An electric bulb is marked '60 W, 220 V'. The resistance of its filament is approximately:
A440 Ω
B807 Ω
C1100 Ω
D13.2 Ω
Answer & Solution
Correct answer: B. 807 Ω
1. From P = V²/R, we get R = V²/P.
2. R = 220² / 60 = 48400 / 60 ≈ 807 Ω.
3. (Actual hot-filament value; cold filament resistance is much lower.)
_Source: NCERT Class 10 Science Ch 11 §11.6.1 Electric Power_
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