I still don't understand current.

[kal.a]

I've read and watched a lot of basic electronic tutorials but I still don't quite get current.

So i bought an Antec power supply to play with. On the label it specifies +5V for a Max. 30A and Min 1.5 A
Does that mean that the minimum current it will output on the +5v is 1.5A?
I hooked it up to a blue LED and measure the current drawn which was about 80 mA. then hooked up the same LED to a "Switching Universal Power Supply" with a 4.5V and 1700mA rating and the current draw was about 8mA. Which didn't make any sense to me.
I wanted to clear up my confusion so I got a 10 Ohm - 5Watt resistor and connected it to each power supply as above. It drew over 400mA from the Antec power supply and 26 mA from the smaller one.

I don't understand where the rest of the current goes? I thought the Min. 1.5A on the label meant that the power supply was going to push that much current through whether needed or not even if it burns out the device?
And what happened to Ohm's law. The 10 Ohm resistor at 4.5V should draw 450mA but that did not happen with the smaller power supply?

 

[Pommie]

A minimum current makes no sense whatsoever. When disconnected it would have to form an arc to pass the minimum. The smaller supply is either broken, a fraud or the output hasn't been set right.

Ohm's law works well. However, it looks like your resistor is 10% high, hence the 400mA. As for the small supply, see above.

Can you post a picture of the rating label with the minimum current on it?

Mike.

 

[spec]

I've read and watched a lot of basic electronic tutorials but I still don't quite get current.

So i bought an Antec power supply to play with. On the label it specifies +5V for a Max. 30A and Min 1.5 A
Does that mean that the minimum current it will output on the +5v is 1.5A?
I hooked it up to a blue LED and measure the current drawn which was about 80 mA. then hooked up the same LED to a "Switching Universal Power Supply" with a 4.5V and 1700mA rating and the current draw was about 8mA. Which didn't make any sense to me.
I wanted to clear up my confusion so I got a 10 Ohm - 5Watt resistor and connected it to each power supply as above. It drew over 400mA from the Antec power supply and 26 mA from the smaller one.

I don't understand where the rest of the current goes? I thought the Min. 1.5A on the label meant that the power supply was going to push that much current through whether needed or not even if it burns out the device?
And what happened to Ohm's law. The 10 Ohm resistor at 4.5V should draw 450mA but that did not happen with the smaller power supply?

Hi kal.a, welcome to ETO

There is not much in life that is certain except death and taxes, as the saying goes. Well, rest assured, you can add Ohms law to that list. Ohms law is the basis for all electronics and you are wise to get the hang of it as a first move.

You were also wise to drop the LED and move to a pure resistor for your experiments.

The 1.5A to 10A specification for your Antec power supply unit (PSU) does not mean what you think: what it means is that the PSU is not designed to work at less than 1.5 A output current and will produce a maximum current of 10A. If you operate that PSU at less than 1.5A load:

(1) you may damage it
(2) it will not work at all
(3) it will produce a voltage higher than 5V
(4) it will produce a large amount of hash and noise at it's output.

That type of PSU is intended for an application where there is always a current drain, like in a personal computer for example. It is not designed to be a bench power supply.

The answer to your problem is to permanently connect a resistor across the power supply of the right value and wattage to take 1.5A

(1) Oms law says:

(1.1) I=V/R or
(1.2) V=I*R or
(1.3) R= V/I

In this case you need to find the value of R so use formula (3):

R= V/I
Where:
V= 5V
I= 1.5A

Thus R = 5/1.5 = 3.33 Ohms The closest standard resistor value is 3.3 Ohms, so that would be ideal. You can work out how much current that would take from the 5v PSU

(2) The power in Watts (W) dissipated by a resistor is:

(2.1) W= V * I or
(2.2) W=( I*I) * R or
(2.3) W= (V*V)/R

You know the Volts and you now know the resistor value, so use formula (2.3) to find the minimum wattage rating for the load resistor:

W= (5*5)* 3.30= 7.58W

Thus 7.58W is the minimum power resistor that would reliably do the job, providing it is in air at 25 degrees centigrade or lower. Best to go for at least a 10W resistor though. Bear in mind that the resistor will get very hot.

