Power source, internal resistance and Ohm's law? - Page 1
Forum section: electronics / beginners

author: Mr D
author_rank: Regular Contributor
timestamp: June 04, 2019, 10:26:06 pm
content: Hi folks,I'm just messing around EveryCircuit, trying to get my head around the basics.When i create a simple circuit consisting of only a 1 volt battery and a one milliohm resistor, i get a nonsensical current of 1 kiloamp even though that is indeed correct according to Ohm's law (seems impossible to get so much current from a 1 volt power source, no??!!)So i guess this is the result of EveryCircuit not simulating the internal resistance of the battery?Is this normal in every simulator (nonsensical answers at the extremes), or do more serious ones take this into effect?Or am i in fact talking nonsense?

author: Audioguru
author_rank: Super Contributor
timestamp: June 04, 2019, 10:38:35 pm
content: It is an example of non-real math.1) There is no 1.0V battery available.2) There is no 1milli-ohm resistor available.Most Simulation programs do not use common sense. They can get 10A (!) from a little 2N3904 transistor that has a maximum allowed current of 200mA and works poorly at and above 100mA.

author: soldar
author_rank: Super Contributor
timestamp: June 04, 2019, 11:00:36 pm
content: Quote from: Mr D on June 04, 2019, 10:26:06 pmWhat did you set the internal resistance to? Because if you set it to zero then you got a result consistent with that.

author: MarkR42
author_rank: Regular Contributor
timestamp: June 04, 2019, 11:05:06 pm
content: It's totally reasonable, because your software is simulating "ideal" components.An ideal 1V battery has zero internal resistance.Ideal wires have no resistance (or inductance etc)An ideal 0.001 ohm resistor actually exists and is a thing. In real life these components are impossible, or at least, unlikely to work like that.It's like the "spherical horse in a vacuum" joke.

author: ejeffrey
author_rank: Super Contributor
timestamp: June 04, 2019, 11:11:54 pm
content: Quote from: Mr D on June 04, 2019, 10:26:06 pmIt certainly isn't impossible. I've used power supplies that produced hundreds of amps at 1V and they were pretty ordinary. Few batteries can do that, but some can. A lab I used to work next to had some single cell NiCd batteries each the size of a car battery. I don't know what there ratings were (or even what they had been used for) but I imagine they could produce a staggering current.Better simulators will have more sophisticated models available, but they won't necessarily use them unless told to. If you just place a 1V battery you are almost always going to get the result you see because there isn't really any other sensible result for the simulator to produce. Only if you provide a model of a specific battery, or create your own will it be able to provide a realistic current limit. Even then, if you do something like a small signal AC simulation, it is going to ignore the large scale parameters that lead to reduces maximum current.

author: cur8xgo
author_rank: Regular Contributor
timestamp: June 04, 2019, 11:29:13 pm
content: Quote from: Mr D on June 04, 2019, 10:26:06 pmYes its real.To cement it in your head, look up the current that a 12V car battery can actually deliver. Do the math to calculate its internal resistance, and you will end up with numbers in the ballpark of what you are seeing there in the sim as far as currents and resistances. Its very important you completely trust that this is a real thing, there is no doubt about it.P.S. There are certainly situations where you could have a real-world 1V source with a real-world 0.001 resistance and see 1000 amps flow for at least some period of time. Capacitor + short for instance..think spot welder with an SCR as switch.https://hackaday.com/2011/11/06/diy-spot-welder-can-join-anything-together-even-copper/https://www.pittnerovi.com/jiri/hobby/electronics/welder/index.htmlYou really need to accept that ohms law here dictates that yes, a 1V source will drive 1000 amps through a 0.001 ohm resistance. There is absolutely nothing unusual about it. Stop thinking in terms of "1V can't drive 1kamp" because that has absolutely no basis in reality. Move on from that.

author: rstofer
author_rank: Super Contributor
timestamp: June 04, 2019, 11:42:14 pm
content: Resistors smaller than 1 milliohm are common as current shunts. Consider a 100A shunt that should produce 0.05V at full scale:R = E / I = 0.05 / 100 = 0.0005 Ohms => 1/2 milliohmAnd that isn't a particularly large shunt! Here's one for 250Ahttps://www.rammeter.com/ram-meter-21m250a50-250-amp-50-dcmv-dc-current-shunt.php?gclid=EAIaIQobChMI-rrp3v_Q4gIVAcJkCh3XgQkOEAQYAyABEgLqDPD_BwE50 mV shunts are quite common in industrial usage.Ohm's Law is a law, not a suggestion! It's not like a speed limit...

