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. $\frac{1}{10}$ fold
• D. $\frac{1}{11}$ fold
  

Answer 1

Answer: (D)

$\frac{1}{11}$ fold

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

Initially, ${\alpha }_1=\theta /I_g$ (i)

Finally, after the shunt is used,

${\alpha }_f=\theta /I$ (ii)

$\therefore \frac{{\alpha }_f}{\alpha }=\frac{\theta /I}{\theta /I_g}=\frac{I_g}{I}=\frac{1}{11}$

So current sensitivity becomes 1/11-fold.