Question

Explain how a moving coil galvanometer is converted into an ammeter. Derive the necessary formula.

Answer 1

A moving coil galvanometer is converted into an ammeter by connecting a low resistance 'S' in parallel.
Let 'G' is the resistance of the galvanometer and 'S' is the low resistance connected in parallel. Since 'G' and 'S' are in parallel so, potential difference across them will be the same. So,
$i_g\times G=(i-i_g)S$
$\frac{i_g}{i}=\frac{s}{s+G}$

$s=$$\frac{i_{g}G}{i-i_g}$