Question
When three identical capacitors are connected in series, their equivalent (total) capacitance is $1 \mu F$. If they are connected in parallel, what will be their total capacitance? If the capacitors are connected to the same source in both the conditions (combinations), then find the ratio of the stored energy in these two types of combinations.

Answer

Let the capacitance of each capacitor be $C \mu F$. So in the case of series combination,
From $\frac{1}{ C _{ S }}=\frac{1}{ C _1}+\frac{1}{ C _2}+\frac{1}{ C _3}$
$\frac{1}{ C _{ S }}=\frac{1}{ C }+\frac{1}{ C }+\frac{1}{ C }=\frac{3}{ C }$
Given Equivalence $\quad C_s=1 \mu F$
$C =3, C _{ s }=3 \mu F$
$\frac{1}{1}=\frac{3}{ C } \Rightarrow C$
When all three capacitors connected in parallel combination, considered equivalent capacitance (say $C _{ P }$ ).
From $C_P=C_1+C_2+C_3 $
$=3+3+3=9 \mu F$ Ans.
Energy stored in series combination $E_S=\frac{1}{2} C_S V^2$
and Energy stored in parallel combination $E_p=\frac{1}{2} C_P V^2$
$\therefore \quad \frac{ E _{ S }}{ E _{ P }}=\frac{\frac{1}{2} C _{ S } V ^2}{\frac{1}{2} C _{ P } V ^2}=\frac{ C _{ S }}{ C _{ P }}=\frac{1}{9} \quad$ Ans.

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