Author Topic: Some noob questions  (Read 14171 times)

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Online Brumby

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Re: Some noob questions
« Reply #50 on: July 18, 2018, 02:19:59 am »
The first thing to understand is that an 'ideal' multimeter measuring current will have zero resistance across its terminals which means it will allow as much current to pass as the circuit it is in can provide.  Such an uncontrolled flow can be a source of danger.  In a real multimeter, however, this resistance will not actually be zero but it will be as close to zero as the designer can get - so it is always wise to think of your multimeter (in current mode) as being a simple piece of wire between between one probe and the other.

Now imagine what would happen if you stuck a piece of wire into the active and neutral of a mains power point that was switched on!!!  Some fireworks, a blown fuse or both - as well as a risk of injury to yourself.  This is the sort of problem that can happen when you don't use measuring equipment properly.

In the case of your wall wart, this is exactly the sort of thing you were doing - but you were fortunate in that things were on a smaller scale and the meter was designed to be able to carry that current.  I would also guarantee you were overloading the wall wart and if you had left it connected it would have heated up very quickly and very likely have been destroyed.

To understand why you kept getting 2A, I will offer you the following very simplified diagram.  This represents your wall wart - but only in a very basic sense.  It is missing a lot of detail that you would find in a real device, but it shows why you would get the same current reading.  To make the maths simpler, I will use the 'ideal' current measuring meter that has zero ohms between the probes.



The story goes like this...
As shown, there are four sections of the transformer that generate 3V each - but the wire that is used in each section has a resistance of 1.5 ohms.
 * When you select 3V, there is only 1 section used and it has a resistance of 1.5 Ω.  Using Ohm's Law V=IR, you get a current of 2A.
 * When you select 6V, there are now two sections used and while this gives you 6V, the resistance across the two sections is 3 Ω.  Again, using Ohm's Law, you get a current of 2A.
 * When you select 9V, there are now three sections used and while this gives you 9V, the resistance across the three sections is 4.5 Ω.  Once more using Ohm's Law, you get a current of 2A.
 * When you select 12V, there are now four sections used and while this gives you 12V, the resistance across the four sections is 6 Ω.  Yet again using Ohm's Law, you get a current of 2A.

This resistance is buried inside the wall wart and something you can't really do anything about it, other than include it in your calculations.  This sort of internal resistance has a name: "Equivalent Series Resistance" or ESR and it is found in every electrical and electronic device - unless it happens to be a superconductor.

It is this ESR and how it is distributed through the wall wart that gives rise to the "constant 2A" phenomenon you observed.


For those wanting to challenge the above, I would like to add that there are several ways of providing a multi-voltage wall wart - and the above is only showing one of those in a very basic conceptual form, purely for the purposes of education, not as a construction project.
« Last Edit: July 18, 2018, 02:24:41 am by Brumby »
 

Offline ArthurDent

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Re: Some noob questions
« Reply #51 on: July 18, 2018, 03:36:46 am »
The wall-wart that you have should have a rating label on it that will tell you what the voltage and safe current rating of the supply is. It sounds like this is a supply that no matter what voltage selection you choose, the short circuit current is limited to 2 amps. You are lucky that it is limited because you never should try to measure the output of an unknown supply by setting a DMM to current and measuring directly across the supply output. A lot of wall-warts would be destroyed by what amounts to a direct short across the output.

Imagine what would happen if you tried this by setting the DMM to current and plugging the leads into a wall outlet. I believe your mains are 230 VAC and with enough current to run appliances. If you plugged the DMM leads into a wall outlet it would blow the DMM’s internal 11A fuse, trip a breaker, or create an arc and burn the probe ends; or some combination of the above. 

As to the confusion about why 2 amps isn’t adequate for everything, what does the work is power, not current or voltage alone. A hair dryer has far different power requirements than a LED nightlight but both may be designed to operate from the same mains voltage.

An automotive bulb may be designed to operate at 12V/1A and an incandescent bulb for your house may be designed to operate at 230V/1A. Both bulbs are designed to load the proper supply at 1A but if you try to run the 230V/1A bulb on 12V it will have too high a resistance at 12V to even glow and will draw very little current while the 12V/1A auto bulb will brightly flash once and be destroyed on 230V because its resistance is far too low for the mains voltage and it will try to draw a very large current and burn out.

