Can you help me understand the relationship between current, voltage and heat generated?

I tried two sets of heated gloves. One uses a battery which generates 7.4 volts x 2 amperes (14.8 watts of power). The other uses an 11.1 volt x 1.4 ampere battery (15.54 watts of power). So, the power generated is similar, but the 11.1 volt gloves heat my hands much more. Why is that?

My memories of high school physics are distant, but I remember one can think of electricity in analogy with a water system, whereby volts are the water pressure, current (measured in amperes) is the flow of water, resistance is the size of the pipe, and power (watts) = volts x current. I also remember the amount of heat generated is proportional to the square of the current, but I’m not sure how to put all the pieces together and explain why the 11.1 volt gloves generate much more heat.

  • $\begingroup$ What do the labels on the gloves or the instructions for the gloves say about the rating of the gloves? $\endgroup$
    – Farcher
    Commented Apr 7, 2016 at 9:55

4 Answers 4


More the power, more the heat generated.

$$P(\textrm{Power})=V\cdot I$$ $$H(\textrm{in joules})\propto V\cdot I\cdot t\,(\textrm{in s})$$ $$ \therefore H(\textrm{in joules})\propto P\cdot t (\textrm{in s})$$

The 11.1 volt gloves does not give greater heat just because it has more Potential difference, but because $\textrm{Power}$ is greater in it.

  • $\begingroup$ So all that counts is the power generated? Two batteries running at different voltage but generating the same power will generate the same heat? How does this reconcile to the Joule effect (am I remembering it correctly?) whereby heat is proportional to the square of the current? $\endgroup$ Commented Apr 6, 2016 at 15:27
  • $\begingroup$ $$V=I*R$$ $$P=(I*R)*I$$ $$\therefore P=I^2R$$ You can do the same using this formula, and the same answer will come. The actual heat generated will be the same if power is the same, but the heat we feel depends upon the material. $\endgroup$ Commented Apr 6, 2016 at 15:29
  • $\begingroup$ So en.wikipedia.org/wiki/Joule_heating is irrelevant in this context? Why? $\endgroup$ Commented Apr 6, 2016 at 15:34
  • 1
    $\begingroup$ It is not irrelevant in this context at all! $$I^2R=\frac{V^2}{R}=VI=P(Power)$$ Heat is proportional to Power, which in turn, is equal to all the above. You can just re-write the formula given in wikipedia to know that Heat is diretly proportional to Power and time $\endgroup$ Commented Apr 6, 2016 at 16:07

As you have confirmed the markings on the batteries to be, on one hand, 7.4V 2000mAh and on the other, 11.1V 1400mAh, and also considering that I have no knowledge about heated gloves, you can think of it as follows:

The 7.4V 2000mAh battery can deliver, say, 200mA for 10 hours at approximately 7.4V. I say can because what determines the actual current, is the load connected to the battery. And approximately because the voltage will drop somewhat during discharge.

The contained energy in a fully charged 7.4V 2000mAh battery can also be expressed as 14.8Wh (watt hours), and if the load (the heating elements) is designed for ten hours use, the supplied power is 1.48W power for those ten hours. If the load is designed for 5 hours use, the battery can deliver 2.96W for that time.

Similarily, the 11.1V 1400mAh battery contains 15.54 Wh, and again if the gloves are designed for ten hours use, the available power for those ten hours is 1.554W. If designed for 5 hours use, the power for that time is 3.1W

The 11.1V gloves heat your hands more because the battery can deliver slightly more power (if designed use time is equal) or because of a lot of other factors in the design of the gloves.

  • $\begingroup$ The OP said the second set heats much more, so a less than 0.5% difference won't do. $\endgroup$
    – L. Levrel
    Commented Apr 6, 2016 at 19:04
  • $\begingroup$ @L.Levrel We don't know what power the two sets of gloves are designed for, or any other design factors as I wrote. $\endgroup$ Commented Apr 6, 2016 at 20:54

Are you sure the batteries are labelled "2 A" and "1.4 A" rather than "2 Ah" and "1.4 Ah"?

