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# The equation of AC voltage is $E=220\; \sin (wt+\large\frac{\pi}{6})$ and AC current is $I= 10 \sin (wt+\large\frac{\pi}{6})$ The average power dissipated is

$(a)\;150\;W \\ (b)\;550\;W \\(c)\;250\;W \\(d)\;50\;W$

$Z= \large\frac{E_0}{I_0}$
$E_0=220$
$I_0 =10$
$Z= \large\frac{220}{10}$
$\quad= 22 \;ohms$
$\phi =\bigg[ \large\frac{\pi}{6} -(- \frac{\pi}{6} ) \bigg]$
$\quad= \large\frac{\pi}{3}$
Power $= \large\frac{E_0}{\sqrt 2} \times \large\frac{I_0}{\sqrt 2} $$\cos \phi \qquad= \large\frac{220}{\sqrt 2} \times \frac{10}{\sqrt 2}$$ \ cos \large\frac{\pi}{3}$
$\qquad= 550\;W$
Hence b is the correct answer.