Physics - Current Electricity Question with Solution | TestHub

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_{\mathrm{G}} = 1 \mathrm{A}$

$\therefore \mathrm{S}=\frac{1\times 22.8}{20-1}=\frac{22.8}{19}=1.2\mathrm{\Omega }$