resistor power

Resistor Power Rating

When an electrical current passes through a resistor due to the presence of a voltage across it, electrical energy is lost by the resistor in the form of heat and the greater this current flow the hotter the resistor will get. This is known as the Resistor Power Rating.

Resistors are rated by the value of their resistance and the electrical power given in watts, (W) that they can safely dissipate based mainly upon their size. Every resistor has a maximum power rating which is determined by its physical size as generally, the greater its surface area the more power it can dissipate safely into the ambient air or into a heatsink.

A resistor can be used at any combination of voltage (within reason) and current so long as its “Dissipating Power Rating” is not exceeded with the resistor power rating indicating how much power the resistor can convert into heat or absorb without any damage to itself. The Resistor Power Rating is sometimes called the Resistors Wattage Rating and is defined as the amount of heat that a resistive element can dissipate for an indefinite period of time without degrading its performance.

The power rating of resistors can vary a lot from less than one tenth of a watt to many hundreds of watts depending upon its size, construction and ambient operating temperature. Most resistors have their maximum resistive power rating given for an ambient temperature of +70oC or below.

Electrical power is the rate in time at which energy is used or consumed (converted into heat). The standard unit of electrical power is the Watt, symbol W and a resistors power rating is also given in Watts. As with other electrical quantities, prefixes are attached to the word “Watt” when expressing very large or very small amounts of resistor power. Some of the more common of these are:

Electrical Power Units

Unit Symbol Value Abbreviation
milliwatt mW 1/1,000th watt 10-3 W
kilowatt kW 1,000 watts 103 W
megawatt MW 1,000,000 watts 106 W

Resistor Power (P)

We know from Ohm’s Law that when a current flows through a resistance, a voltage is dropped across it producing a product which relates to power.

In other words, if a resistance is subjected to a voltage, or if it conducts a current, then it will always consume electrical power and we can superimpose these three quantities of power, voltage and current into a triangle called a Power Triangle with the power, which would be dissipated as heat in the resistor at the top, with the current consumed and the voltage across it at the bottom as shown.

The Resistor Power Triangle

resistor power triangle


power triangle relationship


The above power triangle is great for calculating the power dissipated in a resistor if we know the values of the voltage across it and the current flowing through it. But we can also calculate the power dissipated by a resistance by using Ohm’s Law.

Ohms law allows us to calculate the power dissipation given the resistance value of the resistor. By using Ohms Law it is possible to obtain two alternative variations of the above expression for the resistor power if we know the values of only two, the voltage, the current or the resistance as follows:

[ P = V x I ]      Power = Volts  x  Amps

[ P = I2 x R ]      Power = Current2  x  Ohms

[ P = V2 ÷ R ]      Power = Volts2  ÷  Ohms


The electrical power dissipation of any resistor in a DC circuit can be calculated using one of the following three standard formulas:

resistor power equations
  • Where:
  •       V  is the voltage across the resistor in Volts
  •       I  is in current flowing through the resistor in Amperes
  •       R  is the resistance of the resistor in Ohm’s (Ω)

As the dissipated resistor power rating is linked to their physical size, a 1/4 (0.250)W resistor is physically smaller than a 1W resistor, and resistors that are of the same ohmic value are also available in different power or wattage ratings. Carbon resistors, for example, are commonly made in wattage ratings of 1/8 (0.125)W, 1/4 (0.250)W, 1/2 (0.5)W, 1W, and 2 Watts.

Generally speaking the larger their physical size the higher its wattage rating. However, it is always better to select a particular size resistor that is capable of dissipating two or more times the calculated power. When resistors with higher wattage ratings are required, wirewound resistors are generally used to dissipate the excessive heat.

Type Power Rating Stability
Metal Film Very low at less than 3W High 1%
Carbon Low at less than 5W Low 20%
Wirewound High up to 500W High 1%

Power Resistors

Wirewound power resistors come in a variety of designs and types, from the standard smaller heatsink mounted aluminium body 25W types as we have seen previously, to the larger tubular 1000W ceramic or porcelain power resistors used for heating elements.

wirewound power resistor

Typical Power Resistor

The resistance value of wirewound resistors is very low (low ohmic values) compared to the carbon or metal film types. The resistive range of a power resistor ranges from less than 1Ω (R005) up to only 100kΩ’s as larger resistance values would require fine gauge wire that would easily fail.

