I was ordering some resistors online, and I saw that 0 Ω resistors have a power rating. Why is that? Power through a resistor is calculated with the equation \$P = UI\$ or \$P = RI^2\$. Since \$R = 0\ Ω\$, \$P=0\ W\$.
According to this post (How to calculate Power Rating for Zero Ohm Resistors?), a 0 Ω resistor has no power rating... But Farnell tells me the opposite:
While it may be true that distributors don't want to check every single part individually, in this case it is not down to laziness that the 0Ω resistor has a specified rated power of 125mW.
As pointed out by @BumsikKim's answer, the datasheet for the series does in fact specify this rating - the distributor product page is correctly representing the manufacturers specifications.
From Page 5, we have the following table entry:
Notice how for the entire RC0805 size series, there is a specified rating of 0.125W (1/8W). This includes the 0Ω resistors in that series.
There is also however crucially another specification - Jumper Criteria. This column specifies the rated current for an 0805 jumper (i.e. 0Ω resistor). We can see from the table your jumper is rated for 2A, with an absolute maximum of 5A (presumably short pulse).
So why might a "zero ohm" resistor have such ratings? Simple, it's not a 0Ω resistor. Unless the manufacturer of the resistor you are using have secretly made a room temperature superconductor, the jumper is actually still a resistor, just a very small one. According to the datasheet it is specified to be ~50mΩ or less.
Because the resistance is non-zero, some power will be dissipated. If we plug in the provided numbers, we actually find that the power rating is real and sensible:
$$P = I^2R = 2^2\times0.05=0.2W$$
So in the worst case resistance of 50mΩ, and at the rated current of 2A, it will be dissipating more than the 125mW rating.
Still think the rating is silly?
In a power supply design I had the pleasure of surge testing, the designer had added an 0805 0Ω resistor in series with a 24V DC input, just prior to a TVS diode. During the test, we charged a 10mF capacitor up to 200V and then connected the capacitor to the input of the power supply.
Naturally the TVS started conducting, and the 0Ω resistor turned quite literally into a firework...
It is not really 0Ω. According to the datasheet, page 5, the resistance of the jumper (0Ω resistor) is less then 50mΩ, not the perfect 0Ω.
The most likely explanation is that the resistor is part of a product series, and all product pages on Farnell have the same information for all values in the series.
I mean, if you're Farnell, you aint gonna pay someone to manually create each product entry for the E96 series into your database.
You would have a software tool which would create the product records according to a template. Like, enter the common data from the datasheet only once (brand, series, power, package, photo, etc), and then automatically create all values in the resistor series using these common datasheet values.
Since I once saw a mistake in a resistor manufacturer's part#, I guess the part# would be entered manually for each value too.
Now, 0R resistors aren't exactly 0 ohms, more like a couple tens milliohms, so yes, they do have a max current and max dissipation power.
It's really stating the power rating of the resistor family it belongs to.
Some 0R resistors are in place of a different value in future. If you place this 0R part on a board, that position will be able to accept any resistor in that family.
0ohm resistor are not perfect. You can take 1mohm as value for you calculation. This will lead you to a very low power. You shouldn't bother much about it.
As wikipedia says:
The resistance is only approximately zero; only a maximum (typically 10–50 mΩ) is specified.[*] A percentage tolerance would not make sense, as it would be specified as a percentage of the ideal value of zero ohms (which would always be zero), so it is not specified.
In the ideal world the 0ohm is the ideal wire. In this case the power is calculated as:
In the real world neither the ideal wire neither the actuall 0ohm resistor exists. That means some (little) power is consumed in current-driven applications.
That's why there are different 0ohm resistors with different power ratings; they do dissipate heat so they can be overloaded and burnt.
A physicist's perspective on a resistor \$R = 0 \ \Omega\$:
Applying a constant current source with finite current \$I\$, there is zero power dissipated.
In this case, \$P = I^2 R = 0\$, because \$I\$ is finite. Note that \$V = I R\$ is zero, and so \$V^2 / R\$ isn't infinite, even though \$R\$ is zero.
Applying a constant voltage source with finite voltage \$V\$, there is infinite power dissipated.
In this case, \$P = V^2 / R = \infty\$, because \$V\$ is finite. Note that \$I = V/R\$ is infinite, and so \$I^2 R\$ isn't zero, even though \$R\$ is zero.
More practically, if \$R\$ is small but nonzero, then by similar arguments:
The point isn't whether or not the resistance is exactly zero, but that applying a constant voltage source to a small [zero] resistance results in large [infinite] current in such a way that the final power dissipated is large [infinite and definitely nonzero].
Value of R should be rounded off and close to zero because that looks nothing like a superconductor. Safe to say that all electronical components have a non zero R value, even wires.
