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  4. If a shunt 1 / 10 of the coil resistance is applied to a moving coil galvanometer, its sensitivity becomes

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Q. If a shunt $1 / 10$ of the coil resistance is applied to a moving coil galvanometer, its sensitivity becomes

Current Electricity

Solution:

$\frac{I_{g}}{I}=\frac{S}{S+G}=\frac{(G / 10)}{(G / 10)+G}=\frac{1}{11}$
Initially, $\alpha_{1}=\theta / I_{g}\,\,\,$(i)
Finally, after the shunt is used,
$\alpha_{f}=\theta / I\,\,\,$(ii)
$\therefore \frac{\alpha_{f}}{\alpha_{i}}=\frac{\theta I}{\theta I_{g}}$
$=\frac{I_{g}}{I}=\frac{1}{11}$
So current sensitivity becomes $1 / 11$-fold.