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, \[\frac{{{i}_{g}}}{i}=\frac{4}{100}=0.04,\,\,S=5\Omega \] \[\therefore \] \[\frac{G}{5}=\frac{1}{0.05}-1=24\] \[\Rightarrow \] \[G=24\times 5=120\Omega \]You need to login to perform this action.
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