How Does a Multimeter Meaure Voltage and. . . .

[DW1961]

When you use a 100 ohm resister across the probes the voltage flows between the positive and negative probe around the resistor and the battery (see image). But if you don't use the resistor and you just use the probes, I assume it goes back to the meter? If so, why does it go through the resistor when connected as in the picture, and not just right back to the multi-meter and bypass the resistor?

Edit:
I should have made it clear that using a resistor with the meter is because the meter has no battery test function, thus no internal test resistance.

[ledtester]

In voltage mode your multimeter has a very high impedance between the probes - typically greater than 1 megaohm. So some current will pass through the meter but compared to the 100R resistor it will be insignificant.

[IanB]

the voltage flows between the positive and negative probe around the resistor and the battery (see image)

Voltage does not flow. So the whole premise of your question is flawed.

[Brumby]

Voltage does not flow. So the whole premise of your question is flawed.

That's a somewhat pedantic response.

Considering the nature of the question and the place it was presented, I would be inclined to give the OP some slack and/or provide some education.  Just saying they are wrong is not helpful.

[xrunner]

When you use a 100 ohm resister across the probes the voltage flows between the positive and negative probe around the resistor and the battery (see image). But if you don't use the resistor and you just use the probes, I assume it goes back to the meter? If so, why does it go through the resistor when connected as in the picture, and not just right back to the multi-meter and bypass the resistor?

Well ... Your wording of the problem is a little strange. Are you suggesting the meter is in the DC voltage mode? You need to model the situation to understand it. There are two more "resistors" in the circuit as shown. The battery has internal resistance, and the meter is also placing a high resistance in parallel to the battery and resistor shown.

[IanB]

That's a somewhat pedantic response.

Considering the nature of the question and the place it was presented, I would be inclined to give the OP some slack and/or provide some education.  Just saying they are wrong is not helpful.

I suppose you are right. But I find it hard to think of someone who has been on the forum for three years with over 600 posts as being a beginner.

[Brumby]

When you use a 100 ohm resister across the probes the voltage flows between the positive and negative probe around the resistor and the battery (see image).

To be more accurate, it is current that flows.  In simple terms, voltage is just the pressure that makes current want to flow.

But if you don't use the resistor and you just use the probes, I assume it goes back to the meter? If so, why does it go through the resistor when connected as in the picture, and not just right back to the multi-meter and bypass the resistor?

Remember this: Electricity will flow in each and every part of a circuit that it is able.

In your example, when you have the resistor and meter connected, some current will flow through the resistor and some will flow through the meter.  How much flows through each will depend on the nature of each part.

If we just use simple resistance^** in our calculations, the resistor is 100 ohms and a typical digital multimeter is at least 10,000,000 ohms.  Use ohms law to work out how much current flows through each.


^** Actual circuit parameters in the real world can get really, really tricky, especially as the frequency of things increase - but it is not necessary to worry about that with the example you are trying to understand.

[DW1961]

the voltage flows between the positive and negative probe around the resistor and the battery (see image)

Voltage does not flow. So the whole premise of your question is flawed.

Electricity doesn't flow either, but we gotta talk about it somehow. Would it help if I had said "electrons bumping into each other?"

[IanB]

When you use a 100 ohm resister across the probes the voltage flows between the positive and negative probe around the resistor and the battery (see image). But if you don't use the resistor and you just use the probes, I assume it goes back to the meter? If so, why does it go through the resistor when connected as in the picture, and not just right back to the multi-meter and bypass the resistor?

To elaborate on my previous response, voltage is not something that flows, it is a potential. Measuring voltage is a bit like measuring the weight of an object on a scale.

An ideal voltmeter does not allow any current to flow through it when measuring voltage, in much the same way that an ideal scale does not move when you place a weight upon it.

[DW1961]

In voltage mode your multimeter has a very high impedance between the probes - typically greater than 1 megaohm. So some current will pass through the meter but compared to the 100R resistor it will be insignificant.

Then the meter measures voltage between the probes and resistor?

[Brumby]

That's a somewhat pedantic response.

Considering the nature of the question and the place it was presented, I would be inclined to give the OP some slack and/or provide some education.  Just saying they are wrong is not helpful.

I suppose you are right. But I find it hard to think of someone who has been on the forum for three years with over 600 posts as being a beginner.

Hmmm....   Fair point.

[IanB]

Electricity doesn't flow either, but we gotta talk about it somehow. Would it help if I had said "electrons bumping into each other?"