Low ohmic, low power value resistors are generally used for current sensing applications were, using ohm’s law the current flowing through the resistance gives rise to a voltage drop across it.

This voltage can be measured to determine the value of the current flowing in the circuit. This type of resistor is used in test measuring equipment and controlled power supplies.

The larger wirewound power resistors are made of corrosion resistant wire wound onto a porcelain or ceramic core type former and are generally used to dissipate high inrush currents such as those generated in motor control, electromagnet or elevator/crane control and motor braking circuits.

Generally these types of resistors have standard power ratings up to 500W and are connected together to form resistance banks.

Another useful feature of wirewound power resistors is in the use of heating elements like the ones used for electric fires, toaster, irons etc. In this type of application the wattage value of the resistance is used to produce heat and the type of alloy resistance wire used is generally made of Nickel-Chrome (Nichrome) allowing temperatures up to 1200oC.

All resistors whether carbon, metal film or wirewound obey Ohm´s Law when calculating their maximum power (wattage) value. It is also worth noting that when two resistors are connected in parallel then their overall power rating is increased. If both resistors are of the same value and of the same power rating, then the total power rating is doubled.

Resistor Power Rating Example No1

What is the maximum power rating in watts of a fixed resistor which has a voltage of 12 volts across its terminals and a current of 50 milliamperes flowing through it.

Given that we know the voltage and current, we can substitute these values into the following equation: P = V x I.

resistor power rating example

Resistor Power Rating Example No2

Calculate the maximum safe current that can pass through a 1.8KΩ resistor rated at 0.5W.

Given that we know the resistor power rating and resistance, we can now substitute the values into the power equation of: P = I2R.

resistor power


All resistors have a Maximum Dissipated Power Rating, which is the maximum amount of power it can safely dissipate without damage to itself. Resistors which exceed their maximum power rating tend to go up in smoke, usually quite quickly, and damage the circuit they are connected to. If a resistor is to be used near to its maximum power rating then some form of heatsink or cooling is required.

Resistor power rating is an important parameter to consider when choosing a resistor for a particular application. The job of a resistor is to resist current flow through a circuit and it does this by dissipating the unwanted power as heat. Selecting a small wattage value resistor when high power dissipation is expected will cause the resistor to over heat, destroying both the resistor and the circuit.

Thus far we have considered resistors connected to a steady DC supply, but in the next tutorial about Resistors, we will look at the behaviour of resistors that are connected to a sinusoidal AC supply, and show that the voltage, current and therefore the power consumed by a resistor used in an AC circuit are all in-phase with each other.


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  • S
    Steve Kennedy

    This may seem like a dumb question (probably because it is!), as it reflects an appreciation of electronics still very much in its infancy. One book describes a resistor like ‘brakes for electric current’, and speaks of something like friction inhibiting the flow of current. In that analogy, the power dissipated by a resistor must be akin to the drop in wattage from the supply side of the resistor to the wattage of the output side. Car brakes heat up more when bringing a car travelling at 100kph down to 10kph, than brakes would to bring a car from 20kph down to 10kph. So a 12VDC 2A supply reduced to a 100mA output using a 120 ohm resistor, means a reduction of 22.8W (24W – 1.2W). However, all the calculations of power ratings for resistors only take into account the output side of the resistor. So, in the above example, I’d only use a 120 ohm resistor with a power rating of 1.2W or higher, and forget about the 22.8W. Is the analogy confusing then? Or have I miss some fundamental concept altogether?

    And I guess a superordinate question is this: Does it really matter what ampage my DC power produces as long as it is higher than the circuit load?

    • Wayne Storr

      Your power supply is rated at 12VDC, 2 Amps, that is 2 Amps maximum. The load connected to it will only draw from the supply the amount of current it needs so if your load is rated at 100mA then that’s what it will take from the supply. Even if you had a power supply rated at 12VDC, 5 Amps, the load would still only consume 100mA, then effectively in this case, 4.9 Amps of rated current is unused. Then NO to your question, it does not matter what output amperage your power supply is (within reason) as long as it is higher than the circuit load.

