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Current Electricity question

2021 · 27 Aug · Shift 2 · Q50
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Current Electricity question

2021 · 27 Aug · Shift 2 · Q50

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :
  1. A
    1 Ω\OmegaΩ
  2. B
    5 Ω\OmegaΩ
  3. C
    4 Ω\OmegaΩ
  4. D
    2 Ω\OmegaΩ
View written solutionFree

Correct answer: D

  1. Given data

    • Full scale deflection =50= 50=50 divisions
    • Voltage required for full scale deflection =50 mV=50×10−3 V= 50\,\text{mV} = 50 \times 10^{-3}\,\text{V}=50mV=50×10−3V
    • Current sensitivity =2 div/mA= 2\,\text{div/mA}=2div/mA

  2. Find the current required for full scale deflection

    Current sensitivity means: deflectioncurrent=2 div/mA\frac{\text{deflection}}{\text{current}} = 2\,\text{div/mA}currentdeflection​=2div/mA

    So for 505050 divisions, required current is: Ig=50 div2 div/mA=25 mAI_g = \frac{50\,\text{div}}{2\,\text{div/mA}} = 25\,\text{mA}Ig​=2div/mA50div​=25mA

    Converting into ampere: Ig=25×10−3 AI_g = 25 \times 10^{-3}\,\text{A}Ig​=25×10−3A

  3. Use Ohm’s law to find galvanometer resistance

    Rg=VgIgR_g = \frac{V_g}{I_g}Rg​=Ig​Vg​​

    Substituting values: Rg=50×10−325×10−3=2 ΩR_g = \frac{50 \times 10^{-3}}{25 \times 10^{-3}} = 2\,\OmegaRg​=25×10−350×10−3​=2Ω

  4. Match with the options Rg=2 ΩR_g = 2\,\OmegaRg​=2Ω

    Therefore, the correct option is D.

  5. Comparison with stored answer

    Stored correct answer: D

    Our derived answer: D

    Hence, the answer agrees with the stored correct answer.

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