Question

Two moving coil meters ${ M }_{ 1 }$ and ${ M }_{ 2 }$ have the following particulars
\[\begin{align}
  & {{R}_{1}}=10\Omega ;{{N}_{1}}=30;{{A}_{1}}=3.6\times {{10}^{-3}}{{m}^{2}};{{B}_{1}}=0.25T \\
 & {{R}_{2}}=14\Omega ;{{N}_{2}}=42;{{A}_{2}}=1.8\times {{10}^{-3}}{{m}^{2}};{{B}_{2}}=0.50T \\
\end{align}\]
The spring constants are identical for the two meters. What is the ratio of current sensitivity and voltage sensitivity of ${ M }_{ 1 }$ and ${ M }_{ 2 }$?
$\begin{align}
  & A)1.4,1 \\
 & B)1.4,0 \\
 & C)2.8,2 \\
 & D)2.8,0 \\
\end{align}$

Answer 1

Hint: Current sensitivity and voltage sensitivity of a galvanometer are directly proportional to the number of turns in the galvanometer coil, magnetic field induction and the area of galvanometer coil. At the same time, current sensitivity is inversely proportional to the couple per unit twist whereas voltage sensitivity is inversely proportional to the couple per unit twist as well as the resistance of the galvanometer coil.
Formula used:
$\begin{align}
  & 1){{I}_{s}}=NBA/k \\
 & 2){{V}_{s}}=NBA/kR \\
\end{align}$

Complete answer:
The deflection generated in a galvanometer when unit current flows through it is termed as current sensitivity of galvanometer. The deflection generated in a galvanometer when unit voltage is applied across its terminals is termed as voltage sensitivity of galvanometer. Both of these are directly proportional to the number of turns in the galvanometer coil, magnetic field induction and the area of galvanometer coil. At the same time, current sensitivity is inversely proportional to the couple per unit twist whereas voltage sensitivity is inversely proportional to the couple per unit twist as well as the resistance of the galvanometer coil.
Combining the above explanation with the given question, current sensitivities of ${ M }_{ 1 }$ and ${ M }_{ 2 }$ are given by
$\begin{align}
  & {{V}_{s1}}\quad =\quad {{N}_{1}}{{B}_{1}}{{A}_{1}}/k{{R}_{1}} \\
 & {{V}_{s2}}\quad =\quad {{N}_{2}}{{B}_{2}}{{A}_{2}}/k{{R}_{2}} \\
\end{align}$
where
${{V}_{s1}}$ and ${{V}_{s2}}$ are the voltage sensitivities of ${{M}_{1}}$ and ${{M}_{2}}$, respectively
${{N}_{1}}$ and ${{N}_{2}}$ are the number of turns in ${{M}_{1}}$ and ${{M}_{2}}$, respectively
${{B}_{1}}$ and ${{B}_{2}}$ are the magnetic inductions in ${{M}_{1}}$ and ${{M}_{2}}$, respectively
${{R}_{1}}$ and ${{R}_{2}}$ are the resistances of in ${{M}_{1}}$ and ${{M}_{2}}$, respectively
$k$ is the identical spring constants of both of ${{M}_{1}}$ and ${{M}_{2}}$, as provided
Let this be equation 3.
Using equation 3, ratio of current sensitivities is given by
$\dfrac{{{V}_{s2}}}{{{V}_{s1}}}=\dfrac{{{N}_{2}}{{B}_{2}}{{A}_{2}}{{R}_{1}}k}{{{N}_{1}}{{B}_{1}}{{A}_{1}}{{R}_{2}}k}=\dfrac{42\quad \times \quad 0.5\quad \times \quad 1.8\quad \times \quad {{10}^{-3}}\quad \times \quad 10\quad \times \quad k}{k\quad \times \quad 14\quad \times \quad 30\quad \times \quad 0.25\quad \times \quad 3.6\quad \times \quad {{10}^{-3}}}=\quad 1$
Let this be equation 4.

Therefore, from equations 2 and 4, we can conclude that the correct answer is option $A$.

Note:
From the solution, it is clear that voltage sensitivity of a galvanometer depends upon its resistance whereas current sensitivity of the same does not depend upon its resistance. Hence, one can increase the current sensitivity of a galvanometer by increasing the number of turns of the coil but it is to be noted that this will also increase the resistance of the coil.