Voltage and current in a resistor

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SUMMARY

This discussion centers on the relationship between voltage, current, and resistance in electrical circuits, specifically focusing on resistors. Participants clarify that voltage represents potential energy, while current is the flow of charge. The concept of "voltage drop" across a resistor is explained as the energy dissipated due to resistance, with Ohm's Law (V=IR) serving as a fundamental principle. The analogy of a resistor as a passive element that does not add energy to the circuit is emphasized, along with the importance of understanding the quantification of charge in coulombs.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with electrical concepts such as voltage, current, and resistance
  • Basic knowledge of electrical circuits and components
  • Concept of charge quantification in coulombs
NEXT STEPS
  • Study the principles of AC and DC circuits
  • Explore the concept of power dissipation in resistors
  • Learn about Kirchhoff's Voltage Law and its applications
  • Investigate the relationship between energy and charge in electrical systems
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and anyone seeking to deepen their understanding of fundamental electrical concepts.

  • #31
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.
 
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  • #32
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.
 
  • #33
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.
 
  • #34
daqddyo1 said:
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.
 
  • #35
mohammad_adam said:
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.
 
  • #36
mohammad_adam said:
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.
I would hope so (that you dig deeper into the basics), especially for someone in the fourth year of an EE degree program. Very basic concepts such as the drop in potential (voltage drop) across a resistor are presented early on in electronics, and really aren't very complicated. A voltmeter placed on either side of a resister can be used to measure the difference in potential.
 
  • #37
I think people are confusing my ability to use a voltmeter with my ability to understand the inner workings of "voltage" itself. I'm totally okay with circuit analyses and ohms law and connecting two leads into a voltmeter (in parallel if my memory serves me right - I have done it a few hundred times). Just having troubles visualizing it all, if that's even possible in the first place.

But like sophiecentaur said:

"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."

Perhaps I'm delving too deeply and should do as suggested and just think of these things as "something that flows and has certain properties".

I guess these concepts (EE ones) just have a weirder approach to them than other things because there's so little that can be visualized and so we rely on measurement tools to observe behavior as opposed to trying to picture what's physically happening. Kind of sucky because my brain likes visualization.

Anyways, thank you everyone for the help/input. I greatly appreciate you guys taking the time to try and help me out.
 
  • #38
I think your problem here may just be because you are thinking of Electricity as something different from other aspects of Science. My opinion is that people find it 'strange', simply because they are less familiar with it. We could have ( and you can read plenty of them on PF) a similar conversation about the more familiar Forces, Energy and Mass and we would come up against a wall at the end, because we can never 'understand' it all.
I am a bit 'snobby' about visualisation (a.k.a. analogies) because it can be used as a tool to draw unjustified conclusions. But, Of course, I / we all use them all the time - but privately.
 
  • #39
Yes I totally 100% agree. My problem is overthinking. And the reason I don't overthink the other aspects of science (the more 'familiar' forces etc that you mentioned) is because they're just there and I take them for granted just because they're considered 'visualizable' and less complicated than electricity (even when they're really not).

I do want to continue delving too deeply in my spare time though just because it's fun. I won't delve too deeply to the point where double slits start showing up and I start pondering my own existence haha. I've started to learn not to fall too far into the rabbits hole. That place is way too complicated and not to be understood by someone like me that's for sure.
 
  • #40
Still wouldn't mind if someone could take a stab at explaining voltage drop as opposed to voltage (from say a battery). Looking for a conceptual understanding, not just terminology and tips on how to measure the drop with a voltmeter.
 
  • #41
Point 1:
Remember, potential at a point (measured in volts) is the electrical potential energy of 1 coulomb of charge at that point. That is why 1 volt is defined as 1 joule of energy per coulomb.
If this charge flows through a circuit to another point and let's say through a lamp, then the charge loses some electrical potential energy. (The lost energy is released in the form of heat and light).
Hence we say that there is a potential difference that can be measured across the lamp terminals. This is loosely referred to as a "voltage drop".
Point 2:
If you want to talk about the "voltage" or potential at the positive terminal of a 1.5 V battery you have to be very careful of the semantics. This is because you are measuring the potential relative to some other point in the universe. But which point?
Let's say we're going to measure its potential relative to the negative teminal of that battery. What is its potential there?
Well, it doesn't matter. What matters is the potential difference between the two battery terminals as measured by a voltmeter. That instrument will read 1.5 V (even if the negative terminal is at -47V and the positive terminal is at -45.5V because of the battery's location in some circuit).
Always be very clear in your own mind what is meant by "voltage" and "voltage drop" or even voltage rise" by remembering that they are loose synonyms for the proper terms (electric potential and electric potential difference).
 
