By connecting a resistance of 980 Ω in series with a galvanometer, it is converted into a voltmeter of a certain range. When the resistance of 470 Ω is connected in series, the range is halved. The resistance of the galvanometer is (a) 30 Ω (b) 40 Ω (c) 50 Ω (d) 60 Ω

Answer: (b). Voltmeter range V = Ig(G + R), with the same full-scale current Ig. Ig(G + 980) = V and Ig(G + 470) = V/2 Divide: G + 980 = 2(G + 470) ⇒ G = 980 - 940 G = 40 Ω. So the answer is (b).

By connecting a resistance of 980 Ω in series with a galvanometer, it… | MHT CET 2026 (18 Apr, Shift 1)

Q16MHT CET 2026 · 18 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

By connecting a resistance of $980\ \Omega$ in series with a galvanometer, it is converted into a voltmeter of a certain range. When the resistance of $470\ \Omega$ is connected in series, the range is halved. The resistance of the galvanometer is

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Answer: (b)

Voltmeter range $V = I_g(G + R)$ , with the same full-scale current $I_g$ .
$I_g(G + 980) = V$ and $I_g(G + 470) = \dfrac{V}{2}$
Divide: $G + 980 = 2(G + 470) \Rightarrow G = 980 - 940$
$G = 40\ \Omega$ . So the answer is (b).

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