The peak value of an alternating current is 5 | vedclass

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(B) The instantaneous current is given by the equation $I = I_0 \sin(\omega t)$,where $I_0$ is the peak current and $\omega$ is the angular frequency.

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$1$/$240$ s

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(B) The instantaneous current is given by the equation $I = I_0 \sin(\omega t)$,where $I_0$ is the peak current and $\omega$ is the angular frequency.To reach the peak value,the sine function must be equal to $1$,which occurs when the phase angle $\omega t = \frac{\pi}{2}$.Substituting $\omega = 2 \pi f$,we get $2 \pi f t = \frac{\pi}{2}$.Solving for time $t$,we get $t = \frac{1}{4f}$.Given the frequency $f = 60 	ext{ Hz}$,we substitute this value into the equation:$t = \frac{1}{4 	imes 60} = \frac{1}{240} 	ext{ s}$.Therefore,the current takes $1/240 	ext{ s}$ to reach its peak value from zero.

The peak value of an alternating current is $5 \text{ A}$ and frequency is $60 \text{ Hz}$. How long will the current,starting from zero,take to reach the peak value?

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