meaning of voltage drop

I've read conflicting definitions for voltage drop. Some sources describe it as 'voltage lost' while others 'voltage difference'. I tend to favor
the latter. My reasoning is as follows: If voltage was lost, then adding say additional resistors to a series circuit would mean that there would be a lack of voltage to continue current.
Why I think voltage drop means 'voltage difference' is for this reason: resistors impede the flow of current, so resistance goes up, while current stays the same as opposed to current staying the same but resistance being very small. The former implies a smaller voltage while the latter does not. Given this smaller voltage, we have a reduction in pressure operating over an electric field. So the time that it takes to push a charge becomes longer. This would then be represented by a voltage drop, or the difference between two points in a circuit over some load. It represents a disparity in voltage over time. However, voltage is not lost. Because the voltage after the resistor, would be represented by the applied voltage.
Does that sound correct? If not, why not?
-- conrad
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conrad wrote:

+10 A----[1 ohm]---+ R1 | B R2 | Gnd C----[1 ohm]---+
I = 5 amps
The voltage measured from A to B is 5 volts. The voltage measured from B to C is 5 volts. Best to call it voltage drop.

That is not correct. The current will continue - it will just be a smaller value.
R1 R2 +10 A----[1 ohm]---[1 ohm]---+ | B | Gnd C----[1 ohm]-------------+
I = 10/3 (~3.33) amps
The voltage measured from A to B is 20/3 (~6.67) volts. The voltage measured from B to C is 10/3 (~3.33) volts.

That is completely bollixed up. Given a fixed voltage, current will _decrease_ when resistance increases. The second clause in your sentence "as opposed to current staying the same but resistance being very small" is completely meaningless.
You've gotten completely confused. Focus on the paragraph below:
A current through a resistor will cause a _voltage drop_ across that resistor. The formula is E = IR Voltage (E) equals current (I) times resistance (R) Learn and use the standard term: _voltage drop_.

No, for the reasons described above. Also, voltage *is* lost.
+ 6 -a-[R1]-b-[R2]-c-[R3]-d-[R4]-e-[R5]-f-[R6]--- Gnd
Assume all resistors above are equal value. If you connect the black lead of your meter to gnd and measure to each of the following points, you will get the voltages indicated:
a 6V b 5V c 4V d 3V e 2V f 1V
You drop 1 volt across each resistor. The electrical energy is lost from the circuit - it is converted to heat energy in the resistor. That heat energy does not convert back to electrical energy in the circuit. Now, that should be the last time you ponder voltage being "lost" across a resistor. From now on, refer to it as voltage drop. It will be more understandable to others, and will help you to avoid confusion.
Ed

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See, I've been told that current remains constant in a series circuit. If I am not mistaken, you are implying that current is variable in a series circuit. Is that correct?
-- conrad
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wrote:

