A) 1 : 2
B) 2 : 1
C) 1 : 5
D) 5 : 1
Correct Answer: C
Solution :
Current will be maximum in the condition of resonance, So, \[{{I}_{\max }}=\frac{V}{R}=\frac{V}{10}A\] So, energy stored in the coil, \[{{E}_{L}}=\frac{1}{2}L{{({{I}_{\max }})}^{2}}=\frac{1}{2}L{{\left( \frac{E}{10} \right)}^{2}}\] \[=\frac{1}{2}\times {{10}^{-3}}\left( \frac{{{E}^{2}}}{100} \right)=\frac{1}{2}\times {{10}^{-5}}{{E}^{2}}J\] Also, energy stored in the capacitor, \[{{E}_{c}}=\frac{1}{2}C{{E}^{2}}=\frac{1}{2}\times 2\times {{10}^{-6}}{{E}^{2}}J\] \[={{10}^{-6}}{{E}^{2}}J\] \[\therefore \] \[\frac{{{E}_{c}}}{{{E}_{L}}}=\frac{{{10}^{-6}}{{E}^{2}}}{1/2\times {{10}^{-5}}{{E}^{2}}}=\frac{1}{5}\]
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