The instantaneous values of alternating curre | vedclass

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(B) Comparing the given equations with the standard forms $i = I_0 \sin(\omega t)$ and $e = E_0 \sin(\omega t + \phi)$,we get:$I_0 = \frac{1}{\sqrt{2}

Vedclass · Accepted Answer

$\frac{1}{8}$

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(B) Comparing the given equations with the standard forms $i = I_0 \sin(\omega t)$ and $e = E_0 \sin(\omega t + \phi)$,we get:$I_0 = \frac{1}{\sqrt{2}} 	ext{ A}$$E_0 = \frac{1}{\sqrt{2}} 	ext{ V}$$\phi = \frac{\pi}{3}$The $RMS$ values are calculated as:$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}$The power factor is $\cos \phi = \cos(\frac{\pi}{3}) = \frac{1}{2}$.The average power consumed is given by $P_{av} = E_{rms} I_{rms} \cos \phi$.$P_{av} = (\frac{1}{2}) 	imes (\frac{1}{2}) 	imes (\frac{1}{2}) = \frac{1}{8} 	ext{ W}$.

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

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