A battery doesn't have a fixed intensity. It has an approximately fixed voltage, and the intensity depends on the resistance of the connected load. So usually batteries bear two numbers: voltage and capacity (usually expressed in ampere-hour, 1 Ah=3600 coulombs).

If this is really "2 A" and "1.4 A", is this written on the gloves rather than the batteries? They would be the nominal/typical intensities at which the gloves function. Then your numbers are correct, and there remains your question: why does set 2 heat more?

Well, heat perception is really subjective. Unless you placed your gloves in a calorimeter to measure heating power, what you should ask is: why does set 2 feels like it heats more?

  • Maybe the heat capacity of set 2 is lower, so that its temperature raises faster at power-up: they get warm quicker, but in the stationary state they will provide less power

  • Maybe the outer insulation of set 2 is better, so that it loses less heat to the environment, and it indeed heats you more in spite of consuming less electrical power.

  • $\begingroup$ The labels on the batteries report 1400 mAh and 2000 mAh. The specs on the respective websites report 1.4 ampere and 2 ampere. $\endgroup$ Commented Apr 6, 2016 at 15:32
  • $\begingroup$ @Pythonistaanonymous, those facts strongly suggest that whoever wrote the text for the web site had no idea how the heaters actually work. Of course it could just be coincidence, but then if those specs are right, you should not expect either pair of gloves to stay warm for more than an hour. That's not very many runs on a ski slope. It would not be the first time that the published "specs" in the advertising for some product were completely bogus. Only way to know for sure is to hook up a DMM and test it yourself. $\endgroup$ Commented Apr 6, 2016 at 15:46
  • $\begingroup$ @james large , according to the respective websites, the 12V glove would stay on for about an hour at maximum power, while the 7.4 glove would stay on for ca. 2.5 hours at maximum power. Since the power generated is very similar, I am puzzled. Of course, these numbers could be totally wrong for all I know! $\endgroup$ Commented Apr 6, 2016 at 15:55
  • $\begingroup$ Also, in a slightly broken English, the manufacturer of the 12v gloves writes that ( gerbing.eu/en/products/12v-products/12v-batteries/… ): "Sometimes customer asks us why the runtime of the 7volt heated gloves are longer than the batteries of the 12 volt heated gloves. The answer to that is simple. Because the 12 volt battery must be able to power a 12 volt glove, there can be put less amps into a hybrid battery for the hybrid heated gloves." $\endgroup$ Commented Apr 6, 2016 at 15:55
  • $\begingroup$ @Pythonistaanonymous If the heater with the nominal 1400 mAh battery really draws a constant 1.4 A for one hour (in spite of the battery voltage declining the whole time), then it you should expect it to last for one hour. If the heater with the 2000 mAh battery draws 2 A for one hour, then it too will last one hour. Power and voltage play no role in that equation. Ampere hours is just current times time. $\endgroup$ Commented Apr 6, 2016 at 16:04

I read some of your comments.

Explaining why $H = I^2R $ is not working out

If same power is supplied for same time, equal heating would take place.

$H = I^2R $ is not working here because R in two cases are different.

$V = IR_{1}$ $7.4 = 2 R_{1}$

$R_{1} = 3.7 ~\Omega$

Similarly ,

$R_{2} = 11.1/1.4 = 7.928 ~\Omega$

Now, if you use $H = I^2Rt$ , you get correct answer.

Now, to explain why One glove is heated more.

V marked on cell is EMF.

I marked is max current that can be supplied.(This is due to internal resistance of cell)

P marked is max power that can be supplied.

So, actual current supplied in two cases are actually different.

$V =IR$


$ I_{1}= 7.4/R$


$ I_{2}= 11.1/R$

Clearly, $I_{1}≠I_{2}$


So, heat supplied is actually different. And *11.1*V battery supply more energy.

  • $\begingroup$ @MAFIA36790 Thank you for your edit. I was wondering how to write ohm. $\endgroup$ Commented Apr 6, 2016 at 17:18

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