question_answer

When a shunt of resistance \[12\,\,\Omega \] is connected with a galvanometer, its deflection reduces from 50 div to 10 div. The resistance of the coil of galvanometer is:

A)  \[48\,\,\Omega \]         

B)  \[24\,\,\Omega \]

C)  \[56\,\,\Omega \]          

D)  \[36\,\,\Omega \]

Correct Answer: A

Solution :

 Key Idea: Potential difference across shunt and galvanometer resistance is same. Let G be resistance of galvanometer and \[{{i}_{g}}\] the current through it. Since, G and S are in parallel, we have Ammeter \[{{i}_{g}}\times G=(i-{{i}_{g}})\times S\] \[\Rightarrow \] \[\frac{{{i}_{g}}}{i}=\frac{S}{S+G}\] \[\therefore \] \[\frac{10}{50}=\frac{12}{12+G}\] \[\Rightarrow \] \[12+G=60\] \[\Rightarrow \] \[G=48\,\Omega \]