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Physics · Current Electricity

JEE Main 2026 — 24 January, Morning Shift — Question 28

Two resistors of 100Ω100 \Omega each are connected in series with a 9 V battery. A voltmeter of 400Ω400 \Omega resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ____\_\_\_\_ V.

  1. Option A:

    3

  2. Option B:

    4.5

  3. Option C:

    4

    Correct
  4. Option D:

    2

Answer: C

Step-by-step solution

Current in circuit. I=EReq⁡I=\frac{E}{\operatorname{Req}} Req =100+400×100400+100=180Ω=100+\frac{400 \times 100}{400+100}=180 \Omega ∴I=9180=120 A\therefore \mathrm{I}=\frac{9}{180}=\frac{1}{20} \mathrm{~A} Reading of voltmeter =V=I×80=120×80=4 V=\mathrm{V}=\mathrm{I} \times 80=\frac{1}{20} \times 80=4 \mathrm{~V}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Current Electricity
Topic
Combination of Resistors and cells, Wheatstone Bridge

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