Hello, I was just going through an old quiz and found part of this question quite confusing.

I understand that the perfect voltmeter has infinite resistance and has no current passing through it but I don't really understand the option "it has no voltage drop across it"

Does voltage drop across a voltmeter? If the definition of a perfect meter is that it does not disturb the circuit then wouldn't a voltage drop be a disturbance?

Cheers!

 

This is a quiz to test your knowledge, right?
So, check off the boxes that apply.
To help you understand why the some apply and some do not apply, ask yourself what is a "perfect' ammeter.

 

This is a quiz to test your knowledge, right?
So, check off the boxes that apply.
To help you understand why the some apply and some do not apply, ask yourself what is a "perfect' ammeter.

yes! I did that already and I know the answers are it has infinite resistance and it has no current passing through it, I just found the "no voltage drop across it" interesting. So if i simulate a circuit an place a voltmeter across a resistor, is voltage "dropping" across the voltmeter? It just seems to contradict the point that it is not supposed to disturb the circuit. In my head the reason it doesn't disturb it is because in parallel circuits the voltage remains the same and the voltmeter would be parallel to the resistor but this feels like i still might be misunderstanding something.

 

You are correct, both perfect ammeter and voltmeter are not supposed to disturb the circuit.
So rather than overthink the problem, look at the question from the perspective of which two answers are characteristic of an ideal voltmeter and which two are characteristic of an ideal ammeter.

 

You are correct, both perfect ammeter and voltmeter are not supposed to disturb the circuit.
So rather than overthink the problem, look at the question from the perspective of which two answers are characteristic of an ideal voltmeter and which two are characteristic of an ideal ammeter.

Ok thank you! I often find myself over thinking these things and ending up confused. The perfect ammeter is easier for me to understand as if it has no resistance then it behaves like a wire and no voltage drops across a wire.

So when I read the value on a voltmeter does that mean that voltage IS dropping across the voltmeter? I thought it was just a reference value based on 2 nodes. I thought voltage dropping across something meant there was some kind of resistive element to it?

 

You are misinterpreting the meaning of the phrase "voltage drop".

We tend to use the phrase when analyzing a circuit with components in series, for example, a voltage divider consisting of two resistors in series. If it would help, you can remove the phrase "voltage drop" from your electronics vocabulary and replace it with the correct phrase "voltage difference".

Here is some additional food for thought.
An ideal ammeter would have zero "voltage difference" between its two terminals. But we know that in order for current to flow through any circuit there must be a voltage difference. So how can there be any current flowing?
This apparent contradiction can be understood if we look at Ohm's Law:

I = V / R

In an ideal ammeter, V = 0 and R = 0.
Hence I = 0 / 0 is mathematically indeterminate. The current I in this case will be determined by factors external to the ammeter itself.

 

You are misinterpreting the meaning of the phrase "voltage drop".

We tend to use the phrase when analyzing a circuit with components in series, for example, a voltage divider consisting of two resistors in series. If it would help, you can remove the phrase "voltage drop" from your electronics vocabulary and replace it with the correct phrase "voltage difference".

Here is some additional food for thought.
An ideal ammeter would have zero "voltage difference" between its two terminals. But we know that in order for current to flow through any circuit there must be a voltage difference. So how can there be any current flowing?
This apparent contradiction can be understood if we look at Ohm's Law:

I = V / R

In an ideal ammeter, V = 0 and R = 0.
Hence I = 0 / 0 is mathematically indeterminate. The current I in this case will be determined by factors external to the ammeter itself.

Ok thank you for that, I like voltage difference more because it is always in reference to some other value and that is what i think of.

In your example for the ideal ammeter, I am a bit confused because it seems like an ideal ammeter is just a wire that can tell you what current is flowing through it. If an ammeter is a 2 terminal device and current flows through it shouldn't there be a voltage drop or is it similar to something like steady state in an inductor? But an ideal component is just theoretical right so like you say it has to operate the way it does due to the other components in the circuit.

I think ideal components are quite confusing because when you try to simulate or build circuits they don't behave like the ideal components!

 

The ideal ammeter would have no voltage drop across the sensor. It would detect the magnetic field of the moving charges in the conductor instead of needing a voltage drop potential in series with the conductor as a proxy for the direct magnetic field measurement. A version of these 'ideal' (low load energy consumer during measurement) sensors are called 'Hall' devices.

 

The ideal ammeter would have no voltage drop across the sensor. It would detect the magnetic field of the moving charges in the conductor instead of needing voltage drop potential in series with the conductor as a proxy for the direct magnetic field measurement. A version of these 'ideal' sensors are called 'Hall' devices.

thanks for the info, I will look in to the 'Hall' devices!

 

Ok thank you for that, I like voltage difference more because it is always in reference to some other value and that is what i think of.

In your example for the ideal ammeter, I am a bit confused because it seems like an ideal ammeter is just a wire that can tell you what current is flowing through it. If an ammeter is a 2 terminal device and current flows through it shouldn't there be a voltage drop or is it similar to something like steady state in an inductor? But an ideal component is just theoretical right so like you say it has to operate the way it does due to the other components in the circuit.

I think ideal components are quite confusing because when you try to simulate or build circuits they don't behave like the ideal components!

Hello there,

Now that you know what is going on i guess we can discuss this further.

In your example for the ideal ammeter, I am a bit confused because it seems like an ideal ammeter is just a wire that can tell you what current is flowing through it. If an ammeter is a 2 terminal device and current flows through it shouldn't there be a voltage drop or is it similar to something like steady state in an inductor? But an ideal component is just theoretical right so like you say it has to operate the way it does due to the other components in the circuit.

