Basic current question

The little lost angel kirjoitti:

No. Voltage is same in one potential, so the voltage is 10V over the both resistors.

And then a picky comment but... voltage never flows, it stands still. Current flows.

No, 10A through both, since the voltage is 10V.

Yes, that's right :-)

a?n?g?e? snipped-for-privacy@lovergirl.lrigrevol.moc.com (The little lost angel) wrote @ Mon, 12 Jun 2006 16:11:58 GMT:

Nah, you will have "10V voltage difference across the resistors". That (the 'pressure', and the resistance determines the current that will flow. IR1 = U/R = 10V/1 Ohm = 10A IR2 = U/R = 10V/1 Ohm = 10A (same case) Then the total current (through A) will have to be IR1 + IR2 = 20A

In a real world circuit, all the wires act as very low resistance resistors (dependant on the material, size and length). In schematics of simple electronics, you don't have to pay much attention to that. The schematic will probably not be drawn showing the how the components will be placed anyway.

Andreas B.

I'll discuss a few simplifications that makes considering these things a little easier.

Unless there is a known reason otherwise, the resistance in a wire in a diagram should be mentally treated as a perfect conductor.

Second, any wire to voltage supplies, again unless there is a known reason otherwise, can be removed and replaced with a simple marker to say what it is connected to. That might help, when thinking about this particular problem, too.

So let me redraw your example as I understand it.

This is the same schematic, I think. Technically, you will still need to connect the bottoms of R1 and R2 together and their tops, too, whenever you get around to actually wiring it up to a 10V supply. But for now, thinking about it this way, you don't see wire connecting them together and the connection won't confuse your thinking about it.

Here, you see that they are two relatively independent circuits (or subcircuits, anyway.) R1 doesn't "care" what R2 is doing, and visa versa. They each are directly exposed to a supposed solid 10V supply and there will be 10V/1ohm or 1 amp of current through each, separately. A total of 20 amps, from the point of view of the power supply providing the 10V (assuming, of course, that the supply can actually maintain the 10V regulation while also supplying 20 amps.)

Sometimes, it helps to redraw things this way.

As to the tiny resistance in the wires, it usually can be entirely ignored. As with every rule, there are exceptions. But where it does matter, it usually only matters for a few nodes in the circuit and not the rest of them. Some of the places where it can matter are in high frequency circuits (where it's not the resistance, so much, but the reactances that count), in high current circuits where the resistance can matter, in high voltage circuits where some physical distance may help, or in isolating a digital section from a sensitive analog section. But for thinking purposes _about_ the circuit generally, you usually can conceptualize the wire as perfect.

Jon

Speaking of nits to pick, current doesn't flow - current _is_ flow: flow of charge. :-) (Think of the current in a river - it consists of water flow, no? :-) )

Cheers! Rich

Actually yes, I was trying to figure out if connections matters in a long chain of components when total current draw is low, say 2A vs 5A.

Yup, that does make it much clearer when I was confusing myself. The silly thing is, I actually got the thinking right at first, then had a brain fart when I started putting in numbers :/

How high is high? I was wondering as said, if putting something further down the chain could cause it to fail simply due to the voltage drop. i.e. in a typical PC, you get connectors in a chain of parallel connections, and usually when people run out, they start adding splitters. So I figured that if you used too much wire (nothing huge say just 20cm per extension), things at the end of the chain will simply not get enough voltage esp since more things = more current = higher IR losses right?

Somebody told me it's the resistance of the connectors, especially if the mating is poor, that is probably going to cause problems rather than wire length. But I couldn't find any information on how to determine that, searching a site like

was just PITA and spec pages don't appear to tell how much resistance the pins have.

--
A Lost Angel, fallen from heaven 
Lost in dreams, Lost in aspirations, 
Lost to the world, Lost to myself

Yes, although the extra wire resistance can be an issue in certin instances. Since you're aware of this think about current passing through pcb traces ( an often overlooked issue )

Graham

Ah. Well, "low" or "high" depends on what's going on. For my kinds of circuits, 2A and 5A are both _very_ high. For a circuit card, that kind of current will take a little thinking about trace widths and/or depths.

I tend to redraw every circuit I'm interested in understanding in order to remove busing of power around. The wires don't help me understand them, in fact they make it harder for me, so I get rid of them. Then I try and divide it into sections such that there is a minimum of signal lines going between the sections. (You can usually, without really understanding anything, see what parts can be teased out leaving only a few lines coming in and out and it is usually the case that when you do that you will accidentally and in ignorance luckily also be breaking out the functional groups, as well.)

