10.(b) 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 voltage 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

we know V=IR

Therefore, \phi k=\frac{B_{1}A_{1}N_{1}V}{R_{1}}

Voltage sensitivity of coil M=\frac{B_{1}A_{1}N_{1}}{kR_{1}}

Similarly for coil M2 Voltage sensitivity = \frac{B_{2}A_{2}N_{2}}{kR_{2}}

Their ratio of voltage sensitivity of coil M2 to that of coil M1 

 =\frac{B_{2}A_{2}N_{2}R_{1}}{B_{1}A_{1}N_{1}R_{2}}

=1.4\times \frac{10}{14}

\\=1

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