Current Electricity

JEE Main level, single correct, medium

359 of 380

Question 359 of 380

A galvanometer, whose resistance is 50ohm50ohm , has 2525 divisions in it. When a current of 4×10−44\times {10}^{-4} A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5V2.5V it should be connected to a resistance of:

Show answer and solution

Answer: (B) 200ohm200ohm

Given that the galvanometer has 2525 divisions and the current required for deflection one division is 4×10−4A4\times {10}^{-4}A
current for full scale deflection I=4×10−4×25AI=4\times {10}^{-4}\times 25A
=10−2A={10}^{-2}A

To use this galvanometer as a voltmeter for a maximum voltage of 2.5V2.5V, we need to add a resistance RR in series to the resistance of the galvanometer of G=50ΩG=50\Omega such that,
I=2.5R+G⇒R=2.510−2−50=200ohmI=\dfrac{2.5}{R+G} \Rightarrow R=\dfrac{2.5}{{10}^{-2}}-50 =200ohm

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Circuit builder simulationDrag the sliders and watch the numbers in this question change.Revise Current Electricity firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Current Electricity (Electric Instruments), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.