Physics · Current Electricity

JEE Main 2024 — 9 April, Shift 1 — Question 45

A galvanmeter has a coil of resistance 200Ω200 \Omega with a full scale deflection at 20μ A20 \mu \mathrm{~A}. The value of resistance to be added to use it as an ammeter of range (0−20)mA(0-20) \mathrm{mA} is:

  1. Option A:

    0.40Ω0.40 \Omega

  2. Option B:

    0.20Ω0.20 \Omega

    Correct
  3. Option C:

    0.50Ω0.50 \Omega

  4. Option D:

    0.10Ω0.10 \Omega

Answer: B

Step-by-step solution

G=200Ω\mathrm{G}=200 \Omega

ig=20μ A\mathrm{i}_{\mathrm{g}}=20 \mu \mathrm{~A} i=ig(GS+1)i=i_{g}\left(\frac{G}{S}+1\right)

⇒20×10−3=20×10−6(200 S+1)\Rightarrow 20 \times 10^{-3}=20 \times 10^{-6}\left(\frac{200}{\mathrm{~S}}+1\right)

⇒200 S=999\Rightarrow \frac{200}{\mathrm{~S}}=999

⇒S≈0.2Ω\Rightarrow \mathrm{S} \approx 0.2 \Omega

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Physics
Chapter
Current Electricity
Topic
Electrical Measuring Instruments
A galvanmeter has a coil of resistance 200 Ω with a full scale… | JEE Main 2024 PYQ with Solution · DhiX AI