A capacitor with capacitance$6.00\times {10}^{-5}F$ is charged by connecting it to a 12.0V battery. The capacitor is disconnected from the battery and connected across an inductor with L= 1.50 H. At the time 0.0230 s, how much electrical energy is stored in the capacitor and how much is stored in the inductor?

The charge on the capacitor at any instant is given by

$q=Q\cos \left(\omega t\right)$

Where, q is the charge on the capacitor at any time t , is the maximum charge over capacitor ,$\omega$is the angular frequency .

The current at any instant in LC circuit

The current at any instant is given by

$i=-\omega Q\sin \left(\omega t\right)$

Where, i is the current in the circuit at time ,Q is the maximum charge over capacitor and$\omega$ is the angular frequency.

The energy stored in the inductor

It is the energy stored in magnetic field of the inductor .The source of this energy is external EMF that supplies current in the circuit.

The energy stored in the inductor is given by

$U=\frac{1}{2}L{i}^2$

Where, U is the energy stored in the inductor.

The energy stored in the capacitor

It is a type of potential energy which stored in capacitor in form of electric field.

The energy stored in the capacitor is given by

$U=\frac{q^2}{2C}$

Where U is the energy stored in the capacitor.

0.0230s

The charge on the capacitor at time t=0.0230s

Using

$q=Q\cos \left(\omega t\right)$

Now put the values of constants in above equation

$\begin{array}{r}\left.q=7.20\times {10}^{-4}\mathrm{Ccos}\left(\right(105.4\mathrm{rad}/\mathrm{s}\left)0.023\mathrm{s}\right)\right)\\ q=-5.42\times {10}^{-4}\mathrm{C}\end{array}$

Thus, the charge of the capacitor at time t=0.0230s

The current in the circuit at time t=0.0230s

Using

$\mathrm{i}=-\omega Q\sin \left(\omega t\right)$

Now put value of constants in above equation

$\begin{array}{r}\mathrm{i}=-\left((105\mathrm{rad}/\mathrm{s})\left(7.20\times {10}^{-4}\mathrm{C}\right)\sin (105\mathrm{rad}/\mathrm{s}\times 0.0230\mathrm{s})\right)\\ \mathrm{i}=-0.050\mathrm{A}\end{array}$

Thus,the current in the circuit at time t = 0.0230s.

Using

${\cup }_c=\frac{q^2}{2C}$

Now put the values constants in above equation

$\begin{array}{r}{U}_C=\frac{{\left(-5.42\times {10}^{-4}\mathrm{C}\right)}^2}{2\left(2.45\times {10}^{-3}\mathrm{F}\right)}\\ U_C=2.45\times {10}^{-3}\mathrm{J}\end{array}$

Thus, the electrical energy is stored in the capacitor at time 0.0230s is $2.45\times {10}^{-3}\mathrm{J}$

The calculation of electrical energy stored in the inductor

Using

${\mathrm{U}}_L=\frac{1}{2}L{\mathrm{i}}^2$

Now put the values of constants in above equation

$\begin{array}{r}{U}_L=\frac{1}{2}\left(1.50\mathrm{H}\right)(0.050\mathrm{A}{)}^2\\ U_L=1.87\times {10}^{-3}\mathrm{J}\end{array}$

Thus, the electrical energy is stored in the inductor at time 0.0230s is $1.87\times {10}^{-3}\mathrm{J}$.