This resistor would be perfect, especially if bolted to a heatsink (sheet of aluminium or PSU case possibly): https://uk.rs-online.com/web/p/panel-mount-fixed-resistors/0160708/

 

[ronsimpson]

On the label it specifies +5V for a Max. 30A and Min 1.5 A

"Min 1.5A" Your supply does not meet some specification at less than 1.5A. It was not designed to work with less than 1.5A. This is typical of switching power supplies.
Example: If your load was 10A part of the time and 5A part of the time, your supply would work well.
BUT; If the load went from 10A to 0A it is likely the out put voltage would get above 5.05V for a short time.
Going from 30A to 0 is hard on some supplies. also 0 to 30 is hard on some supplies the voltage would likely drop to 4.9V for a short time.

Example-2: race cars were never designed to drive slow. At walking speeds they are "junk". At full speed they are beautiful.

 

[spec]

I hooked it up to a blue LED and measure the current drawn which was about 80 mA. then hooked up the same LED to a "Switching Universal Power Supply" with a 4.5V and 1700mA rating and the current draw was about 8mA. Which didn't make any sense to me.

A LED is not a resistor. If you are talking about connecting a bare blue LED across a 5V supply you are lucky the LED did not explode. A LED must be fed with a constant current, or at least a circuit to limit the current. To illuminate a blue LED from 5V you would need a resistor in series with the LED of around 150 Ohms

I wanted to clear up my confusion so I got a 10 Ohm - 5Watt resistor and connected it to each power supply as above. It drew over 400mA from the Antec power supply and 26 mA from the smaller one.

That does not make sense as you say but, most probably, the small PSU cannot supply much current.

I don't understand where the rest of the current goes? I thought the Min. 1.5A on the label meant that the power supply was going to push that much current through whether needed or not even if it burns out the device?

Absolutely not. You could make a PSU like that but it would be for a very specialist and odd application

And what happened to Ohm's law. The 10 Ohm resistor at 4.5V should draw 450mA but that did not happen with the smaller power supply?

Nothing ever happens to Ohms Law. It is err.... a law and is carved in stone.

Just a few words about measurements: I do not know how you are measuring the values you are talking about but it is important, as pommie says, to take into account tolerences when making measurements. Take a simple example:

(1) You have a multimeter which has +- 5% accuracy
(2) You buy a power resistor that is +- 20% accuracy (not uncommon for power resistors)

That means, in the worst case, your calculations will be +- 25% in error.

If you had a PSU with an exact output of 5V the meter could read anything between 4.75V and 5.25V.

In practice digital multimeters on the DC volt ranges are typically +- 2% for a low end model to +-1% for an average model, and +- 0.1% for a high end (and very costly) model.

When you are starting out with electronics a good multimeter is essential: the 1% category will be fine.

It is also useful to have an accurate voltage source just to act as an absolute voltage reference. It is dead easy to build a 2.5V +-0.1% voltage reference with a few cheap components. If you would like a circuit just post.

PS, just to illustrate an extreme example of tolerance, although +- 20% is more common these days, at one time high-value electrolytic capacitors typically had a tolerance of -50% +100% (not typing error). So when you bought a 10000 uF capacitor you may only get 5000 uF. On the other hand, if you were lucky, you may get a 20000 uF capacitor. And the errors don't stop there: the capacitance value is also dependent on internal temperature, frequency, ripple current and so on.

 

[kal.a]

Thanks a lot guys. A lot of good info to read.
I can't write much now but I will have a couple more questions when I get back. Hopefully I will finally understand current.




I wired the 10 Ohm resistor again and measured it to get post exact number it draws 465mA with the Antec 5 Volts supply


https://images10.newegg.com/NeweggImage/productimage/17-103-917-03.JPG

 

[spec]

Your PSU is an ATX type intended for a desktop personal computers. https://www.newegg.com/Product/Product.aspx?Item=N82E16817103917

If you don't configure it properly it will do all sorts of stange things. But with a bit of work thay make excellent bench PSUs. There is heaps of data on the internet showing how to do this. The first thing to sort is the turn on signal- I used to know the approach but have forgotton. If you get stuck I will investigate further if you like.

