JEE Main 2023PhysicsCurrent ElectricityMediumMCQ

JEE Main 2023 — Current Electricity Question with Solution

JEE Main 2023 (24 Jan Shift 2)

Question

A cell of emf 90 V is connected across series combination of two resistors each of 100 Ω resistance. A voltmeter of resistance 400 Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be:

Choose an option

Show full solutionCorrect option: A
Correct answer
A40 V

Step-by-step explanation

As we know, voltmeter is connected in parallel to the element being measured.

The equivalent resistance of the above circuit is Req=400×100400+100+100=80+100=180 Ω

Current through the circuit is i=90180=12 A

Reading of the voltmeter is VR=i400×100400+100=12×80=40 V

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Current Electricity chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2023, 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.