Electric current flows. Electricity has two aspects, voltage and current. Current is the thing that flows, and voltage is the driving force that makes current happen.

[DW1961]

Well ... Your wording of the problem is a little strange. Are you suggesting the meter is in the DC voltage mode? You need to model the situation to understand it. There are two more "resistors" in the circuit as shown. The battery has internal resistance, and the meter is also placing a high resistance in parallel to the battery and resistor shown.

It's a meter without a battery tester, so no resistance other than what is necessary to get the meter to work. If I use a meter with a dedicated battery test setting, I get a different voltage reading than the other meter w/o that setting.

I should have made it clear that using a resistor with the meter is because the meter has no battery test function.

[Brumby]

Ah .... so you want to make a meter into a battery tester by adding a load.  Well, that IS the simple way to do it - but for a valid measurement, you will need the right load for the particular battery.

[DW1961]

I suppose you are right. But I find it hard to think of someone who has been on the forum for three years with over 600 posts as being a beginner.

Your personal opinions do not give you a license to disrespect me (or others) for any reason. It's unacceptable, sophomoric, and abusive.


I apologize to the community for this response, but it's a matter of self respect, and I don't need it or want these types of fruitless disrespectful comments. These types of down-talking and disrespectful comments never help anyone -- ever.

[TimFox]

The cell has an internal resistance in series with its chemical electromotive force, which resistance should be much smaller than the very high input resistance of the voltmeter.
The current through the 1 or 10 M\$\Omega\$ voltmeter input from a 1.5 V cell is only 150 nA to 1.5 uA.
To see the effect of the cell's internal resistance, which increases dramatically as the cell discharges, a suitable and relatively small resistor will draw enough current to drop the voltage at the voltmeter input.
100\$\Omega\$ will draw 15 mA from a 1.5 V cell.
Notice that I wrote "cell":  a "battery" is a series connection of cells (despite the labeling on commercial cells).

[DW1961]

When you use a 100 ohm resister across the probes the voltage flows between the positive and negative probe around the resistor and the battery (see image).

To be more accurate, it is current that flows.  In simple terms, voltage is just the pressure that makes current want to flow.

But if you don't use the resistor and you just use the probes, I assume it goes back to the meter? If so, why does it go through the resistor when connected as in the picture, and not just right back to the multi-meter and bypass the resistor?

Remember this: Electricity will flow in each and every part of a circuit that it is able.

In your example, when you have the resistor and meter connected, some current will flow through the resistor and some will flow through the meter.  How much flows through each will depend on the nature of each part.

If we just use simple resistance^** in our calculations, the resistor is 100 ohms and a typical digital multimeter is at least 10,000,000 ohms.  Use ohms law to work out how much current flows through each.


^** Actual circuit parameters in the real world can get really, really tricky, especially as the frequency of things increase - but it is not necessary to worry about that with the example you are trying to understand.

I do get that. But how do the meter know what is passing between the resistor and probes?

[DW1961]

That's a somewhat pedantic response.

Considering the nature of the question and the place it was presented, I would be inclined to give the OP some slack and/or provide some education.  Just saying they are wrong is not helpful.

I suppose you are right. But I find it hard to think of someone who has been on the forum for three years with over 600 posts as being a beginner.

Hmmm....   Fair point.

Is it? Talking down to someone is fair because they have been on a forum for x years?

The error in his logic is that I have been here for 3 years so I should know this stuff. In reality, I visit when I have a question and then go away sometimes for weeks or months even, w/o even thinking about electronics. My total time on this forum typing and replying is minuscule, in actuality. Add up how long it takes to type and reply 600 times average, and you get total time on the board. Perhaps, maybe, several hours or so.

I mean, what he is saying is that since I have been her for so long my question was to low for him to reply nicely to.

It makes the hair on the back of neck stand up when I think people think that sort of reply to another human being is okay.

[DW1961]

Ah .... so you want to make a meter into a battery tester by adding a load.  Well, that IS the simple way to do it - but for a valid measurement, you will need the right load for the particular battery.

Isn't 100 ohms correct for 1.5 to 9v?

[Jwillis]

When you use a 100 ohm resister across the probes the voltage flows between the positive and negative probe around the resistor and the battery (see image). But if you don't use the resistor and you just use the probes, I assume it goes back to the meter? If so, why does it go through the resistor when connected as in the picture, and not just right back to the multi-meter and bypass the resistor?

Edit:
I should have made it clear that using a resistor with the meter is because the meter has no battery test function, thus no internal test resistance.

