A) The peak voltage of the source is 100 volts
B) The peak voltage of the source is 50 volts
C) The peak voltage of the source is \[100/\sqrt{2}\]volts
D) The frequency of the source is 50 Hz
Correct Answer: B
Solution :
\[V=50\times 2\,\sin \,100\pi t\,\cos \,100\pi t=50\sin 200\pi t\] \[\Rightarrow \,\,{{V}_{0}}=50\,Volts\] and \[\nu =100Hz\]You need to login to perform this action.
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