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Current Electricity question

2024 · 29 Jan · Shift 1 · Q67
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  5. /2024 · 29 Jan · Shift 1 · Q67

Current Electricity question

2024 · 29 Jan · Shift 1 · Q67

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A galvanometer having coil resistance 10Ω10 \Omega10Ω shows a full scale deflection for a current of 3 mA3 \mathrm{~mA}3 mA. For it to measure a current of 8 A8 \mathrm{~A}8 A, the value of the shunt should be:
  1. A
    3.75×10−3Ω3.75 \times 10^{-3} \Omega3.75×10−3Ω
  2. B
    3×10−3Ω3 \times 10^{-3} \Omega3×10−3Ω
  3. C
    4.85×10−3Ω4.85 \times 10^{-3} \Omega4.85×10−3Ω
  4. D
    2.75×10−3Ω2.75 \times 10^{-3} \Omega2.75×10−3Ω
View written solutionFree

Correct answer: A

  1. Given data
  • Galvanometer resistance: G=10 ΩG = 10\,\OmegaG=10Ω
  • Full scale deflection current: Ig=3 mA=3×10−3 AI_g = 3\,\text{mA} = 3 \times 10^{-3}\,\text{A}Ig​=3mA=3×10−3A
  • Desired current range: I=8 AI = 8\,\text{A}I=8A

We need the value of shunt resistance SSS so that the galvanometer can measure up to 8 A8\,\text{A}8A.

  1. Current through shunt

At full-scale reading of the ammeter:

  • Current through galvanometer =Ig= I_g=Ig​
  • Remaining current goes through shunt:
Is=I−Ig=8−0.003=7.997 AI_s = I - I_g = 8 - 0.003 = 7.997\,\text{A}Is​=I−Ig​=8−0.003=7.997A
  1. Voltage across galvanometer and shunt is same

Since galvanometer and shunt are in parallel,

IgG=IsSI_g G = I_s SIg​G=Is​S

So,

S=IgGIsS = \frac{I_g G}{I_s}S=Is​Ig​G​

Substitute values:

S=(3×10−3)(10)7.997S = \frac{(3 \times 10^{-3})(10)}{7.997}S=7.997(3×10−3)(10)​ S=0.037.997approx3.75×10−3 ΩS = \frac{0.03}{7.997} approx 3.75 \times 10^{-3}\,\OmegaS=7.9970.03​approx3.75×10−3Ω
  1. Match with options
S≈3.75×10−3 ΩS \approx 3.75 \times 10^{-3}\,\OmegaS≈3.75×10−3Ω

So the correct option is A.

  1. Verification with stored answer

Stored correct answer: A

Our derived answer: A

They match.

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