For an AC circuit the potential difference and current P are given ...
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For an AC circuit the potential difference and current
P are given by \( V=10 \sqrt{2} \sin \omega t \) (in V) and
W \( l=2 \sqrt{2} \cos \omega t \) (in A) respectively. The power dissipated in the instrument is \( \)
(1) \( 20 \mathrm{~W} \)
(2) \( 40 \mathrm{~W} \)
(3) \( 40 \sqrt{2} \mathrm{~W} \)
(4) Zero
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