question_answer

In an ammeter, \[4%\] of the main current is passing through galvanometer. If the galvanometer is shunted with a \[5\Omega \] resistance, the resistance of the galvanometer is:

A) \[120\Omega \]                               

B) \[20\Omega \]

C) \[5\Omega \]                                    

D) \[4\Omega \]

Correct Answer: A

Solution :

Key Idea: Potential difference across galvanometer resistance and shunt is same. Let \[{{i}_{g}}\] be the current across galvanometer and \[i-{{i}_{g}}\] across shunt, then Potential difference across \[G=\] potential difference across\[S\] \[i.e.,\] \[{{i}_{g}}\times G=(i-{{i}_{g}})\times S\] Given,   \[\frac{{{i}_{g}}}{i}=\frac{4}{100}=0.04,\,\,S=5\Omega \] \[\therefore \]  \[\frac{G}{5}=\frac{1}{0.05}-1=24\] \[\Rightarrow \]               \[G=24\times 5=120\Omega \]