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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

When a 12 \Omega resistor is connected in parallel with a moving coil galvanometer, its deflection reduces from 50 divisions to 10 divisions. What will be the resistance of the galvanometer?

A

24 \Omega

B

36 \Omega

C

48 \Omega

D

60 \Omega

Step-by-Step Solution

  1. Principle: The deflection (ϕ\phi) in a moving coil galvanometer is directly proportional to the current flowing through its coil (Ig∝ϕI_g \propto \phi) .
  2. Current Divider Rule: When a shunt resistance (SS) is connected in parallel with the galvanometer (resistance GG), the main current (II) divides between them. The current flowing through the galvanometer (IgI_g) is given by: Ig=I(SS+G)I_g = I \left( \frac{S}{S+G} \right).
  3. Data Interpretation: The initial deflection corresponds to the full current II passing through the galvanometer (before shunting, or if the shunt wasn't diverting current). Let deflection ϕ1=50\phi_1 = 50. So, I∝50I \propto 50. After connecting the shunt S=12 ΩS = 12\ \Omega, the deflection reduces to ϕ2=10\phi_2 = 10. This deflection corresponds to the new current IgI_g flowing through the galvanometer. So, Ig∝10I_g \propto 10.
  4. Calculation: Ratio of currents: IgI=1050=15\frac{I_g}{I} = \frac{10}{50} = \frac{1}{5}. Substitute into the shunt formula: 15=1212+G\frac{1}{5} = \frac{12}{12 + G}. 12+G=12×512 + G = 12 \times 5 12+G=6012 + G = 60
  • G=48 ΩG = 48\ \Omega.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMresistorconnectedparallelmovinggalvanometer

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