A) 0.17 A
B) 0.15 A
C) 0.34 A
D) 0.25 A
Correct Answer: A
Solution :
LC circuit \[\Rightarrow {{V}_{C}}+{{V}_{L}}=0\] and \[{{I}_{C}}-{{I}_{L}}\] \[{{V}_{2}}=\frac{{{L}_{d}}{{I}_{L}}}{dt}..............(1)\] \[\Rightarrow q=C{{V}_{C}}\] \[\Rightarrow \frac{{{d}_{v}}}{{{d}_{t}}}=\frac{d{{V}_{C}}}{dt}\] \[\Rightarrow {{I}_{C}}=\frac{{{C}_{d}}{{V}_{c}}}{dt}.............(2)\] Now, \[{{I}_{c}}={{I}_{L}}=I\] So, Salving (1) and (2), \[\Rightarrow \frac{{{d}^{2}}I}{d{{t}^{2}}}+w_{0}^{2}I=O\] \[\Rightarrow I={{I}_{0}}\sin ({{w}_{o}}t)\]where \[{{w}_{o}}=\frac{1}{\sqrt{LC}}\] \[\Rightarrow {{V}_{L}}=L\frac{dI}{dt}=L{{I}_{o}}{{w}_{o}}{{w}_{s}}({{w}_{o}}t)\] \[Att=0,\,\,{{V}_{L}}=10V.\] Hence\[L{{I}_{o}}{{w}_{o}}=10V\] At some time, where \[{{V}_{L}}=5v\] \[t\] is such that \[\cos ({{w}_{o}}t)=\frac{1}{2}\] \[\Rightarrow {{\cos }^{2}}(wot)+{{\sin }^{2}}(wot)=1\] \[\Rightarrow \sin (wot)=\sqrt{1-{{Y}_{4}}}=\frac{\sqrt{3}}{2}\] Now, \[L{{I}_{o}}{{w}_{o}}=10\] \[\Rightarrow {{I}_{o}}=\frac{10}{{{w}_{o}}L}=10\sqrt{\frac{C}{L}}=0.2\] So at that time \[I={{I}_{o}}\frac{\sqrt{3}}{2}\] \[0.2\times \frac{\sqrt{3}}{2}=0.17\]
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