If the rms current in a 50 text{ Hz} AC circu | vedclass

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(B) Given: $I_{rms} = 5 	ext{ A}$,frequency $f = 50 	ext{ Hz}$.At $t=0$,$I=0$,so the equation for current is $I = I_0 \sin(\omega t)$.First,calculat

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$5 \sqrt{\frac{3}{2}}$

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(B) Given: $I_{rms} = 5 	ext{ A}$,frequency $f = 50 	ext{ Hz}$.At $t=0$,$I=0$,so the equation for current is $I = I_0 \sin(\omega t)$.First,calculate the peak current $I_0 = \sqrt{2} 	imes I_{rms} = 5\sqrt{2} 	ext{ A}$.Next,calculate the angular frequency $\omega = 2\pi f = 2\pi 	imes 50 = 100\pi 	ext{ rad/s}$.Now,substitute $t = \frac{1}{300} 	ext{ s}$ into the equation:$I = 5\sqrt{2} \sin(100\pi 	imes \frac{1}{300})$$I = 5\sqrt{2} \sin(\frac{\pi}{3})$Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$,we get:$I = 5\sqrt{2} 	imes \frac{\sqrt{3}}{2} = 5 \sqrt{\frac{2 	imes 3}{4}} = 5 \sqrt{\frac{3}{2}} 	ext{ A}$.

If the rms current in a $50 \text{ Hz}$ $AC$ circuit is $5 \text{ A}$,and at time $t=0$,the current $I$ is $0$. The value of current $I$ at $t=\frac{1}{300} \text{ s}$ is . . . . . . $A$.

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