Physics - Current Electricity Question with Solution | TestHub

Let the resistance of the galvanometer be $G$ and the main current be $I$.

When the galvanometer is shunted with $S_1=2\Omega$, the deflection is reduced to one-fourth. This means the current through the galvanometer is $I_{g1}=\frac{I}{4}$.

Using the formula for current division, $I_{g1}=\frac{S_1}{G+S_1}I$

$\frac{I}{4}=\frac{2}{G+2}I\Rightarrow G+2=8\Rightarrow G=6\Omega$

When the galvanometer is further shunted with a $1\Omega$ resistance, this new resistor is connected in parallel with the existing $2\Omega$ shunt. The new equivalent shunt resistance $S_2$ is:

$S_2=\frac{2\times 1}{2+1}=\frac{2}{3}\Omega$

The new current through the galvanometer $I_{g2}$ is:

$I_{g2}=\frac{S_2}{G+S_2}I=\frac{\frac{2}{3}}{6+\frac{2}{3}}I=\frac{\frac{2}{3}}{\frac{20}{3}}I=\frac{I}{10}$

The further reduction in deflection is proportional to the decrease in the galvanometer current:

$\Delta I_g=I_{g1}-I_{g2}=\frac{I}{4}-\frac{I}{10}=\frac{5I-2I}{20}=\frac{3I}{20}$

This reduction is given as a fraction $x$ of the deflection observed when shunted with $2\Omega$ ($I_{g1}$).

$x=\frac{\Delta I_g}{I_{g1}}=\frac{\frac{3I}{20}}{\frac{I}{4}}=\frac{3}{20}\times 4=\frac{3}{5}$

Answer: $\frac{3}{5}$