Class 12 Physics Ch-4 Q4.10 | Moving Coil Galvanometer | Current & Voltage Sensitivity Explained

Описание: 📘 Question 4.10 – NCERT Class 12 Physics (Moving Charges and Magnetism):
Two moving coil meters, M₁ and M₂, have the following particulars:

R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³ m², B₁ = 0.25 T
R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10⁻³ m², B₂ = 0.50 T

(The spring constants are identical for both meters.)
Determine the ratio of:
(a) Current sensitivity of M₂ and M₁
(b) Voltage sensitivity of M₂ and M₁


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⚡ Concept Used:

For a moving coil galvanometer —

Current Sensitivity (Iₛ) ∝ N × B × A

Voltage Sensitivity (Vₛ) ∝ (N × B × A) / R


Since spring constant k is same for both:

\frac{I_{s2}}{I_{s1}} = \frac{N₂B₂A₂}{N₁B₁A₁}

\frac{V_{s2}}{V_{s1}} = \frac{N₂B₂A₂/R₂}{N₁B₁A₁/R₁} 


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🧮 Step-by-Step Calculation:

(a) Current Sensitivity Ratio

\frac{I_{s2}}{I_{s1}} = \frac{42×0.50×1.8×10^{-3}}{30×0.25×3.6×10^{-3}} = 1.4

(b) Voltage Sensitivity Ratio

\frac{V_{s2}}{V_{s1}} = 1.4 × \frac{R₁}{R₂} = 1.4 × \frac{10}{14} = 1.0


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💡 Concept Simplified:

Current sensitivity tells us how much deflection per unit current the galvanometer shows.

Voltage sensitivity tells us how much deflection per volt it shows.

Agar resistance badh jaye to voltage sensitivity gir jaati hai ⚡


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