To measure current a DMM is generally placed in series with a load like a 12V/1A auto bulb to measure the actual current the bulb draws. A DMM set to current has a very low resistance between the leads so it doesn’t have a noticeable effect on the voltage being delivered to the load. To accomplish this, internally a DMM measures the voltage drop across a short heavy calibrated wire called a ‘shunt’ and the voltage drop across this shunt is displayed as current, and the total series resistance of the DMM leads, the shunt, and fuse is probably a fraction of an ohm. If you are measuring a 1A load, the voltage drop across the DMM is probably a couple of tenths of a volt. For instance IxR=E (the voltage drop across the DMM) might be 1Ax.2ohm=0.2V drop so if you have a 12V supply connected to a 1A load with the DMM reading current placed in series with the load, the load will see 12V-0.2V or 11.8V and the DMM will display about 1A or whatever the actual current at 11.8V is.
 

Offline Jwillis

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Re: Some noob questions
« Reply #52 on: July 18, 2018, 04:30:28 am »
With the analogy, flow is defined as an amount over time. Speed is not a factor in the "flow" of electrons .We'll use yet another analogy to explain .Think of a Newtons cradle and that best represents how electrons move down a wire.Although in reality its much more chaotic with many other factors . Analogies are used to to best explain a process in the simplest terms.Its much easier to picture in the mind something that may be familiar.
It's not an attempt to insult your intelligence in any way.Just a simple way to explain the relationship between voltage , current and resistance .
 

Offline KL27x

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Re: Some noob questions
« Reply #53 on: July 18, 2018, 04:49:01 am »
The water analogy isn't sorta kinda a half-ass analogy. It is pretty close to perfect.

A high pressure washer and your kitchen faucet might produce the same volume of water per time, but the water coming out of the high pressure washer has more velocity. Thus, it has more kinetic energy. Thus, it can do more work.

In the water analogy, it is most common to refer to the height of the water as a measure of its potential energy. But if you were to pour it to "ground," that turns into kinetic energy. In absence of "resistance," 1 gallon of water poured from 100 feet high would be moving faster by the time it reaches the ground than 1 gallon of water poured from 3 feet.

So, this is why hydroelectric generators are positioned at the bottom of tall water falls, not just stuck in the middle of a big, slow moving river that passes 1000x the volume of water per unit time.

Your adjustable PSU (most likely) has a linear regulator. All of the water it can pour starts out higher than 12V. When you set it under this level, the regulator absorbs the excess "velocity," and turns that energy into heat. But when you pass it through your 0.1ohm current meter DMM, the actual voltage you would record across your DMM leads (if you put another DMM on there on voltage setting) would be close to zero (maybe 1V or 2V tops)*, no matter what voltage you set it to. Almost all the energy is just turned into heat in your PSU. There are other sources of resistance in the PSU that are higher in sum than the 0.1R shunt in your DMM which are busy turning this energy into heat.

*ALL real PSU have internal resistance called impedance in this specific instance. 12V PSU with lower impedance (stronger PSU, essentially) will get that reading higher than one with higher output impedance. This is not necessarily proportional to the max current output, but there's usually a correlation. This stronger PSU would make your DMM do more of the job of turning this energy to heat.
« Last Edit: July 18, 2018, 05:24:20 am by KL27x »
 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #54 on: July 18, 2018, 10:41:28 pm »
Ok, so i'm starting to get an inkling of understanding about this, after playing with my MM and some resistors.

re: Ohm's law: 1 ohm of resistance is almost no resistance, right? Or is it in fact no resistance whatsoever? If that's not the case, how many ohms is, say, a 10cm piece of thin copy wire?
 

Offline rstofer

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Re: Some noob questions
« Reply #55 on: July 18, 2018, 11:14:47 pm »
Ok, so i'm starting to get an inkling of understanding about this, after playing with my MM and some resistors.

re: Ohm's law: 1 ohm of resistance is almost no resistance, right? Or is it in fact no resistance whatsoever? If that's not the case, how many ohms is, say, a 10cm piece of thin copy wire?

