How to Determine Wattage

There are a few methods to quickly determine wattage on a part that doesn’t really list the parameter in the product attributes or the data sheet. Many parts have this rating because they absolutely need it in the description due to it being critical to their design. A few examples include power supplies, resistors, most AC or DC fans, most AC or DC motors, and anything else that deals with power directly. However, there are many electronics parts out there without the rating and this may cause concern. The answer will depend on the type of item asked about. There are two fundamental formulas for power that should be utilized.

P = I * V=I^{2}*R

Where “I” is the current measured/calculated, V is the voltage measured/calculated, and R is any resistance or resistor on the path that current is traveling on.

Three Types of Power Ratings

There are three types of power ratings to consider. There is power generated (supply side), power consumed (load side), and maximum power rating which is the absolute maximum a part can handle before it burns out which can be either load side or supply side depending on the type of part. For example, a power converter will have maximum amperage input requirements and maximum amperage output specifications. If the input amperage is exceeded the load of the power supply will be overloaded and burn out. If the output amperage required is exceeded by the load, the load will not function properly and the power supply may not last as long.

Passive Components

Believe it or not, capacitors and inductors are not perfect devices. The reason that there are ESR equivalents and DC Resistance equivalents is that there is a factor called leakage. Leakage is based on the fact that all materials in devices will have some sort of resistive loss based on how resistance is determined. Devices will typically list what their equivalents are in the datasheets.


Our list of capacitors all have voltage ratings, but no current or power ratings. Capacitors are highly dependent on the maximum voltage used in the circuit design. The context for missing current data is based on the fundamental operation of a capacitor. If the capacitor is used in a non-oscillating, there will be hardly any power drawn by the component because leakage is so low. Determine the parasitic power loss through the following formula:


Where “I” is the current measured/calculated on the wire and RESR is the series equivalent resistance. Remember Ohm’s law is V = I x R and can be manipulated using algebra (so can the power equation). It is easier to determine power loss in an AC application or oscillating DC application for combinations of these three parts working together based on resistive values. Technically, the current isn’t passing through the capacitor based on electric field isolation between layers of a capacitor, but current can be conducted on both sides because the voltage is changing rapidly enough to cause a difference in potential on either side.


Our inductors list current ratings, but no voltage or power ratings. Inductors are dependent on the maximum current used in the circuit design. To determine how much power an inductor can handle before burning out, take the current rating times the voltage used in the circuit. To determine power loss from resistance in the inductor, take the DC Resistance (DCR) and use it in the P=I^2*R formula to determine if there is a significant loss.

Cables and Wires

The special note about wire is that the maximum current is determined by the wire gauge used. Here is a table for single core UL-CSA tested cables for amperage for 4/0 to 24 AWG at 30º C. Data is bound to change based on how wires are tested.

Wire Gauge Ampacity Chart
AWG Size Ampacity (Amps)
4/0 325
3/0 275
2/0 225
1/0 200
1 180
2 170
3 154
4 120
6 95
8 75
10 52
12 34
14 24
16 20
18 9.5
20 6.0
22 5.0
24 3.5

After the current is either measured or compared to charts or calculated, just multiply the rated voltage times the current for the rated power the cables can handle.

Active Components

The same formulas apply to active components to find wattage. The only thing I should mention is most items give maximum current and maximum voltage ratings. So if the wattage is calculated this way, it will be maximum wattage consumed or generated (depends if it is a load or supply). Also, some components like motors or fans have a variable driving current and voltage because of the conditions that the fan is operating in, typically manufacturers will give a nominal value and one can calculate a nominal power. This will be around what the fan/motor should consume in optimal conditions, expect a change if the conditions are not met. Items that require heat-sinks also may have varying voltage or current and it is recommended to use the appropriate heat-sink or there will be variance.

I will cover an example of how to find wattage for COB LEDs in a separate post because there is more detail to that example. Here is the post:

RMS Ratings

The final topic for wattage are items that have Root Mean Square (RMS) ratings for voltage and current. These apply to alternating current applications and some oscillating DC circuits. The power can still be calculated from these values, but the answer will also be in RMS terms. Most manufacturers will only be concerned about the RMS values because finding absolute power gets complicated in elaborate AC grids.


“The context for missing current data is based on the fundamental operation of a capacitor. If the capacitor is used in a non-oscillating, non-voltage changing DC application, there will hardly be any current drawn by the component. Determine the parasitic power loss through the following formula:”

False. The ESR is leakage resistance, and frequency is irrelevant. Resistors (resistance) are not affected by frequency.

" It is easier to determine power loss in an AC application or oscillating/voltage changing DC application for capacitors because the current has the greater potential to pass through a capacitor in those situations."

False, Current DOES NOT pass thru capacitors. Unless they are defective, or, leaky. Current appears to pass thru, but it does not. The dielectric contains E fields, not current flow.

Fortunately, it is easier to determine power loss over an inductor because we have current ratings. Take whatever voltage is used in the design and multiply it by the rated current for the inductor."

False. Rated voltage and curent does NOT equate to power loss. Power loss is resistive (copper losses in the wire) and is a function of current-squared and the ohmic resistance (on a DC or RMS basis) or the integral of the current for AC/RF.

Voltage in the design x inductor current is the power in the rest of the circuit, NOT the inductor.

We cannot make up our own laws of physics.

@daveca A capacitor can still draw very little power in a pure DC operation due to the leakage resistance, the amount is almost negligent so designers don’t have to worry about the ESR in most cases. All items have some sort of parasitic loss (there is no such thing as perfect capacitance, perfect resistance, and perfect inductance). The power equation applies to any voltage and any current and any resistance value (not just resistors only).

As for your statement on current not passing through capacitor you are correct…assuming negligent ESR ratings and pure DC applications, other wise current can still conduct properly on both sides. Anything that changes the voltage fast enough, current won’t care that there is a capacitor in series:

The image above is an LTSpice simulation of a DC Sweep from 0 to 5V in .01V steps. There is a very small minimal current passing through the circuit based on ESR leakage (a matter of femtoAmps) which the total circuit would see as practically zero. The capacitor has an ESR of 500 milliOhms in the diagram. So virtually, the circuit will consume no wattage. In reality, it actually consumes a maximum of 0.0125fW. Consider this application though:

This is the same circuit with a pretty slow triangle wave up to 1.1seconds. The voltage is changing fast enough to cause a much higher current draw on the line, up to plus and minus 25 microAmps, which increases power consumption to 124microWatts. This is a multiplication factor of 9.92 TeraWatts difference compared to the femtoWatt number, which is way more significant of a power loss in a high precision design at higher frequencies. I corrected my terminology to reflect how capacitors work.

Power actually can be calculated based off the current a device can be rated for and the voltage supplied. In turn, the power calculated will be the power the device can handle before breaking down or burning out completely. P=IV and P=I^2R. You are right that the inductor does not consume that amount (that depends if there is a DC Resistance equivalence).