The voltage of an `AC` source varies with time according to the equation,`V=100 sin 100 pi I cos 100 pi t`.Where `t` is in second and `V` is in volt.Then:

Options

A. the peak voltage of the source is `100` volt
B. the peak voltage of the source is `(100//2)` volt
C. the peak voltage of the source is `50` volt
D. the peak voltage of the source is `50 Hz`

Answer

C

Explanation

`V=100 sin 100pi t cos 100 pi t`
`V=50 sin 200 pit`
here `V_(0)=50` &`omega=200 pi , f=100 Hz`