A galvanometer of resistance 100 Ω when connected in series with 400 Ω measures a voltage of upto 10 V. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A is x × 10⁻² Ω. The value of x is: (a) 2 (b) 20 (c) 800 (d) 200

Answer: (b). Full-scale current: Ig = (10)/(100 + 400) = 0.02 A. Shunt for 10 A: S = (Ig G)/(I - Ig) = (0.02 × 100)/(9.98) S ≈ 0.2 Ω = 20 × 10⁻² Ω x = 20. So the answer is (b).

A galvanometer of resistance 100 Ω when connected in series with 400… | JEE Main 2024 (5 Apr, Shift 2)

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Kirchhoff's Laws & Measuring Instruments

A galvanometer of resistance $100\ \Omega$ when connected in series with $400\ \Omega$ measures a voltage of upto $10\text{ V}$ . The value of resistance required to convert the galvanometer into ammeter to read upto $10\text{ A}$ is $x \times 10^{-2}\ \Omega$ . The value of $x$ is:

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Answer: (b)

Full-scale current: $I_g = \dfrac{10}{100 + 400} = 0.02\text{ A}$ .
Shunt for $10\text{ A}$ : $S = \dfrac{I_g G}{I - I_g} = \dfrac{0.02 \times 100}{9.98}$
$S \approx 0.2\ \Omega = 20 \times 10^{-2}\ \Omega$
$x = 20$ . So the answer is (b).

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