Only in the simplest real circuit does an ammeter drop voltage.
A theoretically ideal ammeter does not drop any voltage for one.
But also, in a more complicated real life circuit you can construct a zero voltage drop ammeter so it is not just in the theoretical. The voltage drop may not be 'perfectly' zero but it will be so close to zero that you would not be able to measure it, and thus in the real life rationale it has zero voltage drop also.
The details of such an apparatus is very interesting BTW maybe we will talk about it at some point.
But i am sure for the lack of that, you can imagine a short wire that has a tiny resistance like 0.001 Ohms and you measure the voltage drop across that with a sensitive voltmeter. It's a tiny voltage drop but the sensitive meter picks up the voltage just fine. The voltage is then converted into a current using Ohm's Law. If you take this farther, you get down to maybe 1 micro ohm which may be hard to read though with the volt meter, so we then turn to control theory to get a true zero voltage drop ammeter.
But you may have seen some of the current shunts available in the marketplace. They are often rated at 50mv drop at some current like 10 amps. So when your volt meter reads 50mv you have 10 amps flowing.
We could talk about the controlled zero voltage drop ammeter once you get past this problem.

 

thanks for the info, I will look in to the 'Hall' devices!

There will still be a tiny tiny voltage drop, but in most applications it is small enough to ignore.

 

There will still be a tiny tiny voltage drop, but in most applications it is small enough to ignore.

Sure, nothing is perfect but the loss won't come from series resistance with a non-insertion type current sensor. The extracted energy is from the magnetic field so the sensor is effectively in parallel at that spot extracting a tiny bit of electrical energy from the circuit potential energy source.

 

Sure, nothing is perfect but the loss won't come from series resistance with a non-insertion type current sensor. The extracted energy is from the magnetic field so the sensor is effectively in parallel at that spot extracting a tiny bit of electrical energy from the circuit potential energy source.

That's funny i could have sworn i said "a tiny tiny voltage drop".
But if you want to state it as an energy loss that is fine.
Also, i was thinking more of the kind you buy that is really a hall effect current sensor not just a hall effect device alone because they are readily available from parts distributors. They have a small amount of conduction metal and are self contained. They are inserted in series with the wire to be measured. They are very very close to zero Ohms though even with that conduction metal portion.

 

That's funny i could have sworn i said "a tiny tiny voltage drop".
But if you want to state it as an energy loss that is fine.
Also, i was thinking more of the kind you buy that is really a hall effect current sensor not just a hall effect device alone because they are readily available from parts distributors. They have a small amount of conduction metal and are self contained. They are inserted in series with the wire to be measured. They are very very close to zero Ohms though even with that conduction metal portion.

I used energy because it's a more exact term to describe what's being measured.

The metal conductor mounting part on the sensor device is not needed for the Hall effect (and other contact-less current sensors) to measure current so you can eliminate even that 'tiny tiny voltage drop' by using a sensors like these that sample magnetic energy.

https://forum.allaboutcircuits.com/...c-controlled-battery-array.32879/post-1482098

 

I used energy because it's a more exact term to describe what's being measured.

The metal conductor mounting part on the sensor device is not needed for the Hall effect (and other contact-less current sensors) to measure current so you can eliminate even that 'tiny tiny voltage drop' by using a sensors like these that sample magnetic energy.

https://forum.allaboutcircuits.com/...c-controlled-battery-array.32879/post-1482098

I think it is still arguable that any real energy extraction must cause a voltage drop no matter how small. Sure it may be smaller than with a metal strip for example, and it may be much smaller.

There is a cure for them all though if you want to really talk about a zero Ohm zero energy detector. I'll hold off on that for now though.

 

I think it is still arguable that any real energy extraction must cause a voltage drop no matter how small. Sure it may be smaller than with a metal strip for example, and it may be much smaller.

There is a cure for them all though if you want to really talk about a zero Ohm zero energy detector. I'll hold off on that for now though.

Sure it causes some effect because all measurements require a quantum state being measured of some sort. It's possible to limit the back-action but most measurements cause much stronger back-action than is required by the laws of quantum mechanics.
https://en.wikipedia.org/wiki/Back_action_(quantum)

 

Sure it causes some effect because all measurements require a quantum state being measured of some sort. It's possible to limit the back-action but most measurements cause much stronger back-action than is required by the laws of quantum mechanics.
https://en.wikipedia.org/wiki/Back_action_(quantum)

Well i agree that the effect will be much less without resorting to quantum physics. Even in classical physics you never get anything for nothing

With measurements it is a little different though and we have not yet exhausted all the possibilities. I think it is possible to make a measurement of current that requires so very little of the power in the wire that in theory it could be made smaller than a single particle but maybe to stay real we might be able to get it down to one particle lost per millisecond or something like that

 

But also, in a more complicated real life circuit you can construct a zero voltage drop ammeter so it is not just in the theoretical. The voltage drop may not be 'perfectly' zero but it will be so close to zero that you would not be able to measure it, and thus in the real life rationale it has zero voltage drop also.
The details of such an apparatus is very interesting BTW maybe we will talk about it at some point.

That sounds really interesting, can you please advise me where I could learn more about this at a beginners level?

But you may have seen some of the current shunts available in the marketplace.

I hadn't heard of these until now, I've been reading about them this morning, there seems to be lots of uses for them, I will most likely have to learn about these in my course at some point.

 

I don't need to use any quiz. The perfect voltmeter is the Fluke 289.... *love* that meter

 

That sounds really interesting, can you please advise me where I could learn more about this at a beginners level?



I hadn't heard of these until now, I've been reading about them this morning, there seems to be lots of uses for them, I will most likely have to learn about these in my course at some point.

Hi,

What is it exactly you want to learn about?