A galvanometer of resistance \(G \Omega\) is converted into an ammeter of range 0 to IA. If the current through the galv…

\begin{aligned}
& \text{Given: Galvanometer resistance } = G\ \Omega, \quad \text{Full-scale current } = I\ \mathrm{A}, \\[3pt]
& \text{Current through galvanometer } = 0.1\% \text{ of } I = \frac{0.1}{100}I = \frac{I}{1000}. \\[6pt]
& \text{Let shunt resistance } = S. \text{ Then the same potential difference acts across } G \text{ and } S. \\[3pt]
& \text{So, } I_g G = I_s S, \quad \text{where } I_s = I - I_g = I - \frac{I}{1000} = \frac{999I}{1000}. \\[4pt]
& \Rightarrow S = \frac{I_g G}{I_s} = \frac{\frac{I}{1000} G}{\frac{999I}{1000}} = \frac{G}{999}. \\[6pt]
& \text{The equivalent resistance of the ammeter is } R_A = G \parallel S \\[4pt]
& R_A = \frac{G \times S}{G + S} = \frac{G \times \frac{G}{999}}{G + \frac{G}{999}} = \frac{G^2 / 999}{G(1 + 1/999)} \\[4pt]
& R_A = \frac{G / 999}{1 + 1/999} = \frac{G / 999}{1000/999} = \frac{G}{1000}. \\[6pt]
& \boxed{R_A = \frac{G}{1000}} \\[4pt]
& \text{Hence, the resistance of the ammeter is } \boxed{\tfrac{G}{1000}\ \Omega.}
\end{aligned}