Revered Members,
I can't understand the term Drop in Voltage Drop. Drop means decrease. Is Voltage decreases when current flows through the resistor? Why the name voltage drop instead of simply Voltage?

 

Say 1A of current is flowing through a 1Ω resistor. The voltage at the positive facing side will be 1V higher than the voltage at the end on the negative facing side. So, the resistor "dropped" 1 volt

The volt was there going into the resistor, but it wasn't coming out, so the resistor must have dropped it.

 

Exactly - great explanation.

The voltage at the "input" of the resistor is always higher than the voltage at the "output" of the resistor, with respect to ground. The voltage was reduced (or indeed, dropped), because of the resistor.

I'm sorry to have to say that this is not quite right. The example you were originally given was a good clear one, but your generalisation of it is at best over-simplified. It can only apply to very simple DC circuits, fed by positive DC supplies referenced to earth. These are common in elementary circuit theory problems, but other arrangements exist.

You might start to see the problem better if you think about the following issues.

• Does "higher" mean having a more positive value, or a greater absolute value, neglecting the sign?
• What happens with a negative supply, or if there is more than one supply, some of which may be negative?
• What happens in an AC circuit? (Voltage drops also occur with AC: for example the filament lamps in old-fashioned sets of Christmas tree lights are often wired in series.)

Depending on whether the supply or supplies feeding a resistor is/are positive or negative or AC, and on other voltages that may be present in a complex circuit, the voltage at either end of a particular resistor may be "higher" with respect to ground. This does not necessarily tie in with which end of the resistor might be considered the "input" or "output".

 

I'm sorry to have to say that this is not quite right. The example you were originally given was a good clear one, but your generalisation of it is at best over-simplified. It can only apply to very simple DC circuits, fed by positive DC supplies referenced to earth. These are common in elementary circuit theory problems, but other arrangements exist.

You might start to see the problem better if you think about the following issues.

• Does "higher" mean having a more positive value, or a greater absolute value, neglecting the sign?
• What happens with a negative supply, or if there is more than one supply, some of which may be negative?
• What happens in an AC circuit? (Voltage drops also occur with AC: for example the filament lamps in old-fashioned sets of Christmas tree lights are often wired in series.)

Depending on whether the supply or supplies feeding a resistor is/are positive or negative or AC, and on other voltages that may be present in a complex circuit, the voltage at either end of a particular resistor may be "higher" with respect to ground. This does not necessarily tie in with which end of the resistor might be considered the "input" or "output".

And this is why I'll never become a school teacher. I'll remove my original entry.

To your point though: "Does "higher" mean having a more positive value, or a greater absolute value, neglecting the sign?". My professor has always taught me that a higher voltage is always considered with respect to the common reference, which would mean that a higher voltage always corresponds to a higher numerical value. Ie. +100V is, naturally, a higher voltage than -50V, because |+100| > |-50|.

 

And this is why I'll never become a school teacher. I'll remove my original entry.

Too bad. Your original explanation was a good one for DC.
What Adjuster is referring to is a more generalized one to take into account both DC and AC. I don't think his concept is any better than yours.

 

Too bad. Your original explanation was a good one for DC.
What Adjuster is referring to is a more generalized one to take into account both DC and AC. I don't think his concept is any better than yours.

That is not quite what I meant: the reference to AC was an afterthought, which on reflection might have better not been made to avoid confusing matters. What I was really objecting to was the statement about the voltages referenced to ground.

The voltage at the "input" of the resistor is always higher than the voltage at the "output" of the resistor, with respect to ground.

It may be better to say the end of the resistor receiving a positive conventional current flow will be at a more positive potential than will the other end. These voltages may or may not bear a defined relationship to ground.

Perhaps this is a question of semantics, but in one sense a voltage of say -10kV is "higher" than a voltage of +5V.

 

And this is why I'll never become a school teacher. I'll remove my original entry.

To your point though: "Does "higher" mean having a more positive value, or a greater absolute value, neglecting the sign?". My professor has always taught me that a higher voltage is always considered with respect to the common reference, which would mean that a higher voltage always corresponds to a higher numerical value. Ie. +100V is, naturally, a higher voltage than -50V, because |+100| > |-50|.

@tommydyhr: Don't worry about it. I never have been and never will be a teacher either, I'm just a sick old man sitting at home making pedantic little points while I wait for the Grim Reaper. I certainly cannot speak Danish, whereas you clearly have very good English.

The point about what constitutes a "high" voltage is one of the things I was trying to get at. For circuit theory problems you must of course use the accepted convention, but in some contexts (such as electric shock hazard) one might regard say -10kV with respect to ground as a "higher" voltage than +5V. In such cases, it is the absolute voltage that counts.

The other thing that I meant to point out (but failed to) is that the voltages in some circuits do not necessarily have any defined relationship to ground.

Perhaps next time I will think about it for longer before trying to be clever.

 

I attempted to clear up the confusion of "voltage at" and "voltage drop" some months ago for another beginner. I believe the discourse was a successful one.

A voltage measurement will always by definition be a "voltage drop". It is the potential difference measured between two points. This is independent of the amplitude and direction on any implied current.

When we use the expression "voltage" or "voltage at" we generally accept this to mean with reference to EARTH potential.

 

I attempted to clear up the confusion of "voltage at" and "voltage drop" some months ago for another beginner. I believe the discourse was a successful one.

A voltage measurement will always by definition be a "voltage drop". It is the potential difference measured between two points. This is independent of the amplitude and direction on any implied current.

