# Two moving coil meters,

Question:

Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10 Ω, N1 = 30,

A1 = 3.6 × 10–3 m2B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 × 10–3 m2B2 = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

Solution:

For moving coil meter M1:

Resistance, R1 = 10 Ω

Number of turns, N1 = 30

Area of cross-section, A1 = 3.6 × 10–3 m2

Magnetic field strength, B1 = 0.25 T

Spring constant K1 = K

For moving coil meter M2:

Resistance, R2 = 14 Ω

Number of turns, N2 = 42

Area of cross-section, A2 = 1.8 × 10–3 m2

Magnetic field strength, B2 = 0.50 T

Spring constant, K2 = K

(a) Current sensitivity of M1 is given as:

$I_{\mathrm{s} 1}=\frac{N_{1} B_{1} A_{1}}{K_{1}}$

And, current sensitivity of $M_{2}$ is given as:

$I_{\mathrm{s} 2}=\frac{N_{2} B_{2} A_{2}}{K_{2}}$

$\therefore$ Ratio $\frac{I_{s 2}}{I_{\mathrm{sl}}}=\frac{N_{2} B_{2} A_{2} K_{1}}{K_{2} N_{1} B_{1} A_{1}}$

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

Hence, the ratio of current sensitivity of M2 to M1 is 1.4.

(b) Voltage sensitivity for M2 is given as:

$V_{\mathrm{s} 2}=\frac{N_{2} B_{2} A_{2}}{K_{2} R_{2}}$

And, voltage sensitivity for M1 is given as:

$V_{s 1}=\frac{N_{1} B_{1} A_{1}}{K_{1} R_{1}}$

$\therefore$ Ratio $\frac{V_{s 2}}{V_{s 1}}=\frac{N_{2} B_{2} A_{2} K_{1} R_{1}}{K_{2} R_{2} N_{1} B_{1} A_{1}}=\frac{42 \times 0.5 \times 1.8 \times 10^{-3} \times 10 \times K}{K \times 14 \times 30 \times 0.25 \times 3.6 \times 10^{-3}}=1$

Hence, the ratio of voltage sensitivity of M2 to M1 is 1.