Question

A LC circuit contains a $0.60\text{H}$ inductor and a $25\mu \text{F}$ capacitor. What is the rate of change of current (in ${\text{As}}^{-1}$) at the moment, when charge on the capacitor is $30\mu \text{C}$?

• A. $2$
• B. $3$
• C. $4$
• D. $6$
  

Answer 1

The correct option is A $2$

Given, $L=0.60\text{H};C=25\mu \text{F}q=30\mu \text{C}$


Using KVL we get,

$L\frac{di}{dt}=\frac{q}{C}$

$0.60\frac{di}{dt}=\frac{30\times {10}^{-6}}{25\times {10}^{-6}}$

$\therefore \frac{di}{dt}=2{\text{As}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.