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Ohm's Law discussion

Discussion in 'Discussions, Opinions & Editorials' started by KE5FRF, Apr 26, 2007.

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  1. KE5FRF

    KE5FRF Ham Member

    In the past week, I have encountered two examples of people having a misconception of Ohm's Law and how it works. The theme is the same in both instances, but I wanted to discuss the issue I saw today at work.

    About a month ago, a piece of lab equipment we call a "hotplate" had a switch failure. It is a "rocker" type of switch rated for 20 amps. The hotplate, which is basically a coil type heater inside a ceramic top plate designed to warm a fliud in a beaker or other container, is rated for 220 Volts at 15 Amps. When this failure occured a month ago, we replaced the switch, and lo and behold, the failure occured again. What essentially is occuring is the contact points of the switch are not making at rest position. One has to apply pressure to the switch to get the circuit to close and maintain it. Once you let up, the contacts again relax and break the circuit. I suspect that the contacts are getting dirty due to arching or too hot and warping.

    In the past month, we replaced the switch 4 times.

    Today, my coworker, a man with 30 years experience, and the apprentice we are training, a completely inexperienced young fellow, went about replacing the factory switch with a beefier 30 amp toggle switch. In passing, I mentioned that this might be a "band-aid" fix but might mask the real problem. Certainly, the 30 amp switch will probably last forever, but why were the factory switched going bad when they are rated 5 amps above the operating rating of the device? Possibly, a bad batch of switches from the factory, but that isn't likely.

    Now, here is where the fun begins. I mentioned to the senior fellow of our three man group that we may want to put an amp-probe on the circuit to check if there is a short somewhere, even a partial one. He flatly told me "Nah, its the switch...The wire terminals are dirty an adding resistance to the circuit, and the added resistance is making it get hot.". I'm in a predicament here, because I see this as an opportunity to train the new guy on the use of the test equipment and the theory of Ohm's Law. I also know that the preceding statement "the added resistance is making it get hot" was dead WRONG and totally demonstrated a lack of understanding of Ohm's Law. (Some of you guys might even argue the same thought as he, but the real Elmers here will know that this statement is 180 degrees wrong.

    Now, The first thing I said to my coworkers was, "Indeed, the switch most likely IS the problem, but it might also be a short somewhere". I said this in as polite a way as possible to prevent the appearance of argiung with the other technician in front of the trainee. AND INDEED, I MEANT WHAT I SAID. I know that the switch may be the problem, but not for the REASON that he stated, which is the point!

    Again, he said "No, you are wrong, its the switch, dirty connections will cause resistance and the resistance will cause a voltage drop and make it get hot. Resistors consume voltage and get hot." Again, I politely said that the switch may be causing the problem, and it might be the dirty contacts, but NOT because of resistance. At this point, I was more determined to properly educate the "young skywalker" on Ohm's Law rather than be diplomatic. You see, my coworker is one who flies on autopilot. He knows how to fix a lot of things, but he doesn't always understand the reasons for the failure. He has never been interested in learning anything past the no-brainer approach, and relies totally on past experience. When confronted with a problem that lies outside of this "box", he has trouble. My coworker is my friend, and I always try to be polite and nonconfrontational, but on occasion, we agree to disagree. However, I do not want the new guy to learn things incorrectly.

    So, here is what I contend. The connections on the switch were kind of dull, but not extremely so. They were not particularly pitted, just discolored. What I contend is that arcing is occuring at one or more of those contacts. They are spade type connections and you know how they get loose over time. But of this, I am not totally sure. I also mentioned that the heater element might be warping and thus changing its operating impedance or partially shorting to ground. I also mentioned that the solid state relay may be intermitently shorting to "full on" instead of pulsing the heat as its supposed to do, and thus, the avarage current and power it has to dissapate is more at times. This is not likely, but a possibility to consider. Also, the control circuit might have a component that is faulty and creating a partial short.

