Q. We have a galvanometer of resistance $25\Omega.$ It is shunted by a $2.5\ \Omega$ wire. The part of total current that flows through the galvanometer is given as
A) $\frac{I}{I_{0}} = \frac{1}{11}$
B) $\frac{I}{I_{0}} = \frac{1}{10}$
C) $\frac{I}{I_{0}} = \frac{3}{11}$
D) $\frac{I}{I_{0}} = \frac{4}{11}$
Explanation
(a) $\frac{i}{i_{g}} = \frac{G + S}{S} \Rightarrow \frac{i_{g}}{i} = \frac{S}{G + S} = \frac{2.5}{27.5} = \frac{1}{11}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.