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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)=(220xx220)/(200)=(22xx22)/(2) =242 Omega`
(ii) The rms value of current `= (P)/(V) = (200)/(220)=(10)/(11)A=0.9A`
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