According to the sources, a galvanometer or millivoltmeter can be converted into an ammeter by connecting a low resistance called a shunt ($r_s$) in parallel . The shunt resistance is calculated using the relation $r_s = \frac{I_g R_g}{I - I_g}$, where $I_g R_g$ is the potential difference across the instrument for full-scale deflection ($V_g$), $I$ is the total current to be measured, and $I_g$ is the current through the instrument . In this problem, $V_g = 25 \text{ mV} = 25 \times 10^{-3} \text{ V}$ and the desired range $I = 25 \text{ A}$. Since the current through the millivoltmeter ($I_g$) is typically negligible compared to the total current ($I$), the formula can be approximated as $r_s \approx V_g / I$. Substituting the values: $r_s = (25 \times 10^{-3} \text{ V}) / 25 \text{ A} = 10^{-3} \, \Omega = 0.001 \, \Omega$.