In a moving coil galvanometer, two moving coils M₁ and M₂ have the following particulars: R₁ = 5Ω, N₁ = 15, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T R₂ = 7Ω, N₂ = 21, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of M₁ and M₂? (a) 1: 1 (b) 1: 3 (c) 1: 2 (d) 1: 4

Answer: (a). Voltage sensitivity Vs = NBA/kR. M₁: 15 × 0.25 × 3.6×10⁻³5 = 2.7×10⁻³ (divided by k). M₂: 21 × 0.50 × 1.8×10⁻³7 = 2.7×10⁻³ (divided by k). So the answer is (a) 1:1.

In a moving coil galvanometer, two moving coils M₁ and M₂ have the… | JEE Main 2025 (2 Apr, Shift 2)

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Moving Coil Galvanometer, Ammeter & Voltmeter

In a moving coil galvanometer, two moving coils $\text{M}_1$ and $\text{M}_2$ have the following particulars: $\text{R}_1 = 5\Omega,\ \text{N}_1 = 15,\ \text{A}_1 = 3.6 \times 10^{-3}\text{ m}^{2},\ \text{B}_1 = 0.25\text{ T}$

$\text{R}_2 = 7\Omega,\ \text{N}_2 = 21,\ \text{A}_2 = 1.8 \times 10^{-3}\text{ m}^{2},\ \text{B}_2 = 0.50\text{ T}$

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_1$ and $M_2$ ?

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Answer: (a)

Voltage sensitivity $V_s = \frac{NBA}{kR}$ .
$M_1$ : $\frac{15 \times 0.25 \times 3.6\times10^{-3}}{5} = 2.7\times10^{-3}$ (divided by $k$ ).
$M_2$ : $\frac{21 \times 0.50 \times 1.8\times10^{-3}}{7} = 2.7\times10^{-3}$ (divided by $k$ ).
So the answer is (a) $1:1$ .

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