JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 10th January Evening Slot
Question
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 -
Choose an option
Show full solutionCorrect option: A
Correct answer
A11 10–5 W
Step-by-step explanation
P = I2R
4.4 = 4 106 R
R = 1.1 106
P' = = 106 = 11 105 W
4.4 = 4 106 R
R = 1.1 106
P' = = 106 = 11 105 W
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This is a previous-year question from JEE Main 2019, 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.