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  4. An electric bulb is rated 60 W, 220 V. The resistance of its filament is

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Q. An electric bulb is rated $60 \,W, 220 \,V.$ The resistance of its filament is

AIPMTAIPMT 1994Current Electricity

Solution:

Power $(P) = 60 \,W$ and voltage $(V) = 220$ volts.
Resistance of the filament, $ R=\frac{V^2}{P}=\frac{(220)^2}{60}= 807\, \Omega.$