A) \[10\,\cos \,(500\,t)\]
B) \[-10\,\cos \,(500\,t)\]
C) \[10\,\sin \,(500\,t)\]
D) \[-10\,\sin \,(500\,t)\]
Correct Answer: B
Solution :
[b] In a pure inductive circuit current always lags behind the emf by\[\frac{\pi }{2}\]. If \[v(t)={{v}_{0}}\,\sin \,\omega t\] then \[I={{I}_{0}}\,\sin \,\left( \omega t-\frac{\pi }{2} \right)\] Now, given v(t) = 100 sin (500 t) and \[{{I}_{0}}=\frac{{{E}_{0}}}{\omega L}=\frac{100}{500\times 0.02}\] \[[\because \,L=0.02\,H]\] \[{{I}_{0}}=10\,\sin \,\left( 500t-\frac{\pi }{2} \right)=-10\,\cos \,(500\,t)\]
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