I am looking for a heating resistor (strip), but I would like to keep the current as low as possible. I know that heat is (created and) directly proportional to current and not voltage but on the other hand the power disipated is calculated from current * voltage.

Help me wrap my head around this please.

If I have the option to use 24V, 20W resistor which will run at 0.8 Amps and 12V, 20W heater which will run at 1.6 Amps, will they actually output same (approx) heat. From the power rating I would expect so, but the current says otherwise.

The form factor is the same.

  • 1
    \$\begingroup\$ Just interested, why would you want to minimize current? \$\endgroup\$
    – floppydisk
    Mar 6 at 9:55
  • \$\begingroup\$ I am upgrading a 3D printer and I want to plug into its power supply. \$\endgroup\$ Mar 6 at 12:14
  • 1
    \$\begingroup\$ In this case you should just go with the heater that matches the voltage of your power supply (typically 24V if it's a 3D printer) and has the power that you want. If you connect 2 x 12V 20W heaters in series you'll get 40W of power. If you connect 2 x 24V 20W in parallel you'll also get 40W. However you cannot 'cheat the system' and get more heat for less watts. Simply put, how many watts you burn = how much heat you get. \$\endgroup\$
    – floppydisk
    Mar 6 at 15:18
  • \$\begingroup\$ Define high temperature? \$\endgroup\$ Mar 6 at 15:49
  • 4
    \$\begingroup\$ Did you forget to take into account that the resistance is different? \$\endgroup\$ Mar 6 at 19:42

6 Answers 6


I know that heat is (created and) directly proportional to current and not voltage

This is your mistake. Heat in a resistor is directly proportional to (in fact, equal to) the power. Current is related to power, but it's the power that matters, not the current.

In answer to your comment

Ok... so why, when I have a thin wire, prefer to use high voltage and low current in order not to exceed the cable's rating (which is limited by heat).

I think you're getting confused by the the fact that you always use low current and high voltage for power transmission, where possible. That's a different matter: when we say "high voltage" in this context, we're referring to the voltage between the two power lines. The voltage across a power line, between the source end and the load end, will be lower if the current is lower.


simulate this circuit – Schematic created using CircuitLab

Consider this circuit, which represents a typical 1500 W space heater (the simplest of loads) plugged into a typical 120 V North American mains circuit. The actual current works out to 12.2 A, considering the added resistance of the power line. Each leg of the power line then has 1.22 V (12.2 A * 100 mΩ) dropped across it (that is, between nodes A and B, or between nodes C and D), and dissipates 14.88 W (12.2 A * 1.22 V) of power. The space heater, the actual load, dissipates 1430 W.

Now if we instead use the exact same power cord for the same power load in Europe (I know most of Europe uses 230 V, but using 240 V here makes the numbers work out nicer),


simulate this circuit

Because we want the power to be the same, but the voltage is higher, we use a higher resistance load. The current here works out to be 6.22 A, meaning the voltage drop in each leg of the power line is 0.622 V--approximately half what it was on the 120 V circuit. The power dissipated in each leg of the line is now 0.622 V * 6.22 A, or 3.87 W. The actual load dissipates 1485 W.

Notice how, despite using a higher voltage, the voltage between nodes A and B (and likewise between C and D) went down for the same power load. That is what people talk about when saying that higher voltage means lower current and lower losses: given a load of fixed power, higher voltage between nodes A and D (and thus higher voltage between nodes B and C) means lower voltage between A and B (and between C and D), lower current through the whole circuit, and lower losses in the power lines.

  • \$\begingroup\$ Ok... so why, when I have a thin wire, prefer to use high voltage and low current in order not to exceed the cable's rating (which is limited by heat). \$\endgroup\$ Mar 6 at 17:05
  • \$\begingroup\$ See physics.stackexchange.com/a/92504 \$\endgroup\$
    – Drise
    Mar 6 at 17:25
  • \$\begingroup\$ Yes, that is exactly my point. The answers here suggest that it does not matter if I exchange current for voltage since P = U * I will still be the same. But according to Joule heating, the conductor produces less heat with less current. \$\endgroup\$ Mar 6 at 17:53
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    \$\begingroup\$ @LukášŘádek Because when you're doing that, you're talking about low current and lower voltage, because the resistance is constant. The voltage in the formula P=VI is the voltage *across the resistive element. In a power line, that means the voltage between the source and the load, not the voltage across the load. You might have a nominal voltage of 13.8 kV, but only a fraction of a volt from one end of the line to the other. \$\endgroup\$
    – Hearth
    Mar 6 at 20:07
  • \$\begingroup\$ @LukášŘádek I've expanded the answer to hopefully address your confusion. \$\endgroup\$
    – Hearth
    Mar 7 at 14:51

Heat is directly proportional to power, not current. The heat IS the power dissipated in the resistor.

