Talk:Voltage drop

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Comments[edit]

To all of the contributors of this page, Many thanks. Bare with me as I try to flesh out the article a litte and blend it with other electrical articles. As a mewbie, I would appreciate a few remarks on how best to use the TALP pagess.

The NEC does NOT specify voltage drop! It only recommends such. It does specify ampacity for certain installations using some cables. Sorry, but I am not familiar with practices outside of the United States. Charles E Muhleman, Huntsville IN.

Correct: Art 90.5 explains the "enforcability" of the various language: (A) Mandatory Rules contain the words shall or shall not. (B) Permissive Rules contain the words shall be permitted or shall not be required. (C) Explanatory Material such as includes references to other standards, related other rules of the Code, etc are contained in fine print notes (FPNs). All references to voltage drop in the NEC are found in FPNs ONLY!

Measurement of Voltage drop[edit]

My electrician scoffed at me when I measured voltage drop on a 15 amp circuit (14 awg wire, 100 ft long) using a hairdryer as mentioned in this article. Are there any expert sources that can confirm this as a reliable measure of voltage drop? —The preceding unsigned comment was added by Cynoscion (talk • contribs) 16:30, 13 December 2006 (UTC). Cynoscion 17:22, 13 December 2006 (UTC) Cynoscion[reply]

Voltage Drop in DC circuit[edit]

Present Text Unclear: "A current flowing through the non-zero resistance of a practical conductor necessarily produces a voltage across that conductor." I always thought that voltage is produced by the DC supply, and that the supplied voltage generates the current. In addition, such a "practical conductor" is a passive element of the circuit, and thus produces neither voltage nor current. I propose that the first paragraph be incorporated into the second paragraph, which would be restated as follows into three new paragraphs:

Consider a circuit with a nine-volt DC source; three resistors of 67 ohms, 100 ohms, and 470 ohms; and a light bulb--all connected in series. The DC source, the conductors (wires), the resistors, and the light bulb (the load) all have resistance; all dissipate supplied energy to some degree. Their physical characteristics determine how much energy. For example, the DC resistance of a conductor depends upon the conductor's length, cross-sectional area, type of material, and temperature.

If you measure the voltage between the DC source and the first resistor (67 ohms), you will measure slightly less than nine volts. The current passes through the conductor (wire) from the DC source to the first resistor; as this occurs, some of the supplied energy is "lost" (unavailable to the load), due to the resistance of the conductor. Voltage drop exists in both the supply and return wires of a circuit. If you measure the voltage across each resistor, you will measure a significant number. That represents the energy dissipated by the resistor. The larger the resistor, the more energy dissipated by that resistor, and the bigger the voltage drop across that resistor.

You can use Ohm's Law to verify voltage drop. In a DC circuit, voltage equals current multiplied by resistance. E = IR. Also, Kirchhoff's circuit laws state that in any circuit, the sum of the voltage drops across each component of the circuit is equal to the supply voltage. --ThatAMan (talk) 22:20, 19 August 2012 (UTC)[reply]

I found your explanation of voltage drop in DC circuits a bit confusing and convoluted. It is probably best understood by those that already know what voltage drop means. I suggest a simplified example with a circuit that a layman (like me) can understand. Consider a series circuit with a 9 volt power supply and 3 resistors being 67 ohms, 100 ohms, 470 ohms respectively. If you measured the voltage accross each individual resistor in the circuit you would get the voltage drop for each resistor. Kirchoff's law states that the sum of the voltage drops will total the supplied voltage of 9 volts. You would expect the strongest resistor of 470 ohms to show the largest voltage drop. Why? Explain it. This circuit would be simple enough for most to understand.

The diagram assumes that no voltage is dropped on the return current path as it is earthed at the load. In practice volts are dropped in the neutral conductor as well. — Preceding unsigned comment added by 213.130.123.157 (talk) 12:09, 8 February 2012 (UTC)[reply]

Suggest you rewrite your version with the basic series circuit just described before someone else does....good luck. —Preceding unsigned comment added by Howartthou (talk • contribs) 10:31, 3 March 2008 (UTC)[reply]

Drawing/Graphic[edit]

It is, to me, entirely unclear what the point of the unlabelled, unexplained, unprofessional and hardly referenced drawing in the article is. I believe it is more confusing than helpful. —Preceding unsigned comment added by 82.113.106.16 (talk) 02:00, 6 March 2008 (UTC)[reply]

I thought the diagram was wonderfully colourful, but I'm inclined to agree that I can't work out what the purpose of it is. What's really needed is a simple example "a 10-metre length of 2.5 mm² carrying 15 amps at 12 volts will drop XYZ". —Sladen (talk) 22:21, 3 March 2009 (UTC)[reply]

The diagram IS informative, and in it's current form is certainly more beneficial than confusing. I am inclined to agree with the above comments that it isn't really in keeping with wikipedia style, and really needs to be re-made in some diagam producing package as an SVG image. 87.194.84.113 (talk) 10:35, 8 February 2010 (UTC)[reply]

What is "voltage drop"?[edit]

Proposed new first paragraph: Voltage drop describes how the supplied energy of a voltage source is reduced as electric current moves through the passive elements (elements that do not supply voltage) of an electrical circuit. Voltage drops across internal resistances of the source, across conductors, across contacts, and across connectors are undesired; supplied energy is lost (dissipated). Voltage drops across loads and across other active circuit elements are desired; supplied energy performs useful work. Recall that voltage represents energy per unit charge. --ThatAMan (talk) 22:42, 19 August 2012 (UTC)[reply]

211.27.214.134 insertion is rational since there is voltage drop across the load as well (the current definition "Voltage drop is the reduction in voltage in an electrical circuit between the source and load" excludes the load). If we say "Voltage drop is the voltage across a part of circuit", it will include the load as well. Voltage drop is partial voltage; there is no sense if we talk about the voltage drop across the whole circuit since this is the voltage of the source. Maybe, we have to exclude the additional sources (if there are such) connected in this part of the circuit although the opposite voltage of a source connected contrary to the main source may be considered as voltage drop. In this case, we would keep only the passive elements ("not sources"). From this viewpoint, a transistor is not active but passive element. Am I right? Please, discuss. Circuit dreamer (talk, contribs, email) 16:27, 11 November 2010 (UTC)[reply]

Should mention voltage drop across semiconductor junctions eg in LEDs[edit]

Since Light-emitting_diode#Colors_and_materials links to this article it should mention voltage drop across semiconductor junctions, eg in LEDs, since it is near constant rather than proportional to current as in a resistive element ? - Rod57 (talk) 18:05, 29 October 2016 (UTC)[reply]