Answer:

Step-by-step explanation:
The given parameters of the circuit are;
The voltage applied to the circuit = 240 volt
The types of wire in the circuit are Main wire, wire 1, wire 2, and wire 3
The maximum current that can flow through each type of wire is given by the max power of the machine powered by the wire
Power, P = Voltage, V × Current, I
∴ I = P/V
The maximum power of the machine powered by wire 1 = 2,200 W
The current flowing through wire 1, I₁ = 2,200 W/(240 V) = 55/6 A =

The maximum power of the machine powered by wire 2 = 1,200 W
The current flowing through wire 2, I₂ = 1,200 W/(240 V) = 5 A
The maximum power of the machine powered by wire 3 = 600 W
The current flowing through wire 3, I₃ = 600 W/(240 V) = 2.5 A
Therefore, Kirchhoff's current law, we have;
The current that the main wire can carry, I = I₁ + I₂ + I₃
∴ I = 55/6 A + 5 A + 2.5 A = 50/3 A = 16.
A
The current that the main wire can carry, I = 16.
A
The highest American Wire Gauge (AWG) that can be used for each type of wire are listed as follows;

Therefore, we have;
