Question

In a series circuit $C=2\mu F,L=1mHandR=10\Omega$, 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

The correct option is D 1 : 5

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