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In an AC circuit $V$ and $I$ are given below, then find the power dissipated in the circuit
$V=50 \sin (50 t) v I=50 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$
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$V=50 \sin (50 t) v I=50 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$
Solution:
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Verified Answer
The correct answer is:
$0.625\ W$
$\mathrm{P}=\mathrm{E}_{\mathrm{rms}} \mathrm{V}_{\mathrm{rms}} \cos \theta=\frac{E_0 V_0}{2} \cos \phi$
So, $\quad P=\frac{50 \times 50 \times 10^{-3} \times 1}{2 \times 2}=0.625 \mathrm{~W}$
So, $\quad P=\frac{50 \times 50 \times 10^{-3} \times 1}{2 \times 2}=0.625 \mathrm{~W}$
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