Voltage and current in a resistor

  • #1
So I know that voltage is a sort of pressure due to an imbalance of charges. And current is the resulting flow of charges. So when it comes to a resistor, does the resistance draw a current from the curcuit as well as a voltage? And what is meant by "voltage drop" across a resistor/element. Please use analogies in your answers please and thanks.
 
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Answers and Replies

  • #2
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Voltage is a potential. ie biggest dude ever stands in front of you. If you don't make him mad no harm no foul (lets call him 480v). He could potentially kick your butt, but he probably won't. Now if you make him mad and he takes a swing at you (better run this guy can punch a few amps), misses and hits air, same thing, no harm no foul, he looks like a douche, we all go home. But if your face happens to be there, bang welcome to resistance. So you're a 100 ohms and buddy hits you with 4.8 amps, and your still standing, like a trooper. All of a sudden buddy decides to round house you and you and your friend is standing by your side, You get hit, and your friend gets hit. You're still 100 ohms but, your friend happens to be a big fat guy so he's got like 200 ohms of resistance, So instead of you feeling the full brunt of potential, you only get 1/3 and your friend has 320 Volts dropped on him. However, the remaining 160V and 4.8 amps was enough to drop you to the ground, and buddy, seeings how you made him so mad decides to stomp your face. So you are lying parallel to the concrete, your face at 100 ohms the concrete at 100. He lays his boot down and 'whack' you're out cold 480V on your face, 2.4 amps to you and 480V and 2.4 amps to the concrete. And you know what, buddy's toe hurts, cause Power in= Power out.
 
  • #3
I don't quite understand the last part with the concrete. But lol on the analogy, thanks for that btw..... So what I'm struggling with is this: voltage is potential, if the potential finds a pathway to start actualizing the potential (conductors), then it throws out current, and if someone's face (resistance) is in the way then depending on how fat the face is, a higher voltage will be dropped on/across it? But less current???

And what's meant by drop of current across a load anyway. Is that how much of the total potential (from the source) the element "eats up" from the total potential?
 
  • #4
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What exactly are you studying right now. AC/ DC I could be broad, but it would help me narrow the field a little.
 
  • #5
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For general terms V=IR
 
  • #6
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So I know that voltage is a sort of pressure due to an imbalance of charges. And current is the resulting flow of charges. So when it comes to a resistor, does the resistance draw a current from the curcuit as well as a voltage? And what is meant by "voltage drop" across a resistor/element. Please use analogies in your answers please and thanks.
Do you know how to use wikipedia?
 
  • #7
In my 4th year of electrical engineering. Just never really thought of these concepts fundamentally. I know how to use Wikipedia lol thanks for asking but I thought the point of this forum was to be able to talk on a more personal level with ppl who know these concepts in and out so that I don't just end up with half-baked knowledge about things and end up just rolling with the terms and equations of things I don't really truly understand (like far too many ppl do, at least at my university)
 
  • #8
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In my 4th year of electrical engineering. Just never really thought of these concepts fundamentally.
In that case it is very strange you didn't think of Ohm's law before. Resistor is a passive element and doesn't add energy to the circuit. Hence the "voltage drop" expression.
 
  • #9
Right so what exactly is the "drop"? Because a source (dc in this example) is providing steady voltage so it's not like the "drop across" the resistor is a chunk of that voltage being expended over the resistor. So what is it then?? Sorry if I'm not articulating myself well enough. I just don't understand the expression and how it ties into the whole big pic
 
  • #10
NascentOxygen
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If you connect a 6 Ω resistor across a 12 volt battery, 2A of current flows, and all the 12 V is dropped across that resistor. If you were to replace the 6 Ω by a 4 Ω and a 2 Ω in series, then again a 2A current will flow. Now, the voltage drop across the 4 Ω resistor will be 8 V, in accordance with Ohm's Law.
 
  • #11
I fully agree. But what does it exactly mean that 8v is dropped across the 4 ohm and 4v "dropped across" the 2 ohm?
 
  • #12
Maybe it's a language thing that's messing me up ?
 
  • #13
NascentOxygen
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You could say 8 V is lost across the resistor, or that 8 V appears across it, or that 8 V can be measured across it. "Dropped" has connotations of energy loss, as opposed to energy gain.
 
  • #14
I guess it makes sense. The resistor, depending on its resistance takes the amount of voltage it needs out of the tub of voltage supplied and has it "dropped" across itself in accordance with ohms law. I know I sound like a 5th grade retard here so bear with me lol and thanks for doing so. So with that somewhat settled.... My next question is: how is voltage a measurable quantity? I mean I know that booking a voltmeter across two terminals produces a reading, but what how is the electric "pressure" actually being determined and quantified?
 
