Question

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

• A. $10$ fold
• B. $11$ fold
• C. $1/10$ fold
• D. $1/11$ fold

Answer 1

Answer: (D)

$\frac{I_g}{I}=\frac{S}{S+G}=\frac{(G/10)}{(G/10)+G}=\frac{1}{11}$
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.