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The alternating current in a circuit is given by `I = 50sin314t`. The peak value and frequency of the current areA. `I_(0)`=25 A and f=100HzB. `I_(0)`=50 A and f=50 HzC. `I_(0)` = 50 A and f=100 HzD. `I_(0)`=25 A and f=50 Hz |
Answer» Correct Answer - B From standard equation, we have `I= I_(0)sinomegat`...............(i) Given, I=50sin314t ..............(ii) Comparing Eqs. (i) and (ii), we get `I_(0)`= 50 A, `omega = 2pif` = 314 `rArr f=314/(2 xx 3.i14) = 50 Hz` |
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