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Physics · Current Electricity

JEE Main 2024 — 27 January, Shift 2 — Question 38

Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B , their readings are V1, V2\mathrm{V}_{1}, \mathrm{~V}_{2} and V3\mathrm{V}_{3}. Choose the correct option.

Question figure
  1. Option A:

    V1=V2V_{1}=V_{2}

  2. Option B:

    V1≠V3−V2V_{1} \neq V_{3}-V_{2}

  3. Option C:

    V1+V2>V3V_{1}+V_{2}>V_{3}

  4. Option D:

    V1+V2=V3V_{1}+V_{2}=V_{3}

    Correct

Answer: D

Step-by-step solution

From KVL,

V1+V2V3=0⇒V1+V2=V3\mathbf{V}_{\mathbf{1}}+\mathbf{V}_{\mathbf{2}}\mathbf{V}_{\mathbf{3}}=\mathbf{0}\Rightarrow V_{1}+V_{2}=V_{3}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Physics
Chapter
Current Electricity
Topic
Circuit Analysis, Kirchhoff's Law and Nodal Analysis

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