Question

The voltage of an A.C. source varies with time according to the equation $V=50\, \sin\,100\pi\,t\,\cos\,100\pi\,t$ . Where 't' is the sec and 'V' is in volts. Then

• A. The peak voltage of the source is 100 V
• B. The peak voltage of the source is $100 / \sqrt2$
• C. The peak voltage of the source is 25 V
• D. The frequency ofthe source is 50 HZ

Answer 1

Answer: (C)

The peak voltage of the source is 25 V

Answer 2

$V = 50 \times 2 \, \sin \, 100 \pi \, t \cos \, 100 \pi \, t = 50 \sin 200 \pi t$ $\Rightarrow \, \, V_0 = 50 \, Volts$ and $\nu =100 \, Hz$