Discussion Forum

The sensitivity of the moving coil galvanometer is 'S'.If  a shunt of ($\frac{1}{8}$) the of the resistance of galvanometer is connected to moving coil galvanometer, its sensitivity becomes

Answer:

Explanation:

Given, S= $\frac{G}{8}$ ............(i)

As, S=$ \frac{G}{n-1}$     ............(ii)

Comparing Eqs.(i) and (ii), we get

 $\frac{G}{8}=\frac{G}{n-1}$

 $\therefore$   8=n-1

$\therefore$    n=8+1=9

 Since the range of the galvanometer is increased by nine times, therefore its sensitivity reduces to $\frac{1}{9}$

$\therefore$   S'= $\frac{S}{9}$