In a circuit for finding the resistance of a galvanometer by half deflection method, a 6   V battery…

We are given a galvanometer connected in the first circuit in the above figure, where $E=6 \mathrm{V}$, $R = 11 \mathrm{k\Omega }$ and the deflection $\theta = 9$ divisions. The galvanometer has a resistance of $G$. The current in the circuit,

$I=\frac{E}{R+G} ...\left(1\right)$

Also, since the figure of merit is $60 \mathrm{\mu A}/\mathrm{division}$

$I = 60\times 9 \mathrm{\mu A} = 540 \mathrm{\mu A} ...\left(2\right)$

Substituting equation $\left(2\right)$ in $\left(1\right)$,

$$\begin{gathered} 540\times {10}^{-6} = \frac{6}{11,000+G} \\ \Rightarrow G = \frac{6}{540\times {10}^{-6}}-11,000 \\ \Rightarrow G = 111 \mathrm{\Omega } ...\left(3\right) \end{gathered}$$

In the second circuit, it is given that the deflection is half that of the previous circuit. So, current flowing through the galvanometer half the first circuit. Hence,

$$\begin{gathered} \frac{I}{2}= \frac{E}{R+\frac{GS}{G+S}} \times \frac{S}{S+G} \\ \Rightarrow \frac{I}{2}=\frac{ES}{R\left(S+G\right)+GS} \end{gathered}$$

$\Rightarrow S= \frac{RG \times \frac{I}{2}}{E-\frac{(R+G)I}{2}}$

Substituting the values from equations $\left(1\right), \left(2\right)$ and $\left(3\right)$,

$$\begin{gathered} S= \frac{11 \times {10}^3 \times 111 \times {\displaystyle \frac{540}{2}} \times {10}^{-6}}{6-\left(\frac{6}{2}\right)} \\ =110 \mathrm{\Omega } \end{gathered}$$