The equation of alternating current is : |
\[I=50\sqrt{2}\,\sin \,400\pi t\,amp\]. Then the frequency and root mean square of current are respectively |
A) \[200\text{ }Hz,\text{ }50\text{ }amp\]
B) \[400\pi \,Hz,\,\,50\sqrt{2}\,\,\,amp\]
C) \[200\,Hz,\,\,\,50\sqrt{2}\,\,amp\]
D) \[50\text{ }Hz,\text{ }200\text{ }amp\]
Correct Answer: A
Solution :
[a] \[2\pi nt=400\pi t\] \[\therefore \,\,\,\,n=200\] \[{{I}_{0}}=50\sqrt{2}\,amp\] r.m.s. current\[={{I}_{0}}/\sqrt{2}=50\,amp\]You need to login to perform this action.
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