#### richwaskowitz

##### Senior Member

- Location
- Shelby Twp. MI, near Detroit

2. The feeder supplying it utilizes a J class 25 amp fuse with a let-through current of 5.5kA (UL 508A table SB4.2)

3. I want to prove that a 50 foot length of #10 cable supplying power to this 5 kA rated panel will effectively limit the available fault current to a value below 5kA.

Can it be as simple as taking the resistance of 50 feet of #10 at 0.05 ohms times 2 (out and back) for a total of 0.10 ohms, using I=E/R and coming up with 480 V / 0.1 Ohms = 4800 Amps?