# Calculate the maximum current of zero ohm resistors

According to Ohm's Law, if the resistance is zero, the power dissipated by a resistor is unmeaning.

$$P_d = R \times I^2$$

$$I = \sqrt{ \frac {P_d}{R} }$$

Can one calculate the resistance given that power is defined? Only by knowing power, how can one calculate the maximum current of a zero-Ohm resistor?

I'm a bit confused as I've seen there is zero-Ohm resistors with specified power in the market.

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Ohm value R it is not so simple as you mention. Temperature (i.e. superconductors) as well as physical properties of the conductor material is some of the key elements. Ohm's law includes all this and you have to taking into account before simplify. In any case the above formulas should be proved including voltage as well – GR Tech Apr 24 '14 at 12:27

The power rating (and the tolerance) of a so-called 0$\Omega$ resistor is sort of vestigial- it comes from the series of resistors that the jumper resembles.

The actual current rating does not necessarily represent the same power dissipation as a similar resistor (the limit may be something like current density rather than power dissipation). For example, the Rohm MCR03ERTJ000 is part of the MCR03 series rated at 100mW, but the jumper version is 50m$\Omega$ maximum and rated at 1A max, so only 50mW.

So, it is not valid or safe in general to calculate the current rating, you should look it up in the data or contact the manufacturer, especially if high peak currents will occur, or your jumper could act as a fuse.

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Unfortunately there is no such thing as a true zero ohm resistor (outside of superconductors). Zero ohm resistors available on the market (often called 0R link or jumper) will specify a maximum resistance.

This datasheet is for a range of 0402 surface mount resistors, and specifies that 0R jumper resistors have a resistance of less than 0.05 ohms. It also specifies a maximum power dissipation of 1/16W. This implies that (worst case) the maximum current you can safely pass through the part is: $$\sqrt{ \frac {P_d}{R} } = I$$

Which comes out to around 1.1A, for such a small part it sounds reasonable.

You must bear in mind however that for such low resistances, tracks/wires and solder blobs will have a big impact on this value.

If you need a lower resistance and/or more current capacity, parallel the resistors up.

EDIT: From a theoretical point of view, as the resistance decreases the power dissipation tends towards zero. So for a true zero ohm resistor, no power would be dissipated, allowing infinite current. If your interested in the theoretical side, then I suggest you read some Wikipedia articles on superconductors where the resistance truly is zero. Things get a bit weird.

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... allowing infinite current. In fact, an ideal zero ohm resistor has zero power for any (finite) current. If one allows 'infinite current' (whatever that means), one can have power dissipation in a zero ohm resistor. – Alfred Centauri Apr 24 '14 at 12:26
Real superconductors do not allow infinite current - there is a critical current density $J_C$ (which is also a function of the magnetic field and temperature) above which they are no longer superconductors. – Spehro Pefhany Apr 24 '14 at 12:34
A real power supply cannot supply infinite current either, even if you were to short it out with a superconducting wire. When doing calculations, you may treat your power supply as constant-voltage. But in reality, there is no such thing as a perfect constant-voltage supply. – Alex D Apr 24 '14 at 18:11

If resistance is zero, power is also zero. It's as simple as that. The maximum current of a real "zero ohm" resistor will be listed in the data sheet.

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