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  • If a shunt of 1/10th of the coil resistance is applied to a moving coil galvanometer, its sensitivity becomesOption: 1 10-fold<div clas

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

Option: 1

10-fold


Option: 2

11-fold 


Option: 3

\frac{1}{10} - fold 


Option: 4

\frac{1}{11}  - fold


Answers (1)

best_answer

\mathrm{\frac{I_{\mathrm{g}}}{I}=\frac{S}{S+G} \Rightarrow \frac{I_{\mathrm{g}}}{I}=\frac{(G / 10)}{(G / 10)+G}=\frac{1}{11} }
Initially,\mathrm{ a_1=\theta / I_{\mathrm{g}} \quad }........... (i)
Finally, after the shunt is used,
\mathrm{ \alpha_f=\theta / I }........................(ii)

\mathrm{ \therefore \frac{a_f}{a_i}=\frac{\theta / I}{\theta / I_{\mathrm{g}}}=\frac{I_{\mathrm{g}}}{I}=\frac{1}{11} }
So, current sensitivity becomes 1 / 11-fold

Posted by

vinayak

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