A millivoltmeter of $$25\,mV$$  range is to be converted into an ammeter of $$25\,A$$  range. The value (in ohm) of necessary shunt will be

Solution :
The full scale deflection current
$${i_g} = \frac{{25\,mV}}{G}A$$
where, $$G$$ is the resistance of the meter.
The value of shunt required for converting it into ammeter of range $$25\,A$$  is $$S = \frac{{{i_g}G}}{{i - {i_g}}}$$
$$ \Rightarrow S = {i_g}\frac{G}{i}\,\,\,\left( {{\text{as}}\,i > > {i_g}} \right)$$
So that, $$S \approx \frac{{25\,mV}}{G} \cdot \frac{G}{i} = \frac{{25\,mV}}{{25}} = 0.001\,\Omega $$