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CBSE Class 10 Science 2025 Solved Paper All Sets

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Question : 13 of 20
Marks: +1, -0
An electric bulb is rated 220 V; 11 W. The resistance of its filament when it glows with a power supply of 220 V is :
Solution:  
R=  V2P=  220×22011=4400ΩR = \; \frac{V^2}{P} = \; \frac{220 \times 220}{11} = 4400 \Omega
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