Power (and therefore heat) is directly proportional to both voltage and current.

$$Heat = P = U * I$$

As for your example with power lines, the reason it seems like it is just the current that matters is we are no longer holding power fixed, yet the resistance of the wire IS fixed. The voltage in this case is the voltage dropped across the wire, not the system voltage. Just the voltage lost along the power wire.

Ohm's law can show us how that works:

$$P = U * I$$ $$U = I * R$$ substituting: $$P = I^2 * R$$

So, in this case, lowering the current drastically lowers the power dissipated.

This is different than your situation, because in the case of power lines, it is the load, not the wire, that determines the current. Higher voltage will be stepped down at the load, and the load consumes whatever current it needs at that voltage. Since the power is fixed, and the resistance of the line feeding the load is fixed, the higher voltage allows less current and therefore less power (in the line) lost. A great easy trick to lower power loss in those lines.

Your situation, the resistor is the load itself, so you consider it dropping the entire system voltage across it.

  • 4
    \$\begingroup\$ To the downvoter who failed to leave a comment: Please let me know how I can improve this answer. What made you single out my answer for a downvote? \$\endgroup\$ Mar 6 at 2:14
  • 1
    \$\begingroup\$ My guess is that it was considered a duplicate of the answer given 30 seconds later (the reading order of equally rated answers is random). Also the symbol letter for voltage is U rather than V. Once the other answer has a higher rating, it will appear first in order and be more likely to get more upvotes. So my guess is just bad luck and someone else giving essentially the same answer independently at almost the same time. \$\endgroup\$
    – user107063
    Mar 6 at 3:09
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    \$\begingroup\$ Alright :-).. but isn't that true that when having a thin conductor, we prefer to use high voltage / low current in order not to melt the cable? Meaning that we try to lower the current to prevent high temperature, but P = U * I is still the same. According to your answer, the cable would be heated the same, wouldn't it? \$\endgroup\$ Mar 6 at 17:08
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    \$\begingroup\$ @LukášŘádek This is because the resistance of the wire is constant, unlike in your example, so the power dissipated in the wire is lower with high voltage and low current. It is the voltage dropped across the resistance, not the total system voltage that matters. \$\endgroup\$ Mar 6 at 17:56
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    \$\begingroup\$ @user107063 Thank you. I guess I tend to use V instead of U because I was educated in the US. electronics.stackexchange.com/questions/99584/… \$\endgroup\$ Mar 6 at 18:00

You are correct, \$P=IV\$ where \$P\$ is power, \$I\$ is current, and \$V\$ is voltage. But, here's the thing: Your two resistors have fixed values. Ohm's Law says, \$V=IR\$ where \$R\$ is the fixed value of the resistor. That means, for any given resistor, you cannot change \$I\$ without also changing \$V\$ and vice versa. \$I\$ and \$V\$ are proportional to each other.

If you play around with the two formulas given above, substituting one into the other, you can infer that \$P=I^2R\$, or that \$P=\frac{V^2}{R}\$ In words: For a given fixed resistance, \$R\$, the power is proportional to the square of the applied voltage and proportional to the square of the current. It must be that way because, for a given fixed resistance, as the voltage goes up, the current must also go up, and vice versa.

Plugging in values, \$V=12\$ or \$V=24\$ and \$P=20\$, you will find that the 12V heater must have a resistance of 7.2 Ohms, and the 24V heater must have a resistance of 28.8 Ohms.

