energy stored in series combination or parallel combination of capacitor is equal to sum of the energies stored in individuals capacitors.