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Current Electricity question

2024 · 9 Apr · Shift 1 · Q62
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Current Electricity question

2024 · 9 Apr · Shift 1 · Q62

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A galvanometer has a coil of resistance 200Ω200 \Omega200Ω with a full scale deflection at 20μA20 \mu \mathrm{A}20μA. The value of resistance to be added to use it as an ammeter of range (0−20)mA(0-20) \mathrm{mA}(0−20)mA is :
  1. A
    0.40Ω0.40\Omega0.40Ω
  2. B
    0.10Ω0.10\Omega0.10Ω
  3. C
    0.20Ω0.20\Omega0.20Ω
  4. D
    0.50Ω0.50\Omega0.50Ω
View written solutionFree

Correct answer: C

  1. Given data
  • Galvanometer resistance: G=200 ΩG = 200\,\OmegaG=200Ω
  • Full scale deflection current of galvanometer: Ig=20 μA=20×10−6 AI_g = 20\,\mu A = 20 \times 10^{-6}\,AIg​=20μA=20×10−6A
  • Desired ammeter range: I=20 mA=20×10−3 AI = 20\,mA = 20 \times 10^{-3}\,AI=20mA=20×10−3A

To convert a galvanometer into an ammeter, a small shunt resistance SSS is connected in parallel with the galvanometer.

  1. Use current division condition

At full scale of the ammeter:

  • Current through galvanometer = IgI_gIg​
  • Current through shunt = Is=I−IgI_s = I - I_gIs​=I−Ig​

So,

Is=20×10−3−20×10−6I_s = 20 \times 10^{-3} - 20 \times 10^{-6}Is​=20×10−3−20×10−6 Is=0.020−0.000020=0.01998 AI_s = 0.020 - 0.000020 = 0.01998\,AIs​=0.020−0.000020=0.01998A
  1. Equal potential difference across galvanometer and shunt

Since galvanometer and shunt are in parallel,

IgG=IsSI_g G = I_s SIg​G=Is​S

Hence,

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

Substitute the values:

S=(20×10−6)(200)0.01998S = \frac{(20 \times 10^{-6})(200)}{0.01998}S=0.01998(20×10−6)(200)​ S=4×10−30.01998S = \frac{4 \times 10^{-3}}{0.01998}S=0.019984×10−3​ S≈0.2002 ΩS \approx 0.2002\,\OmegaS≈0.2002Ω

Thus,

S≈0.20 ΩS \approx 0.20\,\OmegaS≈0.20Ω
  1. Match with the options

The required resistance is:

0.20 Ω\boxed{0.20\,\Omega}0.20Ω​

So the correct option is C.

  1. Verification with stored answer

Stored correct answer: C

My derived answer: C

They agree.

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