Question

An LC circuit contains a 40 mH inductor and a 25 μF capacitor. The resistance of the circuit is negligible. The time is measured from the instant the circuit is closed. The energy stored in the circuit is completely magnetic at times (in milliseconds)
(A). 0,314,6.28
(B). 0,1.57,4.71
(C). 1.57,4.71,7.85
(D). 1.57,3.14,4.71

Answer 1

Hint: LC circuit consists of inductor (L) and capacitor (C). Use formula of resonant frequency in LC circuit to calculate time period. There is a difference in LCR and LC circuit. Negligible resistance is considered as zero value of resistance

Complete step by step answer:

To solve this kind of question, first understand what question you want to convey.
We have a LC circuit having an inductor and capacitor connected by wire. Given that circuit resistance is negligible. The value of capacitor and inductor is given in the question. As soon as we close the circuit we have to measure a time and at that time the energy stored in the circuit is magnetic.

To calculate time, we can use frequency formula as time is reciprocal of time.
$T=\dfrac{1}{f}$
Given data: $C=25\times {{10}^{-6}}F,L=40\times {{10}^{-3}}H,R=0\Omega $
We know that, formula of frequency is given by
$\begin{align}
  & f=\dfrac{1}{2\pi \sqrt{LC}} \\
 & f=\dfrac{1}{2\pi \sqrt{40\times {{10}^{-6}}\times 25\times {{10}^{-3}}}} \\
 & f=\dfrac{{{10}^{3}}}{2\pi }Hz \\
\end{align}$
So time is given by,
$T=\dfrac{2\pi }{{{10}^{3}}}mili\sec $
Now we want to calculate energy stored in circuits that are completely magnetic at times.
We know that charge stored in capacitor is given by,
$q={{q}_{0}}\cos (\omega t)={{q}_{0}}\cos (\dfrac{2\pi }{T}t)$
And we also know that magnetic energy around L will be maximum when electrical energy in capacitance is zero.
i.e. $q=0$
So if you put the value of q=0 in the above equation then take a cos inverse of 0.
$\dfrac{T}{2\pi }{{\cos }^{-1}}(0)=t$
So the possible value of t is,
$\begin{align}
  & t=\dfrac{T}{4},\dfrac{3T}{4},\dfrac{5T}{4} \\
 & \\
\end{align}$
Put the value of t in the above equation, note that the value of t is already in millisecond.
$\begin{align}
  & t=\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2} \\
 & t=(1.57,4.71,7.85)\times {{10}^{-3}}\sec \\
\end{align}$
Now as we can see in option, option C is the correct option

Note: To calculate energy stored in completely electric at times.
So, $T=0,\dfrac{t}{2},t,\dfrac{3t}{2}$
In the question, notice that the value of time must be in millisecond. Memorize the unit of capacitor, inductor and resistance.