    Bottom line, Ohm's law requires us to understand that LOW SERIES RESISTANCES allow higher current draw, THUS, more total power and heat. High resistances cause less current draw by the load and thus less heat and power to be dissapated. A potentiometer, if one suitable to handle 15 amps were put in series with the heater, would be HOTTER WHEN THE RESISTANCE IS NEAR ZERO OHMS and COOLER WHEN THE RESITANCE IS IN THE HIGH HUNDREDS OR THOUSANDS OF OHMS.

    I explained all of this to the two guys I am working with, and I kept getting arguments about how resistors consume power...which is true, but not in the way this fellow is trying to fabricate in his mind. He also tried to go into the nature of wire gauges, and said the thinner (higher resistance) wires will get hotter than fatter because of the resistance. Again, another misconception...we all know it is the surface area and heat dissipation properties of smaller gauge wires that make them unsuitable for high currents.

    Anyway, I could not detect a short with my amp-probe in the circuit, which is a shame, because it may reinforce my friends misunderstanding and also make me lose credibility with the younger guy. As I said, I never argued that the switch wasn't the problem, and even described HOW it might be causing it. There may even be another explanation that I can't think of, and someone smarter than I can provide.

    But I think this is a very common misunderstanding of Ohm's Law, and I hope that the nature of my posting will help clear some things up with others who may not understand.
  2. N2RJ

    N2RJ Ham Member

    AS much as you try to explain to people, sometimes they are just right, even when they're wrong.

    And I'm afraid electrical ignorance goes way beyond ohm's law.

    For example, one guy tried to tell me once that when you're wiring additional outlets on the same circuit in your house, that they're wired in series.

    That's right, in series.

    He claims that the additional connector on the outlet to wire additional outlets together makes it a series circuit.

    Now, I can understand for GFI's, where the additional outlets are actually in series with the GFI outlet (and the GFI interrupts them all when a fault is detected). But he wasn't talking about GFI's.

    What a maroon, but I'm afraid they're everywhere.
  3. KL7AJ

    KL7AJ Ham Member

    What you need to do is check the current AT THE MOMENT OF SWITCH CLOSURE. Heating elements draw a lot more current cold than when they're up to temperature. (A standard lightbulb has 10x the resistance hot as it does cold!) Heating elements aren't as extreme, but certainly DO draw larger currents at startup.

    I'll bet almost anything you're torching your switch during turn-on.

  4. N2NH

    N2NH Ham Member


    ... and I'll be he don't know Jack. [​IMG]
  5. KB3LIX

    KB3LIX Ham Member

    Not to burst you bubble Heath, but dirty/corroded connections at the switch COULD indeed be the problem.

    The added series resistance would reduce the over-all current flow in the circuit, but the heat developed in and around the switch by the higher than normal series resistance of the switch (and dirty/corroded connections from the wires to the switch) could warp the internal contacts of the switch causing it to open and fail prematurely.

    The idea of checking the total current flow is a sound idea, and may indicate that the element has a higher than normal current, but cleaning the contacts between the conductors and the switch contact points would probably be the first step I would take.

    If there were a problem in the heating element, you probably would have noticed a change in the heating characteristics before the switch failed. Either too much heat, one particular hot spot, or an open breaker on the mains.

    Lots of different ways to skin the cat !

    I just saw Eric's post above, cold surge current could do the same thing.

    Lots of different ways to skin the cat !

    That is why troubleshooting is an art
  6. KE4PJW

    KE4PJW Ham Member

    Actually he is not.

    R1 R2
    Scenario 1
    R1) 1 ohm switch resistance
    R2) 99 ohm heating element
    Apply 100volts, total power consumption 100 watts.
    1 watt dissipated by R1, 99 watts by R2.

    Scenario 2
    R1) 100ohm switch resistance
    R2) 100ohm heating element
    Apply 100 volts. total power consumption 50 watts.
    25 watts dissipated by R1, 25 watts dissipated by R2.

