Wire Gauge, Current Capacity vs Power Capacity? - Page 1
Forum section: electronics / beginners

author: jstarr
author_rank: Regular Contributor
timestamp: October 13, 2014, 11:15:00 pm
content: I see many tables with various current capacities based on copper wire AWG size (solid copper wire versus stranded, how many strands, etc, etc) but none of these tables mention a relative voltage? Why isnt the the current capacity of any copper wire expressed as a Power capacity? For example I want to use an AGM 12v battery (Solar charged) to power a 12vdc motor used intermittently that draws a maximum of 10A on start-up and 6- 8 amps thereafter. Since the battery will put out more than 12v for most of its capacity, I figure 18 AWG stranded is overkill since I don't care about a 1 vdc voltage drop over 10 feet of 2-wire but the Solar power people tell me I am crazy to use anything thinner than 12 AWG.Any opinions??

author: Falcon69
author_rank: Super Contributor
timestamp: October 13, 2014, 11:29:03 pm
content: Correct me if I am mistaken, but I don't think wire cares much about the voltage, only the amperage through it. Although, Voltage and Amps play together. But can't you have 50,000 volts flowing through a 22gauge wire, as long as it doesn't go above 300mA?

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 13, 2014, 11:39:57 pm
content: The voltage capacity of a wire is only limited by its insulation. A lot of wire is insulated to handle a few hundred volts, and it's sometimes possible to exceed this if you space things out and use air as an insulator (but consult local codes and safety regulations for details).Suppose you had a pair of wires capable of carrying 5 amps. Let's say that, over the length of the run, they had a resistance of 0.1 ohm. The power dissipated in the wires would be I^2*R, or 2.5 Watts.How much power could they deliver to a load? At 5V, they could carry 25 Watts. At 12V, they could carry 60 Watts. At 200V, they could carry 1000 Watts. In each case, the current would be 5 amps, and you'd have to put an additional 0.5V (2.5W) into the input side to deliver the full 5 amps at the rated voltage to the load.The higher the voltage, the greater the power that can be delivered with a given size conductor. And the lower the losses are in the transmission line, when figured as a percentage of the load that's delivered.If you play with the math and work through a few examples, you'll quickly understand why relatively high voltage at relatively low current is what's favored for sending power over long distances.

author: Unknown
author_rank: Unknown
timestamp: October 13, 2014, 11:41:21 pm
content: Example only exaggerated for ease of calculation, but correct me if I've done this maths wrong.wire resistance = 1 ohmsource voltage 20 Vend load = 9 ohms2 amps and 40W being used, 10% of which is dissipated in the wire giving 4W of lost powerwire resistance 1 ohmsource voltage 40 Vend load 19 ohms2 amps again, 80W this time, 5% of which is dissipated in the wire which is again 4WSo it seems the voltage is irrelevant with regard to amperage rating of the wire that is. But the higher the voltage goes, the less relative power you lose.EDIT: The previous posters beat me to it, heh!

author: Falcon69
author_rank: Super Contributor
timestamp: October 13, 2014, 11:53:31 pm
content: SO, based on what these guys have said, the guys at the 'solar company' recommends that size to lower the amount of heat dissapated. Larger wire, less heat. That's the way I understand it anyway. Also, I think that the larger the wire, the less the voltage drop the longer length the wire is. (i think)

author: jstarr
author_rank: Regular Contributor
timestamp: October 14, 2014, 12:08:54 am
content: Quote from: Boltar on October 13, 2014, 11:41:21 pmYou lost me, you say the wire resistance is 1 ohm in both examples so why did you change the end load from 9 to 19 ohm? Were you estimating wire thickness ??10' of 12 AWG is about 0.016 ohm and 10' of 18 AWG is about 0.064 ohm about a 5 fold difference?My question is whether the wire thickness used should be calculated based on power dissipation (voltage dependent) or only current (independent of voltage)

author: Unknown
author_rank: Unknown
timestamp: October 14, 2014, 12:15:40 am
content: I changed the load to keep the current the same at the higher voltage. Basically I was trying to show that power lost in a wire is directly proportional to the electrical current flowing in it. So the power dissipation and the current rating are essentially the same thing.

author: IanB
author_rank: Super Contributor
timestamp: October 14, 2014, 12:16:21 am
content: The wire doesn't know anything about load power. If you look at a given length of wire there is a current flowing through it, a voltage drop along its length, and a power dissipation due to resistance of the wire. You choose your wire size based on two factors: what voltage drop is acceptable, and what level of heating and temperature rise of the wire is acceptable?

author: IanB
author_rank: Super Contributor
timestamp: October 14, 2014, 12:18:06 am
content: Quote from: Boltar on October 14, 2014, 12:15:40 amYou want to keep the load power the same for a reasonable comparison. For instance, consider a 60 W bulb running from a 120 V supply, and then consider a 60 W bulb running from a 12 V supply, using the same cable length and thickness in each case.

