# Difference in EMF of 5000 V and a voltmeter reading of 40 V

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1. Oct 1, 2016

### moenste

1. The problem statement, all variables and given/known data
A power supply used in a laboratory has an EMF of 5000 V. When, however, a voltmeter of resistance 20 kΩ is connected to the terminal of the power supply a reading of only 40 V is obtained.

(a) Explain this observation.
(b) Calculate (i) the current flowing in the meter and (ii) the internal resistance of the power supply.

Answers: (b) (i) 2 mA, (ii) 2.48 MΩ

2. The attempt at a solution
(b) (i) I = V / R = 40 / 20 000 = 2 * 10-3 A.

(b) (ii) E = I (r + R) → 5000 = 2 * 10-3 (r + 20 000) → r = 2 480 000 Ω.

(a) I am not sure why EMF and the PD are different. Maybe it has something to do with the differences between the general differences between the EMF and the PD. As far as I know, their difference is in the fact that PD is work done to move a charge between two points in a circuit and EMF is the total work done to move charge throught a complete circuit. And the formula is also different: V = I R, (R = total resistance), E = I (R + r), (R + r) = total external and internal resistance. But not sure whether this is applied to this situation.

2. Oct 1, 2016

### phinds

Forget EMF for a moment and consider this: if the power supply was rated at 5000V without any mention of EMF would your answers be any different?

3. Oct 2, 2016

### cnh1995

You are on the right track. Apply this to the situation in the problem. A circuit diagram will help.

4. Oct 2, 2016

### moenste

It would be then:
So the current would be I = V / R = 5000 / 20 000 = 0.25 A. That is a larger number than 0.002 A calculated.

Maybe the reason in the difference of the EMF of 5000 V and the real 40 V is in the fact that the power supply is not ideal? This is what I got from a different thread. And because the power source (battery) is non-ideal, it has internal resistance, which we look for in (b).

Thank you, but I think you misunderstood me :). I solved (b), but I'm not sure on the theory part (a). Maybe the reason in the difference of the EMF of 5000 V and the real 40 V is in the fact that the power supply is not ideal? This is what I got from a different thread. And because the power source (battery) is non-ideal, it has internal resistance, which we look for in (b).

5. Oct 3, 2016

### moenste

Volate (PD) is work done to move a charge between two points in a circuit and EMF is the total work done to move charge throught a complete circuit. And the formula is also different: V = I R, (R = total resistance), E = I (R + r), (R + r) = total external and internal resistance.

6. Oct 3, 2016

### Staff: Mentor

It is clear that moenste is designating the open-circuit potential as "EMF" and the terminal potential measured by the voltmeter as "P.D." This may be somewhat unconventional, but is allowable.

Thread cleanup: I have removed posts generated by a misreading of this nomenclature.

7. Oct 3, 2016

### Staff: Mentor

That is correct. When current is drawn it causes voltage loss across the internal resistance.

8. Oct 3, 2016

### ehild

Yes, you are on the right track. See figure. The box represents the real power source, an ideal source with EMF E, and the internal resistance r, connected in series. The current flowing through the circuit causes potential drop across the internal resistance, so you get voltage across the terminals which is lower than the EMF.
In this problem, the resistance of the voltmeter is the load. The voltmeter measures the voltage across its internal resistance.

9. Oct 3, 2016

### moenste

Thank you. Forgot to put the "solved" sign :).

I think for this part:
This explanation is correct:

10. Oct 3, 2016

### ehild

It is correct, but not complete. Add that the voltmeter loads the power supply, and the current flowing through the circuit causes voltage drop along the internal resistance of the battery. The voltage across the terminals is less then the EMF of the battery - by how much?

11. Oct 3, 2016

### moenste

Doesn't the voltmeter just show how much voltage is there?

Why does it causes voltage drop along the internal resistance of the battery?

Terminals means on the poles? + and -?

Sorry, no idea.

12. Oct 3, 2016

### cnh1995

Voltmeter should not load the power supply.
It does show voltage but that voltage is less than the actual open circuit voltage (emf) of the battery.
Yes.
Didn't you already do that in your solution?
Draw a diagram and see how internal resistance and voltmeter resistance are connected.

13. Oct 3, 2016

### ehild

The voltmeter shows the voltage across its terminals. It depends on the loading resistance and also the on the internal resistance of the battery. You said that E=I(r+R), and you get the current I=E/(r+R) flowing in the circuit. That current causes voltage on the resistor R, U=IR according to Ohm's Law. That is the voltage measured by a voltmeter. The same current flows through the internal resistance r: the voltage across it is Ir. The voltages add up, and their sum is equal to the EMF, according to your equation E = I(r+R) = Ir + IR. The first term is the voltage across the internal resistance, r, the second is the voltage across the external resistance R. As $I=\frac{E}{r+R}$, the voltage read by the voltmeter is $U=RI=R\frac{E}{r+R}$. You see, it depends on both resistances.

The current flows through the internal resistance of the battery, and a current I flowing through a resistor means that there is voltage drop across the resistor. It is Ohm's Law. Or it is better to say, that the current can flow through the resistor only when there is potential difference between the terminals of the resistor.
See https://en.wikipedia.org/wiki/Terminal_(electronics)
In case of a battery, they are the + and - poles. But the elements of an electric circuits have some legs which can be connected to the other parts of the circuits. Batteries, resistors, capacitors, inductors, bulbs, diodes have two terminals.

The current I flows through the internal resistance r of the battery. What is the voltage across r? Read the first paragraph of my post.

14. Oct 3, 2016

### ehild

An ideal voltmeter does not load the power supply, but a real one, with finite internal resistance does. The battery is loaded with the internal resistance of the voltmeter, and the voltmeter reads the voltage across its own internal resistance.

15. Oct 3, 2016

### moenste

Load means like power? But isn't it the battery that is supposed to do that? And voltmeters just show how much voltage is flowing in the circuit?

In that case I don't quite understand the answer, sorry.

The higher is the (r + R) the higher is the E? E = I (r + R).

16. Oct 3, 2016

### moenste

The voltage across the terminals of the voltmeter (they are the ends of the cables) is less than the EMF of the battery by 5000 - 40 = 4960 V.

17. Oct 3, 2016

### ehild

Load is something you connect to the power supply. A light bulb, a heater, a hair dryer and so on. From electric point of view they are resistances.
The voltmeter shows the voltage across two points. Voltage does not flow. It is the current that flows

Usually you have a power source with a given EMF. If you buy a battery, the EMF is written on it: 1.5 V , or 9 V.... If you connect something that has resistance R, the current flowing in the circuit is I=E/(R+r).
I suggest you to do some experiments with a battery and small light bulbs.

18. Oct 3, 2016

### ehild

Give it in terms of I and r.

19. Oct 3, 2016

### moenste

Well, most of the resistance is supplied by the battery (2 480 000 Ω) and the voltmeter has a resistance of 20 000 Ω.

And the current flowing through the voltmeter is 2 * 10-3 A.

20. Oct 3, 2016

### cnh1995

Yes. I forgot to write "ideal"..