Depends on the resistance of the wire (size and run length) you are working with and how important the quality of the voltage rail is to the circuit being powered, I think.

It's especially true in cases, like coax cable runs with RF, that you lose at least 50% of your power with each and every break/connection. For typical non-RF situations just passing along power, the connectors are still important and they will have some resistance. Over time, their surfaces will oxidize and those surfaces are less conductive and therefore will become more of a problem in high current situations.

A very common example you can see in almost any home with a room far away from the power panel is to hook up a power saw and try to cut a

2x4 piece of wood with the room light on there. You will see the light dim when the saw is working. But this is because typical home power circuits are wired with inadequate wire gauges for long runs (they use the same wire everywhere throughout) and there is enough voltage drop when pulling 10-15 amps to lose 15-25 volts.

The answer is to use sufficient conductor cross-section for what you are trying to achieve and/or avoid series connectors along the way if that is a problem. Best is to use a separate set of adequate conductors without any intermediate connectors, going all the way right back to the main supply for everything (starring.) But that presents its own problems (cost, time, etc.)

Jon

You know, I often treat conductors exactly the opposite way ( as resistors ). Many poor designs could be avoided by doing this.

I can't recall how many times I've said to the layout guy ' always think of a track as a resistor ' ( and inductor too btw ) or they'll make them long and stringy.

Graham

But that's something entirely different. I assume you mean VSWR ?

Go to a PC group and from time to time someone wil post a pic of a mobo connector that has literally burnt out due to high contact resistance.

Graham

"The little lost angel" wrote in message news: snipped-for-privacy@news.singnet.com.sg...

I am quite surprised by this post. You have been posting here for some time now and I have seen a better understanding of basic concepts from you than this post indicates. What happened? Tom

Totally different reasons, yes.

I can imagine it, already.

Jon

I'm discussing trying to _understand_ how a circuit should work. It doesn't help anyone just trying to begin understanding circuits to confuse themselves with such details. Later on, yes.

It's kind of like trying to understand the motion of the planets in our solar system. Yes, all the planets and moons and the asteroids and so on _do_ have an impact on each other. And yes, if you need to launch a satellite that will get to Saturn and take reasonable pictures, you will be taking many, many things like this into account. But if you are just starting out trying to understand the basics, you instead sweep all that out of your mind and focus just on the two-body problem -- between Sol and one planet. It gets you _pretty close_ and good enough for getting an important first understanding of things.

Jon

Actually, little lost angel did originally say 1+x ohms for the 'further' R which I took to be a recognition of conductor resistance.

Graham

brain fart moment :PpP

I was thinking through the problem and had it correct intuitively. Then I started plugging in numbers to do the maths to see how much of an effect the wire length and different currents would make.

Unfortunately I tend to get a mental block when maths come in. Ever met anybody who could explain high school physics theory to classmates but screw up the calculations beyond recognition? that's me. So I ended up with numbers that didn't match my expectations, got very confused and thought I had misremembered my basics or something.

--
A Lost Angel, fallen from heaven 
Lost in dreams, Lost in aspirations, 
Lost to the world, Lost to myself

which

I know. But it was pretty clear from the OPs comments that even the basics needed work. I don't see any value in dwelling on such things until there is improvement on that score.

Jon

yes, in real life there will be a difference due to resistance of the longer leads on R2 but also they can't make 1 ohm resistors with enough precision that the two resistors will be close in value enough give a result that reflects measure that difference.

--

Bye.
   Jasen

as we know current will always take path of low resistance,since,here both resistances have same value. so ,same amount of current flows through it and r2=r1+x is correct but practically x will be equal to

0,since the connecting wire from r1 to r2 will be of low (or) 0 resistance

Yes. And it will always take the path of high resistance, too.

Ed

Losing 50% of your power at every connection? Commercial transmitters would all be on fire.

15 volt drop at 10 amps is WAY HIGH. Hope the insurance guy doesn't find out about it after the house burns down. My house wiring is crappy and it loses less than 1/3 of that.

I disagree about treating the wires as perfect conductors -- even for a newbie. That is how the dreaded ground loops get started. It can make a big difference even on signals as 'trivial' as 1V P-P video into a 75 ohm load. Most newbies treat 'ground' as an afterthought while to me, it is the most important thing to get right from the beginning. Power regulation/distribution comes next.

GG