author: Unknown
author_rank: Unknown
timestamp: October 14, 2014, 12:22:39 am
content: Quote from: IanB on October 14, 2014, 12:18:06 amThat really wasn't what I was trying to demonstrate. I may have the wrong end of the stick, but the OP wanted to know if he should select wire based on its current rating or power dissipation, I was only trying to show that the two are directly related. I.e. the power dissipated in a wire is the same for a constant current no matter what the voltage is.

author: IanB
author_rank: Super Contributor
timestamp: October 14, 2014, 12:27:39 am
content: Quote from: Boltar on October 14, 2014, 12:15:40 amBut it isn't...The power lost in a wire is proportional to the square of the current, according to the law: Power = I2RSo if you double the current, the power loss increases by four times. If you increase the current by 10 (e.g. from 1 A to 10 A) the power loss is 100 times greater (perhaps 100 W instead of 1 W).

author: jstarr
author_rank: Regular Contributor
timestamp: October 14, 2014, 12:29:11 am
content: Isn't this only about Ohms law --- If my motor will draw 10A at 12v I need a wire thickness that will not contribute much to the voltage loss due to wire heating. So 10A at 12vdc dissipates much less at the same resistance (wire thickness) than 10A at 120vdc.

author: Unknown
author_rank: Unknown
timestamp: October 14, 2014, 12:31:01 am
content: Quote from: IanB on October 14, 2014, 12:27:39 amThe point I was making is keeping the current CONSTANT, not increasing it. If the current through a wire is constant, then so is the power dissipation irrespective of voltage. But by all means let me know if the laws of Physics have changed, I wasn't aware of it. If the maths I did earlier that demonstrated this was wrong, please point it out.

author: IanB
author_rank: Super Contributor
timestamp: October 14, 2014, 12:34:33 am
content: Quote from: jstarr on October 14, 2014, 12:29:11 amYou have that backwards. The wire heating is due to the voltage drop, not the other way around.However, you are correct that you need a wire thickness where the voltage loss in the wire is only an acceptably small fraction of the supply voltage.QuoteNo, 10 A dissipates exactly the same at the same wire thickness, regardless of the supply voltage. As noted above, the power loss is given by the square of the current times the resistance. If the wire resistance is 0.05 ohms the power dissipated at 10 A is 100 x 0.05 = 5 W. Notice that the supply voltage does not appear in this equation.

author: IanB
author_rank: Super Contributor
timestamp: October 14, 2014, 12:38:07 am
content: Quote from: Boltar on October 14, 2014, 12:31:01 amThen you need to choose your words more carefully. Because this is what you actually said:Quote from: Boltar on October 14, 2014, 12:15:40 amAnd this statement is wrong. Power loss isnotdirectly proportional to the current.When trying to make a point, it is important to choose words that support the point you are trying to make.

author: ovnr
author_rank: Frequent Contributor
timestamp: October 14, 2014, 12:39:21 am
content: Quote from: jstarr on October 14, 2014, 12:29:11 amNo.The voltage has no effect on the power dissipated by the wire. If your wire carries 12V 10A or 12 kV 10A, it'll still dissipate EXACTLY as much power*. The only thing that matters is the actual wire resistance (i.e. wire gauge).(*: Assuming it doesn't arc over, at which point all bets are off. Different problem!)

author: Unknown
author_rank: Unknown
timestamp: October 14, 2014, 12:42:41 am
content: 1.21 GWGreat Scott!

author: jstarr
author_rank: Regular Contributor
timestamp: October 14, 2014, 02:16:23 am
content: Quote from: ovnr on October 14, 2014, 12:39:21 amI think you are confusing heat dissipation of the wire versus the total current capacity on the circuit. Power = I * E ; so 12v * 10A is not the same as 12KV * 10A.

author: jstarr
author_rank: Regular Contributor
timestamp: October 14, 2014, 02:21:47 am
content: I hit Post too soon. The end to my last post should be.... so the only characteristic is to decide on how much heat loss from the wire is acceptable , and that is purely a function of the wire thickness, resistance... correct? Why are wire fuses then rated by voltage and current characteristics??

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 14, 2014, 02:41:04 am
content: Quote from: jstarr on October 14, 2014, 02:21:47 amBecause fuses have two jobs. One is to act as a conductor with negligible resistance when presented with low current. But the second is to safely open the circuit and act as an insulator with effectively infinite resistance when presented with high current. When a fuse is doing that first part of the job, it doesn't care about how much voltage is present in the rest of the circuit. But when it's called upon to do that second part of the job, the entire voltage of the power supply is suddenly presented across the fuse. A low voltage fuse presented with a very high voltage will just arc over instead of interrupting the circuit.

author: IanB
author_rank: Super Contributor
timestamp: October 14, 2014, 02:44:30 am
content: Quote from: jstarr on October 14, 2014, 02:16:23 amovnr is not confused.All the way through this thread we have been trying to tell you that the voltage carried by the wire has no effect on the power dissipation. Somehow you are still not getting it.

author: Falcon69
author_rank: Super Contributor
timestamp: October 14, 2014, 02:57:41 am
content: Quote from: AG6QR on October 14, 2014, 02:41:04 amHe's right about that. I remember seeing a video on how the PTC fuses work. Really interesting, let me see if I can find it. I know it doesn't pertain to the OP's original question, but since we are talking about insulators/fuses/wire housings and their job when it comes to heat dissipation, i think it fits here.

