While this installation is not code compliant, neither conductor would be overloaded. If you use the AC resistance from Chapter 9 Table 9 you can solve this as parallel resisitor problem. If we make the circuit 100' long and assume a 55 amp load the resistnace of #14 is 0.31 ohms and the the #8 is 0.078 ohms. The total resistance of these two conductors in parallel is:
1/0.31 + 1/0.078 = 1/Rt = 16.046
Rt= total resistance
1/16.046 = 0.0623 ohms = resistance of 100' of #14 in parallel with 100' of #8
Now if we use E=I(R) to solve for the voltage drop we get E= 55(0.0623) = 3.426 volts
We can now use this voltage drop to solve for the current in each of our conductors.
Again E = I(R) 3.426 = I(0.31) = 11.05 amps for the #14.
3.426 = I(0.078) = 43.92 amps for the #8
These two values do not exactaly equal 55 amps due to rounding. They do total 54.97 amps.
Don(resqcapt19)
[This message has been edited by resqcapt19 (edited 09-24-2001).]
