Q1. [1]
An electric bulb is rated 220 V; 11W. The resistance of its filament when it glows with a power supply of 220 V is :
- (a) 4400 Ω
- (b) 440 Ω
- (c) 400 Ω
- (d) 20 Ω
Previously asked in CBSE board exam
2025 31/4/1 Q13
Generated by claude-sonnet-4-6 · 2026-06-15 06:45 · grounding rag
Model Answer
(a) 4400 Ω
Using $R = \dfrac{V^2}{P} = \dfrac{(220)^2}{11} = \dfrac{48400}{11} = 4400\ \Omega$
Explanation
The formula $R = V^2/P$ is derived from $P = V^2/R$. Always use rated voltage and rated power to find filament resistance. Option (a) is correct.
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