Question

In a series circuit, $C=2 \,\mu F \,, L=1 \,mH$ and $R=10 \,\Omega$. 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

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 1 : 5

Answer 1

Answer: (D)

Current will be maximum in the condition of resonance,
so $i_{\max }=\frac{E}{R}=\frac{E}{10} A$
Energy stored in the coil,
$W_{L}=\frac{1}{2} L i_{\max }^{2}$
$=\frac{1}{2} L\left(\frac{E}{10}\right)^{2}$
$=\frac{1}{2} \times 10^{-3}\left(\frac{E^{2}}{100}\right)$
$=\frac{1}{2} \times 10^{-5} E^{2}$ joule
Energy stored in the capacitor,
$W_{C}=\frac{1}{2} C E^{2}$
$=\frac{1}{2} \times 2 \times 10^{-6} E^{2}$
$=10^{-6} E^{2}$ joule
$\therefore \frac{W_{C}}{W_{L}}=\frac{1}{5}$