Q. A galvanometer of resistance $G$ can measure $1\ A$ current. If a shunt $S$ is used to convert it into an ammeter to measure $10A$ current. The ratio of $\frac{G}{S}$ is
A) $\frac{1}{9}$
B) $\frac{9}{1}$
C) 10
D) $\frac{1}{10}$
Explanation
(b) Here, $I_{g} = 1A$
$I = 10A$
[image:3_image183.emf]
From figure
$I_{g}G = (I - I_{g})S$
$\frac{G}{S} = \frac{I - I_{g}}{I_{g}}$
Substituting the given values, we get
$\frac{G}{S} = \frac{10A - 1A}{1A} = \frac{9}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.