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Current Electricity question

2003 · Shift 0 · Q128
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Current Electricity question

2003 · Shift 0 · Q128

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
An ammeter reads upto 111 ampere. Its internal resistance is 0.81ohm0.81ohm0.81ohm. To increase the range to 10A10A10A the value of the required shunt is
  1. A
    0.03 Ω0.03\,\Omega0.03Ω
  2. B
    0.3 Ω0.3\,\Omega0.3Ω
  3. C
    0.9 Ω0.9\,\Omega0.9Ω
  4. D
    0.09 Ω0.09\,\Omega0.09Ω
View written solutionFree

Correct answer: D

  1. Given data

    • Ammeter full-scale current: Ig=1 AI_g = 1\,\text{A}Ig​=1A
    • Internal resistance of ammeter: Rg=0.81 ΩR_g = 0.81\,\OmegaRg​=0.81Ω
    • Desired new range: I=10 AI = 10\,\text{A}I=10A

  2. Current through shunt When the ammeter is converted to read up to 10 A10\,\text{A}10A, the meter itself should still carry only its full-scale current 1 A1\,\text{A}1A.

    So current through shunt is Is=I−Ig=10−1=9 AI_s = I - I_g = 10 - 1 = 9\,\text{A}Is​=I−Ig​=10−1=9A

  3. Voltage across ammeter and shunt Since the shunt is connected in parallel with the ammeter, potential difference across both is same.

    Voltage across ammeter: V=IgRg=1×0.81=0.81 VV = I_g R_g = 1 \times 0.81 = 0.81\,\text{V}V=Ig​Rg​=1×0.81=0.81V

  4. Find shunt resistance Rs=VIs=0.819=0.09 ΩR_s = \frac{V}{I_s} = \frac{0.81}{9} = 0.09\,\OmegaRs​=Is​V​=90.81​=0.09Ω

  5. Match with options Rs=0.09 ΩR_s = 0.09\,\OmegaRs​=0.09Ω So the correct option is D.

Verification with stored answer

Stored correct answer: D

My derived answer: D

They match.

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