Question

Two capacitors $C_1=C_2=\frac{1}{{\pi }^2}\times 10^{-2}F$ and inductor $L=2\times 10^{-2}H$ are connected in series as
shown in the figure. Initially charge on each capacitors are $4\sqrt{3}\mu C.$ At $t=0$ switch $S_1$ is
closed and at $t=\frac{1}{400}$ sec, switch $S_2$ is also closed. The maximum charge on capacitor $C_2$
during LC oscillation is $k\sqrt{2}\mu C$ . Find $k$ .

Answer 1

Answer: 6

Initial total energy is
$U_{initial}=\frac{q^2}{2C_{eq}}=\frac{2q^2}{2C}=\frac{q^2}{C}=\frac{{\left(4\sqrt{3}\right)}^2}{\frac{1}{{\left(\pi \right)}^2}\times {\left(10\right)}^{-2}}=4.8\times {\left(10\right)}^4$
Let's calculate angular frequency in first case when both capacitors are in series $\omega =\frac{1}{\sqrt{L{C}_{eq}}}\omega =\frac{1}{\sqrt{2\times 10^{-2}\times \frac{10^{-2}}{2{\pi }^2}}}\omega =100\pi T=\frac{2\pi }{\omega }=\frac{2\pi }{100\pi }\Rightarrow T=\frac{1}{50}sec$
At $t=\frac{T}{4}=\frac{1}{200}sec$ All energy will be stored in inductor .and $C_1$ is short-circuited
Maximum energy that can be stored in $C_2$ After that will be
$\frac{q^2}{2C}=4.4\times 10^{4}q=\sqrt{4.4\times 10^4\times 2\times \frac{1}{{\pi }^2}\times 10^{-2}}=6.63\sqrt{2}$ 4