Homework Help: Resistance of the voltmeter

1. Sep 30, 2016

moenste

1. The problem statement, all variables and given/known data
When a particular voltmeter of fixed resistance R, which is known to be accurately calibrated, is placed across the 1.8 kΩ resistor in the diagram below it reads 2.95 V. When placed across the 4.7 kΩ resistor it reads 7.70 V.
(i) Why do these two readings not add up to 12 V?
(ii) Calculate the resistance R of the voltmeter.

2. The attempt at a solution
(ii) Let's find the current of the circuit: I = 12 / 4700 = 2.55 * 10-3 A. Then let's find the current of the 1800 Ω resistor: I1800 = 2.95 / 1800 = 1.63 * 10-3 A. So the resistance of the voltmeter is RV = VV / (I - I1800) = 2.95 / (2.55 * 10-3 - 1.63 * 10-3) = 3226.5 Ω. And it doesn't fit the answer...

I did the same calculation for the second position (voltmeter parallel to the 4700 Ω resistor) and got 1531.3 Ω.

What do I miss here? I consider the voltmeter to be connected in parallel. I also consider 2.95 V and 7.70 V as the reading of the voltmeter and the parallel circuit (so the 12 V is the same for both situations). Maybe the problem assumes that the EMF of 12 V changes to 2.95 V and 7.70 V respectively. Not sure.

P. S. This is a part of the problem, other parts are solved and I got the correct answers. But in case they are required for a better understanding, here is a image of the full problem.

Last edited: Sep 30, 2016
2. Sep 30, 2016

phinds

Your picture is in no way representative of the problem you are trying to solve. First, you have not shown the R of the voltmeter and second, you have combined two situation into one situation in a way that is nonsensical.

3. Sep 30, 2016

Staff: Mentor

Redraw the circuit to highlight the first measurement where the voltmeter of unknown resistance R is placed across the 1.8 KΩ resistor.

The meter reads 2.95 V, so that's the voltage across both R and the 1.8 KΩ resistor. What can you say about the total current $I$? How about $i_1$ and $i_2$?

4. Sep 30, 2016

phinds

EDIT: I misread the problem. Your error is only in not showing the R. gneill has it shown correctly for one of the two cases.

5. Oct 1, 2016

moenste

But the picture is given by the problem. And for both situations I imagined a parallel voltmeter above the 1800 Ohm resistor or above the 4700 Ohm resistor.

Ah, I made a large mistake. My calculations should be correct, but I used the total voltage (12 V) in determing the current and not 12 - 2.95 = 9.05 V. So then we have I = V4700 resistor / R4700 resistor = 9.05 V / 4700 Ohm = 1.93 * 10-3 A. The current of the second resistor is 2.95 / 1800 = 1.63 * 10-3 A. And we find the resistance of the voltmeter to be R = 2.95 V / (Current of 4700 resistor less current of the 1800 resistor) = 2.95 / (1.93 * 10-3 - 1.63 * 10-3) = 10 292 Ω.

I think this should be right.

Update: though I am not sure why (i) "Why these numbers don't add up to 12 V". Any suggestions?

6. Oct 1, 2016

Staff: Mentor

Yes, that looks good.

7. Oct 1, 2016

moenste

Thank you.

Just updated:

8. Oct 1, 2016

Staff: Mentor

Apply what you've learned in other threads about the effects of using a low-resistance voltmeter.

9. Oct 1, 2016

phinds

Yes, and I addressed exactly this in post #4.

10. Oct 1, 2016

moenste

Maybe because we have two resistors with different resistances (1800 Ohm and 4700 Ohm) and even though the resistance of the voltmeter remains the same, but because of the parallel connection the voltage changes. Is this logic correct?

11. Oct 1, 2016

phinds

It's not clear to me what you are saying. Can you be more specific. Talk about the actual circuit, not generalities.

12. Oct 1, 2016

moenste

We have a resistance of 10.3 kΩ for the voltmeter and 1.8 kΩ for the first and 4.7 kΩ for the second resistor.

In the case when the voltmeter is put in parallel to the 1.8 kΩ resistor, we have a total current in the circuit equal to (12 - 2.95) / 4700 = 1.9 * 10-3 A. In the second case the total current is (12 - 7.7) / 1800 = 2.4 * 10-3 A. The currents in the parallel circuit are also different. In the first case they are 1.63 * 10-3 A for the 1800 Ω resistor and 2.7 * 10-4 Ω for the voltmeter. In the second case they are 1.64 * 10-3 A for the 4700 Ω resistor and 7.6 * 10-4 A.

I don't know why the numbers should add up to 12 V. These are two different situations with different currents and voltages.

13. Oct 1, 2016

phinds

Your analysis makes no sense to me but who said they SHOULD add up to 12? The question was why don't, not why they do.

Analyse the VOLTAGES across each of the two resistors in each case of applying the voltmeter, NOT the currents. This question isn't about currents.

