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JEE Main 2019 (10 Jan, Shift 2) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from JEE Main 2019 (10 Jan, Shift 2)

Q42JEE Main 2019 · 10 JAN · SHIFT 2Medium

Electrical Power & Heating Effect

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -

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Answer: (a)

  1. R=PI2=4.4(2×10−3)2=1.1×106 ΩR = \dfrac{P}{I^2} = \dfrac{4.4}{(2 \times 10^{-3})^2} = 1.1 \times 10^6\ \Omega.
  2. With 11 V: P=V2R=1211.1×106=1.1×10−4 WP = \dfrac{V^2}{R} = \dfrac{121}{1.1 \times 10^6} = 1.1 \times 10^{-4}\text{ W}.
  3. So the answer is (a) 11×10−5 W11 \times 10^{-5}\text{ W}.
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