# Current when resistance is 0

A short circuit has $\rm 0$ resistance. We use Ohm's Law $\rm V=IR$. If $\rm R=0$, then $\rm V=0$. Doesn't this mean $\rm I$ can theoretically be anything? Thus $\rm I$ can be like $\rm35A$ for example, and there would be current. But how can there be current without electron potential (voltage)?

Additionally, if $\rm R$ is infinite, then $\rm I=0$. $\rm V$ can be anything. In this case, does this mean for sure that no current will pass through?

But how can there be current without electron potential (voltage)?

In the case where there is no resistance, current (once flowing) does not require any voltage to continue flowing. If you start a current flowing in a superconductor, then even with no applied voltage, it continues to flow.

It doesn't take any force to keep a ball rolling if there is no drag. Likewise it doesn't take any voltage to keep current flowing if there is no resistance.

You're correct that if you have a perfect insulator ($R=\infty$), then any applied voltage will still produce zero current. In the real world, there are no perfect insulators. At some point the voltage will be strong enough to move charges across the material.

In circuits with areas of large resistance, we can safely ignore areas with very small resistance because it doesn't affect the circuit behavior significantly. For this reason, simple circuit behavior is modeled with perfect ($R=0$) conductors.

In almost any real world circuit, that isn't true. The wires have positive (if small) resistance. Current might rise high enough to be a danger, but it isn't infinite.

Even if you were to consider a superconductor as a wire (where resistance really is zero), it still has an inductance value. When resistance is very small (or zero), the inductance becomes significant. It will prevent the current from increasing faster than a certain rate.

• Just for my own knowledge how do you get the current moving in a super conductor I understand voltage would do it but is there another way. Can you use a magnet? I don't know much about super conductors. If the super conductor reaches the right Temp will Eddy currents start on there own? – newguy Sep 14 '15 at 21:48
• See this answer. physics.stackexchange.com/questions/179374/… – BowlOfRed Sep 14 '15 at 21:53
• It means, then, that this equation (U = R * I) is not universally correct, it is correct only for the conditions where the resistance is guaranteed, isn't it. Otherwise U = 0, no matter how much I (current) is there: U = 0 * I = 0 always, at least this is what my math teacher used to say. ;) – forsberg Mar 6 at 11:12
• It's correct at steady-state, not in transition. A voltage drop across a zero-resistance wire will never reach steady-state. – BowlOfRed Mar 6 at 16:07

I can theoretically be anything

EDIT: Oh, you mean the current,$I$, not the pronoun I....

Ohm's law applies only to components which could obey Ohm's law. It is an observational, phenomenological law. If R truly equals zero, then something else is governing the relationship between the current in the circuit and the potential drop. A filament in an incandescent light bulb obeys Ohm's law most of the time even though the resistance changes with temperature.

If one cycles the applied voltage slowly enough, however, one can create a negative slope in the V vs I curve indicating a negative resistance. I'm not sure that's what it means because I don't know if the filament is actually an Ohmic resistance at that point.

In superconducting circuits, there is no resistance and there is no potential drop. The current is produced inductively using an alternating magnetic field.

An infinite R can be produced by cutting a wire in a loop circuit. In that case I=0 and one indeed can let V be anything until the electric field across the gap becomes too large for whatever material is in the gap. Then the gap will become conducting. That's how lightning is produced.

In a circuit that is shorted the current is governed by the internal resistance of the power supply . let's say the supply resistance is 1 ohm and the supply voltage is 35 volts the supply will deliver 35 amps to the shorted circuit. You will not measure a voltage drop across the wire shorting the power supply but there will be a voltage drop inside the supply's internal resistance. In most cases this will damage or destroy the power supply unless it has a current limiting circuit. So even though you do not have an external resistor there is still resistance in the circuit. Even battery's have internal resistance. I understand this is not a physics explanation but I'm an electrical engineer so I'm pretty sure that's how it works

An ideal short circuit is that circuit element for which the voltage across is zero for any current through. One can think of the ideal short circuit in two ways:

(1) as an ideal resistor in the limit of $R \rightarrow 0$

(2) as an ideal voltage source where $V_S = 0$

In both cases, there is current without voltage. This isn't necessarily a contradiction just as there can be flow without pressure (or motion without force).

Likewise, an ideal open circuit is the circuit element (dual to the ideal short circuit) for which the current through is zero for any voltage across. One can think of the ideal open circuit in two ways:

(1) as an ideal resistor in the limit of $R \rightarrow \infty$

(2) as an ideal current source where $I_S = 0$

Every form of energy has inertia. Have a mass $m$ with speed $v$. It won't simply stop unless there is a force that makes it stop. Same way, have a current $I$. It won't stop unless there is a "force" applied.

In our nice Earth, friction stops the mass $m$. And, resistance stops the current $I$. But in space, there is no friction, and mass $m$ can't stop. And in superconducting materials, there is no resistance and current $I$ can't stop. (By the way, this is exactly why MRI equipment generates powerful magnetic fields for so long time: by having a current in zero resistance superconducting materials without a voltage source).

The inertia associated with the current relies in the magnetic energy in the magnetic field associated with the current. Because, the current is the source of the magnetic field.

About $R\to\infty$, there is no current. And that is what happens in electrostatic. You have arbitrary voltages and electric fields, but approximately no current.

"Short" answer: You can't really have a short circuit as such in the first place.

Superconductors aside, everything has resistance. Even free electrons at rest have a little bit of mass and will resist applied voltage just a wee bit. Put those electrons inside something and resistance can only go up. (Again, superconductivity aside. It doesn't just happen by accident.)

Circuits are designed to operate in a specific envelope, where each individual part is working correctly. That means that current in every wire needs to be low enough that voltage across the wire is negligible. Such terms are neglected for the sake of comprehending what the circuit does.

Complex circuits are often simulated by computer-aided design software, working on a complete physical description of the final product. The size and shape of all the wires is typically taken into account, leading to a "schematic" where no parts are directly connected except to resistors. If such a simulation turns up a problem, the designer might track it down to a wire that failed to be an ideal short. (And the solution is usually to get a bigger wire.)

Likewise, air gaps can become conducting given a high enough electric field. I've not encountered a simulator including such automatic analysis, but an engineer designing high-voltage electronics should always pay special attention to insulation.

Lines drawn between circuit elements are a fiction, predicated on all the other assumptions that went into the design of the circuit. Failure modes may simply not be described by the schematic at all. When representing excessive current through a wire, on a schematic or a system of equations, you must remove the wire with your eraser and substitute a resistor instead.