Hi
Many apologies - this is probably the noobiest question ever asked.
I have been trying to measure really simple circuitry (battery, breadboard, wires and resistor) with a multimeter.
I cannot get the figures I am measuring to add up to the expected values.
So I looked at the following page for help https://www.autodesk.com/products/e...standing-voltage-current-resistance-ohms-law/
If you look at the bottom of the page then the writer plugs the numbers into Ohm's Law and says "R = 9V/0.016A which equals 473.68"
Now unless I have missed something totally, 9 divided by 0.016 equals 562.5, not 473.68
Please can someone tell me what I am doing wrong here ?
Thanks
Ben

 

hi Benny,
Welcome to ACC.
The writer of that equation, subtracted the Vled voltage from the 9v supply, then divided by 0.016A
E

I would have chosen Vled as 2v, so [9v -2v]/0.016A = 437.5R [437.68 not 473.68]

 

hi Benny,
Welcome to ACC.
The writer of that equation, subtracted the Vled voltage from the 9v supply, then divided by 0.016A
E
View attachment 151480

Thank you for the welcome and thank you for the incredibly quick reply

OK then I am doing something wrong with my measurements.
I have a breadboard, connecting wires, a 470 Ohm resistor and a 9V battery
When I test the Voltage of the battery just by touching the terminals, I get 8.4V. When I test in parallel across the battery with the circuit connected up, I also get 8.4V. I understand why this is 8.4 as opposed to 9 as the battery is not brand new. Then I test the current and get 0.016A. I know the resistor is 470 Ohm as I have tested that too.
However, those values give me V = I (0.016) * R (470) = 7.5s, which is almost a Volt less than the voltage I have measured.

Again - I am likely to be doing something stupid so I won't be offended if you laugh

Ben

 

hi B,
What type of current meter are you using to measure the current thru the 470R.?

E
Do you have a couple of 220R resistors in your bits box.?

 

Hi @ericgibbs
I'm using the meter in the attached photo
And yes I have many 220R resistors
I have to go out for a while now but will pick up any response in an hour or 2

 

hi,
OK.
Try the left hand circuit, measure across the battery and then across R1 and then R2, post what you measure.
Vb=
R1=
R2=

The other circuit is for the original LED question you posted.
E

 

hi,
OK.
Try the left hand circuit, measure across the battery and then across R1 and then R2, post what you measure.
Vb=
R1=
R2=

The other circuit is for the original LED question you posted.
E

@ericgibbs
Thank you so much for helping out with this. Results are .....

Vb=8.75
R1=4.17
R2=3.81

Photo of circuit attached

​

 

hi,
That is a strange result?
Adding 4.17v and 3.81v = 7.98v , which is 0.77v less than the 8.75v applied voltage.?

Do you have second DVM [voltmeter] that you cross check those readings.?

BTW: when you measure mA currents, use the meters mA current range setting.
I do not read the resistor code banding as 220R, check the resistance with your ohm-meter

E

 

hi,
That is a strange result?
Adding 4.17v and 3.81v = 7.98v , which is 0.77v less than the 8.75v applied voltage.?

Do you have second DVM [voltmeter] that you cross check those readings.?

BTW: when you measure mA currents, use the meters mA current range setting.
I do not read the resistor code banding as 220R, check the resistance with your ohm-meter

E

@ericgibbs I tested both resistors with the meter. Both were well within 1% tolerance of 220R
Unfortunately, I don't have a second meter. I can borrow one on Monday though. Is it likely that my meter is the cause of the problem ?

