Question regarding Ohm's law and a Digital Multimeter.

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

Dude, you seriously need to chill and stop being so aggressive.

I don't need to do a damn thing.

I wasn't the one who sent him on a wild goose chase, over something that wouldn't possibly affect his ckt.

I was simply trying to educate the guy because he was going all theory, without realizing some of the pitfalls.

You did the same thing. People have been getting so caught up in this "maker" stuff, that they've begun to assume that unless they're "told" something, a person, individually, can't figure it out. There's not much simpler than the OP's ckt, yet your claims of burden voltage, which I might add would never have crossed your mind unless you too were told about it, perhaps by a maker-breaker.

What about the percent impedance/regulation of his supply voltage, or even the effect of worn test leads? Both of which are more of a problem, and real ones at that.

You claimed to "educate" the OP, but didn't even realize yourself that you're suggestion had no bearing, suggesting that you yourself had been mislead at some point. Quit being a maker, and start being a learner and thinker- nothing better than this.

It's 3 AM here now, it was about 1-2 AM when I wrote the previous message and it took some time to find the videos in those authors' channels, so I listed the shunt values from memory and listed all three to be as complete as possible, not to confuse anyone (for example by listing the uA range).
I didn't double check the actual values not because I was lazy but because it is late and not in the mood to go to my workbench and spend 10-20 minutes setting everything up.

You're peeing on my leg. Didn't I suggest not overthinking this? You don't have to "set anything up" to know that the shunt impedance of that DMM is insignifigant compared to the dropping resistor already in the circuit.

Just as you don't have to "set up" and pull the ESR meter out to "confirm" the blown Sacons are bad- if that's not overthinking and "getting all theory," tell me what is.

Again, my point was simply to educate him about the fact real world ... test leads have resistance (insignifigant), current shunt has resistance (insignifigant), potentiometer has resistance (dominant.

So you're suggesting that this fraction of an ohm matters just as much as the pot, simply because "it's in the circuit?"

Why are you failing to make this leap?

The power dissipated can be measured with formula P = IxIxR ... for 5mA and 986 ohm the power would be P = 0.005 x 0.005 x 986 = 0.02465 watts if you use or 24 mW

I guarantee you googled that in the last 20 minutes; you had no inkling of Ohm's law's implications, yet you suddenly show me the I2R power formula?

Do you have hands-on electronics experience beyond "making" things on a breadboard? Don't answer me unless you've actually designed your own circuits, even if it's an LED and resistor, like the OP.

People are being mislead by the "makers," and slowly becoming "breakers," without even realizing it.

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

Ok...


Yes, they're Alkaline cells. Just so I understand this, you're saying when I put a load on them (for example, use them to provide energy to a circuit with a resistor), the voltage that they're providing will actually be less than if I measured the voltage from the cells out of circuit?

Yes. If you have, or get, two meters, you'll see it drop as you load them up. The higher the load current, the worse this gets. Alkaline cells' impedance also increases as they approach full discharge.

Keep in mind, I'm new to this. I have no teacher or anything. Everything I've learned I've learned by reading and trying some experiments. I don't know hardly anything yet.

That's OK! We all started somewhere. I certainly didn't know what I do now, when I started. Every day's a school day, no matter how old you are.

That's why I'm asking the questions.

As well you should.

I'm okay with being wrong. I just like to know when I'm wrong so I can correct my thinking.

And these "free-form," build-it-yourself circuits are an excellent way to learn. The junction drop of that LED is more relevant than the meter or breadboard impedance, see below.

I don't understand the analogy you provided about the transistor radio and a central aircon. I just started learning this stuff maybe a week ago. I've spent probably a good 3 days on it though. Had some other stuff to do around the house. I am going to make mistakes. And I'm sure there will be more "stupid assumptions".

That's good! It won't be long until you're building on this and conceiving circuits that seem beyond your grasp at the moment. Soon, you may want to build a small linear power supply. That's what I did after getting started w/ LEDs and such.

As to my comparison: Let's say a large-ish condensing unit for a central A/C draws 25A at 240V, even though the voltage doesn't matter, and a transistor radio draws 35 mA. This is a current ratio of 700:1. A typical breadboard can take maybe 3A per contact, and your LED ckt pulls a mere 4.31 mA; also 1/700th of the contact rating.

IOW, using your breadboard to power the LED is the same as having a circuit of such ampacity for that central AC, but only using it to run a transistor radio. To put it simply, the resistance/limitations of the breadboard contacts won't, in any way, affect your small circuit. Unless said breadboard is corroded or suffered water damage.

[budm]

Re: Question regarding Ohm's law and a Digital Multimeter.

If you want to know how much the AMP meter resistance is being put in series with the circuit, use another Ohm meter and check the resistance between the RED and the Black probes of the Amp meter, if the AMP meter has two ports for example 1 port for low current reading mode up to 500mA, and another port is for current measurement up to 10A. Check the resistance both of them. Subtract your Ohm meter probes resistance from what you read on the Ohm meter, that reading will be how much resistance you introduce into the series circuit.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

Okay, thank you for example that Kaboom. I understand now. I remember learning wires and traces do cause resistance. Although I learned how to calculate how much resistance they take up, I didn't really grasp how little resistance it is. To me, it made sense that the breadboard would be causing the difference there in the end when I took into account the resistance of the LED and tried to calculate the current of the whole circuit. I wasn't certain though, that's why I posted the statement that I did. In case I was wrong, I figured someone would correct me and I appreciate you correcting me. Thanks.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

If you want to know how much the AMP meter resistance is being put in series with the circuit, use another Ohm meter and check the resistance between the RED and the Black probes of the Amp meter, if the AMP meter has two ports for example 1 port for low current reading mode up to 500mA, and another port is for current measurement up to 10A. Check the resistance both of them. Subtract your Ohm meter probes resistance from what you read on the Ohm meter, that reading will be how much resistance you introduce into the series circuit.

Ahh! Thank you Budm! Tomorrow, I'm going over to my friends house to grab my cheap analog meter. Perhaps I can use that. It might just be too hard to get a good reading. The meter was given to me and the person only paid 5$ for it I believe. I'd like to get one of those Fluke 87V's I think they're called one of these days. I'd also like to get a benchtop multimeter sooner rather than later. I just have to find the right one for the right price.

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

Okay, thank you for example that Kaboom. I understand now. I remember learning wires and traces do cause resistance. Although I learned how to calculate how much resistance they take up, I didn't really grasp how little resistance it is. To me, it made sense that the breadboard would be causing the difference there in the end when I took into account the resistance of the LED and tried to calculate the current of the whole circuit. I wasn't certain though, that's why I posted the statement that I did. In case I was wrong, I figured someone would correct me and I appreciate you correcting me. Thanks.

You're welcome.

If you ever want to build your first power supply, I suggest this book. I think the first thing I soldered together was the 5V linear supply in this book.

I still have that power supply. It's had several improvements since then, and of course, still in use.

As its output floats, it can just as easily be used as a -5V supply; I did this to supply the head amps of two HDDs after a bad SATA power adapter killed the onboard converters, before I rebuilt them.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

Which book are you talking about Kaboom?

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

Whoops, forgot the link:

http://www.amazon.com/Building-Power.../dp/B0006F1DLO

Was "forcing" other threads to load, and forgot to drop the link.

I think BCN has bad caps.