JEE Main 2021 — Current Electricity Question with Solution
From: JEE Main 2021 (Online) 31st August Evening Shift
Question
A resistor dissipates 192 J of energy in 1s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s in _________ J.
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Show full solutionCorrect answer: 3840
Correct answer
3840
Step-by-step explanation
E = i2Rt
192 = 16 (R) (1)
R = 12
E1 = (8)2 (12) (5)
= 3840 J
192 = 16 (R) (1)
R = 12
E1 = (8)2 (12) (5)
= 3840 J
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This is a previous-year question from JEE Main 2021, covering the Current Electricity chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.