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Suppose I connect the two terminals of a battery without any resistor resulting in a short circuit. Doesn't that mean the voltage is zero across the terminals of the battery? From Ohm's law V = IR and when R =0, V = 0.

But how is the voltage zero when one terminal of the battery is clearly at a higher potential? The positive charge is concentrated at one terminal and the negative in the other. So how come the voltage becomes zero in a short circuit?

I also can't wrap my head around the fact that voltage immediately becomes nonzero when you add a resistor. I know it works according to Ohm's law. But I don't just get an intuitive idea.

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    \$\begingroup\$ But it's not at a higher potential. At least, not any more... \$\endgroup\$ – Ignacio Vazquez-Abrams May 9 '18 at 10:37
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    \$\begingroup\$ When competing in F1 you must drive in the same direction as everyone else. When you are using voltage sources you must not short them out. \$\endgroup\$ – Andy aka May 9 '18 at 10:50
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    \$\begingroup\$ @user406653 If I is infinite and R is 0, then V=IR does not have to be 0 \$\endgroup\$ – Dusan Bajic May 9 '18 at 12:57
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    \$\begingroup\$ I think a more interesting version of this question - in terms of educational value - would be, "what happens in this circuit as the resistance (of the shorting wire and the battery itself) approaches zero?" \$\endgroup\$ – dwizum May 9 '18 at 16:28
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    \$\begingroup\$ related (if not duplicate): Short circuit = zero voltage? \$\endgroup\$ – Nick Alexeev May 9 '18 at 16:53
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This can only happen if you have an ideal battery. In the real world, nothing is ideal. You have an internal resistance of a battery, usually in the milliohm region, and you also have the wire shorting the battery out. That also has a resistance. What you have now done is created a circuit with resistance, and therefore current. You end up with something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

This means that you will always have a voltage on the battery. The resistance will be very small, hence why there will be a lot of current flowing, which will mean a lot of power dissipated as heat, which is why it can be dangerous to perform such experiments!

As you can see, there is no such thing as a true short circuit, as there will always be resistance somewhere, no matter how small!

As for wrapping your head around voltage becoming non-zero when resistance is added... Well, you don't have to! For the zero condition to exist, the resistance must be infinite, if there was no resistance, then there would be no circuit!

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    \$\begingroup\$ Excellent answer! To the original poster, the heat dissipated by the shorted battery is evidence that the battery has an internal resistance. Hopefully that helps wrap your head around this notion. \$\endgroup\$ – cburf May 9 '18 at 18:53
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    \$\begingroup\$ Not just dissipated in the shorted battery. Heat dissipated in the shorted wire is why you get electrical fires. Or for an even more impressive example, try dropping a spanner across a car battery. (Hint: Don't try this with a battery or a spanner you would like to survive afterwards. Or near anything else which could be damaged by exploding hot metal, including yourself. Actually just check YouTube instead.) \$\endgroup\$ – Graham May 9 '18 at 19:22
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When shorting, the current is not zero as you presume. Batteries have a small internal resistance of \$r\$. If you short the terminals, a current , \$ I = V/r\$ will flow. Where V = Voltage of battery. The short circuit current is limited by that internal resistance.

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    \$\begingroup\$ @user406653 shorting an ideal voltage source makes your theory explode. \$\endgroup\$ – Ruslan May 9 '18 at 12:52
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    \$\begingroup\$ If you know it is not practically possible to achieve, then is it worth worrying about? We can only calculate it with Ohms Law. the fact it is impossible to achieve may be why you are having difficulty getting an intuitive idea of how it would happen. It is always difficult to imagine the result of an impossible scenario \$\endgroup\$ – MCG May 9 '18 at 13:05
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    \$\begingroup\$ shorting a ideal voltage source creates a singularity, may be a black hole! \$\endgroup\$ – Gorkem May 9 '18 at 13:29
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    \$\begingroup\$ Actually that scenario is even indefinite mathematically .We can say I = V/0 = infinity when shorted . But again since shorted, potential has to be same at both end ...So I = 0/0 form ... Not defined :-D \$\endgroup\$ – Meenie Leis May 9 '18 at 13:54
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    \$\begingroup\$ In addition to the internal resistance (which would be the significant factor), don't forget the cable resistance as well. See BigClive for a practical demonstration (who even uses the voltage drop over a copper pipe as a resistor to estimate the current) youtube.com/watch?v=0tGK1nqXr28 \$\endgroup\$ – Jens May 9 '18 at 15:27
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Other people did give, in a way or another, the right practical answer, i.e. you should always consider the unavoidable internal resistance of the battery and/or that of the wire.

