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Hrishant Goswami Grade: 10
        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.
3 years ago

Answers : (1)

Deepak Patra
askIITians Faculty
474 Points
										Sol. P = 60 W V = 220 V = E

R = v^2/P = 220 * 220/60 = 806.67
ε base 0 = √2 E = 1.414 * 220 = 311.08
I base 0 = ε base 0/R = 806.67/311.08 = 0.385 = 0.39 A
3 years ago
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