Discussion in 'General Electronics Chat' started by Vorador, Oct 15, 2014.

Oct 5, 2012
87
1

Sorry for the poorly drawn diagram...

In this circuit, R2 has one terminal open. Yet, I found by accident that there is a voltage between the open end of R2 and positive terminal of the battery. I don't understand why my meter is giving a measurement at all when that resistor has one end floating and doesn't even form a complete loop?

2. ### sirch2 Well-Known Member

Jan 21, 2013
1,008
351
What is the value of R1 and R2, what is the resistance of your meter, what is the reading?

3. ### ddurgin New Member

Mar 2, 2011
4
1
You don't need a complete circuit loop for voltage to be present; for example, you can measure the voltage of an unconnected battery with a volt meter on the bench correct? If I understand correctly, the left end of R2 is connected to the (-) terminal of the battery and the right end is connected only to the meter. In this case there is no current through R2 and therefore the voltage on (either end) of R2 relative to the (+) battery terminal will equal the battery voltage. Is that what you are reading?

4. ### crutschow Expert

Mar 14, 2008
13,470
3,356
Your meter connection forms a complete loop with the battery.

5. ### paulktreg Distinguished Member

Jun 2, 2008
621
124
I presume your voltmeter is digital so it could have an impedance of 1Mohm or above.

We don't know the values of R1 or R2 but if they are low, let's say both are <1kohm, then you are effectively putting a 1Mohm resistor in series with a low value resistor across your battery and you will read virtually the full battery voltage on your meter. If R1 and R2 are 1Mohm then your meter is effectively putting two 1Mohm resistors across the battery and you'll read half your battery voltage on the meter.

The ideal voltmeter has infinite resistance and should read zero volts because it doesn't cause a current to flow but in the real world this just doesn't happen. Have a look at the technical specification for your meter and you'll see it introduces a resistance and treat it as a voltage divider that reads the appropriate voltage.

6. ### profbuxton Member

Feb 21, 2014
233
68
What you are measuring is the battery volts. Connect your meter on the other side of R2 and if you follow the circuit you will see that your meter is across the battery. Since R2 is "open" there is no "voltage drop" across it and your meter is measuring the full battery volts as your circuit is shown.

7. ### MrChips Moderator

Oct 2, 2009
12,623
3,451
When you connect the voltmeter the open end of R2 is no longer open. The voltmeter has internal resistance and hence completes the circuit.

profbuxton is correct. You are now measuring battery voltage with R2 in series.

In theory, the values of R1 and R2 are irrelevant within certain limits.

8. ### GopherT AAC Fanatic!

Nov 23, 2012
6,298
4,020
If the resistance (impedence) of the volt meter is infinite, both ends of R2 will be at the same voltage. Anything less than infinite, you can calculate some voltage drop and you can consider r2+meter impedence as a parallel loop of your circuit.

9. ### Lundwall_Paul Active Member

Oct 18, 2011
222
19
Why would you place a resistor (R2) on the end of the meter probe?

10. ### MrChips Moderator

Oct 2, 2009
12,623
3,451
Just for fun to see what happens?

Lundwall_Paul likes this.
11. ### MrCarlos Active Member

Jan 2, 2010
400
134
If you measure, as you mention, between point A and point B your meter should not indicate anything.
But if you measure between point A and point C the meter will display the battery voltage.
Else, if you measure between B and C Your meter indicate some reading; depending on the value of R2 and the impedance of your meter

File size:
6.9 KB
Views:
37

Oct 5, 2012
87
1
I didn't do it by design. It was in a slightly more complex circuit where I was checking voltages around and just found this oddity. Then out of curiosity, I just made a circuit same as in the diagram to play around with. The battery voltage read 9.42 volts on my meter and R1 and R2 are both 470k. When I connect my voltmeter as shown in the figure, it gives me 6.39 volts.

I have a digital meter and I remember reading it either has 1 or 2M internal resistance. I'll check the manual once again to make sure later.

I did think it was due to the finite resistance of meter, but I was puzzled why the drop is 6.39 volts specifically. How can I calculate this mathematically? If there's a meter resistance + R2 path across the battery, shouldn't the total voltage of meter+R2 still equal the battery voltage, since they are across from it? I wonder if the meter is reading the voltage across its own resistance i.e 6.39 volts, and the other 3 volts are being taken by R2?

Thanks a lot everyone! I really appreciate all the replies I've got! You're an extremely helpful lot!

Oct 5, 2012
87
1
MrCarlos, I'm really very sorry to tell you that there isn't a break between R1 and R2. They are both connected to the the negative terminal of the battery. I should have drawn better (I was really sleepy at the time).

However, the cases you've mentioned still are true but for different points (B & C are the same points in the "correct" circuit).

Thank you so much, MrCarlos!

14. ### MrChips Moderator

Oct 2, 2009
12,623
3,451
You have enough information to calculate the internal resistance of your voltmeter.
See if you can work it out.

Oct 5, 2012
87
1
I used the voltage-divider rule and got 991.2 kilo-ohms. Is that reasonable?

16. ### MrChips Moderator

Oct 2, 2009
12,623
3,451
Sounds reasonable.

Oct 5, 2012
87
1
Thanks a ton everyone! You've been a great help really! Thanks so much!

18. ### wayneh Expert

Sep 9, 2010
12,368
3,224
My meter is ~1MΩ on its 20V scale, so yes, your estimate is very reasonable.