Q. A galvanometer of resistance $36\ \Omega$ is changed into an ammeter by using a shunt of $4\ \Omega.$ The fraction $f_{0}$ of total current passing through the galvanometer is
A) $\frac{1}{40}$
B) $\frac{1}{4}$
C) $\frac{1}{140}$
D) $\frac{1}{10}$
Explanation
(d) $\frac{i_{g}}{i} = \frac{S}{G + S} = \frac{4}{36 + 4} = \frac{4}{40} = \frac{1}{10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.