# Thread: Kirchhoff rules and Ohms law

1. Hi,

i am having a little trouble with the Kirchhoff rule.

the loop rule says that around any closed loop in a circuit the sum of the potential rises (from a battery e.g.) must equql the sum of the potential drops (e.g. resistor).

Ohms law goes: V=I*R

so lets say we have a 6Volt battery, and a 4 Ohm resistor, then that would be

6/4 amps = I

but assuming there is no resistor then we would have

6V = I * 0 (???) what is the current (I)? where is the potential drop that equqls the 6V rise?

can anybody help me before the black hole will eat us on wednesday thanks  2.

3. Originally Posted by evariste.galois
Ohms law goes: V=I*R

so lets say we have a 6Volt battery, and a 4 Ohm resistor, then that would be

6/4 amps = I

but assuming there is no resistor then we would have

6V = I * 0 (???) what is the current (I)? where is the potential drop that equqls the 6V rise?
A 6V battery does not always provide a voltage of 6V.

It has an electromotive force of 6V, and a certain internal resistance (this too is a simplification, but it's enough for many purposes, including your problem).

So the full formula is As long as the load (the external resistor, light bulb or whatever) has a high resistance compared to the internal one in the battery, you can compute the current the way you did: But 4 Ohm is already a very serious load for a pocket-sized battery. Note that "serious" means a small resistance, because it drains the battery more than, say , a 1000 Ohm resistance would. So the approximate formula might not be quite right. In fact, it may be widely off the mark - the real current will be much less than .

If you short-circuit the battery (i.e. ), the approximate formula leads to a division by zero. In mathematics, division by zero is a big "nono". In physics, it's nature's way of saying "you have forgotten something". That something is mostly .

But that's not all. If you short a battery, it will warm up (the energy you are wasting is being converted into heat). The increase in temperature can have various effects depending on what kind of battery you have - from a change in the value of to damaging the battery (dramatically reducing E either temporarily or definitely) to explosion (a serious hazard in car batteries).
The strong current may also modify the chemistry inside the battery, usually making it weaker.

Moreover, the wire or other conductor you used for your short is not a perfect zero Ohm, unless we are talking about superconductors.

One thing is certain: you will not get an infinite current in a piece of wire. Thank goodness for that, because the resulting EMP would, literally, destroy the universe.  4. And, if you replace the fuse in your multimeter with a piece of brazing rod, then forget to replace the fuse, then set the meter to the milliamp scale and try to measure the voltage at the service entrance, you will be doing an experiment approaching the condition you describe, except with a higher voltage. This is something that William McCormick says he once did. He ended up holding two empty rubber bands, those bands being the insulation on the voltmeter leads after the copper conductor vaporized.  5. Originally Posted by evariste.galois
Hi,

i am having a little trouble with the Kirchhoff rule.

the loop rule says that around any closed loop in a circuit the sum of the potential rises (from a battery e.g.) must equql the sum of the potential drops (e.g. resistor).

Ohms law goes: V=I*R

so lets say we have a 6Volt battery, and a 4 Ohm resistor, then that would be

6/4 amps = I

but assuming there is no resistor then we would have

6V = I * 0 (???) what is the current (I)? where is the potential drop that equqls the 6V rise?

can anybody help me before the black hole will eat us on wednesday thanks
The problem that you are running into is that ciruit theory is a stylized approximation to the physics that is described by Maxwell's equations for electrodynamics. It is subject to certain rules in order to be a valid model.

You simply cannot connect and ideal voltage source across a zero resistance and produce a sensible solution. In reality any voltage source includes and internal resistance that will limit the current.

You will eventually find that there is a problem with opening up a switch that controls current flowing into an ideal inductor also. And similarly you cannot throw a switch and instantly introduce a capacitor in parallel with a voltage source.

You may even see such problems "posed" and "solved" by putting resistors in a circuit and then taking limits as the resistance goes to infinity or to zero. Those techniques simply show that the person doing the solving does not really understand the limitations of circuit theory. I have seen this sort of thing done by people who should know better.

Circuit theory only applies to situiations in which the circuit is physically small with respect to the wavelenghts of the electromagnetic waves involved (the low frequency approximation) and in which the idealized circuit elements used are also valid approximations to reality. For any situation in which resistances are very small, very large, or in which switches are postulated that instantaneously change current through an inductor or voltage across a capacitor, circuit theory breaks down.  6. Originally Posted by Harold14370
And, if you replace the fuse in your multimeter with a piece of brazing rod, then forget to replace the fuse, then set the meter to the milliamp scale and try to measure the voltage at the service entrance, you will be doing an experiment approaching the condition you describe, except with a higher voltage. This is something that William McCormick says he once did. He ended up holding two empty rubber bands, those bands being the insulation on the voltmeter leads after the copper conductor vaporized.
Just thinking about it, I see white spots before my eyes. Ha-ha.

I don't know that a battery has internal resistance. I believe that a battery is a diode. And after you run to much amperage through the battery, the diode starts to break down.

If you measure the maximum amps a battery can output during a dead short, basically near zero volts. You will find a ratio or scale back up to its maximum voltage with only the draw of a galvanometer or voltmeter or oscilloscope and near zero amps.

This is a program I wrote. I wrote it a while back. You can get an idea of what the battery will output in amps to a point. The point the battery can no longer raise the amperage with a decrease in voltage.

http://www.Rockwelder.com/Electricity/setupwir.exe

You do not want to discharge batteries to quickly, or feed them excessive power that might cause them to explode.

Sincerely,

William McCormick  Bookmarks
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