A) \[\frac{{{I}_{g}}}{I}=\frac{4}{11}\]
B) \[\frac{{{I}_{g}}}{I}=\frac{3}{11}\]
C) \[\frac{{{I}_{g}}}{I}=\frac{2}{11}\]
D) \[\frac{{{I}_{g}}}{I}=\frac{1}{11}\]
Correct Answer: D
Solution :
In the circuit G and S are in parallel, the potential difference across them is same, hence \[{{I}_{g}}\times G=({{I}_{0}}-{{I}_{g}})\times S\] \[\Rightarrow \] \[\frac{{{I}_{g}}}{{{I}_{0}}}=\frac{S}{S+G}\] Given, \[S=2.5\,\Omega ,\,\,\,G=25\,\Omega \] \[\therefore \] \[\frac{{{I}_{g}}}{{{I}_{0}}}=\frac{2.5}{25+2.5}=\frac{2.5}{27.5}=\frac{1}{11}\] \[{{I}_{g}}=I\] \[\therefore \] \[\frac{{{I}_{g}}}{I}=\frac{1}{11}\]You need to login to perform this action.
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