Question

In an LC circuit the capacitor has maximum charge ${q_0}$. The value of the ${\left( {\dfrac{{di}}{{dt}}} \right)_{\max }}$ is:
A. $\dfrac{{{q_0}}}{{LC}}$
B. $\dfrac{{{q_0}}}{{\sqrt {LC} }}$
C. $\dfrac{{{q_0}}}{{LC}} - 1$
D. $\dfrac{{{q_0}}}{{LC}} + 1$

Answer 1

Hint: to solve the above question we should know the equation for the charge. The general equation for the charge is \[q = {q_0}\cos \omega t\], where ${q_0}$ is the maximum charge that will flow through the circuit and $\omega $ is frequency and $t$ is the time.

Complete answer:
Here in the question a LC circuit is given. LC circuit is an electrical circuit consisting of an inductor, represented by the letter L, and a capacitor, represented by the letter C, connected together.
so, we will prepare a typical LC circuit as shown below,

Firstly, we will write the equation of the charge as follows,
 \[q = {q_0}\cos \omega t\]
The differentiation of the charge will give us the current,
So current, $i = \dfrac{{dq}}{{dt}} = \dfrac{d}{{dt}}\left( {{q_0}\cos \omega t} \right)$
$ \Rightarrow i = {q_0}\dfrac{d}{{dt}}\left( {\cos \omega t} \right)$
$ \Rightarrow i = {q_0}\sin \omega t \times \omega $
$ \Rightarrow i = {q_0}\omega \sin \omega t$------equation (1)
Since we have got the equation for the current, so now we can calculate the $\left( {\dfrac{{di}}{{dt}}} \right)$ by differentiating the above equation as follows,
$\left( {\dfrac{{di}}{{dt}}} \right) = \dfrac{d}{{dt}}\left( {{q_0}\omega \sin \omega t} \right)$
$ \Rightarrow \left( {\dfrac{{di}}{{dt}}} \right) = {q_0}\omega \dfrac{d}{{dt}}\left( {\sin \omega t} \right)$
$ \Rightarrow \left( {\dfrac{{di}}{{dt}}} \right) = {q_0}\omega \cos \omega t \times \omega $
$ \Rightarrow \left( {\dfrac{{di}}{{dt}}} \right) = {q_0}{\omega ^2}\cos \omega t$
Now for this value to be maximum we should take the maximum value of the $\cos \omega t$.
So, the maximum value of the $\cos \omega t$ is 1. So, putting this value in the above equation we get the maximum value of the $\left( {\dfrac{{di}}{{dt}}} \right)$,
So, ${\left( {\dfrac{{di}}{{dt}}} \right)_{\max }} = {q_0}{\omega ^2} \times 1$
$ \Rightarrow {\left( {\dfrac{{di}}{{dt}}} \right)_{\max }} = {q_0}{\omega ^2}$------equation (2)
Now we also know that the value of the $\omega $is $\dfrac{1}{{\sqrt {LC} }}$, i.e. $\omega = \dfrac{1}{{\sqrt {LC} }}$.
So, putting this value of the $\omega $in the equation (2), we get
$ \Rightarrow {\left( {\dfrac{{di}}{{dt}}} \right)_{\max }} = {q_0}{\left( {\dfrac{1}{{\sqrt {LC} }}} \right)^2}$
$ \Rightarrow {\left( {\dfrac{{di}}{{dt}}} \right)_{\max }} = \dfrac{{{q_0}}}{{LC}}$
Hence, the value of the ${\left( {\dfrac{{di}}{{dt}}} \right)_{\max }}$ is $\dfrac{{{q_0}}}{{LC}}$.

So, the correct answer is “Option A”.

Note:
The LC circuit can be used as an electrical resonator and an analogue of a tuning fork and also used for storing the energy oscillating at the circuit’s resonant frequency. This is the reason the LC circuit is sometimes called a resonant circuit, tank circuit, or tuned circuit. A LC circuit is assuming to be loss free and assumes that there is no loss of the energy due to the resistance. But in practical it is not possible that the resistance is zero. The LC circuit is usually oscillating with the minimal damping, so the resistance is made as low as possible.