  • #15
I know that a volt is a joule/coulomb so the amount of work done by a unit of charge.. Which is a coulomb?? Also I have never understood the coulomb. And read a lot in attempts to understand it but I just get more confused. How is charge quantifiable. This blows my mind. Isn't it just essentially "stuff"???
 
  • #16
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You don't drop current but you do draw it.
Your house has a 100 amp service
That means at any given time a hundred amps can be supplied.
To actually visualize stop thinking of things as resistance and start thinking of them as a load.
A microwave needs 1200W of power to nuke your pizza pop.
That means it is going to draw 10 amps of current at 120v.
Inside the microwave 5 volts are allocated to run the clock (your voltage drop) the other 115V are used to microwave the pizza pop, current is equal through both.
And by golly if you are a forth year engineer I need to go back to school or run for the hills.
 
  • #17
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Oh dude, your going back to flux, flux sucks
 
  • #18
NascentOxygen
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I know that a volt is a joule/coulomb so the amount of work done by a unit of charge.. Which is a coulomb?? Also I have never understood the coulomb. And read a lot in attempts to understand it but I just get more confused. How is charge quantifiable. This blows my mind. Isn't it just essentially "stuff"???
A coulomb is just a bunch of electrons (or protons), quite a lot really. You can't get any charge less than 1 electron; that makes it the basic unit of charge (hand in hand with the positron and the proton).
 
  • #19
I don't remember saying that current was dropped. I've been meaning to figure out how "voltage is dropped across a load/resistance". And isn't saying that a load "draws" current misleading? Because I would picture the load as just being there while current is passed through it. And based on its atomic makeup or whatever, it impedes the flow of electrons thereby limiting the amount of current being forced through it.
With regards to the voltage I found this (which may help u understand what I was asking in the first place):
(Someone answering a similar question on the net)

Larger resistors have larger voltage drops. Why? Before we can answer that we must better understand what exactly a resistor does.

A resistor resists the flow of current. This resistance means that some work must be done to "push" current through the resistor. Whenever work is done on charge, we have voltage. Thus, when current flows through a resistor, there is some voltage across the resistor.

The larger the resistance, the more work required to "push" the current through the resistor, the more work done the larger the voltage drop across said resistor.

End quote

So as I see it (and plz correct me if I'm wrong, that's the whole point here), voltage from a battery is more like a bottled up force (from an imbalance of charges wanting to balance themselves out) and then once that force is allowed a pathway, it causes electrons to flow in the whole circuit due to a sort of chain reaction, and then when that chain reaction (current) is slowed by an impedance to that current, the atoms in the resistor do work on the charges to resist that flow thereby causing another voltage/pressure of its own that can be measured.....?

Btw this whol time I've just been picturing a simple series circuit supplied by a dc source.

As for a coulomb being a bunch of electrons.... It's not just a bunch of electrons in the way that avagadros number is a bunch of molecules/mol correct? Because avagadros number is just a way to simplify numbers.... But a coulomb is not just a number of things it's actually "something" about those things/electrons which is being captured. What is that thing (I know ur gonna say it's charge) and how on earth is it being captured?
Maybe I should just go to bed lol

zzzzzzzzZZZZZZZZZZZZZZ
 
  • #20
NascentOxygen
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As for a coulomb being a bunch of electrons.... It's not just a bunch of electrons in the way that avagadros number is a bunch of molecules
Yes, just a large count of electrons. You could have them all grouped together, or count them one at a time as they pass by.
 
  • #21
Yeah but a coulomb is the CHARGE of so many electrons. Not just a number of so many electrons. So u haven't really answered my question. Or in going insane. Don't know which one of the two
 
  • #22
NascentOxygen
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Yes, a count of electronic charge.
 
  • #23
Lol ur not helping me man!
 
  • #24
sophiecentaur
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"A resistor resists the flow of current."
This is where the historical choice of the word "resistor" tends to mess up the understanding of what goes on. The Mechanical Analogy can colour ones appreciation of Electrical Energy Flow and can lead to 'inappropriate' conclusions, I think. A resistor doesn't 'resist' the flow of current any more than a motor, an antenna or a 50W amplifier (i.e. any other load) in a circuit. When a charge passes along any non-ideal conduction path, Energy will be transferred (i.e. away from the circuit).
Rather than talking about a 'force' against another 'force' just think of a resistor being an element that dissipates electrical energy as charges flow around a circuit. Each of the circuit elements will dissipate their share of energy.
V=IR tells you the energy loss per coulomb and all the V's must add up to the V of the power supply (K2: aka Energy Conservation). I always reckon that, once you have stated a Scientific situation with the appropriate Maths, that is probably the best way of expressing it. Arm waving about the same phenomenon is not very often a better approach. Maths is just an alternative to the spoken language and you shouldn't attribute anything magical to the use of 'words' to explain something - except for the familiarity in some cases. The solution is to get used to using Maths for such situations.
 
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  • #25
I see. Thx for ur reply.
 

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