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GUJCET · Physics · Alternating Current
A bulb of 100 W rating is connected with 220 V supply. The resistance of bulb is ____________ .
- A \(2.2 \times 10^{-3} \Omega \mathrm{~m}^{-1}\)
- B \(484 \Omega \mathrm{~m}^{-1}\)
- C \(2.2 \Omega\)
- D \(484 \Omega\)
Answer & Solution
Correct Answer
(D) \(484 \Omega\)
Step-by-step Solution
Detailed explanation
(D) \(484 \Omega\)
\(P=\frac{V^{2}}{R}\)
\(\therefore \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}\)
\(\therefore \mathrm{R}=\frac{(220)^{2}}{100}\)
\(\therefore \mathrm{R}=484 \Omega\)
\(P=\frac{V^{2}}{R}\)
\(\therefore \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}\)
\(\therefore \mathrm{R}=\frac{(220)^{2}}{100}\)
\(\therefore \mathrm{R}=484 \Omega\)
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