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The voltage of an AC source varies with time according to the equation `V=100sin 100pitcos100pit` where `t` is in seconds and `V` is in volts. ThenA. The peak voltage of the source is `100` voltsB. The peak voltage of the source is `50 `voltsC. The peak voltage of the source is `100sqrt(2)` voltsD. The frequency of the source is `50Hz` |
Answer» Correct Answer - B `V=50xx2 sin 100pit cos 100pit=50sin 200pit` `implies V_(0)=50` volts and `v=10Hz` |
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