1 Ohm is a very small number when dealing with electronics.  It is a very high number when dealing with busbar inside a utility substation.  All numbers are relative.

When I think in terms of electronics, I think 220 Ohms for a bright LED from 5V, 330 Ohms if I want it a little dimmer and 470 Ohms as about as dim as I want to go (RED LED).  I might use 330 Ohms from 3.3V if I want a particularly dim surface mount LED - something that might be used for diagnostics.  Later on you will learn to calculate the resistor value for a particular LED current which will result in a specific amount of luminous intensity.  Later on...

I tend to think in terms of 1k, 2.2k, 4.7k and 10k for most transistor projects and 1 Megohm (plus 1 ufd) for op amp integrators (time constant = 1 second, this will come up later...).

The problem with high resistance circuits is that they are easily swayed by noise.  The problem with low resistance circuits it that they tend to use more power.

Here is a calculator that should answer any specific questions you may have:

https://chemandy.com/calculators/round-wire-resistance-calculator.htm

There are MANY other calculators on the Internet.  Some of them are more appropriate for industrial wiring.
 

Offline james_s

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Re: Some noob questions
« Reply #56 on: July 18, 2018, 11:43:46 pm »
Why is it always necessary to explain via analogy?

This is not meant as a criticism of your attempt to explain it. I just don't understand why it's never explained without resorting to analogy!


Well the water analogy works well for many people because you can see, touch and feel flowing water. Ok I suppose you can feel electricity too if the voltage is high enough, but that's not really a recommended way to learn about it!
 
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Offline Rick Law

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Re: Some noob questions
« Reply #57 on: July 19, 2018, 12:18:02 am »
Ok, so i'm starting to get an inkling of understanding about this, after playing with my MM and some resistors.

re: Ohm's law: 1 ohm of resistance is almost no resistance, right? Or is it in fact no resistance whatsoever? If that's not the case, how many ohms is, say, a 10cm piece of thin copy wire?

1 ohm is very little from the scale of resisters you can buy.  You can get (commonly) resisters as low as 0.01 ohm (and may be lower still, but less common) all the way up to multiple giga ohms in one resistor.

But 1 ohm is a lot if you don't want it - for example, on some of my cheaper switches, the contact is about 0.2 ohm to 0.5 ohm!  That is, if I put the switch to the ON position and I measure the resistance at the solder legs - it is 0.2 ohms!
 
Your DMM's probes has wires and so it has resistance as well.  When you put it in resistance mode measuring ohms, and you short the the tip of your two probes, you are measuring the resistance of your probe's wire and the contact resistance between the probes, sockets, etc.  You would probably read 0.25 ohm range - may be less, may be more.

As to your question about 10 cm sheet of copper  - can't answer that because resistance will depend on how thick.  Take a look at this "wire gauge table" - it will show you how many ohms (or milli-ohms) per unit length of copper wire, as well as the cross sectional area for a specific gauge.

https://en.wikipedia.org/wiki/American_wire_gauge
 

Offline rstofer

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Re: Some noob questions
« Reply #58 on: July 19, 2018, 12:19:30 am »
Suppose I had a 480V source and a load current of 1200 Amps (typical for a small unit substation).  Now, assume I wanted to limit the busbar voltage drop to 1% or 4.8 Volts.

R = V / I or 4.8V / 1200A = 0.004 Ohms or 4 milliOhms.  That is 1/250 of 1 Ohm - a pretty small number.  But it is typical of busbar for a unit substation.  The bolted connections have to be done just right!

This is in the world of electrical, not electronics but it does point out that 1 Ohm is not the lower limit of resistance, 0 Ohms is, although it can't be achieved without involving a superconductor.

It's important to remember that Ohm's Law is a Law, not a suggestion.  You can't treat it like a speed limit!

And, yes, I am aware that there are boundary conditions on the Law but they are not relevant to this discussions.  Google tells all about it but it is still irrelevant at this level.