      • S
        Steve Kennedy

        Thanks, Wayne, for your reply. Do you have any advice or comments on the first question in my post?

        • Wayne Storr

          About the brakes of a car, no.

        • WeatherHill

          Hi Steve.
          In your post above and in comparison between the car’s brakes and the electrical brakes (the resistor), you say that the analogy is confusing because you only take into account the output side and not the input side.
          Well let me share my perspective on that.

          Your 12v powersupply has a max potential of 2A or 24W power.
          Your car has a max potential of running as fast as 240kph/149mph.
          Is it correct to calculate how hot the brakes of the car would be after a 240kph to 0kph, when you infact are only going to brake from 12kph to 0kph?
          I believe you will agree with me that the answer to the question is “no”. The LOAD is the current speed you’re travelling at, in this analogy.

          If you think of it like this, I think maybe the analogy is not that confusing after all.

          Also in your post you say “…a 12VDC 2A supply reduced to a 100mA output…”. A resistor is not reducing the output. It only drops the left-over voltage so that the load does not draw it’s amps at full voltage. When it comes to a LED, the resistor will prevent the LED from drawing the full voltage and thus also preventing it from drawing the full current available.

          As an example: if you were to put a 5mm green LED as load in your circuit without a resistor, the LED would draw all the power your supply has: 2A @ 12v = 24W. It will burn out in an instant. Now the LED in my example has a forward voltage drop of 3.3V and has a maximum forward current of 25mA. To calculate the appropriat resistorvalue for this, you would do the following:

          12V – 3.3V = 8.7V
          8.7V / 0.025A = 348 Ohm
          Now 348 Ohm is not a standard resistor, so let’s pick a 360 Ohm resistor, cause it’s the closest above what we need and cause I have one right here.
          8.7V / 360 Ohm = ~24.2mA (this is the current that the LED will get).
          The powerdissipation of the resitor will be…
          0.0242A * 8.7V = 0.21W (so a 1/4W should do, but a 1/2W would be advisable).
          The power supply will have 1.976A left to supply for other sources.

  • B
    Bruno Pera

    Hi! I’m wondering if i can make a HomeMade power supply to my laptop (20VDC-2A) out of an hold printer power supply (out : 40VDC) just making a current divisor with 2 10 ohms resistors. Should they be wirewound, since it will dissipate 20*2=40W?

    Sorry for my rusty English =P


  • M
    Mark Hodgins

    Hi Wayne,, I will try to keep this short,, would you be able to recommend a resistor wattage based on an application? 🙂
    I am working on a motorcycle and replacing older style 3 ohm ignition coil “packs” with newer 1.5 ohm ignition coils that are integrated into the spark plug boot. These ratings are measured on the 12 volt supply side of the coil.
    So in order to prevent damaging the ignition module, I need create 3 ohms of resistance to prevent damaging the aftermarket module that gives the 12 volt supply. So logic would would tell me a 1.5 ohm resistor will create my needed resistance,, however i don’t know what the minimum wattage I can use? 10? 🙂 thanks! I only know the voltage is 12vdc or less, i would think a very low amp or current, as they are very thin wires that plug into the coil, that, and the idea is for the coil to do the hard work anyway, right?. thanks! Mark

    • Wayne Storr

      Hello Mark, assuming all things equal, your 3 ohm coil connected to a 12 volt battery would draw 4 amps, (12÷3 = 4). If you use a 1.5 ohm resistor in series with your new coil, the total resistance would still be 3 ohms, drawing 4 amps. The I^2R loss for the resistor would therefore be: 4x4x1.5 = 24 watts. Then you would require a 1.5 ohm power resistor, similar to above, with a minimum power rating of 25 watts (nearest preferred wattage value).

  • C

    I am working on a project to build a drawbridge from toothpicks. I was allowed to use a motor so I bought a cheap remote controlled car and took it apart and attached it to the bridge. My problem is it moves too fast…I think using resistors is the answer so I bought some at radio shack, but I dont know how many to use and how to figure out how many to use. Should they be in serial or parallel? Thanks for any help.

