The circle consists of a current source with an EMF of 5.4 V and an internal resistance of 1.5 Ohm and consumers

The circle consists of a current source with an EMF of 5.4 V and an internal resistance of 1.5 Ohm and consumers with supports of 4.5 Ohm and 3 Ohm. What is the voltage at the second consumer if they are connected in series in the circuit.

EMF = 5.4 V.
r = 1.5 ohms.
R1 = 4.5 ohms.
R2 = 3 ohms.
U2 -?
Let us write Ohm’s law for a closed electric circuit: I = EMF / (r + R), where I is the current in the circuit, EMF is the electromotive force of the current source, r is the internal resistance of the current source, R is the external resistance of the circuit.
With a series connection of resistances, the following relations are true: R = R1 + R2, I = I1 = I2.
I = EMF / (r + R1 + R2).
I = 5.4 V / (1.5 Ohm + 4.5 Ohm + 3 Ohm) = 0.6 A.
Let’s write Ohm’s law for the second resistance: I = U2 / R2.
U2 = I * R2.
U2 = 0.6 A * 3 Ohm = 1.8 V.
Answer: the voltage across the second resistance U2 = 1.8 V.