question_answer

A galvanometer has a resistance 50 Q. A resistance of 5\[\Omega \] is connected parallel to it. Fraction of the total current flowing through galvanometer is :

A) \[\frac{1}{10}\]                                

B) \[\frac{1}{11}\]

C) \[\frac{1}{50}\]                                

D) \[\frac{2}{15}\]

Correct Answer: B

Solution :

                 The galvanometer G and shunt S is connected in parallel, hence potential difference is the same. \[\therefore \]  \[{{i}_{g}}\times G=(i-{{i}_{g}})\times S\] \[\Rightarrow \]               \[\frac{{{i}_{g}}}{i-{{i}_{g}}}=\frac{S}{G}=\frac{5}{50}=\frac{1}{10}\] \[\Rightarrow \]               \[10{{i}_{g}}=i-{{i}_{g}}\] \[\Rightarrow \]               \[11{{i}_{g}}=i\] \[\Rightarrow \]               \[\frac{{{i}_{g}}}{i}=\frac{1}{11}\]