JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 10th April Morning Slot
Question
In the given circuit, an ideal voltmeter
connected across the 10 resistance reads 2V.
The internal resistance r, of each cell is:
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: C
Correct answer
C0.5
Step-by-step explanation
For the given circuit
Given that VAB = 2 V
I =
Also I(2r + 2) = 1.5 + 1.5 – VAB
2r + 2 = (3–2)3
Given that VAB = 2 VI =
Also I(2r + 2) = 1.5 + 1.5 – VAB
2r + 2 = (3–2)3
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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.