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(D) The given equations are $V = 10 \sqrt{2} \sin \omega t$ and $I = 2 \sqrt{2} \cos \omega t$.We can rewrite the current equation as $I = 2 \sqrt{2}

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(D) The given equations are $V = 10 \sqrt{2} \sin \omega t$ and $I = 2 \sqrt{2} \cos \omega t$.We can rewrite the current equation as $I = 2 \sqrt{2} \sin(\omega t + \frac{\pi}{2})$.Comparing these with the standard forms $V = V_m \sin \omega t$ and $I = I_m \sin(\omega t + \phi)$,we find the phase difference $\phi = \frac{\pi}{2}$.The power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.Since $\phi = \frac{\pi}{2}$,the power factor $\cos \phi = \cos \frac{\pi}{2} = 0$.Therefore,$P = V_{rms} I_{rms} 	imes 0 = 0 \ W$.

For an $AC$ circuit,the potential difference and current are given by $V = 10 \sqrt{2} \sin \omega t$ (in $V$) and $I = 2 \sqrt{2} \cos \omega t$ (in $A$) respectively. The power dissipated in the instrument is ............ $W$.

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