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.

Currents	$r.m.s.$ values
$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2\sqrt{2}}$

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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