« Last Edit: July 19, 2018, 12:22:58 am by rstofer »
 

Offline james_s

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Re: Some noob questions
« Reply #59 on: July 19, 2018, 05:04:57 am »
You don't need to be working on a substation for 1 Ohm to be substantial. I recently built a controller for the motor in a cordless electric lawnmower that can draw up to 40A from a 24V battery pack. If the mosfet had an on-resistance of 1 Ohm it would be trying to drop 40V which is more than the battery can provide. The result is the mosfet would be dissipating a huge amount of power and the motor would never be able to draw anywhere near 40A with only a 24V battery.

Even if it is only 0.1 Ohm it would drop 4V and dissipate 160 Watts which is far too much. 
 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #60 on: July 19, 2018, 12:57:21 pm »
Many thanks to all who are taking the time, i'm reading through all replies carefully and i believe it's having some positive effect!

How's this?:

Say i have a 3V battery in a simple circuit with only a switch (admittedly not a very sensible circuit, but for argument's sake).

Is it fair to say?: With the switch open, i'll measure around 3 volts across that open switch because (as there's no current flowing), that is the potential charge that could flow if i were foolish enough to close the switch?

So that's why if i remove the battery from the circuit and only measure across the terminals of the battery i'll measure 3 volts (ish), whereas if i short-circuit the terminals (not good, i know) and measure, i'll measure almost 0V as the maximum current possible is flowing at that moment so the potential current is almost zero?

Am i on the right track?
« Last Edit: July 19, 2018, 01:03:51 pm by Mr D »
 

Offline Old Printer

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Re: Some noob questions
« Reply #61 on: July 19, 2018, 01:12:17 pm »
I believe the reason your meter would read zero is because the dead short removes any difference between the prob tips, there is nothing to measure. Educated theory may offer another answer, but this is how I see it.
 

Offline rstofer

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Re: Some noob questions
« Reply #62 on: July 19, 2018, 02:43:16 pm »
Am i on the right track?

More or less...  The short circuit of the switch will have a resistance of essentially (but not really) 0 Ohms.  There will be no voltage drop across the switch because when you multiply by zero, the answer is 0.  This isn't a special case, it follows from E = I * R.  R = 0 so regardless of I, E will be 0.   Ohm's LAW

Better, and safer, experiment:  Connect the battery positive lead to your DMM red lead with the meter set to read milliamps.  Connect the meter black lead to 2 each 1k resistors in series.  You should read 3V / 2k = 1.5 mA.  Now short one of the resistors.  You should read 3V / 1k = 3 mA.

You can expand the experiment into a voltage divider circuit.  You will probably want to measure voltage rather than current.  Take out the meter, set it back to volts and reconnect the series resistors across the power supply.  Clip the black lead to the wire headed to the negative battery terminal.  You should read 1/2 the battery voltage at the center of the two series resistors. 

Now change the  top resistor to 100 Ohms.  The reading should be 0.9 * the battery voltage.  The calculation is easy:  Battery voltage divided by total series resistance (to get the current through the loop) multiplied by the bottom resistor value (to get E = I * R for the bottom resistor and the loop current).  Play with this until you really have a handle on the simple circuit.  Swap the location of the 1k and 100 Ohm.  The result would be 0.1 * the battery voltage.

The formula for the simple two resistor voltage divider is  V = (E / (R1 + R2)) * R2 when R2 is the bottom resistor.  The E / (R1 + R2) calculates the loop current I and then V = I * R2, the bottom resistor.

I like millamps, I don't like amps.  Actually shorting a battery, particularly the more exotic batteries, will produce a LOT of current and it only takes a few seconds for the battery to catch fire.  Then you have to run down the hall to the bathroom and flood the flaming mess with water.  Just sayin'...
« Last Edit: July 19, 2018, 02:58:28 pm by rstofer »
 

Online Brumby

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Re: Some noob questions
« Reply #63 on: July 19, 2018, 03:17:19 pm »
So that's why if i remove the battery from the circuit and only measure across the terminals of the battery i'll measure 3 volts (ish), whereas if i short-circuit the terminals (not good, i know) and measure, i'll measure almost 0V as the maximum current possible is flowing at that moment so the potential current is almost zero?

