The current flowing through an AC circuit is | vedclass

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(D) The given equation for current is $I = I_0 \sin(\omega t)$,where $I_0 = 5 	ext{ A}$ and $\omega = 120 \pi 	ext{ rad/s}$.The angular frequency $\

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$\frac{1}{240} 	ext{ s}$

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(D) The given equation for current is $I = I_0 \sin(\omega t)$,where $I_0 = 5 	ext{ A}$ and $\omega = 120 \pi 	ext{ rad/s}$.The angular frequency $\omega$ is related to the time period $T$ by the formula $\omega = \frac{2 \pi}{T}$.Substituting the value of $\omega$: $120 \pi = \frac{2 \pi}{T} \Rightarrow T = \frac{2 \pi}{120 \pi} = \frac{1}{60} 	ext{ s}$.The current reaches its peak value at $t = \frac{T}{4}$ (one-quarter of the time period).Therefore,the time taken is $t = \frac{1/60}{4} = \frac{1}{240} 	ext{ s}$.

The current flowing through an $AC$ circuit is given by $I = 5 \sin(120 \pi t) \text{ A}$. How long will the current take to reach the peak value starting from zero?

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