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NEET PHYSICSMedium

The current flowing in a coil of resistance 90 Ω90~\Omega is to be reduced by 90%90\%. What value of resistance should be connected in parallel with it?

A

9 Ω\Omega

B

90 Ω\Omega

C

1000 Ω\Omega

D

10 Ω\Omega

Step-by-Step Solution

  1. Concept: To reduce the current flowing through a coil (like a galvanometer), a low resistance called a shunt (SS) is connected in parallel with it. This diverts a portion of the current .
  2. Data Given:
  • Resistance of the coil, G=90 ΩG = 90~\Omega.
  • The current in the coil (IgI_g) is to be reduced by 90%90\%. This means the current remaining in the coil is 10%10\% of the main current (II).
  • Therefore, Ig=0.1II_g = 0.1 I.
  • The remaining current passes through the shunt: Is=I−Ig=I−0.1I=0.9II_s = I - I_g = I - 0.1I = 0.9I.
  1. Calculation: Since the coil and the shunt are in parallel, the potential difference across them is equal. Vg=VsV_g = V_s IgG=IsSI_g G = I_s S (0.1I)×90=(0.9I)×S(0.1 I) \times 90 = (0.9 I) \times S 9=0.9S9 = 0.9 S S=90.9=10 ΩS = \frac{9}{0.9} = 10~\Omega
  2. Conclusion: A resistance of 10 Ω10~\Omega should be connected in parallel.
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