A) 1 : 1
B) 1 : 2
C) 2 : 1
D) 1 : 5
Correct Answer: D
Solution :
Current will be maximum in the condition of resonance. So \[{{i}_{rms}}=\frac{v}{R}=\frac{v}{10}A\] Energy stored in the coil \[{{W}_{L}}=\frac{1}{2}Li_{\max }^{2}=\frac{1}{2}L{{\left( \frac{E}{10} \right)}^{2}}\] \[=\frac{1}{2}\times {{10}^{-3}}\left( \frac{{{E}^{2}}}{100} \right)=\frac{1}{2}\times {{10}^{-5}}{{E}^{2}}\]Joule \[\therefore \]energy stored in the capacitor \[{{W}_{c}}=\frac{1}{2}C{{E}^{2}}=\frac{1}{2}\times 2\times {{10}^{-6}}{{E}^{2}}\]Joule \[\therefore \] \[\frac{{{W}_{c}}}{{{W}_{L}}}=\frac{1}{5}.\] Hence, the correction option is [d].
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