Question

In a given L - C circuit, switch S is closed at t = 0, then

• A. maximum charge on capacitor can be $2C\epsilon$.
• B. when charge on capacitor is half of its maximum charge, current is maximum
• C. Charge on capacitor is $\frac{\epsilon C}{2}$ at $t=\frac{\pi }{3}\sqrt{LC}$
• D. Maximum charge on capacitor can be $\epsilon C$

Answer 1

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

Charge on capacitor is $\frac{\epsilon C}{2}$ at $t=\frac{\pi }{3}\sqrt{LC}$


$\epsilon =\frac{q}{C}+L\frac{di}{dt}$
$\Rightarrow \epsilon -\frac{q}{C}=L\frac{d^{2}q}{d{t}^2}$
$\frac{\epsilon C-q}{LC}=\frac{d^{2}q}{d{t}^2}=\frac{di}{dt}\Rightarrow \frac{d^{2}i}{d{t}^2}=-\frac{i}{LC}$
$\frac{d^{2}i}{d{t}^2}=-w^{2}i=w=\frac{1}{\sqrt{LC}}\&T=\frac{2\pi }{w}$
$i=i_0\sin wt$ or $\frac{di}{dt}=i_{0}w\cos wt$
At t = 0,
$L\frac{di}{dt}=\epsilon \Rightarrow \epsilon =L{i}_{0}w\Rightarrow i_0=\frac{\epsilon }{Lw}=\epsilon \sqrt{\frac{C}{L}}\&i=\frac{dq}{dt}=i_{0}sinwt$
${\int }_0^{q}dq={\int }_0^t{i}_{0}sinwtdt\Rightarrow q=-\frac{i_0}{w}[coswt{]}_0^t$
$q=\frac{i_0}{w}(1-coswt)\Rightarrow q=q_0(1-coswt)$
where $q_0=\frac{i_0}{w}=\epsilon C$
$q_{max}=2q_0=2\epsilon C$
When charge on the capacitor is half of its maximum charge i.e., $q_0$,
$q=q_0(1-coswt)$
$\Rightarrow q_0=q_0(1-\cos wt)\Rightarrow \cos wt=0$
Therefore, $i=i_0\sin wt=i_0\times 1=i_0$, so current takes its maximum value.
At $t=\frac{\pi }{3}\sqrt{LC}$,
$q=q_0(1-coswt)=\epsilon C(1-\cos \frac{\pi }{3})=\frac{\epsilon C}{2}$