  • \$\begingroup\$ Yup, I understand that, thank you. But if I have a resistor (heater) which runs 1A/24V and other that runs 2A/12V, would they produce the same amount of heat? I mean... we try to minimize the current on the expense of voltage in thin conductors. \$\endgroup\$ Mar 6 at 17:35
  • 1
    \$\begingroup\$ Yes, they will heat the same. But you should always definitely use appropriately sized conductors, or there will be fire. Search the web for "sizing conductors calculator", add a safety margin, and if in doubt always use thicker wires. \$\endgroup\$ Mar 6 at 19:28
  • \$\begingroup\$ Yup, that is what I always check, no worries. Thank you. I was just curious why we are prefering higher voltage in thinner cables while the total Power would be the same. \$\endgroup\$ Mar 6 at 23:05
  • \$\begingroup\$ @LukášŘádek, thinner cables cost less. Cost less to procure because they contain less copper, and cost less to install because less effort needed to route them and terminate them. \$\endgroup\$ Mar 7 at 3:09
  • \$\begingroup\$ How is that related to power dissipation and my Q? 😁 \$\endgroup\$ Mar 7 at 14:21

I know that heat is (created and) directly proportional to current and not voltage

Incorrect. The heat made is proportional to Resistance x current squared.

P = I2R.

How does that happen? Take Ohm's Law, V=IR, and use it to replace V in Watt's Law, P=VI The result is P = IR • I or I2R.

So no, heat is not at all directly proportional to current. When you take a resistor of specific ohm value, and cut voltage in half, current cuts in half and power quarters.

Here's a trick usable in a North American home. You want a 120W heater but all you can find is a 240V 480W heater. The 480W heater is drawing 2A @ 240V. Its resistance is 240V = 2A • R or R = 120.

We feed it 120V instead. Current is V=IR so 120 = I • 120 or I=1. Alright. So what power is that? Watt's Law P=VI so 120W = 120V • 1A Voilà!

Another way of looking at it is, use two resistors. You can wire the resistors in series or parallel. Think about how that affects voltage and current.

  • 1
    \$\begingroup\$ @quetzalcoatl Oh, you're right! My brainfart. Yes, that would result in a "reduced service life" for that poor heater :) Thanks. \$\endgroup\$ Mar 6 at 9:25
  • \$\begingroup\$ Heaters tend to use red hot wire which has quite higher resistance than wire at lower temperatures. So if you are banking on it drawing only a quarter of the power (and half the current) at 120V, you may be seriously off. If there is a ventilator included, running it at half voltage may not be sufficient for drawing off the heat fast enough for the heater's health. \$\endgroup\$
    – user107063
    Mar 6 at 18:03
  • \$\begingroup\$ @user107063 That's a good point, but IME in any reasonable producing heat range, it becomes hot enough to become reasonably ohmic. The accessory blower would be a problem, yes. \$\endgroup\$ Mar 6 at 18:56
  • \$\begingroup\$ And a diode with 240V will give you 240W. 480W but for half the time. \$\endgroup\$ Mar 9 at 17:28

There might be some confusion how things work. While in that simple circuit example (heater connected to power source via two wires) there is indeed only one (same) current "I", there is not one (but at least 4) important voltages (as each of them have their own resistance). Let's call them:

  • Us = Voltage provided by the source (battery or what have you)

  • U1 = Voltage drop on first piece of the wire (from one terminal of the battery to heater)

  • Uh = Voltage drop on the heater

  • U2 = Voltage drop on the second piece of the wire (from heater to another terminal of the battery).

Now, each of the voltage drops depends on the resistance "R" of that specific piece (let's call them R1 / Rh / R2). For regular pieces of wire and your heater in normal usage (this is simplification for ideal case), that resistance will be always the same no matter what voltage you apply or what current is passing through.

For such resistors connected in series, total resistance of wires and heater is Rtotal=R1+Rh+R2. Thus the current (the same for all elements) will be I=Us/Rtotal.

Now, let's for example set R1 = R2 = 0.5 ohm and Rh=15 ohm. In that case, Rtotal = 0.5 + 15 + 0.5 = 16 ohm. And if your source is:

  • Us = 12V: I = 12 / 16 = 0.75A, power dissipated at each wire is P = I^2 * R1 = 0.75^2 * 0.5 = 0.28W (cca), and power dissipated at heater is about P = I^2 * Rh = 0.75^2 * 15 = 8.44W
  • Us = 24V: I = 24 / 16 = 1.5A, power dissipated at each wire is P = I^2 * R1 = 1.5^2 * 0.5 = 1.13W (cca), and power dissipated at heater is P = 1.5^2 * 15 = 33.75W

So, doubling the voltage of the source quadrupled the power dissipation (AKA heat) - both in heater and in the wires.

