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