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Current Electricity question

2018 · 16 Apr · Shift 1 · Q48
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Current Electricity question

2018 · 16 Apr · Shift 1 · Q48

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A galvanometer with its coil resistance 25 Ω\OmegaΩ requires a current of 1 mA for its full deflection. In order to construct an ammeter to read upto a current of 2 A, the approximate value of the shunt resistance should be :
  1. A
    2.5×10−3 Ω2.5 \times {10^{ - 3}}\,\Omega2.5×10−3Ω
  2. B
    1.25×10−2Ω1.25 \times {10^{ - 2}}\Omega1.25×10−2Ω
  3. C
    1.25×10−3Ω1.25 \times {10^{ - 3}}\Omega1.25×10−3Ω
  4. D
    2.5×10−2Ω2.5 \times {10^{ - 2}}\Omega2.5×10−2Ω
View written solutionFree

Correct answer: B

  1. Given data

    • Galvanometer resistance: G=25 ΩG = 25\,\OmegaG=25Ω
    • Full-scale deflection current of galvanometer: Ig=1 mA=10−3 AI_g = 1\,\text{mA} = 10^{-3}\,\text{A}Ig​=1mA=10−3A
    • Desired ammeter range: I=2 AI = 2\,\text{A}I=2A

  2. Current through shunt

    The shunt carries the remaining current: Is=I−Ig=2−0.001=1.999 AI_s = I - I_g = 2 - 0.001 = 1.999\,\text{A}Is​=I−Ig​=2−0.001=1.999A

  3. Voltage across galvanometer

    At full-scale deflection, Vg=IgG=10−3×25=25×10−3=0.025 VV_g = I_g G = 10^{-3} \times 25 = 25 \times 10^{-3} = 0.025\,\text{V}Vg​=Ig​G=10−3×25=25×10−3=0.025V

  4. Voltage across shunt

    Since galvanometer and shunt are in parallel, Vs=Vg=0.025 VV_s = V_g = 0.025\,\text{V}Vs​=Vg​=0.025V

  5. Shunt resistance

    S=VsIs=0.0251.999≈0.0125 ΩS = \frac{V_s}{I_s} = \frac{0.025}{1.999} \approx 0.0125\,\OmegaS=Is​Vs​​=1.9990.025​≈0.0125Ω

    Thus, S≈1.25×10−2 ΩS \approx 1.25 \times 10^{-2}\,\OmegaS≈1.25×10−2Ω

  6. Match with options

    This corresponds to Option B.

Final Answer: 1.25×10−2 Ω\boxed{1.25 \times 10^{-2}\,\Omega}1.25×10−2Ω​

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