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, \[\frac{{{i}_{g}}}{i}=\frac{4}{100}=0.04,S=5\Omega \] \[\therefore \] \[\frac{G}{5}=\frac{1}{0.04}-1=24\] \[\Rightarrow \] \[G=24\times 5=120\,\Omega \]You need to login to perform this action.
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