A galvanometer has a current range of 15 mA and a voltage range of 750 mV. To convert this galvanometer into ammeter of range 25 ampere, the shunt required is

Given that the galvanometer has a current range of 15 mA i.e., i_g = maximum current measured 
$=15\times {10}^{-3}$ amp. Maximum voltage measured
= 750 x 10^-3 V
$\therefore$Galvanometer resistance
$G=\frac{750\times {10}^{-3}}{15\times {10}^{-3}}=50\mathrm{\Omega }$
Now $\mathrm{S}=\left(\frac{\mathrm{i}}{\mathrm{i}-{\mathrm{i}}_{\mathrm{g}}}\right)\mathrm{G}$
$=\frac{\left(15\times {10}^{-3}\right)\times 50}{25-\left(15\times {10}^{-3}\right)}\simeq 0\cdot 03\mathrm{\Omega }$