# Understanding and calculating suitable AWG based on current draw?

#### Sasuke Uchiha

Joined Jan 15, 2022
21
Hi.

I have been looking at different AWG charts, and I am now more confused than what I was before looking at different AWG charts.

For starters. I have a battery with at 16000mAh @30C rating. Meaning the battery is capable of providing 480A burst current. The wires it uses however, are AWG10, which according to almost all AWG charts, can handle a maximum of 55A. I understand wire length has a lot to do with how much current a wire can handle. If a 10cm AWG10 cable can handle 55A. It doesn't mean a 10m AWG10 cable can handle the same amount of current without overheating/melting, right? What's the formula for calculating a suitable AWG wire size based on current draw and wire length?

Last edited:

#### Ya’akov

Joined Jan 27, 2019
5,659
I think you are reading your battery specifications wrong. Do you have a link to a datasheet?

#### Ian0

Joined Aug 7, 2020
5,118
. I have a battery with at 16000mAh @30C rating. Meaning the battery is capable of providing 480A for one hour.
No, 16A for one hour not 480A.

#### Sasuke Uchiha

Joined Jan 15, 2022
21
No, 16A for one hour not 480A.
16Ah is the battery capacity. But the C rating is 30, so it will be able to give a burst current of 16A*30. It will discharge a lot faster than in 1 hour. But it is able to give a burst current of 480A. Is this not correct?

https://www.power-sonic.com/blog/what-is-a-battery-c-rating/

To be clear. I didn't mean that you can draw 480A from the battery for a full hour. I should have worded myself correctly. I meant that even though the battery is 16Ah. You can draw 480A from it in one go. That's a lot of current travelling through an AWG10 wire rated at 55A

Last edited:

#### MrChips

Joined Oct 2, 2009
26,089
480A for 2 minutes, not 1 hour.

#### Sasuke Uchiha

Joined Jan 15, 2022
21
480A for 2 minutes, not 1 hour.
English is not my first language, I worded myself incorrectly. I will edit the post to clear up the misunderstanding. But my point is. AWG10 is rated at 55A. However, the battery can give a burst current of 480A which is a lot more than what AWG10 is rated to handle. So what is the formula for calculating a suitable AWG wire based on current draw and cable length. The battery was just an example. Just forget about the battery.

#### MrChips

Joined Oct 2, 2009
26,089
The length of the wire is not used in the amperage calculation. Length is used to determine the resistance of the cable and hence the voltage drop and power loss as a result of the cable.

What is required is the resistance per unit length, for example,

10 AWG of copper is 3.3 ohm/ 1000m
0 AWG of copper is 0.32 ohm/ 1000m

The max amperage is based on ambient temperature and the temperature rise.

Are you sure 30C does not refer to 30°C?

#### crutschow

Joined Mar 14, 2008
29,785
The wire tables give the current capacity based on the wire temperature for various conditions and wire insulation type.
You should not exceed those values for long term current conduction.
The current can go higher for short term but, depending upon the current and time, it will overheat and perhaps melt.
For the absolute maximum, you can look up the fusing current of the wire.

The length of the wire does not affect its current rating as that only determines the voltage drop at the end from the wire resistance at the current it is carrying.
So you pick a wire gauge that drops no more than the load can tolerate for the wire length and load current.
Note that for two wires to the load, you must add the resistance of both wires to calculate the voltage drop.

#### Sasuke Uchiha

Joined Jan 15, 2022
21
So you pick a wire gauge that drops no more than the load can tolerate for the wire length and load current.
Sorry, but could you elaborate? I didn't quite understand.

#### Ya’akov

Joined Jan 27, 2019
5,659
A C rating of 40C suggests the battery can be be discharged at that rate, but it doesn't mean it can be discharged at that rate for the same duration as 1C which is one hour at the battery's mAH rating.

From Battery University:
Charge and discharge rates of a battery are governed by C-rates. The capacity of a battery is commonly rated at 1C, meaning that a fully charged battery rated at 1Ah should provide 1A for one hour. The same battery discharging at 0.5C should provide 500mA for two hours, and at 2C it delivers 2A for 30 minutes. Losses at fast discharges reduce the discharge time and these losses also affect charge times.
Additionally, it is almost certain that despite the theoretical ability to discharge at that rate, the claim is only intended as a peak, instantaneous current, not sustained for the maximum time it should be able to deliver it. The ampacity of w depends on its AWG largely because of heating. Even very small wires can take high current if they are not allowed to overheat.

Very short peaks of high current spaced in time can be tolerated by wire rated for a lot less.

[EDIT: Typo repair, h/t @Lo_volt]

Last edited:

#### crutschow

Joined Mar 14, 2008
29,785
but could you elaborate? I didn't quite understand.
You do understand Ohm's law(?).

The wire resistance will cause an IR voltage drop which is proportion to the size of the wire, which determines its resistance per unit length, and the wire length.
The load at the end of the wire will typically have a minimum operating voltage, so the total resistance of the wire (resistance per unit length times its length) has to be no more than will allow the load voltage to stay above this minimum, (the voltage drop below the supply voltage).

#### Lo_volt

Joined Apr 3, 2014
267
This really only depends on what load you are connecting to the battery. The load will draw the current that it needs. You should size the wire to handle that current. You should add a fuse near the battery to protect the wire. As you have said, the battery is capable of very high current. You should set the current rating of the fuse to a value that is less than the current capability of the wire so that the fuse will open if the current exceeds the capability of the wire.

#### Ya’akov

Joined Jan 27, 2019
5,659
You do understand Ohm's law(?).

The wire resistance will cause an IR voltage drop which is proportion to the size of the wire, which determines its resistance per unit length, and the wire length.
The load at the end of the wire will typically have a minimum operating voltage, so the total resistance of the wire (resistance per unit length times its length) has to be no more than will allow the load voltage to stay above this minimum, (the voltage drop below the supply voltage).
Additionally lower voltage means less current so the 40C is only valid, if it is at all, at the terminals