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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

A galvanometer of 50 Ω50~\Omega resistance has 25 divisions. A current of 4×10−4 A4 \times 10^{-4} \text{ A} gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V25 \text{ V}, it should be connected with a resistance of:

A

2500 Ω2500~\Omega as a shunt

B

2450 Ω2450~\Omega as a shunt

C

2550 Ω2550~\Omega in series

D

2450 Ω2450~\Omega in series

Step-by-Step Solution

  1. Concept: To convert a galvanometer into a voltmeter, a high resistance (RR) is connected in series with the galvanometer coil. This ensures that the total resistance is high enough to not draw significant current from the circuit being measured .
  2. Calculate Full Scale Deflection Current (IgI_g):
  • Current per division = 4×10−4 A4 \times 10^{-4} \text{ A}.
  • Total divisions = 25.
  • Ig=25×4×10−4=100×10−4=10−2 A=0.01 AI_g = 25 \times 4 \times 10^{-4} = 100 \times 10^{-4} = 10^{-2} \text{ A} = 0.01 \text{ A}.
  1. Formula: The potential difference (VV) across the combination is given by Ohm's law: V=Ig(G+R)V = I_g (G + R) Where GG is the galvanometer resistance and RR is the series resistance.
  2. Calculation:
  • Given V=25 VV = 25 \text{ V} and G=50 ΩG = 50~\Omega.
  • Rearranging the formula for RR: R=VIg−GR = \frac{V}{I_g} - G R=250.01−50R = \frac{25}{0.01} - 50 R=2500−50=2450 ΩR = 2500 - 50 = 2450~\Omega
  1. Conclusion: A resistance of 2450 Ω2450~\Omega must be connected in series.

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 MAGNETISMgalvanometerresistancedivisionscurrentdeflection

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