Q. We have a galvanometer of resistance 25 $\Omega$. It is shunted by a 2.5 $\Omega$ wire. The part of total current $i_{0}$ that flows through the galvanometer is given as
A) $\left( i/i_{0} \right) = (1/11)$
B) $\left( i/i_{0} \right) = (1/10)$
C) $\left( i/i_{0} \right) = (1/9)$
D) $\left( i/i_{0} \right) = (2/11)$
Explanation
(c) $i_{g} = \ \frac{iS}{G + S} = \frac{i_{0} \times 2.5}{25 + 2.5} = \frac{i_{0}}{9}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.