Q. A galvanometer of resistance 22.8$\ \Omega$ measures 1A. How much shunt should be used, so that it can be used to measure 20A?
A) $1\Omega$
B) $2\Omega$
C) $1.2\Omega$
D) $2.2\Omega$
Explanation
(b) Shunt is a low resistance used in parallel with the galvanometer to make it ammeter.
The circuit is shown in figure.
Voltage across galvanometer = voltage across shunt
$ie.,\ \ \ i_{g}G = \left( i - i_{g} \right)S\ \ or\ S = \frac{i_{g}G}{i - i_{g}}$
Given, $G = 22.8\Omega,\ \ i = 20A,\ \ i_{g} = 1A\ $
$\therefore\ \ S = \frac{1 \times 22.8}{20 - 1} = \frac{22.8}{19} = 1.2\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.