Question

An electric bulb of 100 watt is connected to a supply of electricity of 220 V. Resistance of the filament is

• A. $484\Omega$
• B. $100\Omega$
• C. $22000\Omega$
• D. $242\Omega$

Answer 1

Answer: (A)

$484\Omega$

Answer 2

$\frac{V^{2}}{R} = P$ ⇒ $R = \frac{V^{2}}{P} = \frac{220 \times 220}{100} = 484\Omega$