In a series circuit, $\mathrm{C}=2 \mathrm{\mu F}, \mathrm{L}=1 \mathrm{mH}\text{ and }\mathrm{R}=10 \mathrm{\Omega }\text{. }$ When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be

Current will be maximum in the condition of resonance, so ${\mathrm{i}}_{max}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{\mathrm{E}}{10} \mathrm{A}$

Energy stored in the coil, ${\mathrm{W}}_{\mathrm{L}}=\frac{1}{2}{\mathrm{Li}}_{max}^2=\frac{1}{2}\mathrm{L}{\left(\frac{\mathrm{E}}{10}\right)}^2$

$=\frac{1}{2} \times {10}^{-3}\left(\frac{{\mathrm{E}}^2}{100}\right)=\frac{1}{2} \times {10}^{-5} {\mathrm{E}}^2 \text{ joule }$

Energy stored in the capacitor,

${\mathrm{W}}_{\mathrm{C}}=\frac{1}{2} {\mathrm{CE}}^2=\frac{1}{2} \times 2 \times {10}^{-6} {\mathrm{E}}^2={10}^{-6}{\mathrm{E}}^2\text{ joule }$

$\therefore \frac{{\mathrm{W}}_{\mathrm{C}}}{{\mathrm{W}}_{\mathrm{L}}}=\frac{1}{5}$