JEE Mains · Physics · STD 12 - 3. current electricity
An electric bulb rated \(50 \mathrm{~W}-200 \mathrm{~V}\) is connected across a \(100 \mathrm{~V}\) supply. The power dissipation of the bulb is _______.
- A \(12.5 \mathrm{~W}\)
- B \(25 \mathrm{~W}\)
- C \(50 \mathrm{~W}\)
- D \(100 \mathrm{~W}\)
Answer & Solution
Correct Answer
(A) \(12.5 \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
Rated power \(\&\) voltage gives resistance \(\mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50}\) \(\mathrm{R}=800\) \(\mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800}\) \(\mathrm{P}=12.5 \text { watt }\)…
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