Am i on the right track?

I'll say "Yes", with one qualification.  The phrase "potential current" isn't correct.  I would use the word "voltage" instead.  Everything else reads like you are understanding correctly.

To understand where the voltage of the battery went, just look back at the internal resistance of the power source I mentioned earlier - ESR.  The battery chemistry not only provides the voltage, but presents a certain amount of resistance as well - so the "equivalent circuit" of a battery is a voltage source and a resistor in series.  As a result, when you short out a battery, the voltage generated within the battery is lost through the resistance of the battery.
« Last Edit: July 19, 2018, 03:19:46 pm by Brumby »
 

Offline james_s

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Re: Some noob questions
« Reply #64 on: July 19, 2018, 04:34:03 pm »
I like millamps, I don't like amps.  Actually shorting a battery, particularly the more exotic batteries, will produce a LOT of current and it only takes a few seconds for the battery to catch fire.  Then you have to run down the hall to the bathroom and flood the flaming mess with water.  Just sayin'...

Years ago I learned rather quickly that putting a 9V battery in your pocket with keys and loose change is a bad idea.
 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #65 on: July 19, 2018, 09:52:43 pm »
Thanks, it's getting ever so slightly clearer.

But now i've come up against the following stumbling block:

Say we have a simple circuit which is a 10v battery and a resistor.

If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Let's make the resistor even worse. What's the worst resistor imaginable? No resistor!!!

Now if we measure either "side" of this non-resistor, we certainly won't see 10v on our MM.

So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

« Last Edit: July 19, 2018, 10:02:42 pm by Mr D »
 

Offline james_s

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Re: Some noob questions
« Reply #66 on: July 19, 2018, 10:52:53 pm »
What do you mean "worse" resistor?

As you decrease the value of the resistor, it will draw more current from the battery which means it will dissipate more power. This breaks down when the resistance becomes so low that the source can't supply as much current as it is trying to draw and the voltage will sag below 10V. If you could have a 0.0000 Ohm resistor across it the voltage would sag down to 0 Volts but in reality you can never have quite zero.
 

Offline rstofer

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Re: Some noob questions
« Reply #67 on: July 20, 2018, 12:07:40 am »
Thanks, it's getting ever so slightly clearer.

But now i've come up against the following stumbling block:

Say we have a simple circuit which is a 10v battery and a resistor.

If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Let's make the resistor even worse. What's the worst resistor imaginable? No resistor!!!

Now if we measure either "side" of this non-resistor, we certainly won't see 10v on our MM.

So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would NOT suggest doing it to the car battery.

We know from Ohm's Law that the internal resistance will drop a voltage E = I * R.  It really is that simple!  The thing is, R is quite low so for a given E, I can be quite high.  Like the car battery - it has a very low internal resistance.

This is the Thevenin Equivalent Resistance I mentioned way back near the top and several others have mentioned.

When you put a light load (1A (10 Ohm 10V) or 2A (5 Ohm 10V) on a source with sufficiently low internal resistance, you will measure essentially the full source voltage.  When you put a very low resistance across the source, the internal resistance dominates and the majority of the voltage is dropped inside the device.  Take that internal voltage drop V and square it.  Then divide by the size of the resistor to get the power dissipated in Watts (E squared over R).  Notice how a resistor less than 1 Ohm drives the power up pretty quick.

That's also why I brought up the voltage divider experiment above.  I was using reasonable resistors such that internal heating of the source wasn't likely to occur.  But the top resistor does exactly the same thing when you slide it inside the battery and call it the Thevenin Equivalent Resistance.  The internal resistance forms a voltage divider with a very low value for the top resistor.

Later on you will find that maximum power transfer to the load occurs when the load resistance is equal to the internal resistance.  This is a consequence of the voltage divider equation given above.

If the external load resistance is higher than the internal resistance, the majority of the voltage drop is outside the battery.  Think 1 Ohm internal resistance and 10k load resistance.  Essentially all of the voltage appears across the load resistance.  Now turn it around:  Think of the 1 Ohm internal resistance and a 0.1 Ohm load resistance.  In this case only 10% (approx) of the voltage drop is at the load and the internal resistance is dropping 90% (approx).  And you can bet the battery is heating up.

