Q. 104.2( 25 Votes )

Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10Ω, N1 = 30,

A1 = 3.6 × 10–3 m2, B1 = 0.25 T

R2 = 14Ω, 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 :

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.


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NCERT - Physics Part-I