Question 492: AC fundamentals

Question

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

Accepted Answer

From standard equation, we have
$\begin{aligned}
& I=I_0 \sin \omega t \quad 	ext{.... (i)}\
	ext{Given,} \quad & I=50 \sin 314 t \quad 	ext{.... (ii)}
\end{aligned}$
Comparing Eqs. (i) and (ii), we get
$\begin{aligned}
& I_0=50 \mathrm{~A}, \omega=2 \pi f=314 \
\Rightarrow \quad & f=\frac{314}{2 	imes 3.14}=50 \mathrm{~Hz}
\end{aligned}$

Answer

$I_0=25 \mathrm{~A}$ and $f=100 \mathrm{~Hz}$

Answer

$I_0=50 \mathrm{~A}$ and $f=50 \mathrm{~Hz}$

Answer

$I_0=50 \mathrm{~A}$ and $f=100 \mathrm{~Hz}$

Answer

$I_0=25 \mathrm{~A}$ and $f=50 \mathrm{~Hz}$

AC fundamentals

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

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