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JEE Main 2021 (26 Aug, Shift 2) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from JEE Main 2021 (26 Aug, Shift 2)

Q34JEE Main 2021 · 26 AUG · SHIFT 2Medium

Electrical Power & Heating Effect

An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

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Answer: (a)

  1. Bulb resistance Rb=V2P=1002500=20 ΩR_b = \frac{V^2}{P} = \frac{100^2}{500} = 20\ \Omega; rated current I=PV=5I = \frac{P}{V} = 5 A.
  2. With 200 V supply, total resistance must be 2005=40 Ω\frac{200}{5} = 40\ \Omega.
  3. R=40−20=20 ΩR = 40 - 20 = 20\ \Omega. So the answer is (a).
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