The section of the electric circuit consists of two parallel-connected conductors with resistances R1 = 3 Ohm

The section of the electric circuit consists of two parallel-connected conductors with resistances R1 = 3 Ohm and R2 = 6 Ohm, the current strength in the non-branched part of the circuit is I = 3A. Determine the power of the current passing through the second conductor.

The total resistance when two conductors are connected in parallel is determined by the formula:
R = (R1 * R2) / (R1 + R2)
R = (3 * 6) / (3 + 6) = 18/9 = 2 [Ohm]
According to Ohm’s law, you can calculate the voltage falling across the resistances using the formula:
U = I * R
U = 3 [A] * 2 [Ohm] = 6 [V]
On the second conductor 6V falls, and its resistance is 6 Ohm, we can calculate the power allocated to it by the formula:
P = U * U / R2, where U * U = U ^ 2, i.e. U squared, R2 is the resistance of the second conductor
P = 6 [B] * 6 [B] / [Ohm] = 36/6 = 6 [W]
Answer. 6W.