Question

A Galvanometer of resistance, G is shunted by a resistance S ohm. To keep main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
(A) $ \dfrac{{{S}^{2}}}{(S-G)} $
(B) $ \dfrac{SG}{(S-G)} $
(C) $ \dfrac{{{G}^{2}}}{(S-G)} $
(D) $ \dfrac{G}{(S-G)} $

Answer 1

Hint : Galvanometer is an electromechanical instrument used for detecting and indicating an electric current by deflection of a moving coil. Whenever the galvanometer is shunted by resistance “s”, then the current capacity increases n times.
Shunt is used to divert a part of current by connecting a circuit element in parallel with another.

Complete step by step answer:
Let us consider a galvanometer having resistance ‘G’ is shunted by resistance ‘S’ ohm. It means the resistance ‘S’ is connected with the resistance ‘G’ in parallel.

To keep the main current in the circuit unchanged, the resistance is to be put in series with the galvanometer and the initial resistance should be equal to the final resistance.
i.e.
 $ \begin{align}
  &\Rightarrow {{R}_{initial}}={{R}_{final}} \\
 &\Rightarrow {{R}_{initial}}=G \\
\end{align} $
 To find the resistance, we will use formula when resistance are in parallel or in series
For parallel combination, the resistance is given as:
 $\Rightarrow \dfrac{1}{{{R}_{P}}}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}} $
For series combination, the resistance is given as:
 $\Rightarrow {{R}_{S}}={{R}_{1}}+{{R}_{2}} $
Now, in our case
G and S are in parallel so the resistance is given by:
 $ $ $ \begin{align}
  &\Rightarrow \dfrac{1}{{{R}_{P}}}=\dfrac{1}{G}+\dfrac{1}{S} \\
 &\Rightarrow \dfrac{1}{{{R}_{P}}}=\dfrac{S+G}{GS} \\
 &\Rightarrow {{R}_{P}}=\dfrac{GS}{S+G} \\
\end{align} $
Now, $ {{R}_{P}} $ is in series with R so total resistance will be:
 $\Rightarrow {{R}_{final}}={{R}_{P}}+R $
To keep current unchanged, the initial resistance should be equal to final resistance
 $\Rightarrow {{\text{R}}_{final}}={{R}_{initial}} $
So,
 $\Rightarrow \text{G=}\dfrac{\text{GS}}{\text{S+G}}\text{+R} $
 $\Rightarrow R=G-\dfrac{GS}{S+G} $
 $\Rightarrow R=G\left[ 1-\dfrac{S}{S+G} \right] $
 $\Rightarrow R=G\left[ \dfrac{S+G-S}{S+G} \right] $
 $\Rightarrow R=\dfrac{{{G}^{2}}}{S+G} $
Therefore option (C) is correct.

Note:
It is very important to remember that when the resistance is in series the final resistance is equal to the simple addition of all resistance but when they are in parallel, conditions change. The final resistance in parallel combination is equal to the addition of the reciprocal of all resistances. Do not mix up this concept, otherwise our whole numerical will give the wrong answer. Shunt means a resistance is connected in parallel to another resistance.