From a purely theoretical POV, instead, when you short an ideal voltage source you get a singular, degenerate circuit, with quantities like the current into the short going to infinity. It's the dual case of an ideal current source with no load attached, i.e. with the terminal left open: you get an infinite voltage across them.

This kind of things happen only because ideal sources are just mathematical abstractions, which may give you infinite quantities when used carelessly.

Keep in mind that ideal circuit elements are useful, and simplified, mathematical models of real things. They are not physical objects, so they don't need to follow the law of physics. In particular, ideal sources may source infinite power and this means infinite energy in any finite interval of time: of course this is physically absurd.

In particular, if you fail to model a physical situation correctly, i.e. you oversimplify the model, you can incur in such mathematical absurdities.

If, on the other hand, you are just putting mathematically ideal elements together, there is no need for them to give you something coherent. You are simply writing equations in a graphical ways. Nothing stops you from writing a system of equations with two incompatible equations. The system will have no solution (or will have absurd, non-finite, solutions).

For your specific case, you are just writing two equations and putting them in a system:

1) the equation of the ideal source, saying that the voltage across its terminal (call them A and B) is given and known and different from 0 (call it Vs);

2) the equation of an ideal wire, i.e. a resistance with R=0, which must follow Ohm's law: i.e. I=Vab/R

There is no solution for such a system. Or you may consider I=∞ a "solution", but that is no real number and for any practical purpose is nonsense.

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You're forgetting the internal resistance of the battery. It's normally much less than the load, but it dominates when the load is 0 ohms.

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This is a very common issue for people who are new to circuitry. You are not alone.

The practical answer is that ideal wires and ideal voltage sources are things that do not exist in reality. Everything exhibits some non-ideal behavior. For the most part, these non-ideal behaviors are of minimal importance. The idealized modeling of the circuit lets you predict the majority of the behavior of the circuit. However, in some degenerate circuits, like the one you mention, these non-ideal behaviors play a large role in the behavior of the circuit. Over time, you will be given guidance as to when you can get away with using the idealized models and when you need to bring in more non-ideal behaviors to characterize the circuit.

Technically, if you short an ideal voltage source with an ideal wire, the resulting circuit is inconsistent. An ideal voltage source means that the voltage across the terminals is some fixed amount (i.e. 5V). An ideal wire across those terminals means the voltage difference between those points is 0V. These two statements cannot be put together, because they conflict. You cannot meaningfully short an ideal voltage source with an ideal wire.

But you can short a real battery with a real wire. The results depends on the non-ideal behaviors, such as the internal resistance of the battery.

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Apparently voltage is zero, and the current is infinite. That is, the resistance of the wire and loss will be the load, so if your battery can provide hundreds/thousands/millions of amps (quite improbable), you wont get a short circuit and the voltage in the terminals will be the nominal voltage of the battery. However, and sadly, batteries provide a limited amount of energy, and your tiny load will request a lot of current, given that and according to Ohms, the voltage will drop down. As smaller is the load, bigger is the current and faster will decrease the voltage (a transient which will be a ramp till zero, being the slope of the ramp proportional to the current requested by the load)

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If you a have perfect voltage where you short the wire the following representation is wrong because you don't have a transfer function between the voltage and the current. You assume that R is completly 0 therefore I is decoupled from the voltage.

schematic

You need to analyze the circuit as the following if you want to represent your situation.

schematic

simulate this circuit – Schematic created using CircuitLab

The voltage source will maintain its voltage at the nominal value because it is a perfect voltage source and the current source represent the infine amount of power that the perfect source can provide because the source don't have any internal impedance.