 

[kal.a]

Hi kal.a, welcome to ETO

There is not much in life that is certain except death and taxes, as the saying goes. Well, rest assured, you can add Oms law to that list. Ohms law is the basis for all electronics and you are wise to get the hang of it as a first move.

You were also wise to drop the LED and move to a pure resistor for your experiments.

The 1.5A to 10A specification for your Antec power supply unit (PSU) does not mean what you think: what it means is that the PSU 1s not designed to work at less than 1.5 A output current and will produce a maximum current of 10A. If you operate that PSU at less than 1.5A load:

(1) you may damage it
(2) it will not work at all
(3) it will produce a voltage higher than 5V
(4) it will produce a large amount of hash and noise at it's output.

That type of PSU is intended for an application where there is always a current drain, like in a personal c0omputer for example. It is not designed to be a bench power supply.

The answer to your problem is to permanently connect a resistor across the power supply of the right value and wattage to take 1.5A

(1) Oms law says:

(1.1) I=V/R or
(1.2) V=I*R or
(1.3) R= V/I

In this case you need to find the value of R so use formula (3):

R= V/I
Where:
V= 5V
I= 1.5A

Thus R = 5/1.5 = 3.33 Ohms The closest standard resistor value is 3.3 Ohms, so that would be ideal. You can work out how much current that would take from the 5v PSU

(2) The power in Watts (W) dissipated by a resistor is:

(2.1) W= V * I or
(2.2) W=( I*I) * R or
(2.3) W= (V*V)/R

You know the Volts and you now know the resistor value, so use formula (2.3) to find the minimum wattage rating for the load resistor:

W= (5*5)* 3.30= 7.58W

Thus 7.58W is the minimum power resistor that would reliably do the job, providing it is in air at 25 degrees centigrade or lower. Best to go for at least a 10W resistor though. Bear in mind that the resistor will get very hot.

This resistor would be perfect, especially if bolted to a heatsink (sheet of aluminium or PSU case possibly): https://uk.rs-online.com/web/p/panel-mount-fixed-resistors/0160708/

Thanks a lot I think this post cleared up some of my

Your PSU is an ATX type intended for a desktop personal computers. https://www.newegg.com/Product/Product.aspx?Item=N82E16817103917

If you don't configure it properly it will do all sorts of stange things. But with a bit of work thay make excellent bench PSUs. There is heaps of data on the internet showing how to do this. The first thing to sort is the turn on signal- I used to know the approach but have forgotton. If you get stuck I will investigate further if you like.

Thanks Spec.

If i'd like to go back to your earlier reply and and my misunderstanding. I understood current to be like an available water volume that is controlled by how much I open the tap. If I open the tap just little bit I get very little water even though the watter pressure in the pipe is constant. In other word the device draws as much current as it needs and not any more. Then that turned out to be wrong and I learned that I have to control the amount of current using resistance to insure that no more current than the device need is being forced upon it. And that led to my experiment with the power supplu VS the transformer. I wanted a current supply source that's high and constant to measure across a resistor or to burn an LED but the fact that did not happen confused me.
And when you wrote that I'm wrong in my understanding that the power supply will always put out 1.5A confused me even more. How is the current put out by a power supply (not mine but a reliable power supply) controlled? What determines how much is ouput?

By the way I'm not sure what you mean by configured properly but I watched a couple of youtube clips and got it to work and put out all the voltages it is supposed to. The power supply is not that important to me though. I just want to understand.

Thanks a lot. Much appreciated

 

[AnalogKid]

While there are thousands of different power supplies on the market, internally there are only about a dozen fundamentally different power supply topologies, with many variations of each. This means that moving a particular load, like your power resistor, from one power supply to another often will produce very different results that have little or noting to do with the load itself. This is complicated when the load is an LED or some other non-linear device. To get a better feel for that whole voltage/current thing, start with some higher value resistors like 1.0K, 2.2K, 3.3K, 4.7K, and 10K.

