Question

An inductor of inductance $2.0\, mH$ 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 $\left|\frac{d I}{d t}\right| ?$

• 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 \phi) $
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 ) \ldots \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|=10000 A / s $