Ohms Law - Page 1
Forum section: electronics / beginners

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 03:32:34 pm
content: Sorry probably a stupid question but I am board at work so I decided to try and mathematically understand resistors.I understand that through a voltage divider the voltage drops due to the relationship between resistance and current.Now if I have a single resistor referenced to ground and there is no other path to travel but through the resistor am I correct in seeing no drop in voltage? I was under the theory that resistors dropped both current and voltage.

author: ejeffrey
author_rank: Super Contributor
timestamp: March 26, 2013, 03:55:52 pm
content: For 2-termial components: current flows between the terminals, and there is a voltage difference (aka drop) between the terminals. A current drop doesn't really make sense. For the case of a resistor, the "voltage drop" is equal to "current flow times resistance" -> V=IR.Really this isn't about Ohm's law, but Kirchhoff's laws. If you really understand kirchoff's laws, the rest will take care of itself.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 04:09:53 pm
content: What if you did not know the current but only the input voltage. For example a voltage of 5 through a 10k resistor grounded would be 0.00005 amps. In order to do this the resistor is parallel giving no drop in voltage correct?But if I wanted the voltage to go through a resistor into say an opamp and I wanted to know what that voltage is I would need kerchoffs like you stated I would assume.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 04:14:59 pm
content: +1 On studying Kirchhoff's laws, they are the basis of circuit analysis and actually very simple to understand. If you're into the math, you'll be happy to know that they are essentially a simplification of Maxwell's equations and pretty much similar in form.You should check out MIT's open courseware.For Maxwell's equations:http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/For Kirchhoff's laws and circuit analysis:http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/Audit both these courses and all your questions will be excellently answered.For your second question, if you've got a single voltage source connected to a single resistor, you should see the entire voltage drop across that resistor.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 04:20:33 pm
content: Humm seems I need to look into this more with some experiments. So in my above edit where I added about the opamp you are saying the opamp would get no voltage?

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 04:31:42 pm
content: Quote from: blewisjr on March 26, 2013, 04:20:33 pmNot at all. I *really* recommend you learn circuit analysis, the second MIT video course will very gently ease you into this, the only requirement is that you know linear algebra, or have even just informally solved a system of equations with more than one variable.If you want to set the voltage supplying an opamp to something other than your supply rail, you need a voltage divider, just putting a resistor in series would only effectively serve to limit the current.I *definitely* recommend experimenting. I'd start with something simpler like a battery, a resistor and a forward biased diode in series. Measure the voltage across all combinations of nodes. Measure the current flowing between each device (hint, they will all be the same). Then build a resistive divider, put the diode across the resistor connected to ground, do more measurements. I think that will help clear your confusion.It really can be confusing, so be patient with yourself.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 04:43:29 pm
content: Will a led work for that experiment as the diode? I think you are right about experimenting with the theory I am learning. I have the practical electronics for inventors third and the theory is what is wracking my brain. I have a different book coming in today maybe it will help ease me into the above mentioned book.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 04:54:45 pm
content: Quote from: blewisjr on March 26, 2013, 04:43:29 pmYes, an LED will work, please use Ohm's law to choose the resistor so that the LED doesn't receive more than the maximum current allowed on its data sheet, if you don't have a data sheet, limit the current to 20mA or so.Honestly, I'd say watching the course videos (if you didn't know they are free and online) from the second link are superior to most books for getting you off the ground with Kirchhoff's laws, ymmv.P.S.I highly recommend watching the videos from the first course link so you can see where Kirchhoff's laws come from.What's your level of math ability? If you feel the math is beyond you (not assuming), don't worry at all, we can fix that :-). If that's the case, the first videos are still good to watch, but the calculus referenced might sound intimidating. For simple circuit analysis, linear algebra is all you really need, and that's not so hard to learn, at least in so far as learning to solve systems of linear equations.

author: croberts
author_rank: Regular Contributor
timestamp: March 26, 2013, 04:57:32 pm
content: Hello blewisjrI think the words "through" and "across" are important here. Current flows through a component and voltage is dropped across a component. When I was learning electronics we were taught electron flow. That is current flows from the negative terminal of a source such as a battery to the positive terminal. Where current flowing through a resistor enters the resistor the polarity is negative and where it leaves the resistor the polarity is positive. This helps you understand the polarity of the voltage dropped across the resistor as a result of the current flowing through it. Ohm's Law of course predicts the voltage drop based upon current flow and the value (resistance) of the resistor..

