In an AC circuit E = 50 sin(500t) V, I = 600 | vedclass

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(C) The given equations are $E = 50 \sin(500t)$ and $I = 600 \sin(500t + \frac{\pi}{3}) 	ext{ mA}$.Peak voltage $E_0 = 50 	ext{ V}$.Peak current $I_

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$5$

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(C) The given equations are $E = 50 \sin(500t)$ and $I = 600 \sin(500t + \frac{\pi}{3}) 	ext{ mA}$.Peak voltage $E_0 = 50 	ext{ V}$.Peak current $I_0 = 600 	ext{ mA} = 0.6 	ext{ A}$.Phase difference $\phi = \frac{\pi}{3} = 60^{\circ}$.The average power dissipated in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$.$E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.$P = \frac{E_0}{\sqrt{2}} 	imes \frac{I_0}{\sqrt{2}} 	imes \cos \phi = \frac{E_0 I_0}{2} \cos \phi$.Substituting the values: $P = \frac{50 	imes 0.6}{2} 	imes \cos 60^{\circ}$.$P = \frac{30}{2} 	imes 0.5 = 15 	imes 0.5 = 7.5 	ext{ W}$.Wait, checking the calculation: $50 	imes 0.6 = 30$. $30 / 2 = 15$. $15 	imes 0.5 = 7.5 	ext{ W}$.Re-evaluating the options provided, there might be a typo in the question's options. Given the standard format, if $I = 600 	ext{ mA} = 0.6 	ext{ A}$, the result is $7.5 	ext{ W}$. If $I = 400 	ext{ mA}$, it would be $5 	ext{ W}$. Assuming the closest logical answer based on standard textbook problems of this type, $7.5 	ext{ W}$ is the calculated value.

In an $AC$ circuit $E = 50 \sin(500t)$ $V$, $I = 600 \sin(500t + \frac{\pi}{3})$ mA. What is the power dissipated in the circuit (in $\text{ W}$)? [$\cos 60^{\circ} = 0.5$]

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