Also, a non-switching power supply like an old wall wart (those black plug-in AC adapters that came with calculators and other small electronics) will behave better with these small load currents. Switching power supplies are strange, and even a simple analysis of how they work requires some understanding of how inductors and capacitors behave; probably a bit ahead of where you are. Suffice to say that the transformer requires a minimum current through it for it to act like a transformer - strange but true. That is why your Antek says it needs a minimum load of 1.5 A; that is to stabilize the internal magnetics. Linear supplies rarely have this requirement because the regulation circuits are fundamentally different, and old wall warts are usually a simple linear supply.

ak

 

[spec]

Thanks a lot I think this post cleared up some of my

Thanks Spec.

Hi kal.a,

No sweat about the info

Good to meet you.
(if you ever like my posts please click the 'like' symbol, bottom right)

If i'd like to go back to your earlier reply and and my misunderstanding. I understood current to be like an available water volume that is controlled by how much I open the tap. If I open the tap just little bit I get very little water even though the watter pressure in the pipe is constant. In other word the device draws as much current as it needs and not any more. Then that turned out to be wrong and I learned that I have to control the amount of current using resistance to insure that no more current than the device need is being forced upon it.

You are right. Current is like water and Voltage is like water pressure. Resistance is like a tap: low resistance (1 Ohm) is the tap right open so much water flows. When the tap is almost closed (1 megga Ohm) little water flows.

All you have to do now is translate that to electronics. Water = electrons, water pressure = volts, the tap is resistance. That is all there is to basic electrical circuits. A power supply is like the oceans it has a vast supply of electrons (water) that will never deplete. To be specific 1 Amp is 1.6 x 10^19. electrons passing a point in a wire in a second (that is 16000000000000000000 electrons). But never mind how much water is in the ocean the water will only flow if there is a height differfence. Just so with electrons, they will only flow if forced to do so by a voltage difference.

A PSU is designed to provide a constant voltage at its terminals regardless of what current you take or don't take. But every PSU has a limit. In the case of your PSU the 5V line has a limit of 30Amps: just think how many electrons that is!

And that led to my experiment with the power supplu VS the transformer. I wanted a current supply source that's high and constant to measure across a resistor or to burn an LED but the fact that did not happen confused me.

I have underlined what is wrong. You didn't want a current supply, you wanted a voltage supply. Practically all supplies are voltage supplies. That is what confused you. Suppose you had a 5 V PSU that could provide a maximum current of 3A and you connected a 2 Ohm resistor across it. How much current would flow through the 2 Ohm resistor = 5V/2 Ohms = 2.5A. Now take the same 2 Ohm resistor and connect it to a 5V PSU which had a current capacity of 1000 amps. How much current would now flow through the resistor? Can you tell me?

About the LED. From 0V to around 3V a LED is open circuit and takes no current, then suddenly just over 3v it turns into a low resistance, of say 10 Ohms so with 5V across it the LED will take a large current and melt. LEDs typically run at a current of only 20mA (0.02 Amps)

And when you wrote that I'm wrong in my understanding that the power supply will always put out 1.5A confused me even more. How is the current put out by a power supply (not mine but a reliable power supply) controlled? What determines how much is ouput?

As already stated a PSU does not force current out of its terminals. It has no idea how much current is flowing from its output terminals. All it knows is that it must keep the output voltage fixed at say 5V. Only when the current reaches the limit for the PSU will it say enough is enough and drop its output voltage to zero so no current flows. This is done to protect the PSU from damage. (note that I have simplified greatly for explanation). Always remember that Voltage is the master, current is the slave, not the other way around.

By the way I'm not sure what you mean by configured properly but I watched a couple of youtube clips and got it to work and put out all the voltages it is supposed to. The power supply is not that important to me though. I just want to understand.

