Q13
1 markMCQSection A

An electric bulb is rated 220 V;11 W220\text{ V}; 11\text{ W}. The resistance of its filament when it glows with a power supply of 220 V220\text{ V} is :

Electricity
Electricity

Options

(a)4400 Ω4400\text{ }Ω
(b)440 Ω440\text{ }Ω
(c)400 Ω400\text{ }Ω
(d)20 Ω20\text{ }Ω
Official Answer

The resistance of the filament can be calculated using the formula R = V^2/P, where V is the voltage and P is the power. Substituting the values, we get R = (220 V)^2 / 11 W = 4400 ohms.

resistancevoltagepower

Marking Scheme

  • 11 mark for correct answer

Hint

Use the formula R = V^2/P to calculate the resistance.

Quick Oral Answer

The resistance of the filament is 4400 ohms.

Analysis & Explanation

This problem requires the application of the formula R = V^2/P to calculate the resistance of the filament. The correct answer is 4400 ohms.

Common Mistakes

  1. 1Incorrect calculation of resistance

Interesting Facts

The concept of resistance is important in electrical engineering and electronics.

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2025?

This question carries 1 mark in the CBSE Class 10 Science 2025 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Electricity" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Electricity" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2025 paper?

This is a MCQ question from Section A in the CBSE Class 10 Science 2025 paper.