Two moving coil meters $M_1$ and $M_2$ having the following particulars :-
$R_1 = 10\,\Omega , N_1 = 30, A_1 = 3.6\times10^{-3}\, m^2, B_1 = 0.25\, T$
$R_2 = 14\,\Omega , N_2 = 42, A_2 = 1.8\times10^{-3}\, m^2, B_2 = 0.50\, T$
(The spring constants are identical for the two meters). Determine the ratio of voltage sensitivity of $M_2$ and $M_1$
$R_1 = 10\,\Omega , N_1 = 30, A_1 = 3.6\times10^{-3}\, m^2, B_1 = 0.25\, T$
$R_2 = 14\,\Omega , N_2 = 42, A_2 = 1.8\times10^{-3}\, m^2, B_2 = 0.50\, T$
(The spring constants are identical for the two meters). Determine the ratio of voltage sensitivity of $M_2$ and $M_1$
Medium
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Using the formula of voltage sensitivity,
$\mathrm{V}=\frac{\mathrm{NAB}}{\mathrm{kR}}$
$\therefore \frac{V_{s_{2}}}{V_{s_{1}}}=\frac{n_{2} B_{2} A_{2} \cdot k_{1} R_{1}}{k_{2} R_{2} n_{1} B_{1} A_{1}}$
$=\frac{42 \times 0.50 \times 1.8 \times 10^{-3} \times \mathrm{k} \times 10}{\mathrm{k} \times 14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}}=1$
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