A) \[2\pi v{{q}_{0}}\,\sin \,2\pi vt\]
B) \[3/5\pi v{{q}_{0}}\,\sin \,\pi vt\]
C) \[4\pi v{{q}_{0}}\,\sin \,2\pi vt\]
D) None of these
Correct Answer: A
Solution :
[a] As we know that, As the displacement current through the capacitor is, \[{{I}_{d}}={{I}_{c}}=\frac{dq}{dt}\] ? (i) As given that, \[q={{q}_{0}}\,\cos \,2\pi vt\] Putting this value of q in Eq (i), we get So, \[{{I}_{d}}={{I}_{c}}=\frac{d}{dt}({{q}_{0}}{{\cos }^{2}}\pi vt)\] \[{{I}_{d}}={{I}_{c}}=-{{q}_{0}}\sin 2\pi vt\times 2\pi v\] \[{{I}_{d}}={{I}_{c}}=2\pi v{{q}_{0}}\sin 2\pi vt\]
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