4.10 Two moving coil meters, M1 and M2 have the following particulars: R1 = 10 Ω, N1 = 30, A1 = 3.6 * 10–3 m2, B1 = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 * 10–3 m2, B2 = 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.

4.10 For moving coil meter M1 Current sensitivity of M 1 is given as I s 1 = N 1 B 1 A 1 K 1 and for M 2 is given as I s 2 = N 2 B 2 A 2 K 2 where K 1 and K 2 are spring constants for both M 1 & M 2 . It is given K 1 = K 2 The ratio of current sensitivity is given as I s 2 I s 1 = N 2 B 2 A 2 N 1 B 1 A 1 = 42 * 0.5 * 1.8 * 10 - 3 30 * 0.25 * 3.6 * 10 - 3 = 1.4 Voltage sensitivity of M 1 a n d M 2 is given by V S 1 = N 1 B 1 A 1 K 1 R 1 and V S 2 = N 2 B 2 A 2 K 2 R 2 The ratio of voltage sensitivity V S 2 V S 1 = N 2 B 2 A 2 K 1 R 1 N 1 B 1 A 1 K 2 R 2 = N 2 B 2 A 2 R 1 N 1 B 1 A 1 R 2 = 42 * 0.5 * 1.8 * 10 - 3 * 10 30 * 0.25 * 3.6 * 10 - 3 * 14 = 1

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

R1 = 10 Ω, N1 = 30,

**A1 = 3.6 * 10^–3 m², B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 * 10^–3 m², B2 = 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.**

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Payal Gupta | Contributor-Level 10

8 months ago

4.10 For moving coil meter M1

Current sensitivity of $M_{1}$ is given as $I_{s 1}$ = $\frac{N_{1} B_{1} A_{1}}{K_{1}}$ and

for $M_{2}$ is given as $I_{s 2}$ = $\frac{N_{2} B_{2} A_{2}}{K_{2}}$ where $K_{1}$ and $K_{2}$ are spring constants for both $M_{1}$ & $M_{2}$ . It is given $K_{1}$ = $K_{2}$

The ratio of current sensitivity is given as $\frac{I_{s 2}}{I_{s 1}}$ = $\frac{N_{2} B_{2} A_{2}}{N_{1} B_{1} A_{1}}$ = $\frac{42 * 0.5 * 1.8 * 10^{- 3}}{30 * 0.25 * 3.6 * 10^{- 3}}$ = 1.4

Voltage sensitivity of $M_{1} a n d M_{2}$ is given by $V_{S 1}$ = $\frac{N_{1} B_{1} A_{1}}{K_{1} R_{1}}$ and $V_{S 2}$ = $\frac{N_{2} B_{2} A_{2}}{K_{2} R_{2}}$