In an AC circuit E = 200 sin(50t) text{ V} an | vedclass

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(D) Given equations are $E = 200 \sin(50t) 	ext{ V}$ and $I = 100 \sin(50t + \frac{\pi}{3}) 	ext{ mA}$.Comparing these with standard forms $E = E_0

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$5$

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(D) Given equations are $E = 200 \sin(50t) 	ext{ V}$ and $I = 100 \sin(50t + \frac{\pi}{3}) 	ext{ mA}$.Comparing these with standard forms $E = E_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$,we get:$E_0 = 200 	ext{ V}$$I_0 = 100 	ext{ mA} = 100 	imes 10^{-3} 	ext{ A} = 0.1 	ext{ A}$Phase difference $\phi = \frac{\pi}{3} = 60^{\circ}$.The average power dissipated in an $AC$ circuit is given by $P = E_{	ext{rms}} I_{	ext{rms}} \cos \phi$.We know that $E_{	ext{rms}} = \frac{E_0}{\sqrt{2}}$ and $I_{	ext{rms}} = \frac{I_0}{\sqrt{2}}$.Substituting these values:$P = \left( \frac{E_0}{\sqrt{2}} ight) \left( \frac{I_0}{\sqrt{2}} ight) \cos \phi = \frac{E_0 I_0}{2} \cos 60^{\circ}$.$P = \frac{200 	imes 0.1}{2} 	imes 0.5 = \frac{20}{2} 	imes 0.5 = 10 	imes 0.5 = 5 	ext{ W}$.

In an $AC$ circuit $E = 200 \sin(50t) \text{ V}$ and $I = 100 \sin(50t + \frac{\pi}{3}) \text{ mA}$. The power dissipated in the circuit is (Given: $\sin 30^{\circ} = \cos 60^{\circ} = 0.5$,$\sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$) (in $watt$)

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