10/5 SO is rated officially for 24 amps, but since one conductor (the ground) is not current carrying, it should be rated for 25 amps as per 10/4 SO?

10AWG wire has 10,380 C.Mil (Circular Mills of circumference measurement)

Voltage Drop Foumula: Voltage Drop x cmils / Amperes x 18.7 (for 3 Phase) = One-way distance in feet. (E) x (10,380cmil) / 25Amp x 18.7 = 400 LF.

(E) x (10,380cmil) / 467.5 = 400 LF (E) x 10,380cmil) = 187,000 (E) = xxxxx Volts With voltage drop.

We given a 120v source should be at about 102V when at a full 25 Amps of load maxing out the cable - Am I correct?

And the 20 amp FOH circuit breakers should trip - right?

Given we are breakered for 20 amps, what's the voltage drop on the max load? (E) x (10,380cmil) / 20Amp x 18.7 = 400LF. (E) x (10,380cmil) / 374 = 400LF. (E) x (10,380cmil) = xxxxx (E) = Volts

Our voltage is at xxxx Volts at least in theory, this does not seem right, but than again, I have not seen any melted FOH feeders so I expect nobody has come up to a full 20 Amps for three phases load before. This is given that the amount of voltage drop increases with the amount of load and length which we can assume is true. 8/5 wire would be better at this length, but 10/5 is perhaps

acceptable given it’s not in use coming up to full amperage.

We are considering either going 15 amp breakers at the FOH distribution box, or leaving it as is and telling the crew chiefs of the max amperage of the cable so they are aware of it. Or at lest that’s the going opinion around here so that there is that “extra headroom” available should it be needed. In theory, since this below is a statement of xxxxxx amps per 3 legs is maxing out the available power, that should be fine for any FOH application unless they are tapping it for a follow spot, moving light etc. Which should explain many things.

Would breakering FOH for 15 amps be better?

At 10 Volts of voltage drop which will give a minimum of 110V at FOH, what amperage would be safe to run thru 10/5 SO wire?

(10E) x (10,380cmil) / I x 18.7 = 400LF.

103,800 / I x 18.7 = 400LF.

I x 18.7 = 259.5

I = xxxxxx