Ncert Exercises & Solutions

4.10 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 the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

NEETprep Answer:

For moving coil meter M1:

Resistance, R₁ = 10

Number of turns, N₁ = 30

Area of cross-section,

A_{1} = 3.6 \times 10^{- 3} m^{2}

Magnetic field strength, B₁ = 0.25 T

Spring constant K₁=K

For moving coil meter M2:

Resistance, R₂ = 14

Number of turns, N₂ = 42

Area of cross-section,

A_{2} = 1.8 \times 10^{- 3} m^{2}

Magnetic field strength, B₂ = 0.50 T

Spring constant, K₂ = K

(a) Current sensitivity of M₁ is given as:

I_{s 1} = \frac{N_{1} B_{1} A_{1}}{K_{1}}

And, current sensitivity of M₂ is given as:

I_{s 2} = \frac{N_{2} B_{2} A_{2}}{K_{2}}

∴ Ratio \frac{I_{s 2}}{I_{s 1}} = \frac{N_{2} B_{2} A_{2}}{N_{1} B_{1} A_{1}} = \frac{42 \times 0.5 \times 1.8 \times 10^{- 3} \times K}{K \times 30 \times 0.25 \times 3.6 \times 10^{- 3}} = 1.4

Hence, the ratio of current sensitivity of M₂ to M₁ is 1.4.

(b) Voltage sensitivity for M₂ is given as:

V_{s 2} = \frac{N_{2} B_{2} A_{2}}{K_{2} R_{2}}

And, voltage sensitivity for M₁ is given as:

V_{s 1} = \frac{N_{1} B_{1} A_{1}}{K_{1} R_{1}}

∴ Ratio \frac{I_{s 2}}{V_{s 1}} = \frac{N_{2} B_{2} A_{2} K_{1} R_{1}}{N_{1} B_{1} A_{1} K_{2} R_{2}} = \frac{42 \times 0.5 \times 1.8 \times 10^{- 3} \times 10 \times K}{K \times 14 \times 30 \times 0.25 \times 3.6 \times 10^{- 3}} = 1

Hence, the ratio of voltage sensitivity of M₂ to M₁ is 1.