The model is quite useless frankly, you shouldn't try to model something that doesn't exist nor respect the basic rule of a model. R =0 is out of the scope of the model of Kirchoff.

KVL assume that R never goes to 0. If R = 0, you are now in another domain where you need to use intrinsic properties of the material to derive the current. Supraconductor lines have 0 resistance but their current isn't infinity.

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    \$\begingroup\$ "KVL assume that R never goes to 0". This is wrong. KVL is an approximation under some assumptions (e.g., quasi-static fields) of Maxwell's equations. KVL is, in other words, conservation of energy around a loop (line integral of E field on a closed loop). R is not taken into account. Assuming KVL as a postulate, as many courses in circuit theory do, KVL can be applied even when a circuit element is a 0 ohm resistor. The problem is that not always such a model makes sense (but most of the times 0 ohm resistors, i.e. ideal wires, are perfectly valid circuit elements). \$\endgroup\$ – Lorenzo Donati -- Codidact.org May 9 '18 at 19:37
  • \$\begingroup\$ @LorenzoDonati check back my answer,a pure short at an ideal source cannot be represent as a resistor that a modelization error. A ideal voltage source will maintain voltage at its poles whatever the load you put. You need to represent the short as a current source in this very particular case and not a resistor. The circuit proposed actually respect all the law of KVL. For the wire, there is no such thing has ideal wire, that is just a model approximation that can be apply some cases. \$\endgroup\$ – MathieuL May 9 '18 at 20:07
  • \$\begingroup\$ I didn't criticize your entire answer. Just that statement about KVL. That the OP is facing a modelization mistake is something we both agree on, but this has nothing to do with the validity or applicability of KVL. \$\endgroup\$ – Lorenzo Donati -- Codidact.org May 9 '18 at 20:17
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The existing answers cover the issues of parasitic resistances and how they limit current, but the underlying cause is battery chemistry, which changes that parasitic resistance and the available voltage.

The chemical reactions in a battery are fairly slow compared to the speed of electrons through metals, and chemistry furthermore tends to happen at the contact point of chemicals, which is particularly true for solid-liquid batteries like lead-acid types where most of the action happens at the surface of the lead plates. This is why, for example, your car battery will:

  1. Appear to have a good voltage to a light load, like your voltmeter even when unable to start your car (a very high load)
  2. Only partially regain its capacity if you charge at a high current (say, >=6A) but regain much more capacity if you charge at lower current (say, <= 2A); at a high charging current most of the charge never penetrates deep enough into the chemicals to last, even though the voltage briefly recovers as charge in the immediate vicinity of the reaction is restored.
  3. Lose voltage over time as current is drawn from it.

Jack Ganssle's study of coin cell batteries, for example, shows some data on how the "internal resistance" model is not a constant resistance, but varies based on how much the battery has already discharged and also varies with the amount of current being drawn: https://www.embedded.com/electronics-blogs/break-points/4429960/How-much-energy-can-you-really-get-from-a-coin-cell-

So, as the chemistry is providing a sort of "internal resistance", the charge depletion is also reducing the voltage, and further changing that "internal resistance", and how bad that effect is depends on how much of the available charge you're depleting (ie how much current you're drawing). So, even on the "other side" of the "internal resistance", the voltage doesn't stay the same if you short out the terminals.

As a follow-up note, you shouldn't expect a capacitor to be able to even out the current draw and extend battery life, either, since although capacitors can provide charge, they also have their own parasitic resistances (though some types are better than others), as Ganssle notes in a later article: https://www.embedded.com/electronics-blogs/break-points/4430050/Using-a-capacitor-to-sustain-battery-life This is tangential to your immediate question, but is good to know if you plan to actually design something.

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