An electric bulb is rated $60 \mathrm{~W}, 220 \mathrm{~V}$. The resistance of its filament is\

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Professor PC Thomas Classes · Answer

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