Answer:
[tex]\begin{array}{ccc}Wire \ Number &Maximum\ \ Current \ (A)& AWG\\&&\\Main \ Wire&16\frac{5}{6} &9\\&&\\Wire \ 1&9\frac{1}{6} &12\\&&\\Wire \ 2&5&14\\&&\\Wire \ 3&2.5&17\end{array}[/tex]
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 = [tex]9\dfrac{1}{6} \, A[/tex]
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.[tex]\overline 6[/tex] A
The current that the main wire can carry, I = 16.[tex]\overline 6[/tex] A
The highest American Wire Gauge (AWG) that can be used for each type of wire are listed as follows;
[tex]\begin{array}{ccc}Wire \ Number &Maximum\ \ Current \ (A)& AWG\\&&\\I&16\frac{5}{6} &9\\&&\\I_1&9\frac{1}{6} &12\\&&\\I_2&5&14\\&&\\I_3&2.5&17\end{array}[/tex]
Therefore, we have;
[tex]\begin{array}{ccc}Wire \ Number &Maximum\ \ Current \ (A)& AWG\\&&\\Main \ Wire&16\frac{5}{6} &9\\&&\\Wire \ 1&9\frac{1}{6} &12\\&&\\Wire \ 2&5&14\\&&\\Wire \ 3&2.5&17\end{array}[/tex]
