Question

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

• A. $870\Omega$
• B. $780\Omega$
• C. $708\Omega$
• D. $807\Omega$

Answer 1

Answer: (D)

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