alleryAn 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: syllabus Science Ch 11 §11.6.1 Electric Power_
Keep practising Effects of Electric Current
Browse all Effects of Electric Current questions for CBSE Class 10, or explore more Science topics.
Related questions
Match each household safety component with its duty:Match each domestic alternating current feature with its Indian standard value:Match each electromagnetic vector rule with the quantity it determines:Match each electromagnetic device/concept with its inventor or principle:Match each thermal and power relationship with its formula:Match each resistor combination rule with its formula:Match each factor affecting resistance of a cylindrical conductor:Match each electrical quantity with its defining formula and SI unit: