Two moving coil meters, M 1 and M 2 have the following particulars: R 1 = 10Ω, N 1 = 30, A 1 = 3.6 × 10 –3 m 2 , B 1 = 0.25 T R 2 = 14Ω, N 2 = 42, A 2 = 1.8 × 10 –3 m 2 , B 2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M 2 and M 1 .

Question

Philoid · Accepted Answer

Given:

For moving coil meter M₁

Resistance of wire, R₁ = 10Ω

Number of turns, N₁ = 30

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

Magnetic field strength, B₁ = 0.25 T

For moving coil meter M₂

Resistance of wire, R₂ = 14Ω

Number of turns, N₂ = 42

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

Magnetic field strength, B₂ = 0.50 T

Spring constant, K₁ = K₂ = K

a) Current sensitivity is given by,

For M₁,

…(1)

By putting the values in equation 1, we have

⇒

For M₂

…(2)

By putting the values in equation 2, we have

⇒

Now finding the ration of I₁ and I₂

⇒

Hence, the ratio of current sensitivities is 1.4.

b) Voltage sensitivity is given by,

For M₁,

⇒ …(3)

For M₂

⇒ …(4)

By dividing equation 3 by 4, we get

⇒

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