Note that in "regular" case (the one you talk about - i.e. power is provided by voltage source like a battery or mobile phone charger) you can only get different battery/charger one providing different voltage -- current is not directly controlled by you (and the one specified on the charger is only maximum current, not the one that will be flowing through the circuit).

Also note that you cannot choose voltage drop (or power dissipation) of the the heater (or of the wires) directly - the only thing you can choose (when buying them) is their resistance; and current and power / heating over them will vary according to rest of the circuit (even such simple one!)

Same problem with labeling as above applies. If heater says it is 20W@24V they are actually trying to say that its resistance is fixed at R = U^2 / P = 24^2 / 20 = 28.8 Ohm (which you could check with ohmmeter!). Your 20W@12V is however R = 12^2 / 20 = 7.2 Ohm. It will only produce exactly 20W of heat if you provide exactly 12V (or 24V for the other one) voltage drop on it -- which is not the same as the voltage of the source (although, depending on your wires, it will probably be close enough). Also note that while you can sometimes put more voltage on them but less current for same power, it is not recommended - other things might break (i.e. trying to connect it to 240V but with very small current so power remains at 20W is not a good idea).

And that is "Why do we prefer lower current and higher voltage for thin conductors in order not to burn them" -- because you will "burn them" if you dissipate too much power on them (P1 and P2). And as the power dissipated rises linearly with voltage drop, but quadratically with current, to keep power dissipation on the wire low, you want P1 = I^2 * R1 to be low -- which means lowering not only R1, but much more I (which, as opposed to R as noted above, you do not control directly, but is is dependent on those other calculations, which need to consider rest of the circuit)

As an exercise for the poster: Measure the resistance of your wires, put it in the formulas above, and tell us how much the wires will heat up for first heater, and how much for second one.

(as side note: in reality, it is never so simple. Resistance will vary ever so slightly according to temperature and even humidity and most notably the quality/price of the element, there are tolerances so thing on the label is just approximate, there are such things as current sources etc. -- but nothing that OP should worry about at this stage).


Heat generation of the resistor (i.e. power dissipation) is due to electrons colliding with molecules in the resistor. Moving electrons is what we call "current". The collisions excite the molecules. This excitation is what we call "heat".

So, If I have two resistors, R1 and R2. And R1 is greater than R2. And if both have an identical current passing through. This means R1 is generating more power than R2. (Same current + higher resistance = more collisions = more power)

How is it possible to do this? Well, R1 will need a higher "driving force" to push the same amount of current through it. This "driving force" is what we call "voltage".

So, if you have 5 amps and 20 volts, then there must be a higher resistance and more power than if you had 5 amps and 10 volts.

  • \$\begingroup\$ Very nice and intuitive explanation. Just how I like to explain things, thanks! Therefore 1A/12V should produce the same amount of heat as 2A/12V? Why do we prefer lower current and higher voltage for thin conductors in order not to burn them? \$\endgroup\$ Mar 6 at 17:38
  • \$\begingroup\$ @LukášŘádek absolutely not: it will produce 200% more heat. 1A/12V will dissipate 12W of heat, and 2A/12V will dissipate 24W of heat. \$\endgroup\$ Mar 6 at 18:05
  • \$\begingroup\$ @Matija Nalis My bad... a typo... I meant 1A/24V. In that case, I guess, the answer is yes. And in that case my second question applies... if I have a cable that is not thick enough to run 2A/12V, how does it help to run 1A/24V, since it produces the same amount of heat? \$\endgroup\$ Mar 6 at 23:14
  • \$\begingroup\$ It doesn't "help" since you put different heater with different resistance -- exactly in order to have same power dissipation/heat with different voltages (they're both 20W so they heat the same!). If you however used the same heater element (thus with same resistance), then it would be quite different if you put it on 12V or 24V, as different current would flow and thus different amounts of power would be used / different amounts of heat produced. See my answer for details. Different voltages only "help" if the resistance remains the same (i.e. if it is the same element) \$\endgroup\$ Mar 7 at 1:35

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