author: ArthurDent
author_rank: Super Contributor
timestamp: June 05, 2019, 12:18:15 am
content: The expression that describes what you are seeing is 'GIGO', for garbage in, garbage out. You have not modeled the battery, the wiring, or the load properly, or in enough detail, and the simulator gave you a predictable answer which is logically correct but actually wrong. When you oversimplify the model of the circuit you get an unrealistic answer.I wouldn't try this with an auto battery but a typical 9 volt transistor battery can be loaded to a point where the output voltage at the terminals is 1/2 or 4.5 volts. The value of the resistor required to drop the output by half is the internal resistance of the battery. You can do 4-wire resistance measurements of the wiring used in your circuit and that has to be factored in as well as the 1 milliohm load resistor.These and several other factors will limit your current in the real world. unless you take all these factors into consideration and enter them into your simulator you will be getting unrealistic answers. From your reaction to the simulation I think it is safe to assume you're not talking about situations where 1000 amps can actually be produced. One test I had to do at a place I worked would generate a 20K amp pulse to simulate the current produced by a lightning strike in the ground on a utility watthour meter. I realize there are situations where this stuff could happen but I didn't get the sense that's what you are talking about.

author: AG6QR
author_rank: Frequent Contributor
timestamp: June 05, 2019, 01:29:11 am
content: We don't often see 1 milliohm resistors, but when we do, they may be called "cables", because the resistance wasn't intentional. Currents of 1 kiloamp aren't super common, but they do happen in some types of equipment.Turn the problem around: when you pump a kiloamp through a milliohm cable, what sort of voltage drop do you expect to see across the cable?The simulation was accurately simulating what you told it to simulate. Whether that corresponds to a realistic set of components is another story.Every component has non-ideal characteristics. Every wire (and every other component) has resistance, capacitance, and inductance. Every wire is an antenna, too. A big part of the art of electronics is knowing when the non-ideal characteristics will influence the behavior you see.If you connect a milliohm resistor across a AA cell, the cell will need to be modelled in more detail than a simple ideal voltage source.

author: Mr D
author_rank: Regular Contributor
timestamp: June 05, 2019, 07:30:37 pm
content: Guys, thanks for taking the time.Quote from: ejeffrey on June 04, 2019, 11:11:54 pmI'm really a beginner and i'm baffled by this.A 1 volt battery is a very small, weak battery, right?1 kiloamp is an absolutely enormous current, enough to kill a herd of elephants a 100 times over, right?(i appreciate that the resistance of a 100 herds of elephants is more than a milliohm, but just to make the point! ...)How can such a tiny battery push such an enormous current down a wire?And if we keep reducing the (already miniscule) resistance further, the current rises by a function of that reduction, right? Until we reach some Zeno's paradox-like infinite current??!!Sorry for the stupidity, i'm just trying to learn!

author: timelessbeing
author_rank: Frequent Contributor
timestamp: June 05, 2019, 08:05:32 pm
content: Quote from: Mr D on June 05, 2019, 07:30:37 pmA couple of car batteries in parallelcanproduce a kiloAmpere, yet you can safely grab the terminals. You could probably even lick them. Skin resistance is too high. You would have to generate in the order of megavolts to put that much current through a human.Infinite current would require either infinite voltage, or absolute zero resistance. We have superconductors, but I'm pretty sure the laws of physics don't allow infinite current.

author: garethw
author_rank: Regular Contributor
timestamp: June 05, 2019, 08:07:22 pm
content: 1V from a small AA battery is not, at least with technology available today, going to output 1000A. This is because the internal resistance of the battery is far higher than 1mOhm. The internal resistance changes with temperature and is probably not linear with current flow due to the chemical reaction taking place.Still, a typical NiMH AA battery can output over 20A through a short piece of wire across its terminals, measured with an current clamp.An ideal voltage source will output as much current as necessary to keep its voltage stable. The simulator is treating the 1V source as if it has infinite current available. Ohm's Law is correct, but real life circuits have lots of hidden gems waiting to screw up your day!Gareth

author: Mr D
author_rank: Regular Contributor
timestamp: June 05, 2019, 08:16:16 pm
content: Ah wait, i think i'm beginning to get it!Is it so:Voltage is the measure of the "push" available.With ideal, zero resistance the concept of "push" is meaningless?Therefore with no resistance, the push becomes infinitely strong (if there's no push, there must be infinite flow??!!)But wait!............. Ohm's law always applies?But the equation for current with zero resistance is i = V / 0, which is meaningless and not even allowed in algebra?!

