# meaning of voltage drop

• posted

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?

• posted

------------ Several things don't sound correct. Voltage drop corresponds to voltage difference across the resistor of concern. It is the product of current and resistance. Higher resistance with a given current results in an increase in voltage drop not a decrease as you imply. Higher current through a given resistance does the same. In both cases, the "pressure drop" increases. For most situations, resistance and current are two independent things. All voltages are voltage differences- that is the voltage at one point with respect to another point. In a resistor the voltage drop is + in the direction of conventional current - that is the voltage at the ingoing terminal is + with respect to the outgoing terminal so the voltage is said to "drop". Note that I am not mentioning charge here-for most circuit analysis it is not necessary to refer to it but stick to current which is defined (conventionally) as the rate of change of +charge with time without giving a hoot as to the actual charge carriers (+ or -) Dont get concerned about the time to push a given charge. The electrical pressure supplied by the source so the pressure typically doesn't change. wiring resistance (assuming constant wire temperature) is constant. The current can and will change and depends on the source "pressure" and the total resistance of load and wiring. Increasing wire resistance for a given load will reduce the current-In a way the time to push a given quantity of charge will be longer but don't deal with it that way- consider current which is the rate of change of charge with time (stick to conventional current, not electron current as it is easier in the long run -particularly when you get to AC where the electrons are simply wobbling about and going nowhere). Don't get hung up with what charge carriers such as electrons are doing.

If you have a constant voltage source then this source voltage is distributed between the load and the wiring. Rather than dealing with charge per se at the present time. look at Kirchoff's Laws. These are the basic circuit equations which will not lead you down the garden path. Try to understand what they mean.

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

• posted

+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

• posted

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?

• posted

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.

• posted

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.

• posted

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.

• posted

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).

• posted

On Mon, 23 Apr 2007 04:31:06 GMT, "no_one" Gave us:

• posted

On 22 Apr 2007 22:14:31 -0700, conrad Gave us:

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.

• posted

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.

• posted

Yeah, this site

basically the same thing I have been taught about current. Living the impression that it is 'constant'.

• posted

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?

• posted

I think that you will find that a "voltage drop" is the generally accepted term to indicate that a voltage is developed across a resistive element when a current is passed through that resistance. The "voltage drop" occurs across the terminals of the resistance, and this "drop" is taken away from other elements in the circuit - eg.

100 Ohms is placed across a 100 Volt source. One ampere will flow. An additional 1 Ohm resistance is connected in series with the 100 Ohm load. The 1 Ohm resistance will develop a (almost) 1 Volt drop across its terminals, leaving less than 100 Ohms to be developed across the load.

Current limiting resistors are often refered to as "dropper resistors", because of what they do.

• posted

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

• posted

---------------- 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.

• posted

On Thu, 26 Apr 2007 04:46:30 GMT, "Don Kelly" Gave us:

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.

• posted

the volatge drop and the voltage difference are two different way of understanding the same thing.. suppose a 5V voltage source and two resistances connected to it in series to it. if the first resistance shows a voltage 2v across it the other resistor will show 3V. this way voltage across the resistors reduces the source voltage drop from 5V to 3V or 2V thus it is termed as voltage drop.(if open circuited, the source wd hv shown 5V intead of 3V or 2V) another way of understanding it is way of voltage difference concept. if we see the same examle, the source has 5V and the second resistor shows 3V the voltage drop or potential difference across the first resistor will be 5-3= 2V

• posted

"SuperM" wrote in message news: snipped-for-privacy@4ax.com...

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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).

• posted

On Fri, 27 Apr 2007 02:45:58 GMT, "Don Kelly" Gave us:

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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