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