The loading effect of voltmeters (DMM and analog)

Thread Starter

KathyMechTech

Joined Jan 19, 2018
6
Hi! I'm very new to All About Circuits but I look here often for help in understanding concepts in electronics. I'm working on the last portion of the one problem I have not been able to figure out and I am hoping that you will be gracious enough to help me out. Yes, I read the rules, however, since I have been working on this for several hours I have not listed all the options that I have tried nor the 20 videos I've watched, or the pages and chapters I've read. That being said, here we go!

This is problem 8-17, part C (in Grob's Electronics, 12e). The problem states: Calculate the DC voltage that would be measured across R2 using a DMM having an R1 of 1k ohms on all DC voltage ranges.
The diagram (see attachments) indicates a Total voltage of 12v; R1 = 1k ohms; and R2 = 1.5k ohms. The voltmeter is across R2.

I know that the answer is 7.2 volts but I do not know how the author came to that conclusion. I can work out all the Ohm's Law equations myself but I need a big nudge in understanding the instructions. R1 already has 1k ohms and limiting the other values to 1k ohms was a failure.

Thank you, in advance, for your willingness to help me. This new journey has been quite the challenge!
 

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bushrat

Joined Nov 29, 2014
209

Knowing 2 values easily helps determine other values.

Picture shows 12 V source and 2 resistors in series. Total resistance is 2.5 k ohms (2500 ohms). With that information, total current is 4.8 miliams (or 0.0048 Amps), just as it is stated in picture.
V / R = A
12 / 2500 = 0.0048 A

First resistor is 1 k ohms, and with total current of 4.8 miliamps, it is possible to calculate how much voltage it drops
I x R = V
0.0048 x 1000 = 4.8 V

If first resistor drops 4.8 V, then the second resistor has to drop the rest of the voltage. This circuit has only 2 resistors, so ti is easy to see that. But lets calculate if it is true. So, there is 0.0048 A of current and it's going thru 1500 ohms resistor
I x R = V
0.0048 x 1500 = 7.2 V
 

Thread Starter

KathyMechTech

Joined Jan 19, 2018
6
Thank you, Bushrat. That is exactly how I solved the first third of the problem (part A) but this specifically says something about a DC voltage range. Am I supposed to limiting something in some way?
 

Thread Starter

KathyMechTech

Joined Jan 19, 2018
6
Thank you, Bushrat. That is exactly how I solved the first third of the problem (part A) but this specifically says something about a DC voltage range. Am I supposed to limiting something in some way?
 

bushrat

Joined Nov 29, 2014
209
Digital multimeters usually have very high resistor build in between probes that would have very very small effect on reading voltages. The resistor is usually in megaohms, usually 10,000,000 ohms and sometimes lower or higher. The ranges (I can assume that you are refering to) are settings on multimeter that you need to choose what to display. Typically, first range displays 0 to 200 milivolt (0.2 V) and will not display any higher voltage. next range is 200 milivolt to 2 V, next one is 2 V to 20 V, next one is 20 V to 200 V. Some multimeters can read higher voltages, but lets not get into that.


If you are expecting voltage to be 7.2 and you have selected 200V range, it will only display "7" for you. if you select range 2V, it will show "overload" or something similar. To properly read the voltage, you would have to select 20V range,
 

Thread Starter

KathyMechTech

Joined Jan 19, 2018
6
This is rather embarrassing. Is it possible to delete this thread? I finally made contact with a classmate and she was going on about 10M ohms and I'm, like, WHAT mega-ohms?!? She said, "It's right there in the problem." I'm like, no way! (Sheesh, my California girlhood days are creeping back into my speech!) Anyway, when I re-wrote the problem onto my notebook paper, I skipped a line of text. Now that I see the problem in its entirety it makes sense and I'm well on my way to getting it done.

Thank you to the many people who looked at this thread and was baffled as to how to help me. I appreciate your desire to help.

Now, with my head down in embarrassment, can I just delete this whole thread?
 

Thread Starter

KathyMechTech

Joined Jan 19, 2018
6
Digital multimeters usually have very high resistor build in between probes that would have very very small effect on reading voltages. The resistor is usually in megaohms, usually 10,000,000 ohms and sometimes lower or higher. The ranges (I can assume that you are refering to) are settings on multimeter that you need to choose what to display. Typically, first range displays 0 to 200 milivolt (0.2 V) and will not display any higher voltage. next range is 200 milivolt to 2 V, next one is 2 V to 20 V, next one is 20 V to 200 V. Some multimeters can read higher voltages, but lets not get into that.


