Question

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

1. current sensitivity and
2. voltage sensitivity of M₂ and M₁.

Answer 1

For moving coil meter M₁:

Resistance, R₁ = 10 Ω

Number of turns, N₁ = 30

Area of cross-section, A₁ = 3.6 × 10^–3 m²

Magnetic field strength, B₁ = 0.25 T

Spring constant K₁ = K

For moving coil meter M₂:

Resistance, R₂ = 14 Ω

Number of turns, N₂ = 42

Area of cross-section, A₂ = 1.8 × 10^–3 m²

Magnetic field strength, B₂ = 0.50 T

Spring constant, K₂ = K

(a) Current sensitivity of M₁ is given as:

`I_(s_1) = (N_1B_1A_1)/K_1`

And, the current sensitivity of M₂ is given as:

`I_(s_2) = (N_2B_2A_2)/K_2`

∴ Ratio `I_(s_2)/I_(s_1) = (N_2B_2A_2K_1)/(N_1B_1A_1K_2)`

= `(42 xx 0.5 xx 1.8 xx 10^-3 xx K)/(30 xx 0.25 xx 3.6 xx 10^-3 xx K)`

= `(37.8 xx 10^-3)/(27 xx 10^-3)`

= 1.4

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

(b) Voltage sensitivity for M₂ is given as:

`V_(s_2) = (N_2B_2A_2)/(K_2R_2)`

And, voltage sensitivity for M₁ is given as:

`V_(s_1) = (N_1B_1A_1)/(K_1R_1)`

∴ Ratio `V_(s_2)/V_(s_1) = (N_2B_2A_2K_1R_1)/(N_1B_1A_1K_2R_2)`

= `(42 xx 0.5 xx 1.8 xx 10^-3 xx K xx 10)/(30 xx 0.25 xx 3.6 xx 10^-3 xx K xx 14)`

= `(378 xx 10^-3)/(378 xx 10^-3)`

= 1

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