In the circuit (Figure) the current is to be measured. What is the value of the current if the ammeter shown:
$(a)$ is a galvanometer with a resistance $R_{G}=60.00 \; \Omega$;
$(b)$ is a galvanometer described in $(a)$ but converted to an ammeter by a shunt resistance $r_{s}=0.02 \; \Omega$;
$(c)$ is an ideal ammeter with zero resistance?

(N/A) Total resistance in the circuit is $R_{total} = R_{G} + 3.00 \; \Omega = 60.00 \; \Omega + 3.00 \; \Omega = 63.00 \; \Omega$.
Using Ohm's law,the current $I = V / R_{total} = 3.00 \; \text{V} / 63.00 \; \Omega \approx 0.0476 \; \text{A} \approx 0.048 \; \text{A}$.
$(b)$ The resistance of the galvanometer converted to an ammeter is $R_{A} = \frac{R_{G} \cdot r_{s}}{R_{G} + r_{s}} = \frac{60.00 \; \Omega \times 0.02 \; \Omega}{60.00 \; \Omega + 0.02 \; \Omega} = \frac{1.2}{60.02} \; \Omega \approx 0.01999 \; \Omega \approx 0.02 \; \Omega$.
Total resistance in the circuit is $R_{total} = R_{A} + 3.00 \; \Omega = 0.02 \; \Omega + 3.00 \; \Omega = 3.02 \; \Omega$.
Using Ohm's law,the current $I = V / R_{total} = 3.00 \; \text{V} / 3.02 \; \Omega \approx 0.993 \; \text{A} \approx 0.99 \; \text{A}$.
$(c)$ For an ideal ammeter,the resistance is zero. Thus,the total resistance is $3.00 \; \Omega$.
The current $I = V / R = 3.00 \; \text{V} / 3.00 \; \Omega = 1.00 \; \text{A}$.