Q. A moving coil galvanometer is converted into an ammeter reading upto $0.03\ A$ by connecting a shunt of resistance $4r$ across it and into an ammeter reading upto $0.06\ A$ when a shunt of resistance $r$ is connected across it. What is the maximum current which can be sent through this galvanometer if no shunt is used
A) $0.01\ A$
B) $0.02\ A$
C) $0.03\ A$
D) $0.04\ A$
Explanation
(b) $\frac{i_{g}}{i} = \frac{S}{G + S} \Rightarrow i_{g}G = \left( i - i_{g} \right)S$
$\therefore i_{g}G = \left( 0.03 - i_{g} \right)4r$ …(i)
and $i_{g}G = \left( 0.06 - i_{g} \right)r$ …(ii)
from (i) and (ii)
$0.12 - 4i_{g} = 0.06 - i_{g} \Rightarrow i_{g} = 0.02A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.