# Resistance Wire (Kanthal A1) vs Copper Wire

#### Guest3123

Joined Oct 28, 2014
404
The Copper Wire (PVC Stranded Wire)
Length : 9.6899 Feet
Resistance : 1.000Ω
Gauge : 30AWG
Ω/Ft : 0.1641

The Kanthal A1 Resistance Wire..
Length : 36.3mm, or.. 1.429 Inches
Resistance : 1.000Ω
Gauge : 30AWG
Ω/Ft : 8.36

The Copper Wire
The longer the Copper wire, the more hotter it gets.. The shortter the copper wire, the less hotter it gets. Shorter runs usually have less voltage drop, etc. etc.

The Kanthal A1 Resistance Wire
The longer the run piece of wire, the cooler it becomes. I should know, I build coils for my RDA for my Box Mod. The Shorter the wire.. WATCH OUT !! Because if it's too short, when you press the button, it will vaporize the wire.

I have a Box mod, and I use this stuff all the time. day in, and day out. I have made decent Windows Apps programed in Visual Basic .NET.

Why does Resistance Wire behave differently than Copper wire?

#### Guest3123

Joined Oct 28, 2014
404
But yet, a Short piece of copper wire wont vaporize, like the Kanthal A1 Resistance wire..

In fact, that's a good thing.

The Resistance of a 1 Inch piece of 30AWG Copper Wire is..
0.1641Ω Per Foot, so simply divide by 12"
0.1641 / 12 = 0.013675Ω, so no resistance.

The Resistance of a 1 Inch piece of 30AWG Kanthal A1 Resistance Wire is..
8.36Ω Per Foot, so simple divide by 12"
8.36 / 12 = 0.696666Ω, so almost 7/10th of an Ohm.

Regardless of what I just put in the above text. The longer the kanthal wire, the cooler it will get when creating a heating element, or shorting it between Positive and Negative on a LiPo 2S, 130C, 3,500mAh Battery.

Regardless of what I just put in the above text. The longer the copper wire, the hotter it will get when creating a heating element, or shorting it between Positive and Negative on the LiPo 2S, 130C, 3,500mAh Battery.

#### crutschow

Joined Mar 14, 2008
32,026
..................
Why does Resistance Wire behave differently than Copper wire?
It doesn't.
The only difference is the length of wire at which your various observations happen.
But for a given resistance, a longer copper wire can carry more current without overheating since it has more area to dissipate the heat generated.

You observations about copper wire don't make sense.
A longer wire (either resistance type or copper) will have more resistance and run cooler for a given voltage across it.

#### Guest3123

Joined Oct 28, 2014
404
It doesn't.
The only difference is the length of wire at which your various observations happen.
But for a given resistance, a longer copper wire can carry more current without overheating since it has more area to dissipate the heat generated.

You observations about copper wire don't make sense.
A longer wire (either resistance type or copper) will have more resistance and run cooler for a given voltage across it.
The Copper wire has more resistance, Ohms law.
1 Inch of 30AWG Copper wire is 0.013675Ω, 1828.15 Watts, 365.63 Amps, 5V DC.

The Kanthal wire has less resistance, Ohms law.
1 Inch of 30AWG Kanthal A1 Resistance wire is 0.696666Ω, 35.88 Watts, 7.17 Amps, 5V DC.

I know the video doesn't have anything to do with what I was asking. I also am aware now that what @CrutShow has said is true. But please tell me what is going on in the video I've listed in the above text. The copper wire has a much less resistance, but yet...

#### Guest3123

Joined Oct 28, 2014
404
If I made a 2.7Ω Kanthal wire, and passed 12vdc threw it, it would heat up. It would be 12vdc, 2.7Ω, 4.44A, 53.33 Watts.
If I made a 0.1Ohm Copper wire, and passed 12vdc threw it, it would heat up. It would be 12vdc, 0.1Ω, 120A, 1,440 Watts (1.44kW)

What is going on..?

Is it because the two wires are in series, and connected to each other? Wouldn't the Copper wire burn the paper first if they weren't connected to each other?

#### SLK001

Joined Nov 29, 2011
1,548
ALL wire will heat up if you pass current through it.

It confuses you that a wire dissipating 1.44 kWatt of power is heating up? Likewise, a wire dissipating 53 Watts also?

What is going on...? It's simple - OHM's law is going on.

