A galvanometer of resistance 100 Ω has 50 divisions on its scale and has sensitivitv of 20 μ 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 (a) (b) (c) (d)

Answer: (c). Ig = 50 × 20 = 1 mA 2 V: total 210⁻³ = 2000 Ω ⇒ R₁ = 2000 - 100 = 1900 Ω 10 V: total 10000 Ω ⇒ R₂ = 10000 - 2000 = 8000 Ω 20 V: total 20000 Ω ⇒ R₃ = 20000 - 10000 = 10000 Ω So the answer is (c) R₁ = 1900, R₂ = 8000, R₃ = 10000 Ω.

A galvanometer of resistance 100 Ω has 50 divisions on its scale and… | JEE Main 2019 (12 Apr, Shift 1)

Q168JEE Main 2019 · 12 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

A galvanometer of resistance 100 $\Omega$ has 50 divisions on its scale and has sensitivitv of 20 $\mu$ 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

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Answer: (c)

$I_g = 50 \times 20\ \mu\text{A} = 1\text{ mA}$
2 V: total $\frac{2}{10^{-3}} = 2000\ \Omega \Rightarrow R_1 = 2000 - 100 = 1900\ \Omega$
10 V: total $10000\ \Omega \Rightarrow R_2 = 10000 - 2000 = 8000\ \Omega$
20 V: total $20000\ \Omega \Rightarrow R_3 = 20000 - 10000 = 10000\ \Omega$
So the answer is (c) $R_1 = 1900$ , $R_2 = 8000$ , $R_3 = 10000\ \Omega$ .

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