The deflection in the galvanometer falls to ${\left(\frac{1}{5}\right)}^{th}$ when 4Ω resistance is connected in parallel with it. What will be the deflection if an additional resistance of 2Ω is connected in parallel with the above shunted galvanometer?

We know that,

$\frac{i_g}{i}=\frac{S}{S+G}$

Where S is shunt resistance, G is the resistance of galvanometer.

Here, $\frac{i_g}{i}=\frac{1}{5}$ and $S=4\Omega$

$\Rightarrow \frac{1}{5}=\frac{4}{4+G}$

4 + G = 20

G = 16$\Omega$

In new condition when 2Ω, resistance added in parallel to 4Ω resistance

$\frac{i_g}{i}=\frac{\frac{4}{3}}{\frac{4}{3}+16}$

$\frac{i_g}{i}=\frac{\frac{4}{3}}{\frac{4+48}{3}}=\frac{4}{52}$

$\frac{i_g}{i}={\left(\frac{1}{13}\right)}^{th}$