# 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 $M_1$

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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