    • Wayne Storr

      Hello Colin, this is a very open question. Most motorised cars and toys in general use a wide range of DC brushed or brushless motors operating from about 3 volts with a pair of AA batteries upto about 24 volts with rechargables for larger cars, toys and racers, with angular velocities up to 30,000 rpm. So the size and power rating of a series resistor you would need depends upon your motor and power supply you are using.

      Another more practical option to control your drawbridge would be to use a suitable Motor Gearbox in which you could set the raising and lowering of your drawbridge to any speed you want.

  • s
    sarfaraz ahmad

    how we calculate the power of wire wound resistor according to size.What is the method for distinguishing 5 watt or 10 watt after seeing size.

  • R

    Hi Wayne,
    Thanks for these wonderful tutorials! I am wondering about putting less current through a resistor, and what would happen. I am on a 12V boat, with a 12V water heater. The heating element is 360W, which calculates to 30 amps (nominal). My question related to when this element is being used as a diversion load from a wind generator. (For the readers who don’t know what I’m referring to, a wind generator through a controller charges the boat’s battery, but as that battery gets charged, some of the current is diverted to the water heater.). The controller may send only 10 amps to the water heater, which is 1/3 the capacity. My question is, how much heat would I get in this case. Using the P=VI formula, I would get 1/3 the watts, but using P=I^2R I would get 1/9 the watts. Which is correct, and why?
    Many Thanks in Advance

    • Wayne Storr

      Hello Rebecca, Assuming all things equal, if your heater draws 30 amps at 12 volts its DC resistance will be 0.4 Ohms (12/30) and the I^2R loss will be 30^2 x 0.4 = 360W. If the heater now passes 10 amps then the IxR voltage drop across it will be 4 volts (10×0.4), Therefore the I^2R power dissipated by the heater at this amperage will be 10^2 x 0.4 = 40W, or VxI = 40W, or V^2/R = 40W regardless of the original power rating. Your charge controller can not give a constant 10 amps at 12 volts into the same heater, Ohms Law will not allow it.

  • P
    Pedro Oliveira

    Hello Wayne,
    I was wondering how do you calculate how many volts are needed in order to warm a small electrical resistance with 5 watts at about 24º celcius.
    I’m doing a project and it’s not my area of expertise.
    Thank you in advance,
    Best regards

    • Wayne Storr

      Hello Pedro, It’s not possible to calculate a voltage only given the wattage. However, as the resistance is only rated at 5W, any small DC voltage will do, 6V, 9V 12V, etc. depending upon your control circuit.

  • R

    Hi Wayne,
    Thanks for the quick response. All makes sense, but I still am having trouble. What happens to the remaining 8 volts? It’s a 12V system, everything is 12V. The wind generator is attached directly to the batteries. The heater is attached to the batteries via a relay. The controller operates the relay at 200hz, opening it sufficient times per second so that the battery gets what it needs to charge, and the rest is ‘diverted’ to the heater. So following up on my previous example, how does it work if the wind generator is producing, say, 12 amps and the battery, being almost fully charged, is only consuming 2 amps, leaving 10 amps for the heater?
    Thanks again,

    • Wayne Storr

      I agree you have a 12 volt system, your wind turbine is 12 volts feeding a charge controller/regulator which is 12 volts which feed the 12 volt batteries. But the dummy resistive load is not connected to the batteries, its connected to a separate output on the regulator. You speak of 200Hz, so I would assume that the regulator is using pulse-width modulated (PWM) DC instead of a constant 12VDC to charge the batteries and divert current to the dummy load to prevent overspeeding.

      A resistor has a fixed value (in your case 0.4 Ohms) it can not change but the voltage across it and the current flowing through it can. The regulator using PWM can turn the dummy load ON and OFF very fast (200Hz) at 12 volts, but the average DC voltage and current output level will depend upon the regulators duty cycle ((i.e. ON duration versus OFF duration). If the battery isn’t full, then the regulator passes all the turbines power to the battery. When the battery gets full, the regulator rapidly turns its output ON and OFF (at 200Hz) to maintain the battery at the fullest state of charge possible while sending the rest to the resistive load. You can always connect a DC voltmeter across the dummy load.

