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NEET 1994 (AIPMT) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from NEET 1994 (AIPMT)

Q100NEET 1994 · AIPMTEasy

Electrical Power & Heating Effect

An electric bulb is rated 60 W, 220 V. The resistance of its filament is

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Answer: (d)

  1. R=V2P=220260R = \frac{V^2}{P} = \frac{220^2}{60}.
  2. R=4840060≈807 ΩR = \frac{48400}{60} \approx 807\ \Omega.
  3. So the answer is (d) 807 Ω807\ \Omega.
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