EDIT: when I say your analysis makes no sense to me, I don't mean that your calculations are wrong (I didn't check them) I mean that I have no idea why you would do what you did in order to answer the question you are trying to answer. The question is about voltage and you gave an analysis of current.

14. Oct 1, 2016

Staff: Mentor

The voltmeter is a test instrument. Ideal test instruments don't disturb the thing they are trying to measure.

If attaching the voltmeter changes the way the circuit works, altering its operating conditions, then you are not going to get "true" readings of the undisturbed circuit. With the given circuit you can easily work out what the potential differences should be across the resistors if the circuit is operating normally. If your measured values don't add up as expected, then your meter has disturbed the circuit.

Look at the diagram in post #3. What happens to $i_2$ if R is made very large (even infinite)? What happens if R is made smaller?

15. Oct 1, 2016

moenste

That's probably because I don't understand why the 2.95 V and 7.7 V readings don't add up. I think these are two different circuits and therefore the voltages shouldn't add up to 12 V.

16. Oct 1, 2016

phinds

Well, gneil just made a good suggestion about how to analyse the situation. You are correct in your thinking that they are two different situations but it seems like pulling teeth to get you to do a proper analysis.

Draw the circuit EXACTLY as show. Now add the non-ideal voltmeter across one of the resistors. Calculate the voltage across EACH of the resistors.

Now do that putting the voltmeter across the other resistor.

17. Oct 1, 2016

moenste

The larger the resistance of the voltmeter, the lower is the current and the other way around.

So you are saying that in a correct situation the values should add upt to 12 V? And in our case we have a meter that has disturbed the circuit?

18. Oct 1, 2016

cnh1995

The values will not add up to 12V. When you connect the meter across 1.8k resistor, it reads 2.95V. What would be the voltage across the 4.7k resistor in this situation? Similarly, when you connect the meter across the 4.7k resistor, it reads 7.7V. What would be the voltage across the 1.8k resistor?

19. Oct 1, 2016

Staff: Mentor

Yes, and how will that affect the voltmeter readings (the "measured" values)? What should be the value of R in order for the readings to add up to 12 V as they should if the circuit were undisturbed?
Yes! If the voltmeter was ideal it would not disturb the circuit. But this voltmeter is clearly disturbing the circuit.

20. Oct 1, 2016

phinds

Yes, that is exactly what I was trying, in post #16, to get him to see.

21. Oct 1, 2016

moenste

This should be it:

22. Oct 1, 2016

phinds

That's nonsense. You have not actually analyzed the circuit, you've just used the correct voltage across the meter to produce an incorrect value across the other resistor in each case. This is NOT an analysis, it's just more of the same misunderstanding.

EDIT: and by the way my statement that your value across the meter is correct is an assumption. I did not check your calculations for that, but it's irrelevant to what I just said.

23. Oct 1, 2016

moenste

So,
if we had an ideal voltmeter, in that case the values of the voltmeter connected to the resistances (in our case 2.95 V and 7.7 V) would indeed add up to the 12 V. However, in our case we have a voltmeter that is disturbing the circuit and that is why the readings do not add up to 12 V. So the answer would be that the voltmeter is not ideal and it disturbs the readings. I would guess that we better use a voltmeter with a larger resistance, since the current one is relatively low and therefore more current is being "taken" from the circuit, while in a perfect situation a voltmeter should distract as less current from the circuit as possible.

I think this should be it.

24. Oct 1, 2016

phinds

Yes.

25. Oct 1, 2016

moenste

I am sorry, but I don't understand you.

We have a voltmeter that is connected in parallel. I did represent that in my two graphs. In each case the voltmeter gives a different voltage in the parallel circuit. Since in a parallel circuit the voltage is the same -- I did that VVoltmeter = Vresistor 1800 Ohm or VVoltmeter = Vresistor 4700 Ohm.

Since we have a total voltage of V = 12 V, and VParallel circuit = 2.95 V, to find the V4700 resistor we do: V = V1 + V2. 12 = 2.95 + V4700 resistor, where V4700 resistor = 12 - 2.95 = 9.05 V.

We then find the current of the circuit I = V4700 / R4700 = 1.93 * 10-3 A. Then we find the current of the 1800 Ohm resistor: I1800 = VParallel / R1800 = 2.95 / 1800 = 1.64 * 10-3 A. Then we find the current that flows to the voltmeter: 1.93 * 10-3 - 1.64 * 10-3 = 2.9 * 10-4 A. Then Rof the voltmeter = Vparallel circuit / Ivoltmeter = 2.95 / 2.9 * 10-3 = 10 292 Ohm (if we use the original values and not rounded up).

Second situation: 12 - 7.7 = 4.3 V of the 1800 resistor. I = 4.3 / 1800 = 2.39 * 10-3 A. Current of the 4700 resistor: I = 7.7 / 4700 = 1.69 * 10-3 A. Resistance of the voltmeter: 7.7 / (2.39 * 10-3 - 1.69 * 10-3) = 10 259 Ohm.

Same as the answer 10 300 Ohm in both cases.