have you ever studied electrical circuits? I suggest that you do that as your assertions are getting a bit foolish. FWIW you need to learn about ohms law and what it implies. The current in a circuit is related by the resistance and the voltage. Thus if the resistance changes for a fixed voltage then so does the current. Forget what you have been told; it is either wrong or you have completely failed to understand it. Get a basic book of electrical theory and learn, or go to a reputable community college (or high school class if you are that young) and get an education.
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I know ohm's law and what it implies. If resistance goes up, and current is constant, then voltage increases. If it goes down and current is constant, voltage decreases. But my book and my electronics teacher state that current in a series circuit is constant. And they also say that voltage in a parallel circuit is constant. You are saying that current in a series circuit is not constant. I think that all this is mostly coated with a bunch of fluffy explanations. Consider this, I have a series circuit with two resistors. So let's say my total resistance is 3. If my applied voltage is say 6, then my current is 2 amps. Now, let's add a resistor. And let's say my total resistance is now 4. 6/4 = 3/2 amps. Why should this proportion keep holding true if I add resistors? That is the essential question that I am trying to figure out. This proportion can be proven through empirical means but what about theoretically. How can it be shown that it should be true? And not just some fluffy approximation that is crafted to 'save the phenomenon' so to speak.
-- conrad
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The current is only constant if your source supply is defined as a 'constant current source' which can exist, but is usually a special case analysis.. Most simple resistor circuit analysis assumes a constant voltage source with the current varying from 0 to the maximum, and dependant upon the total resistance of the circuit.
Formally, if you assume that a voltage source is positive and the resitive drops are negative.
(sum of votage sources) + (sum of resitive drops) = 0
in a series circuit.
This is true, even if the current approaches zero. If the current is zero, then the total resistance must be infinite. The source voltage simply divides across the resistors in proportion to their individual resistances.
Conversely, if the resistance goes to zero, theoretically the current will move to infinity, but in practice this never happens because, in the real world, fuses blow, circuit breakers trip, there is always some internal resistance. A constant voltage supply cannot be maintained against a resistance of zero (a direct short).
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Apparently not. Current IS constant in a circuit of a given fixed voltage input with a given fixed resistance. Change either, and the current changes. Until you understand that relationship intimately (Ohm's Law) you will not understand the concept of voltage drop.
A wire IS a resistor. It just has a very low resistance for shorter lengths. At longer lengths, it is a higher value, and carries with it a higher "voltage drop". All resistances drop voltage.
If you cannot grasp those basic terms, you need to start over and it is as simple as that.
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Yeah, this site http://www.tpub.com/neets/book1/chapter3/1-11.htm says basically the same thing I have been taught about current. Living the impression that it is 'constant'.
-- conrad
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Re read the text; it says the same current flows through all the components, not that the current is constant. This is consistent with what you have been told here that the current is "common". Ohms and Kirchoff's laws apply. You appear to be working too hard with these introductory concepts; is this your first class in the subject?
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I think the "constant" bit means, is that it is the "same" at any point in the circuit. So if you have the simple circuit of a battery, wires and resistor (lightbulb for good ideas) and you measure at all points around the ciruit, at the same moment, then you will find all the measurments of current are the same. Constant in location!
Robert
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---------------- The reference is OK. It uses "same current" which has a different meaning than "constant current". If you add a resistor, the current changes- but wherever you measure it, it is the same value as at any other point in the series circuit.
--

Don Kelly snipped-for-privacy@shawcross.ca
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Yeah... it's kinda a rule around these parts...
It's called Ohm's law, and that's what we have been trying to tell him all along.
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message Gave us:

------------------
No- it's NOT Ohms Law. It is a result of Kirchoff's "Current Law" and is true even where the resistors don't obey Ohm's Law (look up Ohms Law- which is often misquoted).
--

Don Kelly snipped-for-privacy@shawcross.ca
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I just posted the same PDF learning aid I posted in abse some three years ago, and it is certainly not mis-stated.
It is a relationship between the resistance of a circuit, and the voltage applied to it, which results in a specific current through said resistance.
So yes, multiple resistances follow Kirchoff's law, and the voltage on each resistive element (including the wires between them) is referred to as the "drop" on each.
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message Gave us:

Quote: ">>If you add a resistor, the current changes- but wherever you measure it,

That is NOT Ohm's Law which is simply the linear relationship between voltage and current in an "ohmic" element.
Ohm's law has nothing to do with the fact that the current is the same everywhere in the loop. KCL does (even where Ohm's Law is meaningless).
That is all that I was trying to say.
I have no problem with use of Ohm's Law to determine the voltage drop across each element in the circuit -provided that they are linear and actually obey Ohm's Law (q.v).
If all the elements are ohmic, then linearity exists and you can go on, (using the far more basic Kirchoff's Laws) to develop the concept of series and parallel equivalents, Thevenin and all that nice stuff. Without linearity - all these things go out the window leaving Kirchoff's Laws and pain in the butt solutions.
--

Don Kelly snipped-for-privacy@shawcross.ca
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us:

http://www.tpub.com/content/neets/index.htm
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E = I * R
Where
E is Voltage I is Current R is Resistane
It's called Ohm's Law, for the fellow who discovered it.
Google.com can find dozens of tutorials about it.
--
Floyd L. Davidson <http://www.apaflo.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) snipped-for-privacy@apaflo.com
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On Sun, 22 Apr 2007 20:31:51 -0800, snipped-for-privacy@apaflo.com (Floyd L. Davidson) Gave us:

I think he needs to go back to even more basic understandings than that. He should start at the beginning.
http://www.tpub.com/content/neets/index.htm
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