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Thread: problem with parallel resistor formula

  1. #1 problem with parallel resistor formula 
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    Ok, I am having trouble understanding how to calculate the resistance of resistors in parallel as you probably gathered from the name of the post.

    I get resistors in series of course its really simple, and at first glance I thought resistors in parallel seemed simple as well but for some reason my answers are coming up wrong.

    I'm not sure where I am making a mistake so I cant really explain it, instead I hope someone here can give me a very basic explanation of the formula and maybe go step by step through this for me;

    3 resistors in parallel lets say 2000, 3000, and 4000

    Also if its not too much to ask I was wondering exactly how a certain scenario would work say if you had 3 resistors as follows; |=resistor

    ||
    |

    See? So that 2 are in series and 2 are parallel yet there are only 3 resistors. Would I calculate the 2 parallel resistors first then simply add the left over resistors value?

    Thanks in advance for the answers!


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    There are two ways you can do it. If you can more easily remember the formula
    Rparallel=R1R2/(R1+R2)
    then you can do it in two steps. Calculate the parallel resistance of the 2000 and 3000 ohm resistor. Then that resistance is in parallel with the 4000 ohms. So you would have
    2000*3000/5000=1200
    1200*4000/5200=923

    The other way to do it is to realize that the inverse of the resistance is conductance and conductance adds in parallel. This method gives you

    1/2000+1/3000+1/4000=1.083E-03
    1/.083E-03=923

    For the last part of the question you are right. Calculate the parallel resistance of the two in parallel then add that to the other resistor in series.


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    Quote Originally Posted by Harold14370
    There are two ways you can do it. If you can more easily remember the formula
    Rparallel=R1R2/(R1+R2)
    then you can do it in two steps. Calculate the parallel resistance of the 2000 and 3000 ohm resistor. Then that resistance is in parallel with the 4000 ohms. So you would have
    2000*3000/5000=1200
    1200*4000/5200=923

    The other way to do it is to realize that the inverse of the resistance is conductance and conductance adds in parallel. This method gives you

    1/2000+1/3000+1/4000=1.083E-03
    1/.083E-03=923

    For the last part of the question you are right. Calculate the parallel resistance of the two in parallel then add that to the other resistor in series.
    One problem I always had with resistors in parallel. Is that if they are not exactly equal. Something that is almost impossible to do.
    Those formulas may not hold on AC current at high hertz and voltage. I believe that is why they demand transformers isolating Halogen building lights in my area.

    If there was some kind of a repeated spike. The lower resistance bulbs could blow. Sending ARC rays from the molten surface of the metal, directly back into the system. Or even the neutral. At very high hertz rates. Benjamin Franklin did rather interesting work on this.

    I tried to warn local non-union electricians about this. But they could not quite understand what I was saying. A week later the union picketed the job and I stood out there with them. Because there are reasons why there are ballasts in lightning. It is for the protection of the grid.

    I have seen what appears to be transformer less lights at Federal buildings and airports. Unless they isolate them from one central location, however I do not believe so. Some prisons still have their transformer isolated halogen lights.

    Transformers can limit hertz and reduce the spikes. Even though we all know transformers for spiking us, when playing with them. Ha-ha.

    But in welding I know that two different lengths of 4/0 wire connected to the same ground terminal, will create power between them. At high hertz. That is amazing. It can create an actual 20 amps of power, between two 4/0 wires connected to each other at the ground on the machine.

    But I do agree with your formulas entirely, for standard line load ohms readings. And I even got an education, learning to add up all three. One problem though. Does that last formula work with one ohm? I don't see how. Looking at it again I see how.

    I looked at it again, in series. It still does not seem right to me. Maybe you should just add the series resistors?



    Sincerely,


    William McCormick
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    William,

    I hate when you post on this subforum because, since I am the moderator, I have to temporarily take you off my ignore list to read your bloviations. Please try to stick to the topic and don't ramble on and reminisce like you do.

    Yes, of course it works with 1 ohm. Try it. Take a 1 ohm, 2 ohm and 3 ohm resistor in parallel.
    Parallel conductance =1/1+1/2+1/3=1.833 mhos
    Parallel resistance = 1/1.833=0.545 ohms
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    Quote Originally Posted by Harold14370
    William,

    I hate when you post on this subforum because, since I am the moderator, I have to temporarily take you off my ignore list to read your bloviations. Please try to stick to the topic and don't ramble on and reminisce like you do.

    Yes, of course it works with 1 ohm. Try it. Take a 1 ohm, 2 ohm and 3 ohm resistor in parallel.
    Parallel conductance =1/1+1/2+1/3=1.833 mhos
    Parallel resistance = 1/1.833=0.545 ohms
    So then you can only use it for parallel not series? Unless you add the series ohms to end of the formula.

    I thought you were doing two parallel resistors and one in series. It does seem to work in parallel. However I tried it with both parallel and series and it could not work.


