# Question on how to measure a circuit length for sizing AC inverter wire

Registered Users Posts: 38 ✭✭
edited November 2018 #1
Hi. In a little 23' motorhome, I've got a typical 30 amp, 35 foot shore power cord terminating into a junction box in the "shore power compartment" on the side of the vehicle. From there, Winnebago ran orange Southwire SIMpull 10/2 + ground NM-B 5 feet to the Parrallax AC/DC load center.

I propose to splice into the 10/2 orange SIMpull (an inline inverter installation) in the middle of the 5' run from the shore power compartment to the load center, then run out and back 15' each way to a 2,000 watt Xantrex XC2000 inverter. For the purpose of putting in the distance to a calculator, what distance "one way" should I use? I don't know if this distance should include any of the existing wiring (35 external cord + 5 feet of orange 10/2). Ignoring any existing wiring, it's a 15' run one way after my splice, so it seems using what Winebago did is fine (orange 10/2) If the entire length (35+5+15) were included, that's different.

Thoughts one which Southwire NM-B to go; orange or black? Thanks

Edit: 11/15/18: Can't believe I forgot this: I just realized I didn't describe something. The inverter is a Xantrex XC 2000 with a 30 amp transfer switch. While the inverter's 2000 watts outputs 16.7AAC, it has a 30AAC transfer relay rating. In my example above, Winnebago added 5' from the demarcation of the incoming shore power cable and the load center. Splicing the inverter in extends that to a total of about 35' running the full 30AAC shore power. If I input to an AC wire calculator just the one way run to the inverter (15'), using the 10/2 SIMpull 30 amp wire seems a no brainer. If (for reasons I'm questioning) the shore power cable and existing wiring must be included into the circuit length, that's (35 + 15 + 5 = 55) with a 30AAC load (assuming that needs to be 30 x 1.25 or 37.5), and assuming 77 degrees F, 3,600 watts, max 2% voltage drop, the indicated AC wire increases to something a little bigger an 10/2. I'm also uncertain that 2% is the correct voltage drop calculator input. I used 1% on the DC side, because this system serves an essential purpose.

• Registered Users Posts: 174 ✭✭✭
What is the max load in Amps on that wire?. Wire size is based on Amps and the fuse protecting it.
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• Solar Expert Posts: 9,583 ✭✭✭✭✭
At 120V, voltage drop (loss) in 10ga wire is generally negligible. What you have to watch for is proper fusing/circuit breakers for the gauge.
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• Registered Users Posts: 38 ✭✭
Ampster said:
What is the max load in Amps on that wire?. Wire size is based on Amps and the fuse protecting it.
The Southwire SIMpull orange 10/2 is rated NEC at 30 amps max and has a 30 mil jacket on it.
• Registered Users Posts: 38 ✭✭
mike95490 said:
At 120V, voltage drop (loss) in 10ga wire is generally negligible. What you have to watch for is proper fusing/circuit breakers for the gauge.

I just realized I didn't describe something. The inverter is a Xantrex XC 2000 with a 30 amp transfer switch. While the inverter's 2000 watts outputs 16.7AAC, it has a 30AAC transfer relay rating. In my example above, Winnebago added 5' from the demarcation of the incoming shore power cable and the load center. Splicing the inverter in extends that to a total of about 35' running the full 30AAC shore power. If I input to an AC wire calculator just the one way run to the inverter (15'), using the 10/2 SIMpull 30 amp wire seems a no brainer. If (for reasons I'm questioning) the shore power cable and existing wiring must be included into the circuit length, that's (35 + 15 + 5 = 55) with a 30AAC load (assuming that needs to be 30 x 1.25 or 37.5), and assuming 77 degrees F, 3,600 watts, max 2% voltage drop, the indicated AC wire increases to something a little bigger an 10/2.

I also and trying to figure out what voltage drop is acceptable on the AC side. I used 1% on the DC side, because this little system serves an essential and important.

Thoughts?

• Solar Expert Posts: 3,854 ✭✭✭✭✭✭
Assuming the extension is the same gauge, #10, the maximum current carrying capacity is 30A, all conductors must be included in the total length  for calculation purposes . However just because the maximum is 30A it's good practice to derate to 80% or 24A, it is important to note that the temperature of the conductors will have an effect on the maximum current carrying capacity, 30A at 30°C is pretty standard, operating in excess of 30°C , the extension lying in the sun for example reduces it's maximum, this is where the derating compensates to some degree.
The maximum voltage drop  should be less than 3%, on the AC side, if there is a need to increase a part of the circuit with a larger gauge, the over current protection "OCP" must be sized for the smallest conductors in the circuit. The inverters maximum capacity of 2000W along with the 30A transfer rating would indicate that there is a margin of safety built in, a derating if you like, preventing overloading of the contacts.
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