author: c4757p
author_rank: Super Contributor
timestamp: March 26, 2013, 05:02:26 pm
content: Quote from: croberts on March 26, 2013, 04:57:32 pmOh please god no. You'll just confuse him. The convention is that current is a positive charge flowing from a higher potential to a lower potential. Sure, the actual physical particles do the opposite, but unless you're actually designing discrete semiconductors, you could do pretty much everything in EE without even knowing there is such a thing as an electron. Let's let people wrap their heads around the 99.9999%-utilized convention before confusing them.

author: olsenn
author_rank: Frequent Contributor
timestamp: March 26, 2013, 05:07:43 pm
content: Sure there's a drop in voltage... it drops from some positive voltage to ground (which it's referenced to). If you peeled the coating off the resistor and probed the middle of the exposed resistor, you would get half the voltage, as it is now a resistive voltage divider

author: c4757p
author_rank: Super Contributor
timestamp: March 26, 2013, 05:10:49 pm
content: I think I see the confusion here. A voltage "drop" doesn't mean it causes the voltage to fall at a certain point, it means there is a voltage difference from one end of the component to the other. A resistor across a 5V power source has a 5V drop, even though the power source remains 5V.

author: croberts
author_rank: Regular Contributor
timestamp: March 26, 2013, 05:14:06 pm
content: Sorry my friend. I knew as I was typing I was in danger of opening an old wound. I withdraw my comment. Cheers!

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 05:15:54 pm
content: Quote from: c4757p on March 26, 2013, 05:02:26 pmI think this is one awful barrier to beginners :-(, especially with less formal math training. If you want to learn the fundamental physics the convention is positive flow is electrons and when you learn the engineering (which is what you'll actually use) the convention is positive flow is holes. The math of course doesn't care which you choose as long as you're consistent.Really really unfortunate that things developed this way. Honestly though, you've eventually got to understand that is doesn't matter and why it doesn't matter if you really want to understand what you're doing.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 05:18:59 pm
content: As far as math goes I am comfortable up to trig. Calc was never my thing but I aced discrete structures (logic math).I am sure I will get my head around this circuit stuff eventually sometimes things seem a bit backwards so far.

author: c4757p
author_rank: Super Contributor
timestamp: March 26, 2013, 05:23:15 pm
content: No need to apologize, I didn't mean to sound snippy. I've just seen so many people get confused by the conventional/electron current distinction, and needlessly so because it just really doesn't matter a rat's ass. Engineering is hard, sure, but the basics here are a lot easier than people make them out to be.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 05:23:22 pm
content: Quote from: blewisjr on March 26, 2013, 05:18:59 pmDo you know methods of solving systems of linear equations? If so, jump right into Professor Agarwal's lectures:http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/I can't think of a better introduction to circuit analysis.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 05:26:02 pm
content: Quote from: c4757p on March 26, 2013, 05:23:15 pmI wish there was a like button here. That's exactly how I feel too. So much domain specific language, so much that makes the subject look intimidating, but what you're really doing, anyone can understand, no matter what their background. I feel that way about the math too. The way school's teach math is garbage imho, they take something everyone can learn and make into something that seems reserved for "smart (tm)" people :-(.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 05:26:33 pm
content: I am good with both linear algebra and matrices might just need a quick brain jog is all.

author: c4757p
author_rank: Super Contributor
timestamp: March 26, 2013, 05:36:15 pm
content: My Analog Circuits class spent so much time on "ideal" (confusing, electronics-as-math) stuff, it was absolutely ridiculous. I spent half an hour at lab the other week trying to explain to my partners why an op amp circuit gave the wrong output when the input had a few megohms of impedance, because "the professor said op amps have infinite input impedance".I finally managed to break them of their "the professor's always right" trance with "She said the inputs are always equal, right? OK, now I'm applying 1V to this input and 2V to that input. 1V = 2V?"The worst part is that the lab specifically addressed this and the professor couldn't be arsed (yeah, I'm American, but screw it, Ilikethat phrase) to prepare the students for it. She was too busy going over nonsense idealized circuits with perfect behavioral sources and whatnot.