#### Lestraveled

Joined May 19, 2014
1,946
The Copper wire has more resistance, Ohms law.
1 Inch of 30AWG Copper wire is 0.013675Ω, 1828.15 Watts, 365.63 Amps, 5V DC.

The Kanthal wire has less resistance, Ohms law.
1 Inch of 30AWG Kanthal A1 Resistance wire is 0.696666Ω, 35.88 Watts, 7.17 Amps, 5V DC.
Your more and less are reversed. 0.013675Ω is less than 0.696666Ω.

#### BR-549

Joined Sep 22, 2013
4,928
I agree with the video. You are seeing the effect of resistance. That's what resistance does.

It produces heat. The amount of heat depends on the amount of resistance.

#### Lestraveled

Joined May 19, 2014
1,946
There is a simple formula to calculate dissipated heat, Watt = Current squared time resistance. In the video, the two resistances are in series and thus have the same current. The wire with the highest resistance will dissipate the highest heat.

#### crutschow

Joined Mar 14, 2008
32,026
If I made a 2.7Ω Kanthal wire, and passed 12vdc threw it, it would heat up. It would be 12vdc, 2.7Ω, 4.44A, 53.33 Watts.
If I made a 0.1Ohm Copper wire, and passed 12vdc threw it, it would heat up. It would be 12vdc, 0.1Ω, 120A, 1,440 Watts (1.44kW)

What is going on..?

Is it because the two wires are in series, and connected to each other? Wouldn't the Copper wire burn the paper first if they weren't connected to each other?
What's going on is the difference between serial and parallel circuits.
In series the current is limited by the high resistance of the Kanthal or nichrome wire so most of the voltage drop (and power) is across the high resistance wire and not the copper wire (that's how a space heater works in your house).
If you connect the two wires separately (in parallel) to the same battery then the current through the copper wire would be much higher and it would dissipate much more power as per V^2 / R (as would occur if you short circuited the mains socket in your house).

#### Guest3123

Joined Oct 28, 2014
404
What's going on is the difference between serial and parallel circuits.
In series the current is limited by the high resistance of the Kanthal or nichrome wire so most of the voltage drop (and power) is across the high resistance wire and not the copper wire (that's how a space heater works in your house).
If you connect the two wires separately (in parallel) to the same battery then the current through the copper wire would be much higher and it would dissipate much more power as per V^2 / R (as would occur if you short circuited the mains socket in your house).
So it's the path of least resistance. Even though the copper wire was 0.1Ω

0.1Ω is MORE resistance, AND 2.7Ω is LESS resistance.

The 12V DC passed thru the path of LESS resistance.

Even though that Ohms law states that 1440kW should have been dissipated thru the Copper wire, which according to Ohms Law, is 12vdc, 0.1Ω, 120A, and again, should have dissipated 1.44kW.

#### Lestraveled

Joined May 19, 2014
1,946
@Guest3123
No, you are not getting it. I don't think you understand how current works in parallel and series circuits.

- If you have two different resistors in parallel, the voltage on each resistor will be the same but the current will be different.
- If you put the same two resistors in series, the current through each resistor will be the same but the voltage will be different.

#### GopherT

Joined Nov 23, 2012
8,009
So it's the path of least resistance. Even though the copper wire was 0.1Ω

0.1Ω is MORE resistance, AND 2.7Ω is LESS resistance.

The 12V DC passed thru the path of LESS resistance.

Even though that Ohms law states that 1440kW should have been dissipated thru the Copper wire, which according to Ohms Law, is 12vdc, 0.1Ω, 120A, and again, should have dissipated 1.44kW.

Dude,

The current is calculated by Adding the resistance of Kanthal (2.7) and copper (0.1) and call that R. V= 12 volts.
Then current (I) is V/R. 12/2.8 = 4.286 amps

Now, to calculate voltage drop across the copper wire snd the voltage drop across the Kanthal, multiply the 4.286 amps by the resistance of the copper and the resistance of the Kanthal. See answers in pictures below. Notice that the total voltage adds to 12.

Last edited:

#### Guest3123

Joined Oct 28, 2014
404
Watching a video, and learning about it.
I'll be back in about an hour or so.

#### BR-549

Joined Sep 22, 2013
4,928
Gopher meant 4.286 amps., not ohms, in his calculation explanation.

#### Guest3123

Joined Oct 28, 2014
404
Just watched a video..