  • #42
mohammad_adam said:
Still wouldn't mind if someone could take a stab at explaining voltage drop as opposed to voltage (from say a battery). Looking for a conceptual understanding, not just terminology and tips on how to measure the drop with a voltmeter.
Voltage drop = energy per unit charge lost.
An emf = energy per unit charge gained.

In a resistor electrons collide with lattice ions and photons are given off whenever an electron drops from conduction to valence. This is why a wire carrying large current feels warm. Electrons here are losing energy since valence band is a lower energy state than conduction band.

In a battery, or generator, electrons gain energy. For a battery, positive & negative ions are moved against the internal E field. In order to move a positive ion towards the positive terminal, work must be done. The chemical reaction known as "redox" (reduction-oxidation) provides this work. We call this gain in energy per unit charge "emf". For the resistor, we call the loss in energy per unit charge "drop", or "voltage drop".

Did I help?

Claude
 
  • #43
Umm-- isn't simpler to just look at the circuit ( loop) from the voltage source - and back. Each of the elements ( wires, resistors, etc) each "drop" some the voltage. The Sum of all of the "drop"s is equal to the source. -- Not the best language, but in power for example, a long extension cord has a higher resistance than a short one, so it "drops" the voltage, and there is less available to the actual load. There is no need to bring charge, emf, columbs, or any other language into this question. V in Volts, R in Ohms, and I in Amps --- draw a circuit, with 3 resistors, and a DC Voltage...
 
  • #44
Windadct said:
Umm-- isn't simpler to just look at the circuit ( loop) from the voltage source - and back. Each of the elements ( wires, resistors, etc) each "drop" some the voltage. The Sum of all of the "drop"s is equal to the source. -- Not the best language, but in power for example, a long extension cord has a higher resistance than a short one, so it "drops" the voltage, and there is less available to the actual load. There is no need to bring charge, emf, columbs, or any other language into this question. V in Volts, R in Ohms, and I in Amps --- draw a circuit, with 3 resistors, and a DC Voltage...
But the OP wanted a more detailed explanation. To start, I would use the circuit theory approach, i.e. Ohm, KVL, KCL. But for a more in depth view, the resistor is a lattice where charges collide with lattice ions and drop to a lower energy state releasing photons (heat). It just depends on how deep of an analysis is being requested. That was what I was thinking when I replied.

Claude
 
  • #45
mohammad_adam said:
Still wouldn't mind if someone could take a stab at explaining voltage drop as opposed to voltage (from say a battery). Looking for a conceptual understanding, not just terminology and tips on how to measure the drop with a voltmeter.
If you want a good analogue, the Potential Difference (Voltage) the Battery is the equivalent to taking (pumping) water to a reservoir at height h. The water makes its way down to the start level, through a series of water wheels, to represent Electrical Loads (forget the pipes, which are wide enough to ignore any friction and for the actual speed of the water to be negligible). The water emerges from the bottom with (notionally) no kinetic energy so the ' vertical drops'(h1, h2, h3 etc.) across each of the water wheels will deliver the same Power as the total Power out of all the water wheels. h will equal h1+h2+h3 +++. Note - this is NOT the classic water model with thick and thin pipes (which is pretty hopeless because it is based on pressure and speeds and not energy).
Gravitational potential (gh) is Joules per kilogram and Electrical Potential is joules per Coulomb

You have to avoid asking the question about how it's all arranged so that the water comes out with no KE; it just ' adjusts itself' for a series of resistors.

Voltage measurement: You just connect the voltmeter across the two points of interest (it is always two points because you are after the potential Difference), either across the battery or across each resistor. The resistor drops will add up to the battery voltage.
 
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  • #46
"voltage drop" to me is Newton's third law...the fundamental concept from physics that should be understood 2 -5 years before Engineering Classes... if he is asking for analogies - then he is not asking about how electrons move, amount of charge ... when you push on something - it pushes back... that is the concept. If you Push electrons though something -- it pushes back. Unless there is some more complex principal you are trying to comprehend - there is no need to discuss anything else.
Even trying to discuss resistance at an atomic level does not work here, because different materials generate "resistance" different ways ... Wire or semiconductor, inductor - electrons vs holes, whatever V= I * R
Sorry to vent a little - I must be missing what the OP is asking -
 
  • #47
Although you are an electric engineering senior, you are asking high-school questions. I don't mean sarcasm, I just want to help. You should buy a proper physics introductory textbook, and then go for the advanced electricity and engineering textbooks, since you may also have a loose foundation in engineering.
You must spend the next 1~2 years of your life self - studying to rebuild a foundation and be a proper engineer.
 
