With a resistance ' X ' connected in series with a galvanometer of resistance 100 Ω, it acts as a voltmeter of range 0 - 15 V. To double the range, a resistance of 1500 Ω is to be connected in series with ' X '. The value of ' X ' in ohm is (a) 900 (b) 1100 (c) 1400 (d) 1600

Answer: (c). Range V = Ig(G + R): Ig(100 + X) = 15. Double range: Ig(100 + X + 1500) = 30 Divide: 1600 + X = 2(100 + X) ⇒ X = 1400 Ω So the answer is (c) 1400.

With a resistance ' X ' connected in series with a galvanometer of… | MHT CET 2025 (25 Apr, Shift 2)

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Kirchhoff's Laws & Measuring Instruments

With a resistance ' X ' connected in series with a galvanometer of resistance $100\ \Omega$ , it acts as a voltmeter of range $0 - 15\text{ V}$ . To double the range, a resistance of $1500\ \Omega$ is to be connected in series with ' X '. The value of ' X ' in ohm is

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Answer: (c)

Range $V = I_g(G + R)$ : $I_g(100 + X) = 15$ .
Double range: $I_g(100 + X + 1500) = 30$
Divide: $1600 + X = 2(100 + X) \Rightarrow X = 1400\ \Omega$
So the answer is (c) 1400.

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