A bulb rated 60W at 220V is connected across a household supply of alternating voltage of 220V. Calculate the maximum instantaneous current through the filament.
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$\text{P}=60\text{W},\text{V}=220\text{V}=\text{E}$
$\text{R}=\frac{\text{V}^2}{\text{P}}=\frac{220\times220}{60}$
$=806.67$
$\in_0=\sqrt{2}\text{E}=1.414\times220$
$=311.08$
$\text{I}_0=\frac{\in_0}{\text{R}}=\frac{806.67}{311.08}$
$=0.385\approx0.39\text{A}$
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