An alternating current is given by I=100 sin | vedclass

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(A) The given alternating current equation is $I = I_0 \sin(\omega t)$.Comparing this with the given equation $I = 100 \sin(50 \pi t)$, we get angular

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$50$

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(A) The given alternating current equation is $I = I_0 \sin(\omega t)$.Comparing this with the given equation $I = 100 \sin(50 \pi t)$, we get angular frequency $\omega = 50 \pi 	ext{ rad/s}$.We know that $\omega = 2 \pi f$, where $f$ is the frequency.So, $2 \pi f = 50 \pi$, which gives $f = 25 	ext{ Hz}$.This means the current completes $25$ cycles in one second.In one complete cycle of a sine wave, the current becomes zero twice (at the start/end and at the half-cycle point).Therefore, in $25$ cycles, the current will become zero $25 	imes 2 = 50$ times in one second.

An alternating current is given by $I=100 \sin (50 \pi t)$. How many times will the current become zero in one second (in $\times$)?

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