Question

Two moving coil metres $M_1$ and $M_2$ have the following particular
$R_1=10\Omega ;N_1=30;A_1=3.6\times 10^{-3}{m}^2;B_1=0.25T;$
$R_2=14\Omega ;N_2=42;A_2=1.8\times 10^{-3}{m}^2;B_2=0.50T$
The spring constants are identical for the two metres. What is the ratio of current sensitivity and voltage sensitivity of $M_2$ to $M_1$ ?

• A. $1.4,1$
• B. $1.4,0$
• C. $2.8,2$
• D. $2.8,0$

Answer 1

Answer: (A)

For meter $M_1$, $R_1=10\Omega ;N_1=30$,
$A_1=3.6\times 10^{-3}{m}^2;B_1=0.25T;k_1=k$
For meter $M_2,R_2=14\Omega ;N_2=42;$
$A_2=1.8\times 10^{-3}{m}^2;$ $B_2=0.50T;K_2=k$
So, current sensitivity $I_s=\frac{NBA}{k}$
$\frac{I_{s_2}}{I_{s_1}}=\frac{N_2{B}_2{A}_2/k_2}{N_1{B}_1{A}_1{k}_1}$
$=\frac{42\times 0.50\times 1.8\times 10^{-3}/k}{30\times 0.25\times 3.6\times 10^{-3}/k}=1.4$
and voltage sensitivity $V_s=\frac{N_{2}BA}{KR}$
Now, $Now,\frac{V_{s_2}}{V_{s_1}}=\frac{N_2{B}_2{A}_2/(k_2{R}_2)}{N_1{B}_1{A}_1/(k_1{R}_1)}=\frac{N_2{B}_2{A}_2{R}_1{K}_1}{N_1{B}_1{A}_1{R}_2{K}_2}$
$=\frac{42\times 0.50\times \left(1.8\times 10^{-3}\right)\times 10\times K}{30\times 0.25\times \left(3.6\times 10^{-3}\right)\times 14\times K}=1$