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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 through the filament.

Hrishant Goswami , 11 Years ago
Grade 10
anser 1 Answers
Deepak Patra
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
Last Activity: 11 Years ago
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