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MHT CET 2026 (14 May, Shift 2) · Moving Coil Galvanometer, Ammeter & Voltmeter

Moving Coil Galvanometer, Ammeter & Voltmeter question from MHT CET 2026 (14 May, Shift 2)

Q13MHT CET 2026 · 14 MAY · SHIFT 2Medium

Moving Coil Galvanometer, Ammeter & Voltmeter

A galvanometer of resistance $50\ \Omega$ is connected to a battery of $3$ V along with a resistance of $2950\ \Omega$ in series. A full scale deflection of $30$ divisions is obtained in the galvanometer. To reduce this deflection to $20$ divisions, the resistance in series should be

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Answer: (d)

Full scale (30 div): $I = \frac{3}{50 + 2950} = 1$ mA.
For 20 divisions: $I' = \frac{20}{30} \times 1 = \frac{2}{3}$ mA.
Total resistance $= \frac{3}{(2/3) \times 10^{-3}} = 4500\ \Omega$ , so series $R = 4500 - 50 = 4450\ \Omega$ .
So the answer is (d) $4450\ \Omega$ .

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