To a current source with an internal resistance of 1 Ohm, an ammeter and a resistor with a resistance of 2

To a current source with an internal resistance of 1 Ohm, an ammeter and a resistor with a resistance of 2 Ohm were connected in series. In this case, the ammeter showed 1A. What will the ammeter show if you use a 3Ω resistor?

Let us first determine the EMF of the source using Ohm’s law for the complete circuit I = E / (R + r),
where I is the current strength, E is the EMF, R is the external resistance, r is the internal resistance of the source.
E = I * (R + r) = 1 A * (1 Ohm + 2 Ohm) = 3 V.
After replacing the resistance, the current in the circuit (EMF and internal resistance did not change):
I = 3 V / (3 Ohm + 1 Ohm) = 3 V / 4 Ohm = 0.75 A.
Answer: 0.75 A.