Two moving coil meters, M₁ and M₂ have the following particulars: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³m², 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.

For moving coil meter M₁:

Resistance, R₁ = 10 QΩ

Number of turns, N₁ = 30

Area of cross-section, A₁ = 3.6 x 10⁻³ m²

Magnetic field strength, B₁ = 0.25 T

Spring constant K₁ = K

For moving coil meter M₂:

Resistance, R₂= 14 Q

Number of turns, N₂= 42

Area of cross-section, A₂ = 1.8 x 10⁻³ m²

Magnetic field strength, B₂ = 0.50 T

Spring constant, K₂ = K

Ans (a).

Current sensitivity of M₁ is given by

Is₁ = N₁B₁A₁/K₁

and, current sensitivity of M₂ is given by

Is₂ = N₂B₂A₂/K₂

Therefore Ratio  Is₂ /Is₁  = N₂B₂A₂ /N₁B₁A₁    (Since Spring constant, K₂ = K)

=>  Is₂ /Is₁  =  (42  x  0.5 x 1.8 x 10⁻³ x K)/(K x 30 x 0.25 x 3.6 x 10⁻³ )  = 1.4

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

Ans (b).

Voltage sensitivity for M₂ is given by

Vs₂ = N₂B₂A₂/ K₂ R₂

And Volatage sensitivity for M₁ is given by

Vs₁ = N₁B₁A₁/K₁R₁

Therefore Ratio Vs₂/ Vs₁ =  N₂B₂A₂ K₁R₁/ K₂ R₂ N₁B₁A₁

Vs₂/ Vs₁  = (42 x 0.5 x 1.8 x 10⁻³ x 10 x K)/(K x 14 x 30 x 0.25 x 3.6 x 10⁻³ )

=>  Vs₂/ Vs₁ = 1

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