A) \[{{\left( \frac{C(V_{1}^{2}-V_{2}^{2})}{L} \right)}^{1/2}}\]
B) \[{{\left( \frac{C{{({{V}_{1}}-{{V}_{2}})}^{2}}}{L} \right)}^{1/2}}\]
C) \[\frac{C(V_{1}^{2}-V_{2}^{2})}{L}\]
D) \[\frac{C({{V}_{1}}-{{V}_{2}})}{L}\]
Correct Answer: A
Solution :
[a] \[q=C{{V}_{1}}\,\cos \,\omega t\Rightarrow \,i=\frac{dq}{dt}=-\omega C{{v}_{1}}\,\sin \,\omega t\] Also, \[{{\omega }^{2}}=\frac{1}{LC}\] and \[V={{V}_{1}}\,\cos \,\omega t\] At \[t={{t}_{1}}\], \[V={{V}_{2}}\] and \[i=-\omega C{{V}_{1}}\,\sin \,\omega {{t}_{1}}\] \[\therefore \,\,\,\cos \,\,\omega {{t}_{1}}=\frac{{{V}_{2}}}{{{V}_{1}}}\] (-ve sign gives direction) Hence, \[i={{V}_{1}}\sqrt{\frac{C}{L}}{{\left( 1-\frac{V_{2}^{2}}{V_{1}^{2}} \right)}^{1/2}}={{\left( \frac{C(V_{1}^{2}-V_{2}^{2}}{L} \right)}^{1/2}}\]
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