    • R

      Wayne, this is most interesting! Thanks for your time. Yes I am talking about PWM DC (sorry, somehow I forgot to mention that), but when I look closely at the circuit diagram, the setup is a bit different than you describe. The turbine is attached directly to the battery, nothing between it and the battery except a fuse. The resistive load is also attached directly to the battery, but for this circuit the PWM regulator has control, and opens and closes the circuit at 200Hz, closing more often as the battery gets full. (I wish I could send you a drawing, but hopefully I’m being clear.)
      So now back to my original quandary: I think I now understand how the circuit works, but I still don’t know the end result – how much heat is the resistive load producing? Let’s say the battery charge is such that the PWM is closing the resistive load circuit 1/3 of its 200 Hz cycles. Will this result in full heat 1/3 of the time and no heat 2/3 of the time, or do you ‘average’ the power/voltage/amperage (I don’t know what word goes in here), thus getting 1/9 of the heat, as voltage and amperage are squared in the power formulas? To put it another way, and this is what I really want to know, suppose I can set the 200Hz PWM to close the circuit 1/3 of its 200Hz cycles. If was to compare how long it would take to heat the water in this situation versus how long it would take to heat the water if the circuit was closed all the time, what would I find out? Would the 1/3 closed at 200 Hz take 3x as long to heat the hot water, or 9x as long? (ignoring heat losses from the tank to the environment over the time periods).
      Sorry to be so long winded!

      • Wayne Storr

        Hello Rebecca, without seeing your installation I tried to answer your question with the information you provided. If you are now saying that the resistive load is a heating element connected directly to the battery, both being fed by the turbine via a fuse then were is the regulator connected. To finally answer one of your questions, full power is only delivered to the heating element when it is switched ON, Then 1/3 of the time its consuming power and 2/3 of the time it is OFF. Put a voltmeter or ammeter in the circuit to see what’s really going on. Enough said.

  • a

    Would a 240 Ohm Carbon Film Resistor 1/4W 5% be okay to use for a parallel circuit with three LEDs with a 9V battery, 3.4 voltage drop and 20mA? All this is new to me, so I need to know before I spend the money to buy the parts and not worry about things over heating.


    • Wayne Storr

      Let’s see, (9 – 3.4)/20mA = 280 Ohm, nearest preferred highest value equals 300 Ohm +/-5% (E24). At 300 Ohm, current equals (9 – 3.4)/300 = 18.67mA. I^2R loss therefore equals 18.67mA x 18.67mA x 300 = 0.1 watts, therefore 1/4W is more than sufficient.

    • WeatherHill

      Hi Annie.
      I believe Wayne Storr misinterpreted your question (or maybe I did). When you say 3.4 voltage drop, I interpreted this as the voltage to be dropped, and that your three LED’s have already dropped the first 5.6 volts (~1.867V each).

      If that is the case, then:
      3.4V / 20mA = 170 Ohm’s
      20mA * 3.4V = 68mW … which is definitely below the 250mW (1/4W).

      Even if I misinterpreted the question, maybe someone else find this helpful.

  • c


    Can anybody tell how power rating affects the performance of a resistor.

    Suppose i have 2 resistor with different power rating. Which one would be better, the one with more power rating or the one with less power rating…

    Please provide me the logic behind that also…

    • Wayne Storr

      When an electrical current flows through a resistor, the resistor tries to stop or “resist” it, hence its name. The lower the value of the resistor in Ohms the easier it is for current to pass through it. Likewise, the higher its resistance in Ohms the harder it is for the current to pass. The effect of resisting the flow of current is that power is converted from electrical energy into heat, (and light for a filament lamp).

      The power rating, also called wattage rating, of a resistor is the maximum power it can dissipate as heat without damage. For a resistor this power rating is given in Watts and can be calculated as: P = I^2R. In most cases, power rating generally relates to a resistors physical property such as construction, type and size. That is larger resistors can dissipate more power than smaller ones of the same ohmic value without overheating. Then the main limitation of a resistor is its power rating so if you have two resistors of the same ohmic value, it would be better to use the one with the higher power rating as this, in theory, would operate cooler than the smaller one.

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