Question

The voltage of an ac source varies with time according to the equation V = 100 sin(100πt) cos(100πt) 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 $100/\sqrt{2}volts$
• D. The frequency of the source is 50 Hz

Answer 1

Answer: (B)

The peak voltage of the source is 50 volts

Answer 2

The given equation can be written as follows

$V = 50 \times 2\sin 100\pi t\cos 100\pi t = 50\sin 2(100\pi t) = 50\sin 200\pi t$

( sin 2θ = 2 sinθ cosθ )

Hence peak voltage $V_{0} = 50volt$ and frequency

$\nu = \frac{200\pi}{2\pi} = 100Hz.$