Two moving coil meters, M_1 and M_2 have the following particulars:


R_1 = 10 \Omega , N_1 = 30,\\\\ A_1 = 3.6 \times 10^{-3} m^2, B_1 = 0.25 T\\\\ R_2 = 14 \Omega , N_2 = 42,\\\\ A_2 = 1.8 \times 10^{-3} m^2, B_2 = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of  current sensitivity of M_2\ and \ \ M_1

Answers (1)
S Sayak

The torque experienced by the moving coil M1 for a current I passing through it will be equal to \tau =B_{1}A_{1}N_{1}I

The coil will experience a restoring torque proportional to the twist \phi

\phi k=B_{1}A_{1}N_{1}I

The current sensitivity is therefore \frac{B_{1}A_{1}N_{1}}{k}

Similarly, for the coil M2, current sensitivity is \frac{B_{2}A_{2}N_{2}}{k}

Their ratio of current sensitivity of coil M2 to that of coil M1 is, therefore,  \frac{B_{2}A_{2}N_{2}}{B_{1}A_{1}N_{1}}

=\frac{0.5\times 1.8\times 10^{-3}\times 42 }{0.25\times 3.6\times 10^{-3}\times 30}=1.4

 

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