author: c4757p
author_rank: Super Contributor
timestamp: March 26, 2013, 05:40:36 pm
content: Quote from: blewisjr on March 26, 2013, 05:26:33 pmSorry for hijacking your thread to bitch about my schoolYep, I highly recommend this. I've found linear algebra (And matrices? That's like "clothingandpants") quite useful in electronics. I think that an understanding of calculus is nice too - you won't really be solving differential equations and stuff outside of theoretical pursuits, but at least being able to picture what an integralmeanscan help you a lot with quick calculations. You should at least briefly revisit calc.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 05:47:58 pm
content: Any suggestions for calc reviews I blame my professor for making it to hard to pass. 5 question tests are a pain. It is not that I can't do it it is more like I always make something negative that should be positive.As for the class tangent I agree I am in school for information security. Nothing but theoretical perfect scenarios that never exist and huge perfect plan documents that never go right when you need it most

author: c4757p
author_rank: Super Contributor
timestamp: March 26, 2013, 05:55:32 pm
content: Quote from: blewisjr on March 26, 2013, 05:47:58 pmI'm going to leave this one open for others' suggestions. I was lucky enough to have an awesome calc teacher in high school who could probably teach a horse differential equations, so I haven't dug around in other resources much.One suggestion: calculus hasstrongroots in physics, and just about every "theoretical" concept in calculus has a physical analogy. If the resource is severely lacking in visual diagrams and such, it's no good. Calculus is one of the branches of math where if you can't find real-world examples, you are a useless teacher, because it's one of the branches tied most closely to the real world.

author: ejeffrey
author_rank: Super Contributor
timestamp: March 26, 2013, 06:10:42 pm
content: Quote from: madshaman on March 26, 2013, 05:15:54 pmPeople in physics still always use the positive current flow convention.As usual, I will argue that neither convention is "fundamentally" right or wrong. The charge of the carrier is irrelevant to maxwells equations. In the cases where the carrier charge makes a difference, they can be either positive or negative (electrons vs. holes in semiconductors or +/- ions in electrochemistry).

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 06:26:05 pm
content: Quote from: ejeffrey on March 26, 2013, 06:10:42 pmYeah, my statement is wrong. Electrons in physics still have -ve charge, and negative potentials are still an excess of negative particles (normally electrons), but electron movement is emphasised, especially when you do experiments like deflecting electrons in a vacuum or look into why one electrode in a high vacuum glows and the other doesn't. I didn't take physics in uni; this is what I remember from high school. When I was first introduced to electronics, current was described as the movement of holes from a positive potential to a negative potential.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 06:32:37 pm
content: Quote from: blewisjr on March 26, 2013, 05:47:58 pmThis one's a bit tougher, iirc, MIT has some video lectures on differential equations under engineering, that's probably the most useful when it comes to electronics, much of the analysis involves differential equations.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 06:36:20 pm
content: Quote from: c4757p on March 26, 2013, 05:36:15 pmThat's utterly ridiculous! Sure, almost every good source states the two golden rules of opamps, *but* they *always* make a clear distinction between an ideal opamp and a real one. Also, it's usually made very clear that the inputs being at equal potential is entirely dependent on feedback, it's obviously not true when the opamp is run open loop.Also, any good teacher or source will clearly explain how important it is to use a better model than an ideal one including bias currents, parasitics etc.