    Which switch is hotter? The one dissipating the most power.
  7. KE5FRF

    KE5FRF Ham Member

    Go back and read my post. I made it clear that yes indeed, the switch connections could cause the problem, BUT NOT BECAUSE OF RESISTANCE. That's the key here. Its NOT RESISTANCE THAT CAUSES HEAT. Arcing or some other abnormal situation, maybe, but not ADDED SERIES RESISTANCE. See what I mean now?

    BTW, the cold switching of the device is not the problem either. WE SELDOM TURN THE DEVICE OFF. And as a troubleshooting mechanism, we instructed the lab technicians not to turn it off at all.
  8. K7KBN

    K7KBN Ham Member

    Current flowing through switch contacts that are clean, smooth and tight would produce very little heat because of very low contact resistance.

    However, if the switch contacts are dirty, rough and/or loose, resistance increases, and current flowing through a resistance DOES produce heat.

    "Twinkle twinkle little star,
    Power equals I squared R."
  9. KL7AJ

    KL7AJ Ham Member

    Oh, how clever. I thought I'd heard them all. [​IMG]
  10. KD6NIG

    KD6NIG Ham Member

    I don't even think your tech even cared about ohms law or anything, and thats the unfortunate thing.

    He only cared about fixing it.

    Now, that may not be the truth, but in most situations like this nowadays, the fix seems to be whatever makes it work, and whatever is cheapest to do said fix.

    Honestly if this problem is continuing, I would question the safety of that hot plate. Not that I think its gonna explode or anything, but anything is possible, and with the possibility of that, comes the possibility of injury. I would question its integrity.

    But I know if I had been the tech in the situation, I would have been told to replace the switch. Git R Done.

    Now, in this situation it may not apply. But what I'm speaking of here was something I noticed when I was in technical school for computers. The goal was no longer to troubleshoot the actual problem, the goal was to find out what part was bad, yank that board, and put a new one in, and more often than not, toss the old-maybe send it back for credit, but in most cases, toss and carry on. Not that its a bad system per se, but I've seen situations in real life where a tech came out, "fixed" the problem and then a bigger problem arose because the part he fixed had taken the heat for the bigger problem, and after replacing the part that had went bad because of the other part, the problem manifested itself.

    In this situation, the fact that the tech was closed minded to it doesn't tell me he didn't want to deal with it. I just honestly have a feeling thats how he was trained. With something like that, I would have taken the measurement, but I wouldn't have told my boss. But since its a simple diagnostic to take, I would have done it anyway.

    Its just that in these days of lore, it appears to be quicker to just tackle the obvious problem, and handle other problems later when they crop up-instead of checking a little further to make sure other stuff didn't get damaged, etc.

    Case in point-a new building is going in next door where I work. They just hooked up the electrical feed last week. Right after they did it, (the next day in fact) one of the "safeties" (on the pole where the feed goes into the building) blew out. The power company came out and was there and gone in 10 minutes-I'm guessing they didn't check anything, they just replaced the thing that blows on the pole and left.

    My suspicions were confirmed yesterday when I heard another pop again. This time though, the safety didn't blow-it was arcing right where the wires went into the conduit onto the pole (they then go under the street to the building) and the pole was on fire. We had to call the fire dept out to check it out.

    I'm just hoping the guy didn't put a larger safety on the pole and that caused it to do this, but it was arcing nicely, you could hear it popping and stuff down here, and it left a nice scorch mark on the pole. The only lights I see the building using right now are the parking lot ones, so something is up.

    I'm just surprised that they didn't check more, but I guess it works the same across the board. Get it fixed and working.

    They were out there for about 7 hours yesterday. Dunno what it was, but I guess the fire made them take a better look......

    Sorry for my rambling, but just trying to make my point.
  11. KE4PJW

    KE4PJW Ham Member

    Actually Heath, you are incorrect here as well. Let say you have a 10 volt 100 watt lamp. That means that you have a current draw of 10 amps and an internal resistance of 1 ohms. So let compare wiring this lamp up with 100ft of 20 gauge wire and 10 gauge wire.

