Question

Two moving coil meters, $M_1$ and $M_2$ have the following particulars:
$R_1=10\Omega ,N_1=30$
$A_1=3.6\times {10}^{-3}{m}^2,B_1=0.25T$
$R_2=14\Omega ,N_2=42$
$A_2=1.8\times {10}^{-3},B_2=0.50T$
(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

(a)

Current sensitivity of $M_1$ is given as,

$I_{s1}=N_1{B}_1{A}_1/k_1$

$I_{s2}=N_2{B}_2{A}_2/k_2$

where,

N=turns,

B=Magnetic field,

k=spring constant,

A=area

R=Electric resistance.

$I_{s1}/I_{s2}=1.4$

(b)

Voltage sensitivity relation is given as,

$V_{s1}=N_1{B}_1{A}_1/k_1{R}_1$

$V_{s2}=N_2{B}_2{A}_2/k_2{R}_2$

where,

N=turns,

B=Magnetic field,

k=spring constant,

A=area

R=Electric resistance.

$V_{s1}/V_{s2}=1$