Answer:
![\begin{array}{ccc}Wire \ Number &Maximum\ \ Current \ (A)& AWG\\&&\\Main \ Wire&16(5)/(6) &9\\&&\\Wire \ 1&9(1)/(6) &12\\&&\\Wire \ 2&5&14\\&&\\Wire \ 3&2.5&17\end{array}](https://img.qammunity.org/2022/formulas/computers-and-technology/high-school/m2jtltxrb4v8yi85ixg0mfsi7c4ff3758j.png)
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 =
![9(1)/(6) \, A](https://img.qammunity.org/2022/formulas/computers-and-technology/high-school/wyozikz2d93uubz7wnrvu2rfh3r3k3fu1p.png)
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;
![\begin{array}{ccc}Wire \ Number &Maximum\ \ Current \ (A)& AWG\\&&\\I&16(5)/(6) &9\\&&\\I_1&9(1)/(6) &12\\&&\\I_2&5&14\\&&\\I_3&2.5&17\end{array}](https://img.qammunity.org/2022/formulas/computers-and-technology/high-school/82ji6oj2ovntofr79cauq6z8ek8o1ev8y7.png)
Therefore, we have;
![\begin{array}{ccc}Wire \ Number &Maximum\ \ Current \ (A)& AWG\\&&\\Main \ Wire&16(5)/(6) &9\\&&\\Wire \ 1&9(1)/(6) &12\\&&\\Wire \ 2&5&14\\&&\\Wire \ 3&2.5&17\end{array}](https://img.qammunity.org/2022/formulas/computers-and-technology/high-school/m2jtltxrb4v8yi85ixg0mfsi7c4ff3758j.png)