To configure an ATX PSU as abench PSU you need to do certain things depending on the particular supply in question.
(1) set the enable inputs to turn the PSU on
(2) possibly load one or more of the supply lines to ensure correct operation
(3) Ensure safety standard are met. this relates mainly to the case which is not designed to be stand alone.
(4) I vaugely remember another setting but i can't be specific right now.

Have fun with electroncs and if all goes wrong and you just cant get it straight, just keep at it. I gave up in frustration a few time when I was in your position. One final thing, basic electronics is dead easy, providing you understand the fundamental enabling rules and get a feel for how to apply them. After Ohms law, there are just two more simple rules to learn and then you are set.

 

[kal.a]

Hi kal.a,

No sweat about the info

Good to meet you.
(if you ever like my posts please click the 'like' symbol, bottom right)

Thanks for pointing that out to me Spec. There are a lot of people that I wanted to thank. Very helpful and informative forum and great memebrs.

You are right. Current is like water and Voltage is like water pressure. Resistance is like a tap: low resistance (1 Ohm) is the tap right open so much water flows. When the tap is almost closed (1 megga Ohm) little water flows.

All you have to do now is translate that to electronics. Water = electrons, water pressure = volts, the tap is resistance. That is all there is to basic electrical circuits. A power supply is like the oceans it has a vast supply of electrons (water) that will never deplete. To be specific 1 Amp is 1.6 x 10^19. electrons passing a point in a wire in a second (that is 16000000000000000000 electrons). But never mind how much water is in the ocean the water will only flow if there is a height differfence. Just so with electrons, they will only flow if forced to do so by a voltage difference.

A PSU is designed to provide a constant voltage at its terminals regardless of what current you take or don't take. But every PSU has a limit. In the case of your PSU the 5V line has a limit of 30Amps: just think how many electrons that is!

Thanks Spec. I think the little light bulb (maybe even and LED) is starting to get just a little bit brighter.

So for a 5v potential the possible current flow is determined by the resistance it is flowing through. Is the resistance is close to zero (because zero is mathematically challenging) then the current will equal the volts divided by the resistance. For example for a resistance of 0.00000000001 and a voltage of 5v 500000000000Amp. Am I on the right track?
And that would be why we have to limit the current applied to an LED for example or any other device. Though and LED will work with a 5v it's resistance is so low, say 10 Ohms that the current going through it would be 500mA which would fry the LED.
If we put a resistor before it, say 250 Ohm or so, to reduce the current to a level that it can handle then the LED would light and not be damaged.

I have underlined what is wrong. You didn't want a current supply, you wanted a voltage supply. Practically all supplies are voltage supplies. That is what confused you. Suppose you had a 5 V PSU that could provide a maximum current of 3A and you connected a 2 Ohm resistor across it. How much current would flow through the 2 Ohm resistor = 5V/2 Ohms = 2.5A. Now take the same 2 Ohm resistor and connect it to a 5V PSU which had a current capacity of 1000 amps. How much current would now flow through the resistor? Can you tell me?

The answer to your question is:

The same amount of current will flow through the resistor as Ohms law still applies regardless of the available current. The amount of current is determined by the voltage and the resistance. 5v divided by 2Ohms still 2.5 Amps and the resistor would have to be capable of handling that kind of current going through it. 5v x 2.5 A = 5 Watt.

 

[spec]

kal,

You are a quick learner,

Everything you said is correct. You have got the resistor current sorted too - my attempt to lead you astray with the 1000 Amps didn't work

 

[KeepItSimpleStupid]

Warning: The water analogy will break down eventually. For now, beware.
It's probably worth introducing the following concepts:
The ideal voltage source: It has a reistance of zero
the ideal current source: It has an resistance of infinity.
Any non-ideal current source can me modeled as a voltage source in series with a resistor.
Any non-ideal voltage source can be modeled as an ideal current source in parallel with a resistor.

A battery can me "modeled" as a resistor in series with a voltage source. Generally we introduce mathematical models to help us understand the behavior of a device.
This battery could also have capacitances too, but for now it's an ideal voltage source in series with a resistor.

