Question

An inductor of inductance $2.0mH$ is connected across a charged capacitor of capacitance $5.0\mu F$ and the resulting $L-C$ circuit is set oscillating at its natural frequency. Let $Q$ denote the instantaneous charge on the capacitor and $I$ the current in the circuit. It is found that the maximum value of $Q$ is $200\mu C$. When $Q=100\mu C$, what is the value of $\left|\frac{dI}{dt}\right|?$

• A. $10000$
• B. $1000$
• C. $100000$
• D. $100$

Answer 1

Answer: (A)

This is a problem of $L-C$ oscillations.
Charge stored in the capacitor oscillates simple harmonically as
$Q=Q_0\sin (\omega t\pm \varphi )$
Here, $Q_0=$ maximum value of $Q=200\mu C=2\times 10^{-4}C$
$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{\left(2\times 10^{-3}\right)\left(5.0\times 10^{-6}\right)}}=10^4{s}^{-1}$
At $t=0,Q=Q_0$ then
$Q(t)=Q_0\cos \omega t$
$I(t)=\frac{dQ}{dt}=-Q_0\omega \sin \omega t$ and
$\frac{dI(t)}{dt}=-Q_0{\omega }^2\cos (\omega t)\dots \text{ (iii) }$
$Q=100\mu C\text{ or }\frac{Q_0}{2}$
$\cos \omega t=\frac{1}{2}\text{ or }\omega t=\frac{\pi }{3}$
$\left|\frac{dI}{dt}\right|=\left(2.0\times 10^{-4}C\right){\left(10^4{s}^{-1}\right)}^2\left(\frac{1}{2}\right)$
$\left|\frac{dI}{dt}\right|=10000A/s$