Question

The voltage of an ac source varies with time according 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 $50Hz$

Answer 1

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

Given,

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

$\Rightarrow V_0=50$ Volts and

$\nu =100Hz$