JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 9th April Morning Slot
Question
A moving coil galvanometer has resistance 50
and it indicates full deflection at 4mA current.
A voltmeter is made using this galvanometer
and a 5 k resistance. The maximum voltage,
that can be measured using this voltmeter, will
be close to :
Choose an option
Show full solutionCorrect option: D
Correct answer
D20 V
Step-by-step explanation
G = 50
S = 5000
Ig = 4 × 10–3
V = ig (G + S)
V = 4 × 10–3 (50 + 5000)
= 4 × 10–3 (5050) = 20.2 volt
S = 5000
Ig = 4 × 10–3
V = ig (G + S)
V = 4 × 10–3 (50 + 5000)
= 4 × 10–3 (5050) = 20.2 volt
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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.