Two conductors are connected in parallel, the resistance of the second is 9 Ohm, the current in the supply

Two conductors are connected in parallel, the resistance of the second is 9 Ohm, the current in the supply circuit is 0.6A. The strength of the current passing through the first conductor is 0.2 A. What is the resistance of this section and the voltage at its ends?

With a parallel connection, the current in the sections of the circuit is added up:
I = I1 + I2, where I is the current in the supply circuit, I1 is the current in the first conductor, I2 is the current in the second conductor
From here we find the current in the second conductor:
I2 = I – I1 = 0.6 [A] – 0.2 [A] = 0.4 [A]

The voltage at the ends of the section can be calculated according to Ohm’s law:
U = I * R, where I is the current in the section of the circuit, R is the resistance of this section
U2 = I2 * R2 = 0.4 [A] * 9 [Ohm] = 3.5 [V]

Since the conductors are connected in parallel, the voltages at their ends are equal.
U1 = U2 = 3.6 [B], where U1 and U2 are the voltage applied to the first and second conductors

Knowing the voltage on the conductor and the current through it, we calculate the resistance of the entire section:
R = U / I. where R is the total resistance of the circuit section, U and I are the voltage and current of the circuit section
R = 3.6 [V] / 0.6 [A] = 6 [Ohm]

Answer: R = 6 [Ohm]; U = 3.6 [B]