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Current Electricity question

2021 · 31 Aug · Shift 1 · Q63
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Current Electricity question

2021 · 31 Aug · Shift 1 · Q63

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Consider a galvanometer shunted with 5 Ω\OmegaΩ resistance and 2% of current passes through it. What is the resistance of the given galvanometer ?
  1. A
    300 Ω\OmegaΩ
  2. B
    344 Ω\OmegaΩ
  3. C
    245 Ω\OmegaΩ
  4. D
    226 Ω\OmegaΩ
View written solutionFree

Correct answer: C

  1. Given data

    • Shunt resistance: S=5 ΩS = 5\,\OmegaS=5Ω
    • Fraction of total current through galvanometer: 2%=0.022\% = 0.022%=0.02

    So if total current is III, then: Ig=0.02II_g = 0.02IIg​=0.02I Current through shunt is: Is=I−Ig=0.98II_s = I - I_g = 0.98IIs​=I−Ig​=0.98I

  2. Use the property of parallel branches Since galvanometer and shunt are in parallel, potential difference across them is same: IgG=IsSI_g G = I_s SIg​G=Is​S where GGG is galvanometer resistance.

  3. Substitute values 0.02I⋅G=0.98I⋅50.02I \cdot G = 0.98I \cdot 50.02I⋅G=0.98I⋅5

    Cancel III: 0.02G=4.90.02G = 4.90.02G=4.9

  4. Solve for GGG G=4.90.02=245 ΩG = \frac{4.9}{0.02} = 245\,\OmegaG=0.024.9​=245Ω

  5. Match with options 245 Ω245\,\Omega245Ω corresponds to Option C.

Final Answer

The resistance of the galvanometer is: 245 Ω\boxed{245\,\Omega}245Ω​

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