Conversion of galvanometer into ammeter: Ammeter is connected in series in a circuit. So, its resistance should be such that it does not increase the equivalent resistance value of the circuit appreciably. So, to convert a galvanometer into an ammeter a very low resistance value shunt is to be connected in parallel with the galvanometer. If potential difference between points $X$ and $Y$ be $V_x$ $-V_y$, then $\;V_x-V_y\;=I_g \times G = I_s \times S$ $\;\;\text{Or,}\;\;I_g \times G\;=I_s \times S$ $\therefore \;S\;=\;\frac{I_g}{I_s} \times G$ .......(1) $\;\;\text{Again}\;\;I\;=I_s+I_g$ $\therefore \;I_s\;=I-I_g$ Putting in equation (1) $S=\;\frac{I_g}{I-I_g} \times G$