Question

Two moving coil meters, $M_1$ and $M_2$ have the following particulars:
$R_1$ =$10\,\Omega$ , $N_1$ = 30,
$A_1$ = $3.6 × 10^{–3} m^2$, $B_1$ = $0.25\, T$
$R_2$ = $14 \,\Omega$, $N_2$ = 42,
$A_2$ = $1.8 × 10^{–3 }m^2$, $B_2$ = $0.50 T$
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of $M_2$ and $M_1$.

Answer 1

For moving coil meter $M_1$:
Resistance, $R_1$ =$10 \,Ω$
Number of turns, $N_1$ = $30$
Area of cross-section, $A_1$ = $3.6 × 10^{–3 }m^2$
Magnetic field strength, $B_1$ = $0.25 \,T$
Spring constant $K_1 = K$

For moving coil meter $M_2$:
Resistance, $R_2$ = $14 Ω$
Number of turns, $N_2$ =$42$
Area of cross-section, $A_2$ = $1.8 × 10^{–3} m^2$
Magnetic field strength, $B_2$ = $0.50 \,T$
Spring constant, $K_2 = K$
(a) Current sensitivity of $M_1$ is given as:
$I_{s1}$ = $\frac{N_1B_1A_1}{K_1}$
And, current sensitivity of $M_2$ is given as:
$I_{s2} =\frac{N_2B_2A_2}{k_2}$
Ratio$\frac{I_{s1}}{I_{s2}} = \frac{N_1B_1A_1}{N_2B_2A_2} = \frac{42 \times 0.5 \times 1.8 \times 10^{-3} \times K}{K \times30 \times 0.25 \times 3.6 \times 10^{-3 }}= 1.4$

Hence, the ratio of current sensitivity of $M_2$ to $M_1$ is 1.4.

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(b) Voltage sensitivity for $M_2$ is given as:
$V_{s2} =\frac{ N_2B_2A_2}{K_2R_2}$
And, voltage sensitivity for $M_1$ is given as:
$V_{s1}= \frac{N_1B_1A_1}{K_1R_1}$
Ratio $\frac{V_{s2}}{V_{s1}} = \frac{N_2B_2A_2K_1R_1}{N_1B_1A_1K_2R_2} =\frac{ 42 \times 0.5 \times1.8 \times 10^{-3} \times 10 \times K}{K \times 14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}}= 1$

Hence, the ratio of voltage sensitivity of $M_2$ to $M_1$ is 1