You can't get around the necessity of wrapping your head around Ohm's Law.  It explains everything but you need to play with it to get an intuitive feel.  What happens to I when you vary E for constant R.  What happens to I when you vary R with constant E?  What happens to E when you vary R with constant I?  Sit down with a scratch pad and lay down that voltage divider and see how the number actually work.  Same with the voltage divider using the battery internal resistance (assume 1 Ohm just for a number) and various external resistors or even an external 2 resistor divider.

« Last Edit: July 20, 2018, 01:09:46 am by rstofer »
 

Offline james_s

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Re: Some noob questions
« Reply #68 on: July 20, 2018, 12:17:39 am »
As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would suggest doing it to the car battery.


I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying.
 

Offline rstofer

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Re: Some noob questions
« Reply #69 on: July 20, 2018, 01:10:10 am »
As has been mentioned above in several posts, every voltage source will have an internal resistance.  For example, the internal resistance of a AAA battery is quite high when compared to a car battery.  You might survive putting 0 Ohms across a AAA but I would suggest doing it to the car battery.


I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying.

Fixed!  Thanks...
 

Offline ArthurDent

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Re: Some noob questions
« Reply #70 on: July 20, 2018, 02:06:20 am »
"I'm gonna guess you meant to say you would *not* suggest doing that to a car battery. The results are likely to be fairly spectacular but certainly not something I'd recommend trying."

Years ago I owned a MG1100, a inexpensive little car that had a metal ratchet-type device to hold the hood (bonnet) up when you opened it. I was driving on a rough dirt road once and the car just died and smoke was billowing from under the hood. What had happened is the small bolts holding the upper end of the metal device had vibrated out and it had dropped directly across the battery terminals, which had no insulating covers. The sub-ohm resistance of the device forced enough current through the internal cell connectors of the battery so that one melted and the battery was now an open circuit. It got hot but no boom.

Oh, the car was a standard so pushing it and popping the clutch turned the engine over fast enough so residual magnetism in the actual generator allowed the car to start and I got home.
 

Online Brumby

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Re: Some noob questions
« Reply #71 on: July 20, 2018, 04:35:07 am »
So where does this relationship break down? I mean the relationship that says that voltages stay the same measured across a resistor, regardless of the resistor value?

To be very clear - the relationship NEVER breaks down.

You just need to catch up with your understanding - and don't fret ... you ARE getting there.

The answer to your confusion is the same as I (and others) have already mentioned - Equivalent Series Resistance.  You really need to get that concept drilled into your thinking - and then the whole mystery will be solved (well, at least this corner of the world of electronics.)

UNLESS you have a power source that is superconductive, there will ALWAYS be some amount of Equivalent Series Resistance.  For car batteries, this is very low - but it is not zero.  For alkaline batteries it is significantly higher - even though it is still small on the scale of resistance.  Your average 18650 lithium batteries are somewhere in between.

Whenever you look at a power source, you must ALWAYS consider there is a resistor of some value hidden inside it that you can't get around.

With this in mind, you have made an error in the following:
If the resistor is 10 ohms, and we place our MM either side of the resistor, we measure 10v.

If we change the resistor to 5 ohms, and measure again, we still see 10v.

Ok, so 5 ohms is a worse resistor then 10 ohms, but we still measured 10v.

Assuming we have a meter with a 10M input impedance, we can ignore the effect it has on the power source.  (Yes, even this will have an effect on the power source, but it will be very, very small.)  Let's say we measure the voltage (with no other resistors) and you get a reading of 10V.

Now, when you connect you 10 ohm resistor, you will NOT get a reading of 10V.  It will be slightly less.  This is because the current flowing through the circuit goes through TWO resistances.  The obvious one is the 10 ohm resistor you've connected across the power source.  The second one is the hidden internal resistance of the power source - the ESR.  Each of these will drop a number of volts according to Ohm's Law - and the only one you can measure directly is that of the external 10 ohm resistor you used.  You have to calculate the voltage drop caused by the ESR.

