question_answer

We have a galvanometer of resistance 25 n. It is shunted by a 2.5 n wire, the part of total current that flows through the galvanometer is given as

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,   \[S=2.5\,\Omega ,\,\,\,G=25\,\Omega \] \[\therefore \]  \[\frac{{{I}_{g}}}{{{I}_{0}}}=\frac{2.5}{25+2.5}=\frac{2.5}{27.5}=\frac{1}{11}\]                                 \[{{I}_{g}}=I\] \[\therefore \]  \[\frac{{{I}_{g}}}{I}=\frac{1}{11}\]