So, now we have two properties of this battery:
Open circuit voltage and short circuit current.
That modeled resistor we can call "ESR" or "Effective Series Resistance". smaller is better.
In car batteries, they like to use minimum "Cold Cranking Amps" which is really marketing. bIGGER is better.
ESR of a battery is somewhat difficult to measure.

because of that series resistance, the battery can only deliver a certain amount of current. The "real world" makes add some other things to that result too:
The resistance of the cables.
The "contact resistance" of say the battery clamp to the battery; The clamp to the wire, etc.
Thermal effects, if any.
Dissimilar metal effects, if any.
In some very rare cases we have to consider vibration.
In some cases "very rare", we have to consider the earth's magnetic field.

Generally, when teaching we don't overload the brain with all of possible issues. We keep things simple.
We mention "ohm law" but we intentionally "leave out" the fact that it's temperature dependent.

Let's nip this concept early too:
ben Franklin made a mistake. His mistake results in "The world" defining his "current", conventional current and hat "current" "flows" from positive to negative. That "current" is not electrons. They happen to go the "other way". The "other way" matters "sometimes". Your meters measure "conventional current".

The Amp is defined as 1 C/s. A (C) or a Coulomb is a unit of charge. Charge doesn't flow like a fluid flows in a pipe, so it breaks down already.

I may have confused you. I challenged a College course Electronics instructor that my answer was right and he actually agreed, but said "You're not supposed to know that yet!"

One other comment. Units derived from proper names get capitalized. Ω= Ohms, S=Siemens, A=Amps, V=Volts, Hz=Hertz. cm, m, g don't.

 

[kal.a]

Good evening Gentlemen,


I was playing around with a couple of LEDs and a resistor and measuring current and voltage drop across the LEDs and making sure I got a handle on the basics. I had one 1K resistor, one red LED and one 2191U1-24 LED and one of the ways I wired them was as follows:

24.24 VDC with the two LEDs in parallel together and in series with the 1K resistor:
Resistor Vf21.93
Red LED Vf2.29 If 0.021
2191U1 did not light up at all and had 0 Vf


Why did the voltage get all routed to the red LED and nothing went to the 24V LED?


Thanks
Kal

 

[KeepItSimpleStupid]

2191U1 did not light up at all and had 0 Vf

makes NO sense. The definition of parallel would mean they have the "same" voltage across each.

The "same" voltage would have been Vf of the smaller LED. e.g. 2.29 V

the 24 V LED actually consists of a LED with a small Vf, but has a series resistor and a 1N1004 Diode. It can operate between 12 and 24 V, but it actually sees a current source (24 V in series with your 1K resistor.

The circuit contains resistors, diodes, standard LEDs a voltage source in both series and parallel and no longer simple.

 

[kal.a]

Thanks KISS,

I tried it again and the 24V LED actually lights up but so dimly I couldn't see it but as soon as I pull out the red LED the 24V LED lights up quite bright.

I played around a bit more and here's what's happening:

The 24V LED has a red and white leads when I connect it by itself directly to 24 DC supply it works only when the red lead is to the positive and not the other way around.
Now I put a resistor on the + side and the red lead of the 24V LED on the other side of the resistor then the white lead to - (Negative)and it lights up as expected. All good.
Then I put the resistor on the + side and both LEDs positive side to the resistor (meaning the red lead of the 24V LED and the longer lead of the red LED) and ground the other side and that's when the 24V LED doesn't fully light up.
But then I left the red LED wired correctly to the resistor and rewired the 24V LED with the white lead to the resistor (the resistor is still connected to +) and the red lead to + (positive) and the 24LED lights up brightly!!!!
What's going on?

 

[KeepItSimpleStupid]

tried it again and the 24V LED actually lights up but so dimly I couldn't see it but as soon as I pull out the red LED the 24V LED lights up quite bright.

That is the right answer.

There is a diode in that 24 V LED, so it's effectively an open circuit on one direction. The diode allows a high reverse voltage or PRV or PIV.

The LED with the small Vf is in itself a diode, but the max reverse voltage is about 6V. That mechanism may or may not kill the LED. If it's sufficiently current limited it can act as a zener diode.

