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JEE Main 2005 (AIEEE) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2005 (AIEEE)

Q199JEE Main 2005 · AIEEEMedium

Kirchhoff's Laws & Measuring Instruments

A moving coil galvanometer has 150150 equal divisions. Its current sensitivity is 1010- divisions per milliampere and voltage sensitivity is 22 divisions per millivolt. In order that each division reads 11 volt, the resistance in ohmsohms needed to be connected in series with the coil will be -

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Answer: (c)

  1. G=IsVs=10 div/mA2 div/mV=5 ΩG = \frac{I_s}{V_s} = \frac{10\text{ div/mA}}{2\text{ div/mV}} = 5\ \Omega
  2. Full scale current Ig=15010=15 mAI_g = \frac{150}{10} = 15\text{ mA}; full scale voltage needed =150×1=150 V= 150 \times 1 = 150\text{ V}
  3. R=1500.015−5=10000−5=9995 ΩR = \frac{150}{0.015} - 5 = 10000 - 5 = 9995\ \Omega
  4. So the answer is (c) 9995.
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