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Physics · Current Electricity

JEE Main 2024 — 1 February, Shift 2 — Question 31

In an ammeter, 5%5 \% of the main current passes through the galvanometer. If resistance of the galvanometer is G , the resistance of ammeter will be :

  1. Option A:

    G20\frac{G}{20}

    Correct
  2. Option B:

    G199\frac{\mathrm{G}}{199}

  3. Option C:

    199 G

  4. Option D:

    200 G

Answer: A

Step-by-step solution

Given:5% of main current passes through galvanometerResistance of galvanometer: GLet resistance of ammeter be RALet shunt resistance be S\begin{aligned} \text{Given:} \quad &\text{5\% of main current passes through galvanometer} \\ &\text{Resistance of galvanometer: } G \\ &\text{Let resistance of ammeter be } R_A \\ &\text{Let shunt resistance be } S \end{aligned} Current distribution:Ig=0.05IIs=I−Ig=0.95I\begin{aligned} \text{Current distribution:} \quad &I_g = 0.05I \\ &I_s = I - I_g = 0.95I \end{aligned} For parallel combination:IgG=IsS0.05I G=0.95I S0.05G=0.95SS=0.05G0.95=G19\begin{aligned} \text{For parallel combination:} \quad &I_g G = I_s S \\ &0.05I \, G = 0.95I \, S \\ &0.05G = 0.95S \\ &S = \frac{0.05G}{0.95} = \frac{G}{19} \end{aligned} Resistance of ammeter:RA=GSG+S=G⋅G19G+G19=G219G(1+119)=G219G⋅2019=G20\begin{aligned} \text{Resistance of ammeter:} \quad R_A &= \frac{G S}{G + S} \\ &= \frac{G \cdot \frac{G}{19}}{G + \frac{G}{19}} \\ &= \frac{\frac{G^2}{19}}{G\left(1 + \frac{1}{19}\right)} \\ &= \frac{\frac{G^2}{19}}{G \cdot \frac{20}{19}} \\ &= \frac{G}{20} \end{aligned} RA=G20\boxed{R_A = \frac{G}{20}}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Physics
Chapter
Current Electricity
Topic
Electrical Measuring Instruments

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