Question

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

• A. 708 Ω
• B. 870 Ω
• C. 807 Ω
• D. 780 Ω

Answer 1

Answer: (C)

807 Ω

Answer 2

$R = \frac{V^{2}}{P} = \frac{(220)^{2}}{60} = 807\Omega$