author: jstarr
author_rank: Regular Contributor
timestamp: October 14, 2014, 03:49:44 am
content: QuoteSorry for being dense but I guess that why this is the Beginners Section :-) I will go back to my example question, under the conditions of a constant load (10A) at constant voltage (12V) would 18AWG be adequate or do I need 12 AWG. The difference in wire resistance is an increase in total resistance of about 0.05 ohms (for 10 feet). The voltage drop would be about 0.5vdc and the heat loss would be about 5W. Is that too much for 18 AWG to handle safely?

author: IanB
author_rank: Super Contributor
timestamp: October 14, 2014, 04:00:28 am
content: Quote from: jstarr on October 14, 2014, 03:49:44 am12 AWG is recommended. Here's why:If you use 18 AWG a 10 ft run of cable will be a total length of 20 ft (10 ft out and 10 ft back).The resistance of 18 AWG is 6.52 ohms/1000 ft, so the resistance of your 20 ft will be 0.13 ohms. With a load current of 10 A the voltage drop will be 1.3 V. Therefore with a 12 V supply the motor at the other end will see only 10.7 V. That is a voltage loss of 1.3/12 = 11%. Usually an 11% loss in voltage is considered too much.With 12 AWG the resistance is 1.62 ohms/1000 ft. Now the voltage drop is only 0.32 V, or 2.7%. The motor will see 11.68 V. With that voltage drop you should not see much loss in performance of the motor.

author: Falcon69
author_rank: Super Contributor
timestamp: October 14, 2014, 04:01:16 am
content: I would use the 12awg for two reasons.No.1 The solar guys who do this sort of stuff for a living and know what is needed recommended the 12awg.No.2 The large gauge wire will do better as for as less voltage drop, and also allow for better heat dissipation. As well as provide extra room for unforeseen things like extra current through wire if something happens.Stick with what they experts have told you. If It turns out that the 18awg is not good enough for that application, then you have spent more in labor and wire to replace it. Just do it right the first time.

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 14, 2014, 04:02:36 am
content: Quote from: jstarr on October 14, 2014, 02:16:23 amovnr is not confused. You seem to be confusing power dissipated in the wire with power delivered to the load. Power dissipated in the wire is I * E, where I is the current in the wire, and E is the voltage present from one end of the wire to the other end of that individual conducting wire. The voltage across the load does not enter into it. The voltage across the wire is just the current times the resistance of the wire (load resistance doesn't enter into it).Power delivered to the load is also written as I * E, but this time, the E represents the voltage presented across the load. The equation is still P=I*E, but the letters represent different quantities.Consider a circuit consisting of two wires carrying power from a source to a load. The heating in a wire depends only on the current running through that wire, not on the voltage as measured from one wire to some different wire. One individual wire "sees" only the current running through it, and the resulting voltage across the wire, as measured from one end of the wire to the other end of that same wire. It doesn't "know" anything about the voltage of the other wire. On the other hand, the power source and the load both see the voltage presented between the two wires.

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 14, 2014, 04:11:12 am
content: Quote from: jstarr on October 14, 2014, 03:49:44 amIt won't catch on fire and melt, if that's what you're asking. But, assuming the run has a total of 20 feet (two conductors, each 10 feet long), you'll lose almost 10% of your power in that short transmission line. What's the cost of oversizing your solar panel by 10%, versus the cost of increasing your wire gauge to lower the power loss?

author: jstarr
author_rank: Regular Contributor
timestamp: October 14, 2014, 12:58:24 pm
content: BIG thanks for all of your patience in helping me to understand the principles behind this. I have one more question as I did not state the setup accurately. Another beginner mistake.... my apologiesThe motor being supplied by the AGM battery is powered thru a switching supply. The supply input is 12 VDC and the motor takes 5VDC at 10A peak. I have read that switching supplies are near 85% efficiency, so am I correct that the draw on the battery will only be 4.2A * 0.85 = 3.54 A peak??

author: cowana
author_rank: Frequent Contributor
timestamp: October 14, 2014, 01:08:42 pm
content: So if the motor draws 10A at 5v:Assume 100% efficiency converter:10/(12/5) = 4.16AHowever it's not quite that good - say 85% efficiency will increase the 12v current consumption slightly:4.16/0.85 = 4.89A draw at 12v

author: ovnr
author_rank: Frequent Contributor
timestamp: October 14, 2014, 04:06:56 pm
content: Quote from: jstarr on October 14, 2014, 12:58:24 pmSwitching supplies may range between <40% and 98% efficiency. Please tell us which PSU you're actually using.And no, you've got it backwards. If your load is 5V @ 10A over a 85% efficient switcher, your battery will see a 12V @ 4.9A load.Secondly, if possible put the PSU near the motor - you will have four times less power loss in your cables (4.9A vs 10A; assuming your cables are 0.25 ohms, that's 6W vs 25W).