When we use the expression "voltage" or "voltage at" we generally accept this to mean with reference to EARTH potential.

Perhaps you could provide a link to that, I fear my efforts may merely have muddied the waters.

Getting even relatively simple concepts clear in the mind and then expressing them clearly and unambiguously is an art in itself. Despite a comment in an earlier posting, I am no educator. It has always been clear to me that I lack certain necessary skills, perhaps especially where it comes to putting myself in the student's position so that I do not take things for granted that should not be.

When I try my hardest to be clear, the most likely result seems to be over-complication and even greater confusion. This is galling to me as I find myself with a certain amount of experience which I would like to pass on. I am nearing the end of my career (and possibly nearing the end of more than that, as you might guess from my familiarity with certain equipment). Never mind, I shall continue to try, perhaps practice will help.

 

Here is the like to a previous discussion on "voltage drop"

http://forum.allaboutcircuits.com/showthread.php?t=56372

 

If I may belabor the point, here is a demonstration which I had to deliver to a PhD candidate in Physics.

Here is a 9V battery. What is the voltage at the +ve terminal? The answer is not 9V nor is it 0V. This answer is "unknown".

There is always a potential at every point in space, including the terminal of the battery. There is even a potential when measured with respect to ground.

In this case, the black probe is not connected to GROUND. The +ve terminal could very well be sitting at 1000V and we may not realize this.

 

Wonderful. That was one of the points that I wanted to make, but failed to. In an ideal world, could we hope that it might be tackled before postgraduate level?

Edit: Actually. this is a better thing than what I had intended. As you say, every point is at some potential, which can be considered referred to earth potential. In practical situations however, we may be interested in whether the potential with respect to ground is defined in a way that is not readily changed, or whether it is much less predictable, perhaps easily altered by a minuscule flow of charge.

The potentials on the terminals of the battery on the table fall into one of those categories.

 

But now the student thinks, OK, I shall ground the black probe, and then I shall be able to measure the potentials at each terminal (with a real meter, not a perfect thought-experiment one).

He or she does so, and taking the difference of the two potentials to be the battery voltage, concludes that the battery is completely dead.

Why may this be wrong? (This is a beginners' question: preferably to be left for a while for beginners to think about it.)

 

I do get all of the points that's been made in this thread, but I doubt that I'd ever be able to explain them to anyone who doesn't have at least some kind of electrical background.

The more "basic" a subject is, the easier it is to confuse people, by being either too thorough, or simply just overcomplicating things.

I'll just stick to the, quite precise, definition found at answers.com:

(′vōl·tij ′dräp)

(electricity) The voltage developed across a component or conductor by the flow of current through the resistance or impedance of that component or conductor. ​

 

But now the student thinks, OK, I shall ground the black probe, and then I shall be able to measure the potentials at each terminal (with a real meter, not a perfect thought-experiment one).

He or she does so, and taking the difference of the two potentials to be the battery voltage, concludes that the battery is completely dead.

Why may this be wrong? (This is a beginners' question: preferably to be left for a while for beginners to think about it.)

Yes, that is a very good experiment which we should expect a beginner to encounter.
We should be prepared for a proper explanation.

 

I do get all of the points that's been made in this thread, but I doubt that I'd ever be able to explain them to anyone who doesn't have at least some kind of electrical background.

The more "basic" a subject is, the easier it is to confuse people, by being either too thorough, or simply just overcomplicating things.

I'll just stick to the, quite precise, definition found at answers.com:

(′vōl·tij ′dräp)

(electricity) The voltage developed across a component or conductor by the flow of current through the resistance or impedance of that component or conductor. ​

That definition leaves out the part bout "The power lost with that voltage drop is dissipated by heat".

When calculating how big of a transistor you'll need to drive a motor, you calculate or measure the voltage dropped across the transistor, then multiply that value with the current through the transistor to find out how much heat is dissipated, so you know how large of heatsink you will need. Well, larger heatsink or different transistor.

5V @1A is 5 W, when that current goes into a 1 Ohm resistor, there will be 4V@1A on the other terminal, the remaining 1 Volt dropped x 1 Amp of current means 1 Watt of power will be turned into heat by the resistor due to dropping that voltage.

Voltage drop is a very common measurement in electronics, it applies to about everything when looking at Thermal Characteristics (i.e. Why does this thing get so hot?). From diodes to LEDs to transistors, resistors, and even simple wire, over a long distance, will dissipate power through them as heat.

Perhaps if you think of V*I=W, Volts*Amps=Watts, you'll understand more about voltage drops, the voltage doesn't simply vanish, that energy merely changes form to heat.

 

But now the student thinks, OK, I shall ground the black probe, and then I shall be able to measure the potentials at each terminal (with a real meter, not a perfect thought-experiment one).

He or she does so, and taking the difference of the two potentials to be the battery voltage, concludes that the battery is completely dead.

Why may this be wrong? (This is a beginners' question: preferably to be left for a while for beginners to think about it.)

To add more intrigue to the problem, what would you observe if you were to use two voltmeters to simultaneously measure the voltage at each terminal of a 9V battery while both black leads of the voltmeters were connected to EARTH GROUND as shown in the diagram?

 

And this leads to the next step... What would you observe if you removed the GROUND connection but connected the two black leads?

 

It might be useful to say something about the meters, but maybe that would spoil the problem for all but the slowest.

 

So there cannot be any voltage on an airplane??? We had a drawing on the wall in the avionics shop at Midway that showed an airplane towing a bucket of dirt with a "Ground" wire going to it. The verbiage on the drawing said, "What is ground?"