Q124 / 180 · KCET 2021 · Physics
In an oscillating $L C$-circuit, $L=3 \mathrm{mH}$ and $C=2.7 \mu \mathrm{F}$. At $t=0$, the charge on the capacitor is zero and the current is $2 \mathrm{~A}$. The maximum charge that will appear on the capacitor will be
- A$1.8 \times 10^{-5} \mathrm{C}$
- Correct answerB$18 \times 10^{-5} \mathrm{C}$
- C$9 \times 10^{-5} \mathrm{C}$
- D$90 \times 10^{-5} \mathrm{C}$
Solution
The charge is a function of time $(t)$, is given by
$q=q_{0} \sin \omega t...(i)$
where, $q_{0}=$ maximum charge,
and $\omega=$ angular frequency.
Differentiating Eq. (i), we get
$I=\frac{d q}{d t}=\frac{d}{d t}\left(q_{0} \sin \omega t\right)=\omega q_{0} \cos \omega t$
At, $\quad t=0$,
$\begin{aligned} I &=\omega q_{0} \cos (0)=\omega q_{0} \\ \therefore \quad I &=\frac{1}{\sqrt{L C}} q_{0} \text { or } q_{0}=I \sqrt{L C} \quad\left[\because \omega=\frac{1}{\sqrt{L C}}\right] \\ &=2 \sqrt{3 \times 2.7 \times 10^{-6}} \\ &=18 \times 10^{-5} \mathrm{C} \end{aligned}$
Hence, the correct answer is option: 'B'.