JEE Main 2020 — Current Electricity Question with Solution
From: JEE Main 2020 (Online) 4th September Morning Slot
Question
A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C0.10 W
Step-by-step explanation
PR = 0.5 W
i2R = 0.5 W
iR = 2.5
i = 0.2 A & R = 12.5
Also, V = E – ir
2.5 = 3 – (0.2)r
r = 2.5
Power dissipated in internal resistance
= i2r = (0.2)2(2.5) = 0.1 W
i2R = 0.5 W
iR = 2.5
i = 0.2 A & R = 12.5
Also, V = E – ir
2.5 = 3 – (0.2)r
r = 2.5
Power dissipated in internal resistance
= i2r = (0.2)2(2.5) = 0.1 W
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This is a previous-year question from JEE Main 2020, 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.