If you are expecting voltage to be 7.2 and you have selected 200V range, it will only display "7" for you. if you select range 2V, it will show "overload" or something similar. To properly read the voltage, you would have to select 20V range,
Thank you, Bushrat! Evidently, I'm working with just half a deck tonight. I missed the text that gave the value of Rv being 10M ohms -- which fits into your illustration perfectly. Thank you for your explanation.
 

Thread Starter

KathyMechTech

Joined Jan 19, 2018
6
@JoeJester, Cute. My first thought was to divide 77 and 260 in half. Thankfully, my brain did a double-take and I looked up Fluke 77. I saved myself a heap-load of embarrassment there! Otherwise, I'm going to look over the problem tomorrow when I'm not so battered by the day. Why is this your favorite problem? Do you encounter formal problems very often?

One thing I enjoy about schematics is how simple and clean they look. Too bad my breadboard isn't as beautiful when I have all the components together : / So far, this has been an interesting semester as I try to wrap my mind around this new-to-me language. There have been many days when I think I should just finish my communications degree but electronics is incredibly captivating. I'm definitely hooked! Next semester I'll show a little more wisdom and take fewer credits.

I'm attaching the documentation for how I solved the problem, if any other students are interested. There may be an easier way but this seemed very straight-forward right along everything I've been doing for the past month or so.
 

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MrAl

Joined Jun 17, 2014
7,189
Hi! I'm very new to All About Circuits but I look here often for help in understanding concepts in electronics. I'm working on the last portion of the one problem I have not been able to figure out and I am hoping that you will be gracious enough to help me out. Yes, I read the rules, however, since I have been working on this for several hours I have not listed all the options that I have tried nor the 20 videos I've watched, or the pages and chapters I've read. That being said, here we go!

This is problem 8-17, part C (in Grob's Electronics, 12e). The problem states: Calculate the DC voltage that would be measured across R2 using a DMM having an R1 of 1k ohms on all DC voltage ranges.
The diagram (see attachments) indicates a Total voltage of 12v; R1 = 1k ohms; and R2 = 1.5k ohms. The voltmeter is across R2.

I know that the answer is 7.2 volts but I do not know how the author came to that conclusion. I can work out all the Ohm's Law equations myself but I need a big nudge in understanding the instructions. R1 already has 1k ohms and limiting the other values to 1k ohms was a failure.

Thank you, in advance, for your willingness to help me. This new journey has been quite the challenge!
Hi,

While modern digital multimeters usually have a high input impedance (taken to be resistance) analog meters are a different story. Then can vary quite a bit as to there input specs. For example, 20k Ohms per volt. I had one a long time ago that had 20k Ohms per volt on some scales and 50k Ohms per volt on other scales.

The whole idea is to just replace the meter with an equivalent resistance, like 10 megohms or 100k or whatever you calculate the internal resistance to be. Dont be fooled though, 10megohms is not infinite, it affects some readings with higher impedance circuits and even with lower impedance circuits you may want really good accuracy.
 

JoeJester

Joined Apr 26, 2005
4,196
@KathyMechTech

The 10M meter paralled with the 1.5k resistor doesn't reduce the equivalent resistance much (0.08 percent). This causes the voltmeter to read 7.19V as 20V scale is the minimum scale on most 4 digit DMMs, notwithstanding calibration errors and the +1 or two count in the meter specifications.

In the example with a 20k/volt sensitivity, that describes the meter movement of 50 uA full scale. So, to make it read 1 volts you need know the meter resistance and to add a resistor in series to cause no more than 50 uA flows when measuring 1V. That resistor is always in series with the meter. To get the higher scales, you add an additional appropriate resistance. At 10V you need a total of 200k.

It changes the loading. Now the Req would be 1.5k || 200k or 0.75 percent lower, plus or minus calibration error.

Using lower value resistors does NOT dramatically demonstrate meter loading effects. If your analog meter does not have a mirror background, you can introduce paralax errors.

Granted the 20M potentiometer exaggerates the problem, but it creates a image that places meter loading in your mind.

That same problem I used in the previous posting, graphically across the whole adjustment range of the potentiometer can be viewed like:

problem-06.jpg

There are times to be aware of meter loading.
 
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