# Q7.12  An $LC$ circuit contains a $20mH$ inductor and a $50\mu F$ capacitor withan initial charge of $10mC$. The resistance of the circuit is negligible. Let the instant the circuit is closed be $t=0$.            ( c)     At what time is the energy stored             (i) completely electrical (i.e., stored in the capacitor)?

Answers (1)

at any instant, the charge on the capacitor is:

$Q=Q_0cos(w_{natural}t)=Q_0cos(2\pi f_{natural}t)=Q_0cos\left ( \frac{2\pi t}{T} \right )$

Where time period :

$T=\frac{1}{f_{natural}}=\frac{1}{159}=6.28ms$

Now, when the total energy is purely electrical, we can say that

$Q=Q_0$

$Q_0=Q_0cos(\frac{2\pi}{T})$

$cos(\frac{2\pi t}{T})=1$

this is possible when

$t=0,\frac{T}{2},T,\frac{3T}{2}....$

Hence Total energy will be purely electrical(stored in a capacitor) at

$t=0,\frac{T}{2},T,\frac{3T}{2}....$.

## Related Chapters

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
##### Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions