An electric iron with a power of 1 kW is powered from a 220V network. The length of the connecting cord is 1.5 m

An electric iron with a power of 1 kW is powered from a 220V network. The length of the connecting cord is 1.5 m. The power loss in the cord is 0.1% of the power of the iron. Determine the diameter of each of the copper wires in the cord. How do you find the amount of copper wires in a cord?

We translate the values ​​from given to the SI system:
P = 1 kW = 1000 W.
L = 1.5 m.
Pp = 0.001 * P.

1. Let’s find the losses on the cord:
Pp = 0.001 * P = 0.001 * 1000 = 1 W.
2. Losses in the cord are generated as heat, hence:
Pп = I² * R, where I is the current in the circuit, R is the resistance of the element, in our case the wires.
3. To determine the current in the circuit, we use the expression:
P = U * I, where I is the current in the circuit, U is the voltage.
Let us express the current strength from this expression:
I = P / U = 1000/220 = 4.55 A.
4. Now we express the resistance from the expression for the power loss:
R = Pp / I² = 1 / 4.55² = 0.048 Ohm.
5. Expression for determining the resistance of a wire by specific resistance:
R = ρsp * l / Ssech, where R is the resistance of the wire, l is the length of the wire, Ssec is the cross-sectional area, ρsp is the resistivity of the wire, ρ copper = 1.7 * 10–8 Ohm * m.
Let us express the cross-sectional area from this expression:
Ssection = ρsp * l / R.
Our wire consists of two conductors, which means l = 2 * L, then:
Ssection = ρsp * 2 * L / R = 1.7 * 10–8 * 2 * 1.5 / 0.048 = 1 * 10 ^ -6 m².
7. To determine how many conductors are in a wire, you need to know the cross-sectional area of ​​one conductor that makes up the wire and divide our resulting cross-section by this area.
Answer: 1 * 10 ^ -6 m² or 1 mm².