Q. The resistance of a galvanometer coil is R, then the shunt resistance required to convert it into a ammeter of range 4times, will be
A) 4R
B) R/3
C) R/4
D) R/5
Explanation
(b) Ammeter is made by connecting a low resistance shunt S in parallel with galvanometer G. since G and S are in parallel, the potential difference across them is same.
$i_{g} \times G = \left( i - i_{g} \right) \times S$
Given, $G = R,\ i = 4i_{g}$
$S = \frac{i_{g}}{4i_{g} - i_{g}} \times R = \frac{i_{g}}{3i_{g}} \times R = \frac{R}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.