author: Smokey
author_rank: Super Contributor
timestamp: March 26, 2013, 06:49:21 pm
content: Spice is a really good resource for playing with fundamental principals type stuff as a sanity check. The more complex your circuit, the more you need to be aware that spice isn't perfect and can lie to you, but for simple stuff it's really good. It's no replacement for breaking out the breadboard, but it sure is faster to check basic stuff and as a sanity check. Try out LTspice if you don't have a spice of choice yet. It's pretty confusing at first, but there are great tutorials for basic stuff.Toss your circuit down and then you can check the voltage/current at any point you want, even spots that would be hard to measure in an actual physical circuit. You never are out of stock of some wacky value component with spice.To get into the education part of things, you really need that theoretical ideal background. It's important to learn how to think like an engineer, which is what you are really going to school for. It's not usually the lecture that is the problem, it's the labs. I get how it's hard sometimes to relate those abstract ideal concepts to real world circuits, which is why I think every engineering class should be based around lab sessions with a large goal-type project in mind. Some big actual physical thing that gets built up circuit-block by circuit-block as you learn about each sub-circuit one lab at a time. Each lab builds up and tests a circuit that eventually goes into the whole of the project. That way each circuit is not only reinforced by being built and tested and analyzed, but you see it's relation to a bigger project. The example I usually come up with is an audio power amplifier. A simple amplifier is built up of a bunch of fundamental circuit blocks that pretty much snap together and can be built and tested independently. You have all kinds of transistor biasing, opamps, current mirrors and sources, differential inputs, feedback concepts, exposure to lots of different components. Lots of good stuff. That building-blocks-of-a-larger-project concept was how some of my programming class labs were structured and I know those are the ones where I remembered the most stuff. Unfortunately all my basic EE labs were of the "build this circuit, take a measurement, and tear it apart" kind of deals.That is all independent of the issue of not being able to understand the lab TA because he doesn't speak your language very well though

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 07:15:39 pm
content: Quote from: Smokey on March 26, 2013, 06:49:21 pmSpice has saved me a couple times from something I didn't notice, I'd see behaviour utterly different from what I expected and was forced to figure out why.Quote from: Smokey on March 26, 2013, 06:49:21 pmI think that's why people who have a real interest (beyond acquiring s decent career) in their subject have an easier time learning. I've been programming for over 30 years and I started when I was 7; the theory and math I learned in uni was indispensable, but my learning was always driven by the desire to accomplish something. Same thing with electronics, when I want to build something or gain experience with a realm I'm unfamiliar with, there's always a real project I really want to build which causes me to buy books, discuss with others and most importantly, put real circuits together that I *want* to work.Each such exploration leads to new realms I never even knew existed, that might spark an interest and cause me to go deeper or broader. It might depend on learning style, but I've always found self-motivated and directed learning to be the most effective.This is totally different from the academic approach where you might not give a crap about what you're doing other than to get your grade.Quote from: Smokey on March 26, 2013, 06:49:21 pmI'd go one further and suggest mentorship and independent projects should *replace* most of a formal engineering education, whatever the field. With the internet, and so many people with such a wide skill set, we're almost at the point where that's feasible for anyone.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 26, 2013, 07:20:42 pm
content: Quote from: blewisjr on March 26, 2013, 05:26:33 pmThen definitely jump right on those MIT videos, other than good books like "Art of Electronics" it's hard to find a better coverage of the basics.Another book I can recommend, (I'll have to double-check the title and author when I get home) is (iirc) "Basic Circuit Theory". It was written in the 60s or something but it's just fantastic. I bought it used and not sure if it's still in print though.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 26, 2013, 07:38:18 pm
content: I agree with you when you say it is easier to learn when you are interested in it. I started programming at 11 when I got my first computer because I wanted the answer to how and why the programs did what they did. Now at 27 the same thing drove me to start electronics as a hobby. The same thing drove me to my major in computer security. If ever at any point I can not ask why or how anymore there is no reason to continue because there is no more drive to learn something new.

author: MacAttak
author_rank: Unknown
timestamp: March 26, 2013, 08:00:13 pm
content: Quote from: blewisjr on March 26, 2013, 03:32:34 pmI think of it like this... and hopefully I can convey my mental image of it adequately...In the simplest of circuits, you have a voltage source and a resistive load. Even if there is just a bare wire connecting the voltage source poles, that wire still has some non-zero resistance.A B/--------\| |$ -$ =$ -$ =| |\---------/D CSince the wire length between A and B and the length between C and D can be anything, let's make it simple and say that they are infinitely short - which means zero resistance between them. It should be obvious that if A is infinitely close to B (and C is close to D) that the voltage measured at A will be identical to B and C is identical to D.In simpler terms, thinking of this as a typical resistor component and a AA battery, the top lead of the resistor is exactly the same voltage as the top lead of the battery. And the bottom lead of the resistor is exactly the same voltage level as the bottom lead of the battery. That's just another way of saying that the voltage difference between the two resistor leads (the "drop") is exactly the same as the voltage produced by the source battery.Or even more simply stated - the voltage dropped by all components in the circuit exactly counteracts the voltage of the source. Therefore, the resistor in this example *must* be expressing a voltage drop.Or you can think of it in another way. Let's suppose that the wire between A and B is long enough to have some resistance - which is actually true anyways. This means that we effectively have a voltage divider here - even if only one "resistor" component is in the circuit, due to the intrinsic resistance of the wire itself. So really, even the most simple of possible circuits has a voltage divider. Even a straight short will be a voltage divider if you measure the voltage at various points along the length of the shorting wire.

