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Sensitivity of moving coil galvanometer is $s$. If a shunt of $\left(\dfrac{1}{8}\right)^{th}$ of the resistance of galvanometer is connected to moving coil galvanometer, its sensitivity becomes:
\[\begin{align}
  & \text{A}\text{.}\dfrac{s}{3} \\
 & \text{B}\text{.}\dfrac{s}{6} \\
 & C.\dfrac{s}{9} \\
 & D.\dfrac{s}{12} \\
\end{align}\]

Answer
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Hint: A moving coil galvanometer is an electromagnetic device which is sensitive to the flow of small current in the circuit and is used to measure small current in the circuit. If follows the principle, that when a current carrying coil is kept in the proximity of a magnetic field it experiences a torque.

Formula used: $s=\dfrac{G}{n-1}$ where $G$ is a constant, which is the character of the galvanometer and $n$ is the range of the galvanometer.

Complete step by step answer:
A galvanometer is said to be sensitive, if it can show large deflections even when a small current flows through the circuit and thus sensitivity of the galvanometer is the ratio of change in deflection $d\theta$ to change in the current \[dI\]. This can be represented mathematically as $S=\dfrac{d\theta}{dI}$ .
We also know that, if $s$ is the sensitivity of the galvanometer, then $s=\dfrac{G}{n-1}$ where $G$ is a constant, which is the character of galvanometer and $n$ is the range of the galvanometer, which is the number of divisions the pointer moves to detect the small current.
Then, we can write, $S=\dfrac{G}{n-1}$, also given that $s=\dfrac{G}{8}$
Then, $\dfrac{G}{8}=\dfrac{G}{n-1}$
Or, $n-1=8$
Or $n=9$
Clearly, the range of the galvanometer increased to $9$.i.e. for small current, it points away $9$ divisions, then the sensitivity is said to become $S=\dfrac{s}{9}$

So, the correct answer is “Option C”.

Note: More deflection on the galvanometer, more the sensitivity of the galvanometer. The torque experienced by the current conductor in the magnetic field causes the pointer in the galvanometer to show deflection in the reading of the galvanometer.