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(D) Given: $i = \frac{1}{\sqrt{2}} \sin(100\pi t) 	ext{ A}$.Comparing with $i = i_0 \sin(\omega t)$,we get peak current $i_0 = \frac{1}{\sqrt{2}} 	e

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$\frac{1}{8}$

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(D) Given: $i = \frac{1}{\sqrt{2}} \sin(100\pi t) 	ext{ A}$.Comparing with $i = i_0 \sin(\omega t)$,we get peak current $i_0 = \frac{1}{\sqrt{2}} 	ext{ A}$.Given: $e = \frac{1}{\sqrt{2}} \sin(100\pi t + \frac{\pi}{3}) 	ext{ V}$.Comparing with $e = e_0 \sin(\omega t + \phi)$,we get peak voltage $e_0 = \frac{1}{\sqrt{2}} 	ext{ V}$ and phase difference $\phi = \frac{\pi}{3}$.Calculate $RMS$ values:$i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} 	ext{ A}$.$e_{rms} = \frac{e_0}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2} 	ext{ V}$.Average power consumed in the circuit is given by $P = i_{rms} e_{rms} \cos \phi$.$P = (\frac{1}{2}) (\frac{1}{2}) \cos(\frac{\pi}{3}) = (\frac{1}{4}) (\frac{1}{2}) = \frac{1}{8} 	ext{ W}$.

The instantaneous values of alternating current and voltage in a circuit are given as $I = \frac{1}{\sqrt{2}} \sin(100\pi t)$ and $E = \frac{1}{\sqrt{2}} \sin(100\pi t + \frac{\pi}{3})$. The average power in watts consumed in the circuit is:

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