JEE Main 2017 — Current Electricity Question with Solution
From: JEE Main 2017 (Offline)
Question
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15, it shows full
scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a
voltmeter of range 0 – 10V is:
Choose an option
Show full solutionCorrect option: B
Correct answer
B1.985 × 103
Step-by-step explanation
Given : Current through the galvanometer,
ig = 5 × 10–3 A
Galvanometer resistance, G = 15
Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.
V = ig (R + G)
10 = 5 × 10–3 (R + 15)
R = 2000 – 15 = 1985
= 1.985 × 103
ig = 5 × 10–3 A
Galvanometer resistance, G = 15
Let resistance R to be put in series with the galvanometer to convert it into a voltmeter.
V = ig (R + G)
10 = 5 × 10–3 (R + 15)
R = 2000 – 15 = 1985
= 1.985 × 103
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This is a previous-year question from JEE Main 2017, 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.