When you replace the 10 ohm resistor with a 5 ohm resistor, the current will increase.  You are still facing a circuit with 2 resistances - the external 5 ohm one and the ESR within the power source.  Note that the ESR is the same as it was for the 10 ohm resistor - so with increased current, Ohm's Law tells us that the voltage drop because of the ESR will be greater.  This means the voltage you measure across your 5 ohm resistor will be even lower than it was with the 10 ohm resistor.


If you are any good with algebra, you can perform the above as a real experiment and you should be able to calculate the ESR of the power source.  If you can do that, you will have mastered this fundamental!
« Last Edit: July 20, 2018, 04:37:42 am by Brumby »
 

Online Brumby

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Re: Some noob questions
« Reply #72 on: July 20, 2018, 04:45:16 am »
If you are any good with algebra, you can perform the above as a real experiment and you should be able to calculate the ESR of the power source.

Just be careful about the power rating of the resistors.  To do the above as a real experiment, the 10 ohm resistor will need to be able to handle around up to 10W - and the 5 ohm resistor, up to 40W.  They are going to heat up quite a bit.  If the ESR is very low, then these figures will be pretty close to what you will encounter, however as the ESR rises, the lower the power rating for these resistors will need to be...

BUT

be very aware that the ESR will also cause the power source to heat up - and that could provide some spectacular and dangerous outcomes if left to run for more than a couple of seconds.


PS.  When choosing a power source for the above experiment, don't use a lab power supply.  A lab power supply will have regulated output and possibly current limiting that will thwart any attempt at calculation.
 Use a battery.
« Last Edit: July 20, 2018, 04:53:27 am by Brumby »
 

Offline Mr DTopic starter

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Re: Some noob questions
« Reply #73 on: July 21, 2018, 11:23:03 am »
Thanks, i'm starting to get an intuitive feeling for this (although it's still foggy ;))

But i'm still struggling with some aspects of voltage.

Say we have a 9v battery. We measure for voltage across the terminals and measure 9v(ish).

Now, we touch only the + probe of the MM against the + of the battery while leaving the  MM - probe dangling. We measure 0v as expected.

Now, we lay a 10cm piece of copper wire on the table.

With the +MM still attached to the battery +, we touch the - probe to the end of the copper wire. We measure 0v as expected.

Now we attach the end of copper wire to the - battery terminal. Now if we touch the end of the copper wire with the MM -, which still having the + attached to the battery +, we measure 9v.

Soooooo, my question is: how does the MM "know" what is at the end of the copper wire? How does it "know" that in the last step, i attached the negative battery terminal to the copper wire?

As far as i can see it, there are only two possibilities:

Either a) some information must have travelled along the copper wire from - battery terminal to the MM - probe and into the MM.

Or b) some information must have travelled from the battery + terminal, into the MM+ probe, through the MM, out of the MM - probe, along the copper wire, into the - battery terminal, through the battery, out of the battery + terminal, into the MM+ probe, and into the MM.

Is it one or the other? Or a mixture of both? Or have i (yet again) fundamentally failed to understand something?



« Last Edit: July 21, 2018, 11:27:49 am by Mr D »
 

Offline ArthurDent

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Re: Some noob questions
« Reply #74 on: July 21, 2018, 12:53:29 pm »
"Soooooo, my question is: how does the MM "know" what is at the end of the copper wire? How does it "know" that in the last step, i attached the negative battery terminal to the copper wire?"

Using "know" is not how you should visualize it, the circuit is just following the laws of physics. An exact analogy is a lighting circuit in your house. If you turn a switch on (like connecting a lead to the copper wire you have on the table), current can now flow in the completed circuit and the light controlled by the switch will go on. Turn the switch off to open the circuit and the light goes off.

If you have a table lamp plugged into a wall outlet, the lamp can be on but when you unplug the lamp you break the circuit (remove the lead from the copper wire you have on the table), and the table lamp goes off. This is more obvious because you can actually see the physical separation between between the plug and the wall outlet. The only "know" part is your knowingly completing or opening a circuit to control the flow of current.

Don't overthink this. 
 


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