 

[MrAl]

Hello there Kala,

Most of the questions have been addressed already but there are some other things that are worth noting.

First off, an LED is not a good starting point for discovering what current is, or anything else for that matter. That's because LED's are much more complicated than a device like an ordinary resistor. An LED voltage and current relationship is exponential in nature, while a resistor has a much less complicated relationship. Because of this, a resistor is the best device to start with as you probably already found out.

Second, very often you will find that nothing says more than a direct measurement. A measurement tells us if we really have what we think we have. Various devices like power supplies and batteries behave in a manner which does not conform to the definition of an ideal source, so a measurement or two helps us find out what is really happening as compared to what we think is happening. The number of measurements we need is often the same as the number of variables we are working with.

For a resistor, the relationship is described pretty well with Ohm's Law:
I=V/R

and since we have three variables we probably want to measure three things:
1. The resistance, using a trusted Ohm meter.
2. The voltage, using a trusted volt meter.
3. The current, using a trusted current meter.

If we fail to measure just one of these quantities, we could very well end up with results that dont make sense.
For example, if we 'think' we have a 10 ohm resistor but it's really a 20 ohm resistor and we just measure current and voltage and the current is 2 amps and voltage 20 volts, we will see:
I=V/R
so:
2=20/20

but that's not right (remember we 'thought' we had a 10 ohm resistor). This could easily cause confusion.
Then finally we measure the resistance and find out it is really 10 ohms, so we do it again and get:
2=20/10

and now it all makes sense.

So the lesson to be learned is to be sure to measure as many quantities as you can.

In the case of the power supplies and batteries, they almost never put out the EXACT voltage that they have stamped on the label. It will always be at least a little different. This means that it is mandatory to measure the voltage first, and thet means we have to measure the voltage for each and every load we connect, not just one time only for one particular load. That's because the voltage often changes a little for each load we use. A 10 ohm might draw the voltage down by 0.1v, while a 5 ohm load might draw it down by 0.2v, etc. So always measure the voltage at the same time that you measure the current. The resistance can usually be measured once as long as you dont go over about half the power rating of the resistor. Note this is not the case using an LED either, because the LED internal resistance changes dramatically with different voltages.
Also, dont go over about half the power rating of the resistor, and power can be calculated from any of these:
P=I*I*R
P=V*V/R
P=I*V

Since we would like to limit the power to about half of the power rating of the resistor, we might calculate that using either of:
Phalf=(I^2*R)/2
Phalf=(V^2/R)/2
Phalf=(I*V)/2

As a last note, we dont connect LEDs to a voltage source without a dropping resistor. Connecting an LED directly to 5v could burn it out or just damage it so that it changes characteristics. The LED usually will go dim after that too. The dropping resistor limits the current to the LED.
But for understanding current for the first time, it is better to start with a resistor. Leave the LED's for later.

 

[KeepItSimpleStupid]

This **broken link removed** is a good read. I don't know how much you will understand though.

Some type of DC power supplies are:
unregulated
Linear
Switching

Unregulated is at the mercy of the utility power. Linear has low ripple, but has a poor conversion efficiency. Switchy typically uses frequencies outside of the audio range, but has higher ripple. The higher the frequency used, the smaller the supply. The magnetics can be smaller.

Under the switching category there is buck, boost and buck-boost as well as isolated. there are AC to DC converters as well as DC to DC converters.

Each supply does different things when you overload them. They can have overvoltage and or over-current protection. The simplest overcurrent protection is a fuse.

Lab supplies may fall under CC/CV or constant current/constant voltage or constant voltage/current limiting. Some lab supples fall under the 4 quadrant operation. This means that they operate when V+I-, V-I+, V-,I+ and V-,I-.

A DC supply cannot generally be used to charge a battery without a series diode.

I do want to mention a few types of AC supples. One is called energy limiting. The magnetics are lossy and they will withstand a short circuit. AC supplies can fall into isolated and non-isolated as well.

 

[large_ghostman]

A DC supply cannot generally be used to charge a battery without a series diode.

Eah???