    With the 10 gauge wire, the total resistance of the wire is .2ohms. Total circuit resistance is 1.2ohms. With a 10 volt supply, it would draw 8.33 amps and dissipate 83.3 watts. The power dissipated by the wire would be 13.88 watts.

    With the 20 gauge wire, the total resistance of the wire is 2.08 ohms. Total circuit resistance is 3.08 ohms. With a 10 volt supply, it would draw 3.25 amps and dissipate 32.5 watts. The power dissipated by the wire would be 21.97 watts.

    In this application, the resistance of the wire makes the 20 gauge wire unsuitable, not it's current capacity.
  12. W9RFM

    W9RFM Ham Member

    Look up series arcing faults. Arcing can cause great heat, and also adds impedance.
  13. KE5FRF

    KE5FRF Ham Member


    I'm sure that a theoretical scenario can be dawn on a piece of paper that shows a situation where two equal resistors will dissipate the same amount of power. But I think the crux here is that the RESISTANCE does NOT generate the heat. It is the total power in the system, the current and the voltage, coupled with the low resistance of the heating element, that create the heat. You yourself demonstrated in your pictorial that 100 ohms creates 100 watts in your scenario but 200 ohms creates 50 watts.

    Now, the key here is that the instrument was operating normally up until the point of the failure. The heating element was sufficiently heating the liquids at the setpoint we always expect. There was no evidence of inefficency or having power robbed from somewhere else.

    Again, I admitted from the word go that the switch could be the problem, and even said it likely IS the problem. But it was the way that my coworker said "NO" and argued the other possibilities were wrong.

    But I am glad you brought up this example, because it did offer something deeper that my mind had forgotten and overlooked. I assure you, friend did not think the scenario through at all and has never done the first Ohm's Law calculation in his career.

    It is an interesting twist that you propose, and indeed could be the culprit here. Perhaps this is why some people err in their understanding of Ohms law. Little tricky scenarios like the one you just showed. But again, the key here is that the resistance doesn't CAUSE the heat. If this were the case, would we not use beefy 1 megohm resistors as heating elements?

    But your scenario makes the discussion interesting and makes understanding the rest of it more complicated. Thanks for throwing the monkey wrench into the works.

    Do you understand what I am getting at here? My friend [​IMG]
  14. KL7AJ

    KL7AJ Ham Member

    You have to have SOME resistance to generate heat. In the equation I^2*R, if R is zero, so is the heat (power)!

    All other things being equal (i.e. a fixed voltage source with infinite current capacity) the power dissipated in the load (heat generated) will always be greater with smaller resistance. However, the current also will rise as the inverse of the resistance...hence the SQUARED term.

    On the other hand, if the load resistance is on the same order as the resistance of your power cord, the most power you can achieve is when the load resistance is EQUAL to that of the power cord! You start losing ground at any value of R less than that.

  15. KE5FRF

    KE5FRF Ham Member

    OK, Terry. The problem here is that you are talking about the distribution of power through the circuit and not the cause of the heat. I guess you could say we are both correct here in the facts that we are presenting, but the bottom line is that RESISTANCE DOESN'T CAUSE THE HEAT NOR THE CONSUMPTION OF POWER. The distribution of the resistance only does just that, DISTRIBUTES THE HEAT AND THE POWER. The heat and the power are already there, and given a low total impedance of the circuit, will be greater than occurs when you add resistance.

    Indeed, the switch fault, I concede, might be an example of where power was redisributed due to a voltage drop. But still, the basic precept here is that P=ExI....if totsl "I" is lower, total P will be lower. My friend has ignored this and formulated a generalized misconception that a high resistance will generate heat. Again, if this were the case, would we not be building heater elements from 1 megohm materials? NO. We do not because 1 megohm prevents current flow.

    This has turned into an interesting conversation...I didn't expect it to.

