Question: Q. 16. (i) A parallel plate capacitor $\left(C_{1}\right)$ having charge $Q$ is connected, to an identical uncharged capacitor $C_{2}$ in series. What would be the charge accumulated on the capacitor $C_{2}$ ?
(ii) Three identical capacitors each of capacitance $3 \mu \mathrm{F}$ are connected in turn in series and in parallel combination to the common source of $V$ volt. Find out the ratio of the energies stored in two configurations.
A [O.D. Set I, II, III 2016]
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Solution:
Ans. (i) Both caprectors would come to common potontial
$V_{\text {commen }}=\frac{\text { Total charge }}{\text { Net capacitance }}=\frac{Q}{c_{1} c_{2}}\left(c_{1}+c_{2}\right)$
$\Rightarrow$ charge on second capacitor $\left(q^{\prime}\right)=Q\left(c_{1}+c_{2}\right)$
(ii) Capacitance in series:
$\frac{1}{c_{\text {net }}}=-\frac{1}{c_{1}}+\frac{1}{c_{2}}+\frac{1}{c_{3}} \Rightarrow \frac{1}{c_{5}}=-\frac{1}{3}+\frac{1}{3}+\frac{1}{3}-1 \Rightarrow c_{3}=4$
Parallel combination
$C_{\text {pet }}=c_{1}+c_{2}+c_{3} \Rightarrow c_{p}=3+3+30 q \Leftrightarrow c_{p}=9$ llF
Ratio of energies $=\frac{1}{2} c_{s} v^{2}: \frac{1}{2} c_{p}, 1: 9$
$\therefore$ Ratio of energies in cories ma paralied $=1.9$
[Topper’s Answer 2016]