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    William McCormick
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    Quote Originally Posted by William McCormick
    So then you can only use it for parallel not series? Unless you add the series ohms to end of the formula.
    Yes, that's what you have to do.
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    I remember when I first learned how to calculate parallel resistance. I think I was always looking for an easier answer, that in turn made me resistant to the correct formula. I don't even think about it anymore. I find a spreadsheet is the easiest way, just create a bunch of cells with the correct formulas and plug in the number. I use spreadsheets for pretty much everything these days.
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    Quote Originally Posted by (In)Sanity
    I remember when I first learned how to calculate parallel resistance. I think I was always looking for an easier answer, that in turn made me resistant to the correct formula. I don't even think about it anymore. I find a spreadsheet is the easiest way, just create a bunch of cells with the correct formulas and plug in the number. I use spreadsheets for pretty much everything these days.
    I rarely use ohms. Almost everything is in amps. Because the ohms at rest, in most circuits do not match the run ohms. The ohms are created by powering the equipment. In most of the heater and induction loads.

    Some heater loads are fixed ohm. And I can use that formula.

    Harold's formula converts the ohms to amps using one volt as the input, and the formula volts/ohms=amps, it then adds up or totals the amperes. After you total the amps then the formula divides one volt by amps to get the ohms. Using the formula Volts/amps=ohms.

    You could also use the watts instead. They are the same value as the amps at one volt with a similar ohm input.

    We just do not normally do it like that. Although using a WIRE calculator we do actually do it just like that, we just do not even realize it.


    http://www.Rockwelder.com/Electricity/setupwir.exe

    This is a handy little program, you can create a bunch of copies of it on the desktop and workout circuits. Compare different scenarios. It is free to use.

    I just misunderstood the example.


    Sincerely,


    William McCormick
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  10. #9  
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    William, in my world your ideas would not work. I work with microprocessors and mostly low current analog circuits. Heating and increase in resistance is not something that I often have to consider. The only time I do is when trying to make a temperature compensated voltage reference or if I'm worried about frequency drift.
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    Quote Originally Posted by (In)Sanity
    William, in my world your ideas would not work. I work with microprocessors and mostly low current analog circuits. Heating and increase in resistance is not something that I often have to consider. The only time I do is when trying to make a temperature compensated voltage reference or if I'm worried about frequency drift.

    Which ideas?


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    William McCormick
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  12. #11  
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    Quote Originally Posted by Harold14370
    There are two ways you can do it. If you can more easily remember the formula
    Rparallel=R1R2/(R1+R2)
    then you can do it in two steps. Calculate the parallel resistance of the 2000 and 3000 ohm resistor. Then that resistance is in parallel with the 4000 ohms. So you would have
    2000*3000/5000=1200
    1200*4000/5200=923
    Remember that the formula "Rparallel=R1R2/(R1+R2)" only works for two resistors in parallel. It is a shortcut since most parallel resistance problems only have two resistors in parallel. If you have more parallel resistors, you have to use the more general form that Harold stated later in his post.
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    Quote Originally Posted by jammer
    Quote Originally Posted by Harold14370
    There are two ways you can do it. If you can more easily remember the formula
    Rparallel=R1R2/(R1+R2)
    then you can do it in two steps. Calculate the parallel resistance of the 2000 and 3000 ohm resistor. Then that resistance is in parallel with the 4000 ohms. So you would have
    2000*3000/5000=1200
    1200*4000/5200=923
    Remember that the formula "Rparallel=R1R2/(R1+R2)" only works for two resistors in parallel. It is a shortcut since most parallel resistance problems only have two resistors in parallel. If you have more parallel resistors, you have to use the more general form that Harold stated later in his post.
    Just give me a good spice simulator and I'm happy. LTSpice works pretty good.
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    Quote Originally Posted by (In)Sanity
    Just give me a good spice simulator and I'm happy. LTSpice works pretty good.
    Simulators are good but they dont help give an underlying knowledge of how it works, thats like using a calculator for basic addition. Doing the maths makes you think about it more.

    The poor guys probably been scared off now anyway. Revealing the infinite complexitys of electricity isnt going to help him understand ohms law any better.
    'Aint no thing like a chicken wing'
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    Quote Originally Posted by harvestein
    The poor guys probably been scared off now anyway. Revealing the infinite complexitys of electricity isnt going to help him understand ohms law any better.
    I have to admit, once I realized I had to apply not all that complex but somewhat cumbersome formulas to branch circuits and such and then apply those results to even more formulas to get the total resistance I was a bit taken back.

    Without a doubt spice simulators don't replace basic knowledge. They do however let you test what you've learned without having mosfets blown in to pieces and smoking resistors
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  16. #15  
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    Quote Originally Posted by (In)Sanity
    Without a doubt spice simulators don't replace basic knowledge. They do however let you test what you've learned without having mosfets blown in to pieces and smoking resistors
    Are you crazy?! Thats the best part!!

    By the way when i was referring to scaring the guy off i was referring to Williams outlook on the world of ohms law. Not anything that you had said. :?
    'Aint no thing like a chicken wing'
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