# In a galvanometer there is a deflection of 10 divisions per mA. The internal resistance of the galvanometer is 78 ohm. If a shunt of 2 ohm is connected to the galvanometer and there are 75 divisions in all on the scale of the galvanometer,calculate the maximum current which the galvanometer can read.

$Ig = \frac{75}{10}, and, Imax = \frac{(Rg+Rs) Ig}{Rs} = \frac{(78+2) 75}{(78) 10} = \frac{100}{3}\boldsymbol{A}$
$Imax = \frac{(Rg+Rs) Ig}{Rs} = \frac{(78+2) 75}{(78) 10} = \frac{100}{3}\boldsymbol{A}$