In an ammeter, 4% of the main current is passing through the galvanometer, If shunt resistance is 5 Ω, then resistance of galvanometer will be (a) 60 Ω (b) 120 Ω (c) 240 Ω (d) 480 Ω

Answer: (b). Ig G = (I - Ig)S ⇒ G = ((I - Ig)S)/(Ig) Ig = 0.04I ⇒ G = (0.96 × 5)/(0.04) = 120 Ω So the answer is (b) 120 Ω.

In an ammeter, 4% of the main current is passing through the… | MHT CET 2024 (9 May, Shift 1)

Q43MHT CET 2024 · 9 MAY · SHIFT 1Easy

Kirchhoff's Laws & Measuring Instruments

In an ammeter, $4\%$ of the main current is passing through the galvanometer, If shunt resistance is $5\ \Omega$ , then resistance of galvanometer will be

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Answer: (b)

$I_g G = (I - I_g)S \Rightarrow G = \frac{(I - I_g)S}{I_g}$
$I_g = 0.04I \Rightarrow G = \frac{0.96 \times 5}{0.04} = 120\ \Omega$
So the answer is (b) $120\ \Omega$ .

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