Question

The alternating current in a circuit is given by $I=50\,\sin \,314\,t$ . The peak value and frequency of the current are

• A. $I_{0}=25\,A$ and $f=100\,Hz$
• B. $I_{0}=50\,Hz$ and $f=50\,Hz$
• C. $I_{0}=50\,A$ and $f=100\,Hz$
• D. $I_{0}=25\,A$ and $f=50\,Hz$

Answer 1

Answer: (B)

$I_{0}=50\,Hz$ and $f=50\,Hz$

Answer 2

From standard equation, we have
$I=I_{0} \sin \omega t$..(i)
Given, $I=50\, \sin 314\, t$ ...(ii)
Comparing Eqs. (i) and (ii), we get
$I_{0}=50\, A,\, \omega=2 \pi\, f=314$
$\Rightarrow f=\frac{314}{2 \times 3.14}=50\, Hz$