MCQEasyJEE 2024Ohm's Law & Resistance

JEE Physics 2024 Question with Solution

A galvanometer having coil resistance 10Ω10 \, \Omega shows a full scale deflection for a current of 3mA3 \, \text{mA}. For it to measure a current of 8A8 \, \text{A}, the value of the shunt should be:

  • A

    3×103Ω3\times 10^{-3} \, \Omega

  • B

    4.85×103Ω4.85\times 10^{-3} \, \Omega

  • C

    3.75×103Ω3.75\times 10^{-3} \, \Omega

  • D

    2.75×103Ω2.75\times 10^{-3} \, \Omega

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: galvanometer resistance Rg=10ΩR_g = 10 \, \Omega, full-scale deflection current Ig=3×103AI_g = 3 \times 10^{-3} \, \text{A}, and required current I=8AI = 8 \, \text{A}.

Find: the shunt resistance RsR_s required to convert the galvanometer into an ammeter.

For a galvanometer connected in parallel with a shunt, the potential difference across both is the same. Therefore,

IgRg=IsRsI_g R_g = I_s R_s

where

Is=IIg=80.003=7.997AI_s = I - I_g = 8 - 0.003 = 7.997 \, \text{A}

Detailed Calculation

Substituting the values into the relation,

(3×103)×10=7.997Rs(3 \times 10^{-3}) \times 10 = 7.997 \, R_s 0.03=7.997Rs0.03 = 7.997 \, R_s Rs=0.037.9973.75×103ΩR_s = \frac{0.03}{7.997} \approx 3.75 \times 10^{-3} \, \Omega

Therefore, the value of the shunt is 3.75×103Ω3.75 \times 10^{-3} \, \Omega. The correct option is C.

The solution labels the option as D, but its working gives 3.75×103Ω3.75 \times 10^{-3} \, \Omega, which matches option C in the given options.

Common mistakes

  • Using the total current 8A8 \, \text{A} directly through the galvanometer is incorrect because only the full-scale current IgI_g flows through the galvanometer. The remaining current goes through the shunt. Always use Is=IIgI_s = I - I_g for the shunt branch.

  • Forgetting that the galvanometer and shunt are in parallel leads to using a wrong series relation. Since they are in parallel, the voltage across both is equal. Use IgRg=IsRsI_g R_g = I_s R_s, not addition of resistances.

  • Not converting 3mA3 \, \text{mA} into 3×103A3 \times 10^{-3} \, \text{A} gives a completely wrong shunt value. Convert all currents into SI units before substitution.

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