 

Here's a closer shot of one of the resistors
I read that (backwards) as Red Red Black Black Brown (220R 1%)

 

hi,
Thanks for the clearer image.
Let us know the result of the meter check.
E

 

@ericgibbs I just changed the battery in the multimeter and I am now getting very different results which seem far more accurate. I am beginning to think that there is something seriously wrong with my multimeter. As I said before, I will try another one on Monday and let you know what happens

 

hi,
Thanks for the feedback, look forward to seeing your results.
E

 

Thank you for the welcome and thank you for the incredibly quick reply

OK then I am doing something wrong with my measurements.
I have a breadboard, connecting wires, a 470 Ohm resistor and a 9V battery
When I test the Voltage of the battery just by touching the terminals, I get 8.4V. When I test in parallel across the battery with the circuit connected up, I also get 8.4V. I understand why this is 8.4 as opposed to 9 as the battery is not brand new. Then I test the current and get 0.016A. I know the resistor is 470 Ohm as I have tested that too.
However, those values give me V = I (0.016) * R (470) = 7.5s, which is almost a Volt less than the voltage I have measured.

Again - I am likely to be doing something stupid so I won't be offended if you laugh

Ben

The reason for the error between the calculated current and the measured current is that the meter is adding an additional impedance in the circuit (in this case around 55 ohm).

So with the meter in circuit the battery is presented with a resistance of 470+55 = 525 ohm, resulting in the measured current of 16mA with an applied voltage of 8.4V.

It is very important to understand the loading effects of electrical measuring instruments and how they can affect the expected result.

 

The reason for the error between the calculated current and the measured current is that the meter is adding an additional impedance in the circuit (in this case around 55 ohm).

hi Hymie,
If my test current meter added an extra 55R in series with the circuit path I was measuring. I would say the meter was faulty.
The 'normal' loading effect of taking a measurement is not the cause of the TS's problem.

E

 

hi Hymie,
If my test current meter added an extra 55R in series with the circuit path I was measuring. I would say the meter was faulty.
The 'normal' loading effect of taking a measurement is not the cause of the TS's problem.

E

In my experience one of the short-comings of cheap digital meters is that they have a relatively high resistance when measuring current – causing the error noted.

For most circuits, when measuring a relatively low currents an additional 50-60 ohms in the circuit will make no difference; but when the circuit resistance is low, combined with a low voltage – the 50-60 ohms becomes an issue.

Even good quality meters will have a noticeable resistance when set on a low current measurement range.

 

hi H,
I would agree a cheap meter may not display the correct measurement value.
The decent DVM's I use have a 1R and 0.1R [mA, Amp ranges].

The ~50R internal series that the TS meter has, is well above the acceptable value.
If you consider the simple test circuit I posted,

a 8.75V nominal into 440R load, he got readings of,
[he used 1% resistors, which is a good choice.]

Vb=8.75
R1=4.17
R2=3.81

E

EDIT:
Hi H,
Downloaded the manual for the users Ragu17b DVM, not much help regarding the internal resistances value.?
It does mention that a low battery will cause incorrect readings, shows a low battery warning

 

Hi @ericgibbs
I think I sorted it now - I was using connector wires between my multimeter wire and the breadboard just so I could poke connectors into the breadboard and pull them out whenever I wanted.
I took the wires out of the equation and everything started working normally as expected.
It seems that one of the connector wires has a problem so that's in the bin now.
It explains things to me as the readings were all over the place.
Thanks for your patience and help yesterday - it was much appreciated.
Ben

 

hi Benny,
Thank you for the feedback, it may also help other hobbyists.
Eric

EDIT:
I see that you are interested in learning about electronics, look at this AAC link.

https://www.allaboutcircuits.com/worksheets/

 

Generally, you can't use the same meter to measure both current and voltage because most current meters have an internal resistance. They do probably drop less than 0.6 V, but that number may be significant.

You CAN use your voltmeter for a series circuit and see if Vb=V1+V2+V3....Vn, or the voltage drop of all the components equal the battery voltage. There are also component tolerances. Resistors might be +-5% and a LED voltage drop may nominal be 2.1 V, but there ia a maximum and minimum. The nominal drop varies by the color of the LED.

There are feedback ammeters which have a voltage drop below 1 mV.

And a general note is that the meters affect your measurement. Let's say for measuring voltage, Zin is 10 megohm (common). You cannot expect to measure the voltage across a 10 M resistor because 10 M || 10 M is 5 M; You just change the resistor value to 5 M ohms.