Question

An electric bulb of 100 W is designed to operate on 220 V. Resistance of the filament is

• A. 484 Ω
• B. 100 Ω
• C. 22000 Ω
• D. 242 Ω

Answer 1

Answer: (A)

484 Ω

Answer 2

By using $P = \frac{V^{2}}{R} \Rightarrow R = \frac{V^{2}}{P} = \frac{(220)^{2}}{100} = 484\Omega$SSS