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 (a) current sensitivity and (b) 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"_1"B"_1"A"_1)/"K"_1`

And, current sensitivity of M₂ is given as:

`"I"_("s"_2) = ("N"_2"B"_2"A"_2)/"K"_2`

∴ Ratio `"I"_("s"_2)/"I"_("s"_1) = ("N"_2"B"_2"A"_2"K"_1)/("K"_2"N"_1"B"_1"A"_1)`

= `(42 xx 0.5 xx 1.8 xx 10^-3 xx "K")/("K" xx 30 xx 0.25 xx 3.6 xx 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) = ("N"_2"B"_2"A"_2)/("K"_2"R"_2)`

And, voltage sensitivity for M₁ is given as:

`"V"_("s"_1) = ("N"_1"B"_1"A"_1)/("K"_1"R"_1)`

∴ Ratio `"V"_("s"_2)/"V"_("s"_1) = ("N"_2"B"_2"A"_2"K"_1"R"_1)/("K"_2"R"_2"N"_1"B"_1"A"_1)`

= `(42 xx 0.5 xx 1.8 xx 10^-3 xx 10 xx"K")/("K" xx 14 xx 30 xx 0.25 xx 3.6 xx 10^-3) = 1`

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