Physics Circle home Student Login

Home›Chapters›Moving Charges and Magnetism PYQs›Moving Coil Galvanometer, Ammeter & Voltmeter›Question 28

NEET 2008 (AIPMT) · Moving Coil Galvanometer, Ammeter & Voltmeter

Moving Coil Galvanometer, Ammeter & Voltmeter question from NEET 2008 (AIPMT)

Q28NEET 2008 · AIPMTMedium

Moving Coil Galvanometer, Ammeter & Voltmeter

A galvanometer of resistance 50 Ω\Omega is connected to a battery of 3 V along with a resistance of 2950 Ω\Omega in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

Show answer and solution

Answer: (b)

  1. Full scale (30 div): I=350+2950=1I = \frac{3}{50 + 2950} = 1 mA.
  2. For 20 divisions: I′=23I' = \frac{2}{3} mA.
  3. Total resistance =3(2/3)×10−3=4500 Ω= \frac{3}{(2/3) \times 10^{-3}} = 4500\ \Omega, so series R=4500−50=4450 ΩR = 4500 - 50 = 4450\ \Omega.
  4. So the answer is (b) 4450 Ω4450\ \Omega.
Report a mistake

More questions on Moving Coil Galvanometer, Ammeter & Voltmeter