#### Guest3123

Joined Oct 28, 2014
404
Dude,

The current is calculated by Adding the resistance of Kanthal (2.7) and copper (0.1) and call that R. V= 12 volts.
Then current (I) is V/R. 12/2.8 = 4.286 amps

Now, to calculate voltage drop across the copper wire snd the voltage drop across the Kanthal, multiply the 4.286 ohms by the resistance of the copper and the resistance of the Kanthal. See answers in pictures below. Notice that the total voltage adds to 12.

View attachment 97832 View attachment 97833
Where do you get 100mOhms from? It's 0.1Ohms. That's 1/10th of an Ohm. It's not 1.837 Watts. It's 12V² / 0.1Ω = ((12v x 12v) / 0.1Ω = 1,440 Watts.

But it's not.. It's (0.428V * 0.428V) / 0.1Ω = 1.83184 Watts.

Ok ok.. I got this now.. Your right. lol.

(11.556V * 11.556V) / 2.7Ω = 49.45968 Watts.. Well I'll be damned.. It's true.
Yes I know my precision is off.

Here let's fix that.
4.2857142857142857142857142857143 Amps.

4.2857142857142857142857142857143 Amps x 2.7Ω = 11.571428571428571428571428571429 Vdc
4.2857142857142857142857142857143 Amps x 0.1Ω = 0.42857142857142857142857142857143 Vdc
0.42857142857142857142857142857143Vdc + 11.571428571428571428571428571429Vdc = 12Vdc Exactly.

Calculating the Precise Watts.
0.42857142857142857142857142857143Vdc² / 0.1Ω = 1.8367346938775510204081632653061 Watts.
11.571428571428571428571428571429Vdc² / 2.7Ω = 49.591836734693877551020408163269 Watts.

God damn, it really does pay to study. Thanks for the help guys. I understand now, and I see what it going on.

#### Guest3123

Joined Oct 28, 2014
404
Because like what @Lestraveled said..

- If you have two different resistors in parallel, the voltage on each resistor will be the same but the current will be different.
- If you put the same two resistors in series, the current through each resistor will be the same but the voltage will be different.

So, then I went and watched the video about Resistors in Series.. and VOLLA !! I payed attention, and learned exactly what Gopher posted.

But instead of being given the answers, I actually understand what is going on in the first video, because I taught myself how to understand what the hell was going on. With the help of course of everyone that gave me direction.

#### cmartinez

Joined Jan 17, 2007
7,965
But instead of being given the answers, I actually understand what is going on in the first video, because I taught myself how to understand what the hell was going on. With the help of course of everyone that gave me direction.
And that, my friend, is the very essence of this site.
Glad you were able to figure things out on your own. Welcome aboard!

Word of advice: don't stop here, there's plenty of other exciting stuff you can learn and do fun things with.

#### GopherT

Joined Nov 23, 2012
8,009
Where do you get 100mOhms from? It's 0.1Ohms. That's 1/10th of an Ohm. It's not 1.837 Watts. It's 12V² / 0.1Ω = ((12v x 12v) / 0.1Ω = 1,440 Watts.

But it's not.. It's (0.428V * 0.428V) / 0.1Ω = 1.83184 Watts.

Ok ok.. I got this now.. Your right. lol.

(11.556V * 11.556V) / 2.7Ω = 49.45968 Watts.. Well I'll be damned.. It's true.
Yes I know my precision is off.

Here let's fix that.
4.2857142857142857142857142857143 Amps.

4.2857142857142857142857142857143 Amps x 2.7Ω = 11.571428571428571428571428571429 Vdc
4.2857142857142857142857142857143 Amps x 0.1Ω = 0.42857142857142857142857142857143 Vdc
0.42857142857142857142857142857143Vdc + 11.571428571428571428571428571429Vdc = 12Vdc Exactly.

Calculating the Precise Watts.
0.42857142857142857142857142857143Vdc² / 0.1Ω = 1.8367346938775510204081632653061 Watts.
11.571428571428571428571428571429Vdc² / 2.7Ω = 49.591836734693877551020408163269 Watts.

God damn, it really does pay to study. Thanks for the help guys. I understand now, and I see what it going on.

One piece of advice, after you write the first three digits that are not zeros. You can round everything else off as zeros. Just put the right number of zero until the decimal (or use appropriate prefix for to a avoid most zeros. Nice work plowing through it!