    But again, the underlying principal I was trying to demonstrate is that current flowing through an electrical system, with a sufficient voltage, is what creates the heat in the system. The resistance or impedance of the circuit stands to inhibit current flow, thus decrease the total power consumption. A 100 watts light bulb has a lower impedance than a 50 watt light bulb. The resistance doesn't create the heat, it only redistributes it.

    SHWEW...this is a tangled discussion, [​IMG]
  16. KE4PJW

    KE4PJW Ham Member

    No problem Heath. What I am attempting to convey is prespective.

    If for some reason, say, arching, the switch begins to go up in resistance, it will dissipate more power. Say it dissipates 25 watts. Ok, that's not enough to keep the heating element from operating, but it is more than enough to completely destroy the switch mechanism.

    Let's say the heating element is 2KW @ 110V. It pulls 18.18 amps and has a resistance of 6.05 ohms.

    If the switch has a resistance of 0.075 ohms, current draw would be 17.96 amps and total power would be 1945 watts.

    The switch would dissipate 24.2 watts.

    The higher the resistance in the switch, the more heat the switch dissipates. The [higher] resistance causes the heat because you are dealing with a voltage divider circuit.

    Anyway, hope that helps you understand why he said, what he said. Good luck, and let me know what you find out.
  17. KE5FRF

    KE5FRF Ham Member

    OK, Joshua nailed this right smack dab on the head. My exact point.

    There are a lot of people out there doing tech type jobs that fly on autopilot. Those folks are usually very competant at what they do, so flying on autopilot is not neccessarily a bad thing, because they take their experience as far as they can. Experience is an awesome tool in a troubleshooting arsenal. But to be just a little better than competent, one should strive to understand the principals behind things. I do not claim to be a super-tech...and a lot of things I have learned and forgotten, or never learned at all. But one thing I always try to do is consider all the possibilities.

    With my friend, he refused to investigate other potential problems not because he understands Ohm's law better than I, but rather because he DOESN'T. His technique of going about repairs is to replace parts. The only time he uses a VOM is to do continuity checks or see if a power supply is putting out 12 volts. He has never performed a pont to point check in his life. I'm not disrespecting him...he's a good technician, but he doesn't want to be bothered with other fault checking procedures. Replace the switch and be done with it. What if there WAS a partial short somewhere in the circuit? We would have replaced a 20 amp switch with a 30 amp switch, and had the potential for a fire or a meltdown.

    I wanted to make the new technician understand things a little deeper than just "change the switch". I wanted him to understand the reasons for doing it and for checking other problem possibilities as well.
  18. VA2GK

    VA2GK XML Subscriber

    Ohh man, noooooooooooooooooooooooooooooo [​IMG]

    Dozens of years of studies and dozens of years of experience down the drain [​IMG]

    I was sure an increased resistance was creating more heat, but I was relying on ohm's law, and now it's clear old Gorg was wrong all along.

    You kill me man, I am not worthy [​IMG]
  19. KE5FRF

    KE5FRF Ham Member

    Terry, you are 100 percent correct...except the part where you said..."I hope this helps you understand why he said what he said".

    I ASSURE YOU, I've been working with my friend for almost eight years..My friend does not know why he said what he said....In fact, my friend's explanation for what he said would be totally wrong, as I have outlined. I must admit that my mind has not went where yours went, and this demonstrates an inadequate thought process on my part...for which I am embarassed...but my thought process was based on a foundation of the basics...I assure you that with my friend, it was just flying on autopilot.

    Thanks everyone for your sage input on this thread. I will be taking these examples back to my coworkers and discussing them. Ohm's law seems so simple on the surface, but can be work in odd ways when you dig a little deeper.
  20. N2RJ

    N2RJ Ham Member

    The best part was that he said that the little tab that you break to separate the two outlets (for example if one is a lamp outlet controlled by a switch) is actually a component, which is why it's a series circuit.

    I suspect he said that after realizing how much he looked like a complete ass, but that was comedy at its best.
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