The voltage of an ac source varies with time according to the equation V=100sin⁡100πcos⁡100πt where t is in seconds and V is in volts. Then

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The peak voltage of the source is 100 volts

The peak voltage of the source is 50 volts

The peak voltage of the source is 100/2volts

The frequency of the source is 50 Hz

answer is B.

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Detailed Solution

V=50×2sin⁡100πtcos⁡100πt=50sin⁡200πt⇒V0=50 Volts and v=100Hz

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