Question

Two moving coil meters, M1 and M2 have the following particulars:R1 = 10 W, N1 = 30,A1 = 3.6 × 10–3 m2, B1 = 0.25 TR2 = 14 W, 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) voltagesensitivity of M2 and M1.

Answer 1

Given: The resistance ( R 1 ) of the moving coil meter M 1 is 10 Ω, number of turns N 1 is 30, area of cross-section ( A 1 ) is 3.6× 10 −3   m 2 and magnetic field strength ( B 1 ) is 0.25 T.

For the moving coil meter M 2 , the resistance ( R 2 ) is 14 Ω, number of turns N 2 is 42, area of cross-section ( A 2 ) is 1.8× 10 −3   m 2 , and magnetic field strength ( B 2 ) is 0.50 T.

(a)

Current sensitivity of M 1 is given as,

I 1 = N 1 B 1 A 1 K 1

And, current sensitivity of M 2 is given as,

I 2 = N 2 B 2 A 2 K 2

where, the spring constant for M 1 and M 2 are K 1 and K 2 , both are equal to K.

Ratio of ( I 2 )and ( I 1 ) is given as,

I 2 I 1 = N 2 B 2 A 2 K 2 × K 1 N 1 B 1 A 1 = 42×0.5×1.8× 10 −3 K × K 30×0.25×3.6× 10 −3 =1.4

Therefore, the ratio of current sensitivity of M 2 to M 1 is 1.4.

(b)

Voltage sensitivity of M 1 is given as,

V 1 = N 1 B 1 A 1 K 1 R 1

And, voltage sensitivity of M 2 is given as,

V 2 = N 2 B 2 A 2 K 2 R 2

where, the spring constant for M 1 and M 2 are K 1 and K 2 , both are equal to K.

Ratio of ( V 2 ) to ( V 1 ) is given as,

V 2 V 1 = N 2 B 2 A 2 K 2 R 2 × K 1 R 1 N 1 B 1 A 1 = 42×0.5×1.8× 10 −3 K×14 × K×10 30×0.25×3.6× 10 −3 =1

Therefore, the ratio of voltage sensitivity of M 2 to M 1 is 1.