$A$ multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure $10 \text{ mA}$,$100 \text{ mA}$,and $1 \text{ A}$ using a galvanometer of resistance $10 \text{ } \Omega$ that produces maximum deflection for a current of $1 \text{ mA}$. Find the values of $S_1, S_2$,and $S_3$ that have to be used.

(D) To convert a galvanometer into an ammeter,a low-value resistance (shunt) is connected in parallel. Let $G = 10 \text{ } \Omega$ be the galvanometer resistance and $I_g = 1 \text{ mA} = 0.001 \text{ A}$ be the full-scale deflection current.
For range $I_1 = 10 \text{ mA} = 0.01 \text{ A}$ (between terminals $A$ and $B$): The shunt resistance is $S_1 + S_2 + S_3$. The current through the shunt is $I_1 - I_g = 0.01 - 0.001 = 0.009 \text{ A}$.
$I_g G = (I_1 - I_g)(S_1 + S_2 + S_3) \implies 0.001 \times 10 = 0.009(S_1 + S_2 + S_3) \implies S_1 + S_2 + S_3 = \frac{10}{9} \approx 1.11 \text{ } \Omega$.
For range $I_2 = 100 \text{ mA} = 0.1 \text{ A}$ (between terminals $A$ and $C$): The shunt resistance is $S_2 + S_3$. The current through the shunt is $I_2 - I_g = 0.1 - 0.001 = 0.099 \text{ A}$.
$I_g(G + S_1) = (I_2 - I_g)(S_2 + S_3) \implies 0.001(10 + S_1) = 0.099(S_2 + S_3)$.
For range $I_3 = 1 \text{ A}$ (between terminals $A$ and $D$): The shunt resistance is $S_3$. The current through the shunt is $I_3 - I_g = 1 - 0.001 = 0.999 \text{ A}$.
$I_g(G + S_1 + S_2) = (I_3 - I_g)S_3 \implies 0.001(10 + S_1 + S_2) = 0.999 S_3$.
Solving these equations simultaneously:
$S_3 = \frac{10}{999} \approx 0.0101 \text{ } \Omega$,
$S_2 = \frac{10}{99} - S_3 \approx 0.101 - 0.0101 = 0.0909 \text{ } \Omega$,
$S_1 = \frac{10}{9} - (S_2 + S_3) = \frac{10}{9} - \frac{10}{99} = \frac{100}{99} \approx 1.0101 \text{ } \Omega$.