Question

Two moving coil metres $M_{1}$ and $M_{2}$ have the following particular
$R_{1}=10\,\Omega; N_{1}=30; A_{1}=3.6\times 10^{-3}\, m^{2}; B_{1}=0.25\, T;$
$R_{2} =14\, \Omega; N_{2}=42; A_{2}=1.8 \times 10^{-3}\,m^{2}; B_{2}=0.50\, T$
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 $M_{1}$, $R_{1}=10\, \Omega; N_{1}=30$,
$A_{1}=3.6 \times 10^{-3}\, m^{2}; B_{1}=0.25\, T; k_{1}=k$
For meter $M_{2}, R_{2}=14\, \Omega ; N_{2}=42;$
$A_{2}=1.8 \times 10^{-3}\,m^{2};$ $B_{2}=0.50\, T;K_{2}=k$
So, current sensitivity $I_{s}=\frac{NBA}{k}$
$\frac{I_{s_2}}{I_{s_1}} = \frac{N_{2}B_{2}A_{2}/ k_{2}}{N_{1}B_{1}A_{1}k_{1}}$
$=\frac{42\times0.50\times1.8\times10^{-3}/k}{30\times0.25\times3.6\times10^{-3}/k}=1.4$
and voltage sensitivity $V_{s} = \frac{N_{2}BA}{KR}$
Now, $Now, \frac{V_{s_2}}{V_{s_1}}=\frac{N_{2}B_{2}A_{2} /(k_{2}R_{2})}{N_{1}B_{1}A_{1} /(k_{1}R_{1})}=\frac{N _{2}B_{2}A_{2}R_{1}K_{1}}{N_{1}B_{1}A_{1}R_{2}K_{2}}$
$=\frac{42\times0.50\times\left(1.8\times10^{-3}\right)\times10\times K}{30\times0.25\times\left(3.6\times10^{-3}\right)\times14\times K}=1$