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 \[{{i}_{\max }}=\frac{V}{R}=\frac{V}{10}A\] 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 \[{{W}_{C}}=\frac{1}{2}C{{E}^{2}}=\frac{1}{2}\times 2\times {{10}^{-6}}{{E}^{2}}={{10}^{-6}}{{E}^{2}}\ joule\] \[\therefore \frac{{{W}_{C}}}{{{W}_{L}}}=\frac{1}{5}\]
You need to login to perform this action.
You will be redirected in
3 sec
