When a resistance of 5 Ω is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250 mA, however when 1050 Ω resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ Ω. Numerical answer type. Work it out, then open the answer.

Answer: 50. Voltmeter: Ig(G + 1050) = 25 Ammeter: Ig G = (0.25 - Ig) × 5 ⇒ Ig(G + 5) = 1.25 Divide: (G + 1050)/(G + 5) = 20 ⇒ G + 1050 = 20G + 100 19G = 950 ⇒ G = 50 Ω Answer: 50 Ω

When a resistance of 5 Ω is shunted with a moving coil galvanometer… | JEE Main 2023 (13 Apr, Shift 1)

Q97JEE Main 2023 · 13 APR · SHIFT 1Medium

Kirchhoff's Laws & Measuring Instruments

When a resistance of $5\ \Omega$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $250\text{ mA}$ , however when $1050\ \Omega$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ $\Omega$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 50

Voltmeter: $I_g(G + 1050) = 25$
Ammeter: $I_g G = (0.25 - I_g) \times 5 \Rightarrow I_g(G + 5) = 1.25$
Divide: $\frac{G + 1050}{G + 5} = 20 \Rightarrow G + 1050 = 20G + 100$
$19G = 950 \Rightarrow G = 50\ \Omega$
Answer: $50\ \Omega$

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