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JEE PHYSICS MCQ: CURRENT ELECTRICITY

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A resistor develops 400 J400 \mathrm{~J} of heat in 10 s10 \mathrm{~s} when a current of 2 A2 \mathrm{~A} is passed through it. The resistance and the energy developed in 10 s10 \mathrm{~s}. If the current is doubled is
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Correct option: C
10Ω,1600 J10 \Omega, 1600 \mathrm{~J}
Option C (10Ω,1600 J10 \Omega, 1600 \mathrm{~J}) is correct because the resistance is calculated using H=i2RtH = i^2Rt, and the new energy is calculated with the doubled current.

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