Question

At t=0 switch $s_1$ is closed & $s_2$ remains open. At time $t=\frac{\pi }{2}\sqrt{LC},s_1$ is opened and $s_2$ is closed. Now again counting time from zero, at the event of closing of $s_2$ and breaking of $s_1$ . Current in inductor L as a function of time can be given as $i^{\prime }=i_0^{\prime }sin(w^{\prime }t+\varphi )$

Here $i_0^{\prime }$ is

• A. $\epsilon \sqrt{\frac{C}{L}}$
• B. $2\epsilon \sqrt{\frac{C}{L}}$
  
• C. $\frac{3}{2}\epsilon \sqrt{\frac{C}{L}}$
  
• D. $\frac{\epsilon }{2}\sqrt{\frac{C}{L}}$

Answer 1

The correct option is C $\frac{3}{2}\epsilon \sqrt{\frac{C}{L}}$


When $s_1$ is closed & $s_2$ is open By K.L.L is $ϵ=L\frac{di}{dt}+\frac{q}{C}\Rightarrow L\frac{di}{dt}=ϵ-\frac{q}{c}or\frac{d^{2}i}{d{t}^2}=-\frac{i}{LC}$

$\frac{d^{2}i}{d{t}^2}=-w^{2}i,wherew=\frac{1}{\sqrt{LC}}\dots \left(1\right)$

so $i=i_{0}sinwt-\left(2\right)\Rightarrow \frac{dq}{dt}=i_{0}sinwt$

${\int }_0^{q}dq=i_0{\int }_0^{t}sinwtdt\Rightarrow q=\frac{i_0}{w}(1-coswt)$

Solving for boundry condition i.e. at t=0

$ϵ=L\frac{di}{dt}$ & at q=0 at t=0 also i=0

we get $q_0=ϵc\dots \left(3\right)\&i_0=ϵ\sqrt{\frac{C}{L}}\dots \left(4\right)$

When $s_1$ is opened & $s_2$ is closed dounting same from zero again. Let now current be $i^{\prime }$



By conservation of energy$\Rightarrow \frac{1}{2}L{i}^{\prime 2}+\frac{(q_0+q{)}^2}{2c}+\frac{q^2}{2c}=TotalEnergy$

or

$\frac{d{i}^{\prime }}{dt}=-\frac{(q_0+2q)}{LC}$

$\frac{d^2{i}^{\prime }}{d{t}^2}=-\frac{2i^{\prime }}{LC}\Rightarrow \frac{d^2{i}^{\prime }}{d{t}^2}=-w^{12}{i}^{\prime }$

$w^{\prime }=\sqrt{\frac{2}{LC}}$ so $i^{\prime }=i_0^{\prime }sin(w^{\prime }t+\varphi )$

By boundry condition i.e. at t=0, $q^{\prime }ϵc$

$\&i^{\prime }=ϵ\sqrt{\frac{C}{L}}\text{also}l\frac{d{i}^{\prime }}{dt}=\frac{q_0}{C}\Rightarrow \frac{d{i}^{\prime }}{dt}=\frac{ϵ}{L}$

by solving all we get

$i^{\prime }=\frac{3}{2}ϵ\sqrt{\frac{C}{L}}sin(w^{\prime }t+\varphi )$ where $tan\varphi =\sqrt{2}$