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A light bulb has the rating 200W 220V. F...

A light bulb has the rating 200W 220V. Find (i) resistance of the bulb filament (ii) rms value of current flowing through the filament.

Text Solution

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(i) `R = V^2/P = (220 xx 200)/200 = (22 xx 22)/2 = 242 Omega`
(ii) The rms value of current
`=P/V = 200/220 = 10/11A =0.9 A `
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