Question

A galvanometer of resistance $22.8 \, \Omega $ measures $1 \, A$ . How much shunt should be used, so that it can be used to measure $20 \, A$ ?

• A. $1 \, \Omega $
• B. $2 \, \Omega $
• C. $1.2 \, \Omega $
• D. $2.2 \, \Omega $

Answer 1

Answer: (C)

Shunt is a low resistance used in parallel with the galvanometer to make it ammeter.

The voltage across galvanometer = voltage across the shunt
Given, $G = 22.8 \Omega , \, \, i = 20 A , \, \, $ $i_{G} \, = \, 1 \, A \, \, $
$\therefore S=\frac{1 \times 22 .8}{20 - 1}=\frac{22 .8}{19}=1.2\Omega $