Your multimeter measures a Potential Difference between the positive probe and negative probe. It then compares and calculates that difference from a reference value set in the chip. This gives a readout on the display.
In old analog volt meters the voltage and current would create a magnetic field in the meter coil that has a specific impedance this would cause a deflection . And as you know, with a set impedance (resistance),  when voltage rises so must current. So as voltage rises the meter deflects more.

In your picture, Your measuring the potential voltage across both the resistor and the battery. Any battery will drop in voltage under enough load.  Since the meter "draws' very little current because of a very high impedance. (ussually 1 to 10 mega ohms) the multimeter can't simulate a load high enough to drop the voltage of the battery. The resistor draws more current across it because of it's much lower impedance dropping the voltage. So the meter can measure an accurate voltage because it effects the battery very little compared to the 100 ohm resistor. 
This is a load test of the battery at approximately 15mA. If there is no significant drop in voltage then the battery is good. If it drops to less than ( I believe ) 1V the battery is considered depleted. 
Suppose that's the best way I can describe it.

[DW1961]

The cell has an internal resistance in series with its chemical electromotive force, which resistance should be much smaller than the very high input resistance of the voltmeter.
The current through the 1 or 10 M\$\Omega\$ voltmeter input from a 1.5 V cell is only 150 nA to 1.5 uA.
To see the effect of the cell's internal resistance, which increases dramatically as the cell discharges, a suitable and relatively small resistor will draw enough current to drop the voltage at the voltmeter input.
100\$\Omega\$ will draw 15 mA from a 1.5 V cell.
Notice that I wrote "cell":  a "battery" is a series connection of cells (despite the labeling on commercial cells).

So the resistor loads the battery and the meter reads and calculates the volts from the pos and neg battery connection from the probs to the meter?

[Kleinstein]

Ah .... so you want to make a meter into a battery tester by adding a load.  Well, that IS the simple way to do it - but for a valid measurement, you will need the right load for the particular battery.

Isn't 100 ohms correct for 1.5 to 9v?

The suitable resistor depends on the size of the battery. 100 Ohm is quite low for a coin cell, but quite high for a car battery.
It depends on the battery type if one needs a load resistor to test a battery.

For the alkaline cells one can get away without a load resistor, as the open circuit voltage of the cells goes down. Usually 1.2 V per cell are considered empty, as not much charge is left than. However it depends on the circuit and some work down to less than 1 V while others already complain (stop working) at 1.4 V.
Old style dry cells and Li coin cells like CR2032 need a load to get a  meaningful result. When empty the open circuit still stays nearly the same, but the internal resistance goes up and the voltage thus drops under load. A those cells are usually used with light load one can use a larger resistor, like 1000 or even 10000 ohm for testing. Such a battery may be empty for circuits that need relatively high current like a flash light, but may still work for a low load like a clock.

I noticed the difference between the battery cells with an alarm clock: with a dead alkaline cell the clock stoped / slowed down but the alarm still worked till the end. With a dry cell the clock still worked, but the alarm failed when the battery was empty.

[DW1961]

Regardless of how is measures voltage using the resistor, I just tested my MM using it's built in test mode using an AAA A battery:

100 Ohm resister and MM set to 20VDC = 1.57v
No Resister  and MM set to 20VDC= 1.58v
Built in 1.5v test mode = 1.516v

This indicates to me the battery is not only good, but fully charged. That's why the resistor test and the non resistor test look almost identical. As far as the built in test goes, I dunno why the discrepancy in voltage, except that the built in circuit uses a larger resistor?

What I need now is a low battery to see the difference between the resistor and no resistor.

[Jwillis]

I know we're no really supposed to advertise or promote products here. But i picked up one of these and it really helped to understand ESR and how it changes as the battery depletes.
https://www.aliexpress.com/item/1005004269893522.html?spm=a2g0o.order_detail.order_detail_item.11.7cb2f19cLsM6Ou
With the assistance of the battery data sheet of course you need to select a load resistor that will pull an average normal use current . In the case of a AA 1.5V cell it is somewhere around 15mA. So this is probably why a 100 ohm resistor was used.
 

[BeBuLamar]

That's a somewhat pedantic response.

Considering the nature of the question and the place it was presented, I would be inclined to give the OP some slack and/or provide some education.  Just saying they are wrong is not helpful.

I suppose you are right. But I find it hard to think of someone who has been on the forum for three years with over 600 posts as being a beginner.

I am fine with someone who doesn't know that's what we here for. But as you said I suspect that the OP is trying to pull our legs.