# Smallest possible resistance (highest current) [duplicate]

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I'm new to electronics and I've been familiarizing myself with some basic formulas, specifically Ohm's law. It seems to imply that if resistance is smaller than 1, let's say 0.00001 Ohm, then I can get 100000 A of current with just 1V of voltage. So is it actually possible (not in just theory) to achieve that kind of current by lowering the resistance, or there's some limitations to how low the resistance can be?

## marked as duplicate by Dmitry Grigoryev, Michel Keijzers, Finbarr, Voltage Spike, RoyCApr 4 '18 at 18:03

• it depends on your power source. An AA battery for instance will limit the maximum current achievable to around 1 amp, when new. 100k AA batteries in parallel would be needed to reach your 100kA into a very low resistance, or 1k car batteries. – Neil_UK Mar 12 '18 at 14:13
• Imagine the resistance of steel to steel arc welding of a 6" circular tube being welded from inside to out. As the tube goes around copper electrode wheels, the shunt resistance of welded material goes down while the gapped area needs more current to heatup. This is the only example I have seen using 10k~100kVA ~4Vac. This called Diffusion bonding and it uses a Zirconium shim and is used in Candu Nuclear reactors – Tony Stewart Sunnyskyguy EE75 Mar 12 '18 at 15:58
• Standard thickness copper foil (at standard foil weigh of 1 ounce per square foot) has thickness of 35 microns or 1.4 mils. The resistance at room temperature is 0.000498 ohms. Putting 100 square of foil in parallel yields 5 micro Ohms. What can you do with that? – analogsystemsrf Mar 12 '18 at 17:49
• In the ideal world yes, things like this can happen. In the physical world everything has parasitics (resistance,capacitance and inductance). Even superconductors have nano-ohms of resistance an are not perfect. Circuits a just a model of how things work, you also have to be able to physically build the model for it to be useful. We are limited to 'conductive' materials for current carriers and all materials have resistance. – Voltage Spike Apr 4 '18 at 16:41

By Ohm's Law on it's own the current, theoretically, can go to infinity in a circuit with zero resistance. Of course, in reality you would need a superconducting loop to do that.

However, if you have an infinite current in a loop, you also generate an infinite magnetic field.

Once that field gets large enough, the super-conductor will cease to be super-conductive in an event known as quenching. Then things go BOOM!

Further, even if you can overcome the above, you would run into quantum and relativistic effects.

Yes, that's possible.

Yes, there's a physical limit to how low the resistance of a passive component can be - 0 Ω. We call that superconductivity, and if you, for example, cause a current in a closed superconductor loop, the current will run forevermore.

Notice that your 10 µΩ resistor is not trivial to build – even copper has a specific resistance.

• Is it actually 0 Ohm though or like something very small but still greater than 0? Because if it's 0 we'd be dividing by 0 when using Ohm's law formula (or these things don't apply in physics?) – Greg Mar 12 '18 at 14:04
• @Greg A superconductor has zero resistance. Current flowing through a superconducting coil has been observed to be unchanged at the measurable level (parts per million) over a time span of 10s of years. Although experimentally that only bounds the resistance above by a very, very tiny value, there are good theoretical reasons to believe it's actually zero. – Neil_UK Mar 12 '18 at 14:10
• @Greg by that logic one can conclude that physical bodies cannot be at rest, since $t=\frac{S}{V}$. – Dmitry Grigoryev Mar 12 '18 at 14:14
• @Greg you have to think the other way around. You can't have any voltage across a superconductor. That's what's impossible. Not the superconductivity itself. Your voltage source will just break down trying to supply infinite current. – Marcus Müller Mar 12 '18 at 14:18
• Greg - its not a matter of dividing by zero - with superconductivity ohms law simply doesn't apply. Ohms law define the current/voltage relationship across a given resistance. With Superconductivity there is no resistance, so no voltage/current relationship exists. Think of voltage as simply being the result of current flow across a resistance, hence no resistance, no voltage. – Norm Mar 12 '18 at 17:04

While there is no physical limit to how low a resistance could be, there is indeed a limit on maximum current you can have in a specific conductor. Once the current value is high enough to involve the motion of every last charge carrier at its maximum speed, you will be physically unable to increase the current any further.

Yea. But it also depends on the power supply you use. It has a limit on how much current it can maximum deliver. Also the resistor will have a wattage rating, which limits the maximum power ( $I^2R$), it can dissipate across. Otherwise it will overheat and burn away. Depending on the material used to make the resistance, it will have an intrinsic property called resistivity, $\rho$. Resistance is related to resistivity by $R = \rho L/A$. Resistivities are of order $\mu \Omega m$ usually. Now you can calculate how the dimensions should be for building small resistances.

Your voltage source will have an internal resistance so lowering the external resistance will only increase the current so far.

Even if you have a superconducting wire (zero resistance), there is a critical current and critical field surface (temperature dependent) above which the wire will no longer superconduct.

Finally, even if you had a zero resistance source (which is impossible, except for exactly zero volts) and a superconducting wire, the current would not increase instantaneously, rather it would rise linearly with time because even straight piece of wire has inductance. A few mm of wire might have ~1nH of inductance.

Conductors have a resistance that increases with temperature, T, with a certain positive tempco,(PTC). So when high currents heat up the conductor $Pd=I^2R(T)$ ,the power drops as it heats up (assuming source maintains constant voltage). This applies to welding and light bulbs.

The grid can be very low resistance but at the point of service is raised by distribution transformer design to be 10% or so of the rated V/I(max) so residential breakers are rated for 10kA.

The heat from current , I regardless of whether it is μΩ, mΩ or Ω in the power source and load will be $Temp~ rise ['C] = I^2 \cdot R(T)~\cdot t\cdot R_{th}$ for thermal resistance Rth and R as a function of temp T.

The resistance is a function of resistivity ρ and thickness x so $R=ρ/x$

Batteries also have an effective series resistance, ESR for a ΔV (e.g.= -41%) drop in voltage then ESR = ΔV/I. Thus for a car with cold cranking amp (CCA) rating of 500A rated at 7.5V from 12.5 or ΔV=5V/500A = 10 mΩ.

LiPo 18650 size batteries are in the same range of ESR. (5~25mΩ new) So intuitively you can understand that the quality of the battery is greater with low ESR but is not fixed in time, as it rises with age and depleted charge.

Then high temperature superconductors (zero Ohm) are popular now for research in Fusion power where massive magnetic forces are needed to start the fusion process to convert H2 into He and release the heat needed to power a city from pineapple size tank of Hydrogen. This is the energy of the future if they ever get it working. ( old joke is they never will get it working so it is always the "energy of the future" ) but maybe in my lifetime.