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

[moenste]

Homework Statement

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.

 

[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?

 

[cnh1995]

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.

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

 

[moenste]

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?

It would be then:

A voltmeter of resistance 20 kΩ is connected to the terminal of the power supply a reading of 5000 V is obtained.

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).

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

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).

 

[moenste]

What do you think is the difference between EMF and voltage?

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.

 

[NascentOxygen]

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.

 

[NascentOxygen]

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).

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

 

[ehild]

(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.

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.

 

[moenste]

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.
View attachment 106874

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

I think for this part:

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.

This explanation is correct:

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).

 

[ehild]

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

I think for this part:

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).

This explanation is correct:

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?

 

[moenste]

voltmeter loads the power supply,

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

current flowing through the circuit causes voltage drop along the internal resistance of the battery.

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

voltage across the terminals

Terminals means on the poles? + and -?

The voltage across the terminals is less then the EMF of the battery - by how much?

Sorry, no idea.

 

[cnh1995]

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

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.

Terminals means on the poles? + and -?

Yes.

Sorry, no idea

Didn't you already do that in your solution?

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

Draw a diagram and see how internal resistance and voltmeter resistance are connected.

 

[ehild]

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

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.

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

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.

Terminals means on the poles? + and -?

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.

Sorry, no idea.

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.

 

[ehild]

Voltmeter should not load the power supply.

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.

 

[moenste]

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.

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?

Didn't you already do that in your solution?

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

Draw a diagram and see how internal resistance and voltmeter resistance are connected.

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

 

[moenste]

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?

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.

 

[ehild]

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?

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

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

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.

 

[ehild]

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.

Give it in terms of I and r.

 

[moenste]

Give it in terms of I and r.

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.

 

[cnh1995]

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.

Yes. I forgot to write "ideal"..

 

[ehild]

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.

In general, the voltage across the internal resistance of the battery is Ir, and the potential difference across the terminals of the loaded battery is U=E-Ir, which is equal to the voltage across the external resistance, IR.

 

[ehild]

Yes. I forgot to write "ideal"..

Yes, but it was not an ideal voltmeter.

 

[David Lewis]

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

That’s true in general. In this situation, E is fixed because it is an independent variable (given to you in the problem statement). Conversely, r is a dependent variable (depends on the values of E and R).