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Question
A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.
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Solution
Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, Erms = 220 V
P = V2R,
where R = resistance of the bulb
`therefore R = v^2/P = (220xx220)/60`
=806.67
Peak value of voltage (E_0) is given by,
`i_0 = E_0/R`
`⇒ i_0 = (311.08)/806.67 = 0.39 A `
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