Would a heating element have a very high resistance, or a very low resistance? (All comments in this post are based around the fact that the voltage is the same for each situation) I would have thought that a higher resistance would have resulted in more heat loss, but I've been taught that the higer the current, the more energy is lost to heat. Therefore, a lower resistance would release more heat.

Which one is right? Thanks for any help.

it is difficult to visualize the fact that reduced resistance and increased current generated more heat . If somebody try to explain me without much math because I know what

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    \$\begingroup\$ P=I*V so if your V is fixed you have to increase I. How do you do that? Decrease R. \$\endgroup\$ – Brian Drummond Oct 3 '17 at 10:36
  • \$\begingroup\$ Based on V = I * R... \$\endgroup\$ – Solar Mike Oct 3 '17 at 10:38
  • \$\begingroup\$ But as R approaches 0 such as in a superconductor, won't there be no resistance to create heat? \$\endgroup\$ – vercellop Dec 11 '18 at 7:05

All of this relates to two things:

Ohm's Law: \$ R = \frac{V}{I}\$

Joule Heating \$ P_\text{Heat} = V \cdot I\$

The first tells us that if we keep the voltage \$V\$ a constant, the current will increase when the resistance decreases. This makes sense since the resistance is a measure of how hard it is to have current flow from one node to another - if it resist less, more current can flow.

The second then tells us that the power increase with current and with voltage. If we keep the voltage constant, but increase the current, the power will increase.

In a resistor, all this energy is turned into heat. Thus, if we have more power being dissipated in the resistor, it will get warmer if it has a lower resistance (given it has a constant voltage across its terminals).

  • \$\begingroup\$ see the diagrams in the other answer \$\endgroup\$ – Solar Mike Oct 3 '17 at 11:05
  • \$\begingroup\$ The voltage over the heating element is not constant as the only thing that changes is the heating element. Source impedance is never 0. \$\endgroup\$ – le_top Oct 3 '17 at 13:05
  • \$\begingroup\$ @le_top ofcourse, source impedance is never 0, but one can often assume it is without significant error. You could also say how this ignores any black-body radiation effects, or magnetic effects, or thermal noise currents injected back at higher frequencies but none of those are generally an issue either. The question was "given a constant voltage" and I answered as such. Assuming ideal sources happens all the time. \$\endgroup\$ – Joren Vaes Oct 3 '17 at 13:08
  • \$\begingroup\$ The question is with regards to a heating element. Heating element resistances are normally low, so the source impedance becomes important. Most people will suppose that the voltage is constant regardless of the source. Even if the heating element is powered from a low voltage source (battery or regulated), it is relevant to consider the source impedance. Most applications discussed here are "low voltage" electronics. The lowest resistance we can create is a 0 Ohm resistor and in that case we have near 0 power consumption. \$\endgroup\$ – le_top Oct 3 '17 at 13:49
  • \$\begingroup\$ @le_top This could just as well be the heater element in a TCXO, which would likely be a quite high resistance compared to most regulated supplies output impedance. Ofcourse, source impedance is important, but the question literally states that the voltage is to be kept constant. \$\endgroup\$ – Joren Vaes Oct 4 '17 at 14:32

The math has already been tackled, so I'll go for the visual:

The easiest way to visualise it is with water. Imagine a pipe with water flowing through. imagine this is the flow of electricity. Now we need to add a resistor. lets pretend we have a 10 ohm and a 1k resistor.

To visualise this with the water, imagine the 10 ohm resistor has thinned the pipe, so the water is constricted. It's like putting your finger on the end of a hosepipe, the water has more energy out the other side. Here is a visual aid to help: enter image description here

The higher the resistance, the wider the path. The lower the resistance, the smaller the patch, and hence more current.

  • \$\begingroup\$ a visual answer - neat +1 from me \$\endgroup\$ – Solar Mike Oct 3 '17 at 11:04
  • \$\begingroup\$ This is incorrect when the resistance is lower than the source resistance - a lower resistance will then consume less power! In the drawing this would be close to the resistance being as wide as the original pipe - you are suggesting that this would consume more power! Maybe, if the pump itself is not consuming more. \$\endgroup\$ – le_top Oct 3 '17 at 13:02
  • \$\begingroup\$ Yes, that is true.... I was trying to make it as simple as possible (as the OP seems like a beginner) and going to deep into it may be more confusing. For complete beginners, I feel this is a rather simple way of giving them some sort of visualisation to help them get their head around it \$\endgroup\$ – MCG Oct 3 '17 at 13:26

If somebody try to explain me without much math

There will always be some math....

The power dissipated by a resistor is voltage x current and, if you use ohms law, you can derive two other expressions namely P = \$I^2 R\$ and P = \$V^2/R\$.

Using the first derived formula (P = \$I^2 R\$)....

If the resistor is applied across a fixed voltage and its resistance drops by 10% then current will increase by 10% and, as you can see from \$I^2 R\$, power must increase and hence it gets hotter because \$I^2\$ is 1.21 higher and hence \$I^2 R\$ is 1.1 times higher.

  • \$\begingroup\$ But what if your heating element approaches superconductivity? Surely there won't be any resistance to make heat even though the current is high. That's where it's not intuitive. \$\endgroup\$ – vercellop Dec 11 '18 at 7:04
  • \$\begingroup\$ Superconductors cannot dissipate heat therefore it is no longer a heating element. @vercellop \$\endgroup\$ – Andy aka Dec 11 '18 at 8:45
  • \$\begingroup\$ Yes, exactly - I mean that following your explanation above you might extrapolate that as you keep lowering resistance and increasing current, you'll keep getting more heat, but obviously that doesn't hold. At which point does lowering resistance cease to give you more heat? \$\endgroup\$ – vercellop Dec 12 '18 at 7:26
  • \$\begingroup\$ You need to draw up a scenario to analyse. \$\endgroup\$ – Andy aka Dec 12 '18 at 10:18

None of the replies posted so far mention the internal resistance of the supply.

From mathematical analysis we know that the power absorbed by a load is highest when the load it the complex conjugate of the source impedance \$X_{load} = \overline{X_{source}}\$ .

Now let's consider that the load is only "Resistive" which is approximately the case for a heating element.

For the sake of the example, we suppose that the source voltage is 100V and the source impedance is 1 Ohm.

Look at the extremes:

  • When the load impedance is infinite, the current is 0 and power consumed in the load is 0;
  • When the load impedance is 0, the current is 100A, but the voltage over the load is 0. The power consumed by the load is 0W, while the power consumed by the source is 10000W (100Vx100A).

Look at the theoretical best case:

  • When the load is 1 Ohm, the current is 50A (100/2), the voltage is 50V (100/2), and therefore power consumed by the load is 2500W, which is equal to the power consumed by the supply.

And intermediate cases:

  • When the load is 1.1 Ohm, the current is 47.6A, the voltage over the load is 52.4A and power consumed by the load is 2494W.
  • When the load is 0.9 Ohm, the current is 52.6A, the voltage over the load is 47.4V and the power consumed by the load is 2493W.

As you can see, the theory is confirmed. By deviating just a little bit from the ideal resistance we have lower power consumption in the load. The voltage also changes even though the internal source voltage is constant.


A smaller resistance will produce more heat as long as the new resistance is still higher or equal than the source resistance (if the source is purely resistive).


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