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  • #48
Windadct said:
"voltage drop" to me is Newton's third law....the fundamental concept from physics that should be understood 2 -5 years before Engineering Classes... if he is asking for analogies - then he is not asking about how electrons move, amount of charge ... when you push on something - it pushes back... that is the concept. If you Push electrons though something -- it pushes back. Unless there is some more complex principal you are trying to comprehend - there is no need to discuss anything else.
Even trying to discuss resistance at an atomic level does not work here, because different materials generate "resistance" different ways ... Wire or semiconductor, inductor - electrons vs holes, whatever V= I * R
Sorry to vent a little - I must be missing what the OP is asking -

The analogue for N3 is surely the Back emf as you try to change the current in an inductor. As with the reaction force from a mass when you push it, no Power is necessarily lost. In a component that is dissipating Energy, Power is loss.

There is a value 'R' which is the ratio of PD and Current for all components but it is only constant when you have a linear conductor (metal - Ohm's Law applies). It is risky to use the concept 'R' when dealing with a semiconductor when the VI characteristic is not a straight line. For calculations, you should stick with V and I in any equation you try to form.
 
  • #49
Just dropping in with my own analogy here. Imagine a world where lakes are on islands floating above the ocean. A lake flow into a river, which flow downwards, decreasing the waters potential energy. I we were to put a drain in the middle of a lake, connected directly to the sea below, the water would get there a lot faster. In this analogy, the river is the resistor, impeding the waters wish to get salty.

In these terms, I think of the voltage as the gravitational potential energy stored in the water. As voltage is due to separation of charges, the water goes downwards due to separation from the ocean. As a footnote, the Aral Sea is an open circuit.
 
  • #50
Sophie -not to be contrarian- IMO if the force applied is being equally reacted ( steady state as with drag / friction / resistance ) --- then the F applied = F reaction -- and work is done(power loss) . I agree to the inductor - this is the case where energy is stored in the system - inductors / capacitors... in these cases we are using Impedance - and we are talking about time varying cases - but regardless of the time - the Total force applied (Voltage) at any point in time is equal to the sum of each of the elements "reactive" force. ( really KVL right ?- the total sum =0, although the OP may ONLY see KVL as a mathematical tool - it is a (the) fundamental issue)
Consider a motor - not a resistor - looks like an inductor, then the back emf - is proportional to the mechanical force ( torque) and the power is the EMF ( Volts) * current... if there are resistors capacitors - what ever else in the circuit --- the back EMF becomes one of the SUM of forces - resisting the flow of current ( as I say pushing back).
This is my understanding - I do not consider this an analogy, but the fundamental principal. The same principal can be applied to many ( if not most) classical physical systems. Water in pipes, springs, gravity and friction blocs...
 
  • #51
Windadct said:
Sophie -not to be contrarian- IMO if the force applied is being equally reacted ( steady state as with drag / friction / resistance ) --- then the F applied = F reaction -- and work is done(power loss) . I agree to the inductor - this is the case where energy is stored in the system - inductors / capacitors... in these cases we are using Impedance - and we are talking about time varying cases - but regardless of the time - the Total force applied (Voltage) at any point in time is equal to the sum of each of the elements "reactive" force. ( really KVL right ?- the total sum =0, although the OP may ONLY see KVL as a mathematical tool - it is a (the) fundamental issue)
Consider a motor - not a resistor - looks like an inductor, then the back emf - is proportional to the mechanical force ( torque) and the power is the EMF ( Volts) * current... if there are resistors capacitors - what ever else in the circuit --- the back EMF becomes one of the SUM of forces - resisting the flow of current ( as I say pushing back).
This is my understanding - I do not consider this an analogy, but the fundamental principal. The same principal can be applied to many ( if not most) classical physical systems. Water in pipes, springs, gravity and friction blocs...

Either way round, it is not a good idea to try to equate Potential Difference (an Energy quantity) with Force. Using the right quantities and believing what the Maths delivers is usually the best way forward.
Volts are only allowed to be called a 'force' in the context of "emf", which is an ancient tradition and the term is used with fingers crossed.
 
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