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QJEEM19Ph217d: JEE Main 2019 Physics - Class 12 JEEM19Ph217d Solution: A galvanometer of resistance 100 has 50 divisions on its scale…

A galvanometer of resistance 100 Ω\Omega has 50 divisions on its scale and has sensitivitv of 20 μ\mu A/division. It is to be converted to a voltmeter with three ranges of 0-2V, 0-10 V and 0-20 V. The appropriate circuit to do so is

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/e8247ed39fa86bd66083caeebaf9563ae2150f4c95a0196f6101288a55fd862c.webp
  • B https://ik.imagekit.io/novatra/photos/aa6cc25dedb9bc81f4e56fc4e22016aa4a5d4f0ab4039b7b5cdefebf86e077f6.webp
  • C https://ik.imagekit.io/novatra/photos/e8d995118e7ea72acbfad1397ef128de7dd46e2c0ac64d99e1293e3ac16f29a4.webp
  • D https://ik.imagekit.io/novatra/photos/3270ac0f5ad8e556649f9223efa7e20aa0f7ab68e14e1bed6d3ac4765019e1b7.webp

Step-by-Step Verified Solution

20 × 50 × 10–6 = 10–3 Amp

V1=210−3=100+R1{V_1} = {2 \over {{{10}^{ - 3}}}} = 100 + {R_1}

1900=R11900 = {R_1}

V2=1010−3=(2000+R2){V_2} = {{10} \over {{{10}^{ - 3}}}} = \left( {2000 + {R_2}} \right)

R2=8000{R_2} = 8000

V3=2010−3=10×103+R3=10×103R3{V_3} = {{20} \over {{{10}^{ - 3}}}} = 10 \times {10^3} + {R_3} = 10 \times {10^3}{R_3}

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