The deflection in a moving coil galvanometer is reduced to half when it is shunted with ' $X^{\prime}…
The deflection in a moving coil galvanometer is reduced to half when it is shunted
with ' $X^{\prime} \Omega$ coil. The relation between ' $\mathrm{X}^{\prime}$ and resistance of galvanometer ' $\mathrm{G}^{\prime}$ is
$2 X=G$
$4 X=G$
$X=2 G$
$X=G$
Solution
$\mathrm{I}_{\mathrm{G}} \cdot \mathrm{G}=\mathrm{I}_{\mathrm{x}} \cdot \mathrm{X}$
The total current gets equally divided between the galvanometer and the shunt.
$\therefore \mathrm{I}_{\mathrm{G}}=\mathrm{I}_{\mathrm{X}}$