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.

Answer:

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 \[{{U}_{s}}:{{U}_{p}}=1:9\]