author: garethw
author_rank: Regular Contributor
timestamp: June 05, 2019, 08:17:48 pm
content: With regards to the NiMH battery I mentioned just now, Yes I have tried this out of curiosity when I got my current clamp.As far as I remember the voltage dropped to around 1.25V and drew around 24A during the short circuit. Resistance is Voltage / Current so:1.25 / 24 = 52mOhm.This is close to the estimated figure of 50mOhms from Energiser's datasheet.I definitely wouldn't recommend trying it yourself as it can result in batteries rupturing etc.Gareth

author: garethw
author_rank: Regular Contributor
timestamp: June 05, 2019, 08:20:13 pm
content: Voltage / 0 = Infinity. That is why the simulator assumes the perfect voltage source has infinite current available.

author: timelessbeing
author_rank: Frequent Contributor
timestamp: June 05, 2019, 08:23:47 pm
content: Quote from: Mr D on June 05, 2019, 08:16:16 pmYes. Volts are a measure ofElectromotive ForcePlumbing is a good analogy for electricity.Think of water pressure as electrical potential, and the cross section (for example hose diameter) as resistance. The volume of water moving past a point is like current.Quote from: Mr D on June 05, 2019, 08:16:16 pmAs resistance approaches zero, current diverges to infinity. Obviously no such thing.

author: AG6QR
author_rank: Frequent Contributor
timestamp: June 05, 2019, 08:23:57 pm
content: Quote from: Mr D on June 05, 2019, 07:30:37 pmNot necessarily.The voltage of a cell is determined by the chemicals involved in the reaction, not by the size of the cell. I'm not aware of any common battery chemistries that have a 1V output, but 1.2V is certainly common, for example in rechargeable NiCads that used to be so popular.Any NiCad cell has 1.2V, whether it's a AAA cell or a 500 pound behemoth. Large NiCad batteries usually consist of multiple cells in series, but at least in theory, you could wire a few hundred thousand AA NiCad cells in parallel to create a single 1.2V battery. Such a 1.2V battery would have a minuscule internal resistance and would be capable of many kiloamps.I can't think of a practical reason to do it, but it could be done.

author: cur8xgo
author_rank: Regular Contributor
timestamp: June 06, 2019, 07:31:20 pm
content: heres a much better way to get it in your headvoltage is equivalent to potential energyjust like lifting something in the air and asking "how much work can this thing do if I drop it?"now to get my point across..1V may seem like not very much voltage, so cant "do" very muchby that logic, lifting something 1 inch off the ground seems like it cant "do" very muchreally? how about if you lift 10,000 empire state buildings off the ground 1 inch then ask what work you can do with the energy released by dropping them?

author: Mr D
author_rank: Regular Contributor
timestamp: June 06, 2019, 07:39:15 pm
content: Thanks, but can you clarify your analogy?Is voltage like the weight that's going to fall one inch, or like the distance (1 inch) an arbitrary weight is going to fall?But i like your analogy for another reason:You asked: "how much work can this thing do if I drop it?"If there is nothing for the weight to interact with, it'll fall forever. So it'll do an infinite amount of the least possible amount of work!?Is that like, in an ideal circuit, the current becoming infinitely high when a voltage pushes current down a wire with no resistance?

author: AG6QR
author_rank: Frequent Contributor
timestamp: June 06, 2019, 09:39:26 pm
content: Lifting mass through a gravitational field is somewhat like "lifting" charge through an electric field.The unit of mass is the kilogram. The unit of charge is a Coulomb (about 6x10^18 electrons)On earth's surface, it takes about 10 Joules to lift a 1kg mass 1 meter. It always takes exactly 1 Joule to "lift" 1 Coulomb of charge through a 1 Volt electric field. That's another way of saying that a Volt is one Joule per Coulomb.An amp is the rate at which charge moves through a circuit. An amp is one Coulomb per second.The power equation says that power is Voltage times current.(Joule/Coulomb)x(Coulomb/sec) = Joule/secThe definition of a Watt is a Joule per second, so the units do work out the way they should.The corresponding excercise for gravity is left for whomever wants to do it.

author: cur8xgo
author_rank: Regular Contributor
timestamp: June 06, 2019, 10:35:40 pm
content: Quote from: Mr D on June 06, 2019, 07:39:15 pmThe voltage is potential energy, so its the combination of the weight and the height.Since super conductors are on the edge of physics I'm not really sure how much current flows when a voltage is applied. It would be interesting to look up how super conducting experiments have been run and what the current levels have been, and supplied how.So instead of imagining zero resistance, which you will never, ever see, unless you are literally working with super conductors, lets just say a very small resistance, which is small enough to be "ideal" for all intents and purposes but still non-zero enough to allow understanding at an extreme. Lets say 100 micro ohms.An entire circuit with no resistance is not necessarily "ideal" since it wouldn't really do anything unless you are messing with super conductors. However a section of a circuit between two components could be considered ideal since usually energy wasted in wire resistance is not the goal.If the building doesn't interact with anything as it falls, no work is being done. So think of an object falling towards a planet with no atmosphere. No work is being done, as there are no interactions. But something is clearly happening. Potential energy is being converted to kinetic energy.I _think_ that in a super conductor, something similar is happening. No work is being done, but the current (and/or resultant magnetic field?) represents a kinetic energy of sorts?

