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.

---

(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