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, \[\frac{{{i}_{g}}}{i}=\frac{4}{100}=0.04,S=5\Omega \] \[\therefore \] \[\frac{G}{5}=\frac{1}{0.04}-1=24\] \[\Rightarrow \] \[G=24\times 5=120\,\Omega \]