When current I is passed through galvanometer coil, it is deflected through 16°. But when it is shunted by a resistance of 8 Ω, the same current produces a deflection of 4°. The resistance of the galvanometer is (a) 48 Ω (b) 16 Ω (c) 24 Ω (d) 12 Ω

Answer: (c). Deflection ∝ galvanometer current, so with the shunt Ig = 4/16 I = I/4. Current divider: Ig = S/G+S I ⇒ 8/G+8 = 1/4 G + 8 = 32 ⇒ G = 24 Ω So the answer is (c) 24 Ω.

When current I is passed through galvanometer coil, it is deflected… | MHT CET 2026 (12 May, Shift 2)

Q8MHT CET 2026 · 12 MAY · SHIFT 2Medium

Moving Coil Galvanometer, Ammeter & Voltmeter

When current $I$ is passed through galvanometer coil, it is deflected through $16^\circ$ . But when it is shunted by a resistance of $8\ \Omega$ , the same current produces a deflection of $4^\circ$ . The resistance of the galvanometer is

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Answer: (c)

Deflection $\propto$ galvanometer current, so with the shunt $I_g = \frac{4}{16} I = \frac{I}{4}$ .
Current divider: $I_g = \frac{S}{G+S} I \Rightarrow \frac{8}{G+8} = \frac{1}{4}$
$G + 8 = 32 \Rightarrow G = 24\ \Omega$
So the answer is (c) $24\ \Omega$ .

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Moving Coil Galvanometer, Ammeter & Voltmeter question from MHT CET 2026 (12 May, Shift 2)

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