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Physics · Current Electricity

JEE Main 2026 — 4 April, Morning Shift — Question 11

A voltmeter with internal resistance x Ω can measure up to 20 V. To increase its range to 30 V, the required modification is

  1. Option A:

    connect x2Ω\frac{x}{2}\Omega in series

    Correct
  2. Option B:

    connect x2Ω\frac{x}{2}\Omega in parallel

  3. Option C:

    connect xΩx\Omega in series

  4. Option D:

    connect 2xΩ2x\Omega in parallel

Answer: A

Step-by-step solution

Current for full scale: I = 20/x. For 30 V, total resistance = 30/I = 30x/20 = 1.5x. So series resistance = 1.5x - x = x/2

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Morning Shift
Subject
Physics
Chapter
Current Electricity
Topic
Electrical Measuring Instruments
A voltmeter with internal resistance x Ω can measure up to 20… | JEE Main 2026 PYQ with Solution · DhiX AI