• # question_answer Net capacitance of three identical capacitors in series is $1\mu F.$ What will be their net capacitance, if connected in parallel? Find the ratio of energy stored in the two configurations, if they are both connected to the same source.

If n identical capacitors, each of capacitance C are connected in series combination, then the equivalent capacitance,            ${{C}_{S}}=\frac{C}{n}$ and when connected in parallel, then the equivalent capacitance,            ${{C}_{p}}=nC\Rightarrow \frac{{{C}_{p}}}{{{C}_{S}}}=\frac{nC}{C/n}={{n}^{2}}$             or         $\frac{{{C}_{s}}}{{{C}_{p}}}=\frac{1}{{{n}^{2}}}$                                    ?(i) According to problem, $C=n{{C}_{s}}=3\times 1\mu F=3\mu F$ For each capacitor, in parallel combination,             ${{C}_{p}}=nC=3\times 3=9\mu F$ or ${{C}_{p}}=9\mu F$ Also, for same voltage, energy stored in capacitor is given by             $U=(1/2)C{{V}^{2}}$ [for V = constant]?(ii) or                     $U\propto C$ $\Rightarrow \frac{{{U}_{s}}}{{{U}_{p}}}=\frac{{{C}_{s}}}{{{C}_{p}}}=\frac{1}{{{n}^{2}}}=\frac{1}{{{(3)}^{2}}}=\frac{1}{9}$            [from Eq (i)] $\Rightarrow \frac{{{U}_{s}}}{{{U}_{p}}}=\frac{1}{9}$ or ${{U}_{s}}:{{U}_{p}}=1:9$