author: Nerull
author_rank: Frequent Contributor
timestamp: June 06, 2019, 11:08:31 pm
content: The power rails in some of the high end workstation motherboards can supply hundreds of amps at 1.8V. Meanwhile, neon sign transformers supply tens of thousands of volts but can barely output a few milliamps. Voltage tells you nothing about how much power a supply can output.Voltage is the change in energy of a unit charge moving through an electric field.As an analogy to height, you could think of a change in kinetic energy per unit mass. There isn't a named unit for that to my knowledge, but it could be expressed as J/kg. A 1 kg unit dropped from some specific height will gain 1J of kinetic energy, and this height would be your 'height-volt'.You could drop a 1 tonne weight from the same height. It will gain 1000 times more kinetic energy, but its 'height-volt' measurement would be equal because it is normalized to a unit mass.

author: rstofer
author_rank: Super Contributor
timestamp: June 06, 2019, 11:37:23 pm
content: Ohm's Law holds because it is likely impossible to achieve 0 Ohms of resistance. Even 10-100is not zero. Division by zero is not a realistic problem.There are many Ohm's Law tutorials on the Internet.

author: Brumby
author_rank: Unknown
timestamp: June 07, 2019, 02:13:22 am
content: I really don't know why so many people have gone to such great lengths to give answers to questions that weren't asked.I will address your points:Quote from: Mr D on June 04, 2019, 10:26:06 pmAbsolutely correct.QuoteI am no expert - but I would expect that if you used a native definition of a voltage source, the internal resistance would not be included, i.e. it would be considered to be zero. If you wanted to simulate a real-world voltage source, you could create one by selecting an appropriate voltage source with an appropriate series resistor.If there were any predefined such models, there would need to be a huge variety as not all 1V voltage sources will have the same ESR. For a 1.5V example, take AAAA, AAA, AA, C and D cell batteries - each of those will have different ESR - and each of THOSE will differ between chemistry and brands. Then you have power supplies, solar cells and whatever else can generate a potential - and all of those will have different ESR.Setting up the exact values for a real world source will usually be quite straightforward - and it will save having to search through a potentially large range of predefined models in the hope of finding the right match.QuoteNot at all. It is a very fair question.You now know that you have to be aware of the difference between the ideal world of the circuit simulator and the real world of electronics.A simple example of this difference can be found in the simulation of an astable multivibrator. Four resistors, two capacitors and 2 transistors in the classic structure will not start. The reason is that if you use exactly the same value components in the two halves, then there will not be any way for the simulator to choose which side will turn on first. You can get around this by making one value slightly different. This introduces asymmetry and the simulator can then work out the starting condition where one side will turn on first. After that, the running phase will settle down to what you are familiar with.This problem never happens in a real multivibrator circuit, because the chances of getting four pairs of identical components is basically zero.

author: Mr D
author_rank: Regular Contributor
timestamp: June 07, 2019, 07:25:04 am
content: Thanks Brumby,In my short time on this forum, i've noticed quite a lot of resistance from some people to the idea of using a simulator for learning anything about electronics.Personally i think i've learnt quite a lot a short time from simulators.But some people, instead of addressing the principal that i'm asking about, simply say in "in a real circuit, that's not how it works". So they expect me, as well as grasping the basic principal, to also grasp the subtleties of real-world component interaction at the same time!Ok, so EveryCircuit is a bit of a toy, fair enough.A question:Are there other simulators, like Multisim, that, given enough information about what component you've chosen, will factor into the simulation the real world-ness that would make it more realistic?And do they, if instructed to do so, even add in some fuzzy or randomness (into for example, resistor values) so that an astable multivibrator would simulate more accurately?The reason i'd like to keep learning with simulators is that it's just quicker to throw components around on a screen, create multiple versions to compare, do a quick hour of learning without first covering the kitchen table with multimeters, breadboards etc.

author: Brumby
author_rank: Unknown
timestamp: June 07, 2019, 08:03:48 am
content: Quote from: Mr D on June 07, 2019, 07:25:04 amThat, I'm afraid, is a question I cannot answer - as I do not have that information.

author: timelessbeing
author_rank: Frequent Contributor
timestamp: June 07, 2019, 08:55:53 am
content: You will learn a lot more from building real circuits than from a simulator.