Q. The voltage of an ac source varies with time according to the equation $V = 100\sin{100\pi t}\cos{100\pi t}$ where $t$ is in second 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 $50\ Hz$
Explanation
(b) $V = 50 \times 2\sin{100\pi t}\cos{100\pi t} = 50\sin{200\pi t}$
$\Rightarrow V_{0} = 50\ volts$ and $v = 100Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.