Determine the voltage at the ends of a 100 cm long aluminum conductor with a cross-sectional area

Determine the voltage at the ends of a 100 cm long aluminum conductor with a cross-sectional area of 0.09mm ^ 2, in which the current is 0.9A. The specific resistance of iron is 0.025 Ohm mm ^ 2 / m.

L = 100 cm = 1 m.

S = 0.09 mm2.

I = 0.9 A.

ρ = 0.025 Ohm * mm2 / m.

U -?

According to Ohm’s law for a section of a circuit, the voltage at the ends of the conductor U is equal to the product of the current in the conductor I by its resistance R: U = I * R.

Since the conductor is homogeneous and has a cylindrical shape, its resistance R is expressed by the formula: R = ρ * L / S, where ρ is the resistivity of the material from which the conductor is made, L is the length of the conductor, S is the cross-sectional area of ​​the conductor.

U = I * ρ * L / S.

U = 0.9 A * 0.025 Ohm * mm2 / m * 1 m / 0.09 mm2 = 0.25 V.

Answer: at the ends of the aluminum conductor, the voltage is U = 0.25 V.