Voltage and current in a resistor

[mohammad_adam]

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.

 

[dotmatrix]

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.

 

[mohammad_adam]

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?

 

[dotmatrix]

What exactly are you studying right now. AC/ DC I could be broad, but it would help me narrow the field a little.

 

[dotmatrix]

For general terms V=IR

 

[zoki85]

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?

 

[mohammad_adam]

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)

 

[zoki85]

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.

 

[mohammad_adam]

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

 

[NascentOxygen]

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.

 

[mohammad_adam]

I fully agree. But what does it exactly mean that 8v is dropped across the 4 ohm and 4v "dropped across" the 2 ohm?

 

[mohammad_adam]

Maybe it's a language thing that's messing me up ?

 

[NascentOxygen]

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.

 

[mohammad_adam]

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?

 

[mohammad_adam]

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"?

 

[dotmatrix]

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.

 

[dotmatrix]

Oh dude, your going back to flux, flux sucks

 

[NascentOxygen]

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).

 

[mohammad_adam]

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 please 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 going to say it's charge) and how on Earth is it being captured?
Maybe I should just go to bed lol

zzzzzzzzZZZZZZZZZZZZZZ

 

[NascentOxygen]

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.

 

[mohammad_adam]

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

 

[NascentOxygen]

Yes, a count of electronic charge.

 

[mohammad_adam]

Lol ur not helping me man!

 

[sophiecentaur]

"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.

 

[mohammad_adam]

I see. Thx for ur reply.

 

[mohammad_adam]

I feel so lost when it comes to trying to understand the fundamentals. I have a hard time just running with things from an overview perspective. And then the more I inquire about the fundamentals the more confused I get. It's rather disheartening. End rant. I'll eat some ice cream and I'll be fine. Thx guys. And didn't mean to be rude to the person that suggested Wikipedia. Cheers.

 

[sophiecentaur]

I see. Thx for ur reply.

It would be appreciated, on PF, if you could use 'proper' words in your contributions. Firstly, it is a matter of courtesy and secondly, in a technical subject, non-word abbreviations are used in a technical context. "ur" can mean many things; "your" means only one thing.

 

[mohammad_adam]

I shall keep it mind for next time. Thanks.

 

[sophiecentaur]

I feel so lost when it comes to trying to understand the fundamentals.

You have all our sympathy in this. The way to deal with that problem has to be to stick with it and to start at the beginning. Many people try to run before they can walk. Doing simple example exercises can be a massive help - like doing scales every day when learning the piano. Things can become much easier when you are really familiar with the initial numerical relationships. boring boring, I know but it will produce results.

 

[sophiecentaur]

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

The concept of 'charge' came along long before anyone was aware of electrons. There is absolutely no need to think 'particles' in most of everyday Electricity problems. Think of it a s 'something' that flows and has certain properties and that can be measured by some means. I suggest the reason that you are struggling with is could simply be lack of familiarity. The same question ("what is?") can be asked about all the quantities we deal with (mass, time etc). It's all very abstract, when you delve deeply into it.

 

[mohammad_adam]

Yes I totally agree. It's a shame that in universities things are taught at such high pace with such heavy course loads that students rarely get time to ponder these things in great detail. I plan to make it a sort of hobby to dig deep into the basics. I know it's not necessary for engineering purposes but interesting nonetheless.

 

[sophiecentaur]

My point is that the basics that you learn for Engineering will help you with in-depth 'understanding', if you get those basics as second nature. The two levels are in agreement with each other, remember - so one can help the other.

 

[daqddyo1]

The semantics in this thread is confusing. Instead of talking "voltage" everywhere, let's talk about electric potential (V) (measured in volts or joules per coulomb) and potential difference (deltaV).
For a current to flow from point A to point B, there must be a potentil difference between the 2 points and a pathway for the chargs to flow. The pathway could be a light bulb and wires connected between A and B
Although I hate to do it, let's talk about the flow of positive charge in the following example:
Let's say the potentail at A is +20V and the potential at B is 0V. (This means that point B could be grounded). Hence the potential difference (deltaV) between the points A and B is -20V.
If 1 coulomb of (+) charge (Q) flows from A to B, the energy dissipated (deltaE) by the charge will be 20 joules.
{deltaE= QxdeltaV}
The light bulb will release most of this energy as heat because of its resistance, BUT the connecting wires also have a little resistance, so they will each dissipate a small amount of heat energy as well.
So, between A and B, we have wire1...lamp...wire2.
Yoiur voltmeter might show the following maesurements: across the ends of wire1, 0.1 volt, across the lamp,19.7 volts.and across wire2, 0.2 volt.
As an exercise, if the current flow through the circuit is 2 amperes, calculate the reistance of each of the parts of the circuit using Ohms Law.

 

[sophiecentaur]

The semantics in this thread is confusing. Instead of talking "voltage" everywhere, let's talk about electric potential (V) (measured in volts or joules per coulomb) and potential difference (deltaV).
For a current to flow from point A to point B, there must be a potentil difference between the 2 points and a pathway for the chargs to flow. The pathway could be a light bulb and wires connected between A and B
Although I hate to do it, let's talk about the flow of positive charge in the following example:
Let's say the potentail at A is +20V and the potential at B is 0V. (This means that point B could be grounded). Hence the potential difference (deltaV) between the points A and B is -20V.
If 1 coulomb of (+) charge (Q) flows from A to B, the energy dissipated (deltaE) by the charge will be 20 joules.
{deltaE= QxdeltaV}
The light bulb will release most of this energy as heat because of its resistance, BUT the connecting wires also have a little resistance, so they will each dissipate a small amount of heat energy as well.
So, between A and B, we have wire1...lamp...wire2.
Yoiur voltmeter might show the following maesurements: across the ends of wire1, 0.1 volt, across the lamp,19.7 volts.and across wire2, 0.2 volt.
As an exercise, if the current flow through the circuit is 2 amperes, calculate the reistance of each of the parts of the circuit using Ohms Law.

Why do you hate to do it? The rest of what you write is absolutely fine. Who cares about the charge of the particles that happen to carry the charge? ("A minus times a minus is a plus", is well enough known by anyone who is even tinkering with EE). We are talking Classical Electricity here, which was developed way before anyone came across the electron. Imo, it's when people (like you), who are clearly OK with the subject, start to make some sort of distinction between conventional current and where electrons go, that the less well informed start to get uncomfortable. You really should not 'admit' / suggest that there's any real confusion between the two approaches; it only makes things worse.

 

[Mark44]

Lol ur not helping me man!

Besides what sophiecentaur said, the use of "text-speak" isn't permitted here at PF.
From the rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/):

Language:
All posts must be in English. Posts in other languages will be deleted. Pay reasonable attention to written English communication standards. This includes the use of proper grammatical structure, punctuation, capitalization, spacing, and spelling. In particular, "I" is capitalized, there's a space after (but not before) a comma, a period, and other punctuation. Multiple exclamation marks are also discouraged. SMS messaging shorthand ("text-message-speak"), such as using "u" for "you", "please" for "please", or "wanna" for "want to" is not acceptable.