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JEE Main 202427 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
27 January
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2024 · 27 January · Evening Shift · Physics · Current Electricity

Three voltmeters, all having different internal resistances are...

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

Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential

Options

  1. AV1=V2{V}_{1} = {V}_{2}
  2. BV1≠V3−V2{V}_{1} \ne {V}_{3} - {V}_{2}
  3. CV1+V2>V3{V}_{1} + {V}_{2} > {V}_{3}
  4. DV1+V2=V3{V}_{1} + {V}_{2} = {V}_{3}

Correct answer

D. V1+V2=V3{V}_{1} + {V}_{2} = {V}_{3}

Step-by-step solution

Let's consider the following diagram:

Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential

According to the Kirchhoff's voltage law, the algebraic sum of the voltages in a closed loop is zero.

From the above figure, it can be written that

V1+V2−V3=0⇒V1+V2=V3{V}_{1} + {V}_{2} - {V}_{3} = 0 ⇒ {V}_{1} + {V}_{2} = {V}_{3}

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