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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

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प्रश्न

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.

बेरीज
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उत्तर

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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पाठ 39: Alternating Current - Exercises [पृष्ठ ३३०]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 39 Alternating Current
Exercises | Q 3 | पृष्ठ ३३०

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