currentpowerresistorsvoltage
I have a 0 Ohm resistor here.
Can someone tell me how the power dissipation will be calculated for this type of 0 ohm resistor.
The part says it has a power rating of 0.1W.
But, if we use either V squared over R formula or the I squared R formula, in both of the case, we would either get 0 or infinity for the power dissipation of the resistor.
So, how to calculate? If we cannot calculate, then why does the manufacturer provide a power rating value?
The voltage rating is for the resistor series typically and specifies the maximum peak voltage you can apply without danger of damaging the resistor due to corona, breakdown, arcing, etc.
The power rating is completely independent of the voltage rating. It specifies the maximum steady state power the package is able to dissipate under given conditions.
You have to conform to both specs. If placing the maximum voltage across the resistor results in more power than the spec allows you have to reduce the voltage until you meet the spec. Likewise you can't increase the voltage above the rating just because you're not hitting the maximum power limit.
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...
Best Answer
Even nominally "0 ohm" resistors have some resistance.
The datasheet for the part you linked says that zero-ohm resistors have "< 20 mOhm" of resistance.
I don't know how you're using this part, but in most cases for a resistor that small, the exact value of the resistor won't affect the current flowing through it very much, so you can conservatively assume that the resistor value is actually equal to 20 mOhm and use P = I^2 R to find the maximum rated current.