Electronics > Beginners

Some noob questions

<< < (15/28) > >>

ArthurDent:
"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.

Brumby:

--- Quote from: Mr D on July 19, 2018, 09:52:43 pm ---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?

--- End quote ---

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:

--- Quote from: Mr D on July 19, 2018, 09:52:43 pm ---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.

--- End quote ---

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!

Brumby:

--- Quote from: Brumby on July 20, 2018, 04:35:07 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.

--- End quote ---

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.

Mr D:
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?



ArthurDent:
"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. 

Navigation

[0] Message Index

[#] Next page

[*] Previous page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod