Question

Initially the capacitor is charged to a potential of $5\text{V}$ and then connected to position $1$ with the shown polarity for $1\text{s}$. After $1\text{s}$, it is connected across the inductor at position $2$.

Choose correct alternative(s)

• A. The potential across capacitor after $1\text{s}$ of its connection to position $\left(1\right)$ is $5[2-\frac{1}{e}]\text{V}$
• B. The maximum current flowing in the $L-C$ circuit when the capacitor is connected across the inductor is $10[2-\frac{1}{e}]\text{A}$
• C. The frequency of $LC$ oscillation is $\frac{100}{\pi }\text{Hz}$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The frequency of $LC$ oscillation is $\frac{100}{\pi }\text{Hz}$

Position $\left(1\right)$-

For RC circuit,
Using KVL,
$E-IR-\frac{q}{C}=0$
$E-\frac{dq}{dt}R-\frac{q}{C}=0$
$\frac{dq}{dt}R=E-\frac{q}{C}$
$\frac{Rdq}{E-\frac{q}{C}}=dt$
Integrationg above equation within limits
${\int }_{q_0}^q\frac{Rdq}{E-\frac{q}{C}}={\int }_0^{1}dt$

$R{\left[\frac{ln(E-\frac{q}{C})}{\frac{-1}{C}}\right]}_{q_0}^q=[t{]}_0^1$

$ln(E-\frac{q}{C})-ln(E-\frac{q_0}{C})=-\frac{t}{RC}$
Here $q_0$ is initial charge, given that initially capacitor is charged to $5\text{V}$, hence $\frac{q_0}{C}=5\text{V}$
Also, $E=10\text{V},C=10\text{mF},R=100\Omega ,t=1\text{s}$
Substituting the values we get,
$ln(10-\frac{q}{C})-ln(10-5)=-\frac{1}{100(10\times {10}^{-3})}$
$ln(2-\frac{q}{5C})=-1$
$2-\frac{q}{5C}=\frac{1}{e}$
$\frac{q}{C}=5[2-\frac{1}{e}]$
Hence potential across capacitor at $t=1$ is $V=\frac{q}{C}=5[2-\frac{1}{e}]\text{V}$

Position$\left(2\right)-$

$\to$During $L-C$ oscillations, when current in it is maximum. Total energy will be stored in the inductor $\left(\frac{1}{2}L{I}^2\right)$,

Using conservation of energy,
$\frac{1}{2}C{V}^2=\frac{1}{2}L{I}^2$
$\frac{1}{2}(10\times {10}^{-3}){\left[5\right(2-\frac{1}{e}\left)\right]}^2=\frac{1}{2}\times 2.5\times {10}^{-3}\times I^2$
$\Rightarrow I_{max}=10(2-\frac{1}{e})\text{A}$
Frequency of $LC$ oscillation
$f=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi }\frac{1}{\sqrt{2.5\times 10\times {10}^{-6}}}\Rightarrow f=\frac{100}{\pi }\text{Hz}$
$a,b,c$ are correct