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Physics Question on The Moving Coil Galvanometer

A galvanometer has a coil of resistance 200 Ω200 \, \Omega with a full scale deflection at 20 μA20 \, \mu A. The value of resistance to be added to use it as an ammeter of range (0–20) mA is:

A

0.40 Ω0.40 \, \Omega

B

0.20 Ω0.20 \, \Omega

C

0.50 Ω0.50 \, \Omega

D

0.10 Ω0.10 \, \Omega

Answer

0.20 Ω0.20 \, \Omega

Explanation

Solution

Step 1: Formula for shunt resistance The shunt resistance RsR_s is given by:

Rs=IgRgI−Ig,R_s = \frac{I_g R_g}{I - I_g},

where:

  • Ig=20 μA=20×10−6 AI_g = 20 \, \mu A = 20 \times 10^{-6} \, \text{A} (full-scale deflection current),
  • Rg=200 ΩR_g = 200 \, \Omega (resistance of the galvanometer),
  • I=20 mA=20×10−3 AI = 20 \, mA = 20 \times 10^{-3} \, \text{A} (ammeter range).

Step 2: Substitute the values

Rs=(20×10−6)⋅200(20×10−3)−(20×10−6).R_s = \frac{(20 \times 10^{-6}) \cdot 200}{(20 \times 10^{-3}) - (20 \times 10^{-6})}.

Simplify the numerator:

(20×10−6)⋅200=4×10−3.(20 \times 10^{-6}) \cdot 200 = 4 \times 10^{-3}.

Simplify the denominator:

(20×10−3)−(20×10−6)=19.98×10−3.(20 \times 10^{-3}) - (20 \times 10^{-6}) = 19.98 \times 10^{-3}.

Thus:

Rs=4×10−319.98×10−3≈0.20 Ω.R_s = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.20 \, \Omega.

Final Answer: 0.20 Ω.0.20 \, \Omega.