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Current Electricity question

2019 · 9 Apr · Shift 2 · Q68
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Current Electricity question

2019 · 9 Apr · Shift 2 · Q68

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A ?
  1. A
    0.02 ohm
  2. B
    0.2 ohm
  3. C
    0.002 ohm
  4. D
    0.5 ohm
View written solutionFree

Correct answer: B

  1. Given data
  • Galvanometer resistance: G=50 ΩG = 50\,\OmegaG=50Ω
  • Maximum current through galvanometer: Ig=0.002 AI_g = 0.002\,\text{A}Ig​=0.002A
  • Desired ammeter range: I=0.5 AI = 0.5\,\text{A}I=0.5A

To convert a galvanometer into an ammeter, we connect a small shunt resistance in parallel with the galvanometer.

  1. Current through shunt

The total current is I=0.5 AI = 0.5\,\text{A}I=0.5A, out of which only Ig=0.002 AI_g = 0.002\,\text{A}Ig​=0.002A should pass through the galvanometer.

So current through shunt is

Is=I−Ig=0.5−0.002=0.498 AI_s = I - I_g = 0.5 - 0.002 = 0.498\,\text{A}Is​=I−Ig​=0.5−0.002=0.498A
  1. Use parallel branch voltage condition

Since galvanometer and shunt are in parallel, potential difference across both is same.

Voltage across galvanometer:

V=IgG=0.002×50=0.1 VV = I_g G = 0.002 \times 50 = 0.1\,\text{V}V=Ig​G=0.002×50=0.1V

Let shunt resistance be SSS. Then

V=IsSV = I_s SV=Is​S

So,

S=VIs=0.10.498S = \frac{V}{I_s} = \frac{0.1}{0.498}S=Is​V​=0.4980.1​ S≈0.2008 ΩS \approx 0.2008\,\OmegaS≈0.2008Ω
  1. Final answer

Thus the required resistance is approximately

S≈0.2 ΩS \approx 0.2\,\OmegaS≈0.2Ω

So the correct option is B.

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