Question

A 2.1-µF capacitor is charged to a potential difference of 16.0 V and then connected across a 0.44-mH inductor. What is the current in the circuit when the potential difference across the capacitor is 8.0 V? (Give the magnitude.)
____________ A

Answer 1

Charge on capacitor will be

$q_0=CV=2.1\times16=33.6\mu C$

When Capacitor connected with inductor, there will be charge oscillation. At t=0, charge will be maximu . now frequency of oscillation will be

$\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2.1\times10^{-6}\times0.44\times10^{-3}}}=3.29\times10^4$

Hence

$I=I_0\cos(\omega t)$

Potential across the capacitor is negative of potential across the inductor as

$V=-L\frac{dI}{dt}\Rightarrow \frac{q}{c}=-L\frac{dI}{dt}=\sqrt{\frac{L}{C}}I_0\sin(\omega t)=\sqrt{\frac{L}{C}}I_0\cos(\omega t-\frac{\pi}{2})$

Now

$Z=\sqrt{X_L^2+X_C^2}=\sqrt{L^2\omega^2+\frac{1}{C^2\omega^2}}=20.47$

$Z=\sqrt{X_L^2+X_C^2}=\sqrt{L^2\omega^2+\frac{1}{C^2\omega^2}}=20.47\\ I_0=\frac{V}{Z}=\frac{16}{20.47}=0.782$

i.e.

$V=V_0\cos(\omega t-\frac{\pi}{2})\Rightarrow \cos(\omega t-\frac{\pi}{2})=\frac{8}{16}\\ \Rightarrow \omega t-\pi/2=\pi/3\Rightarrow t=7.96\times10^{-5}$

Now

$I=I_0\cos(\omega t)=0.789\times\cos(3.29\times10^4\times 7.96\times10^{-5})=-0.6823$