author: IanB
author_rank: Super Contributor
timestamp: March 27, 2013, 04:16:34 am
content: Quote from: blewisjr on March 26, 2013, 03:32:34 pmThis is really basic high school level circuit theory. ejeffery gave you some accurate information at the start, but then the thread got into linear algebra, calculus, MIT courseware and who knows what. That may be good stuff, but it is far more advanced than anything you needed to know to answer your questions.Get back to the real basics here.Electric current is a flow of "stuff" (electric charge) between two points.Electric current flows in circuits, which are closed loops that start and end at the same point. If there is no closed loop, there is no current.Electric current flows between two points in a circuit when there is a potential difference between those two points to provide a driving force. Potential difference is measured as voltage, and it is like a kind of "pressure" that makes electric current flow along.If you start at some point in a circuit and step around it, measuring the changes in voltage (potential differences) as you go, then when you get back to the starting point all the potential differences will add up to zero.There has to be something in a circuit like a battery to provide a motive force to make the current flow at all, or all the potential differences around the circuit will be zero and no current will flow.In a single closed circuit the current will be the same everywhere around the loop. Current flow cannot be created or destroyed, it always balances.Between any two points in a circuit where there is a potential difference and a current flowing between them, the resistance of that part of the circuit is given by the potential difference divided by the current. Ohm's law says that the current flowing through a resistor is proportional to the potential difference between each end. If you double the potential difference you double the current.Take a simple example of a 1.5 V battery and a 10 ohm resistor connected between the terminals. Starting at the battery negative and stepping to the battery positive, the voltage potential goes up by 1.5 V (the potential difference is +1.5 V). Stepping from the battery positive across the resistor back to the battery negative, the potential goes down by 1.5 V (the p.d. is -1.5 V). The total potential difference around the circuit is (+1.5 V) + (-1.5 V) = zero, as it must be. If you went round the block to and from the same point, and yet found yourself at a different place from where you started it would be a strange universe.Secondly, the current flowing into one end of the resistor must be the same as the current flowing out of the other end, as there is nowhere else for the current to flow. This must also be the same as the current flowing through the battery for the same reason.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 27, 2013, 05:33:38 am
content: Sorry if we went on a tangent IanB. Honestly though, your description exactly contains Kirchhoff's laws, and at some point you have to apply them to understand anything beyond the most trivial circuit and for that you have to solve systems of linear equations (which *is* high school stuff). There are shortcuts to the brute force application of Kirchhoff's laws, but again, you need knowledge of linear systems to understand how/why they work.To the OP/blewisjr: yeah, if you're feeling overwhelmed definitely take it slow, but since your studying compsci, I don't think you'll have trouble with the second set of linked MIT course videos at all. In fact, MIT hires excellent teachers, so I'd venture it's actually easier to learn from their lectures than from another school, ymmv.All that aside, linear algebra, imho shouldn't be portrayed as difficult, and the OP said he/she is decent at it, but I hope we didn't make the OP feel overwhelmed. I've of the mind that anyone can learn anything as long as others are there to give them a hand up.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 27, 2013, 09:34:25 am
content: Actually I really like the tangents that get generated on the forum. I got way more out of this thread so far then I thought I would.IanB thanks for that you actually worded that really well. I understand the basics of current flow "how it moves", my real issue is I was miss understanding the term voltage drop. Your example cleared that right up. Thank you very much.I will for sure check out those MIT courses they look pretty good. I really like how these universities do this. I actually learned a lot from one of the Stanford CS courses online when I was trying to understand how to create generic container storage in C a few years ago. Who would have thought Stanford had a whole course on that.Keep in mind it was a dumb question I originally posted it really was. That is what I get for not thinking properly through the concept of voltage drop. I have high school level physics and chemistry under my belt but that was 10 years ago just failed to dust off the cobwebs with the voltage drop thing. Makes total sense why the Drop would take us down to 0.Now back on topicSo with a voltage divider if we put in 12v as the input we would get like below. At what point would the potential go negative. From my understanding thevoltages through a divider must when added together = the input voltage.+ ------- 10k|----------- 6V10k- ----------|----------- 6V

