Question

The voltage of an ac source varies with time according to the equation $V=100sin100\pi tcos100\pi t$ where t is in seconds and V is in volts. Then

• 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 $\frac{100}{\sqrt{2}}$ volts
• D. The frequency of the source is 50 Hz

Answer 1

The correct option is B The peak voltage of the source is 50 volts

$V=50\times 2sin100\pi tcos100\pi t=50sin200\pi t$

Comparing it with the standard expression for the AC voltage, $V\left(t\right)=V_{m}sin\left(2\pi ft\right)$
Therefore the peak value of the AC voltage is $V_m=50Volts$.

The frequency of the AC voltage is $100Hz$.