author: AndrejaKo
author_rank: Frequent Contributor
timestamp: March 27, 2013, 11:11:53 am
content: It's sort of right, but I'd present it a bit different. Let's first connect the multimeter black probe to the battery negative terminal. We'd get this:+ ---------|-------------12 V10k|----------- 6 V10k- ----------|----------- 0 VTop tap is red probe right before the first 10k resistor, middle tap is between resistors and lower tap is after second 10k resistor.Let's take a look at your example:+ ------- 10k|----------- 6V10k- ----------|----------- 6VTo get top 6 V, you'd have to measure like this: First connect the red probe before top 10k resistor and black probe between the two resistors. This will give you 6 V. Here we moved the multimeters ground so that it's at different voltage than the battery negative terminal.To get the negative voltage, we should connect the red probe of the meter to bottom tap after second 10k resistor while the black probe is connected to the tap between resistors.Right now, I don't see a simple way of getting the bottom 6 V reading without inverting the voltage reading on the meter.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 27, 2013, 12:32:28 pm
content: I read somewhere that in the divider the voltages should add to the input voltage but this could be false info because I think kerchoffs law says all the voltages is a series add to 0 so the -6v seems to make more sense in that case.

author: madshaman
author_rank: Frequent Contributor
timestamp: March 27, 2013, 01:04:38 pm
content: Quote from: blewisjr on March 27, 2013, 12:32:28 pmI don't know, but maybe gravity as a force field example will help. Imagine you're on some hilly terrain with various paths you can walk. Let's say you choose a route through the paths that takes you back to your starting position.We know that your potential energy vs gravity goes up when you're climbing and down when you're descending. We also know that when you arrive back at your starting position your net change in potential energy is zero no matter what path you choose.Let's say at various points on your walk (maybe whenever your cross an intersecting path) you measure the change in your potential energy from your previous measuring point. In fact, with something a large as the earth, you can treat it as an infinite plane and assume your potential energy is linear with your altitude, so we'll just measure altitude instead.Whenever you take a measurement, call your altitude at the previous point "red" and your altitude at the current point "black" and call the difference between the two (red minus black) your "voltage drop". No matter what path you choose, the sum of your "voltage drops" will be zero.If you start at the top of the heighest hill, travelling only downwards, taking measurements as you go, you'll have a series of positive voltage drops. Let's call the lowest point in this terrain "the ground", when you've climbed down to ground, let's say you choose a path that leads directly back to the top, let's call it "battery road" or maybe "voltage source". When you reach your start position at the top, your final "voltage drop" measurement will be equal to the sum of all your previous measurements, but negative/opposite-in-sign.That's Kirchhoff's voltage law in a nutshell and also it's corollary in Maxwell's equations. The measurements in the example is what the meter is showing.Hope that makes it more clear.

author: c4757p
author_rank: Super Contributor
timestamp: March 27, 2013, 01:12:39 pm
content: Quote from: blewisjr on March 27, 2013, 12:32:28 pmThe voltagedifferences. It's just Kirchhoff's voltage law: the sum of all the voltage differences in a loop equals zero. Look at the attached circuit. If you move clockwise from ground (at the bottom left corner), you go up 30, down 10, down 10, down 10. That's (+30) + (-10) + (-10) + (-10) = 0.

author: blewisjr
author_rank: Frequent Contributor
timestamp: March 27, 2013, 01:48:24 pm
content: Ok got it makes sense great analogy and example. Now I for sure understand what is going on with voltage drop. Thanks for all the amazing help and giving your time to try different routes of explanations. If I was not working I would have had time to experiment and you would have had an easier time explaining I think but 12 hr days limit what I can do when I am on my rotation. Again much thanks.