The Ohm's law calculations of the parts do not agree with the whole












1












$begingroup$


In the circuit below, the Ohm's law calculations of the parts do not agree with the whole:




  1. the voltage drops of the parts do not add up to the total voltage, and

  2. the current does not calculate to be the same throughout the entire circuit.


Why? Actual VOM readings are in red and done with a digital VOM.



Circuit anomaly










share|improve this question









New contributor




Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$








  • 1




    $begingroup$
    What instrument was used to measure the voltages?
    $endgroup$
    – TimWescott
    9 hours ago










  • $begingroup$
    I assume you realize that I should have asked "Why?" before adding the last sentence, but I can see how that might be confusing. Maybe I can still edit my original comment and put "why?" in the correct place. The device I used is a E SUN DT830 Digital Multimeter.
    $endgroup$
    – Sanity Check
    9 hours ago








  • 1




    $begingroup$
    Is there a resistance or loss between the components? Meaning if you check from the lead of 1 bulb to the other lead of the other is there a loss there? Is there a voltage drop between the last bulb and the battery? Voltage doesn't just disappear there is something that is measured when you measure both that is not being measured when measuring individually. If not then you have a bad meter.
    $endgroup$
    – Robert Fay
    9 hours ago












  • $begingroup$
    Mine was a poorly-phrased request for instrument type and model number -- because a meter with an offset could give you results such as you see. Voltage loss in the wires, or shooting at a moving target as the battery voltage decreases, is much more likely.
    $endgroup$
    – TimWescott
    9 hours ago
















1












$begingroup$


In the circuit below, the Ohm's law calculations of the parts do not agree with the whole:




  1. the voltage drops of the parts do not add up to the total voltage, and

  2. the current does not calculate to be the same throughout the entire circuit.


Why? Actual VOM readings are in red and done with a digital VOM.



Circuit anomaly










share|improve this question









New contributor




Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    What instrument was used to measure the voltages?
    $endgroup$
    – TimWescott
    9 hours ago










  • $begingroup$
    I assume you realize that I should have asked "Why?" before adding the last sentence, but I can see how that might be confusing. Maybe I can still edit my original comment and put "why?" in the correct place. The device I used is a E SUN DT830 Digital Multimeter.
    $endgroup$
    – Sanity Check
    9 hours ago








  • 1




    $begingroup$
    Is there a resistance or loss between the components? Meaning if you check from the lead of 1 bulb to the other lead of the other is there a loss there? Is there a voltage drop between the last bulb and the battery? Voltage doesn't just disappear there is something that is measured when you measure both that is not being measured when measuring individually. If not then you have a bad meter.
    $endgroup$
    – Robert Fay
    9 hours ago












  • $begingroup$
    Mine was a poorly-phrased request for instrument type and model number -- because a meter with an offset could give you results such as you see. Voltage loss in the wires, or shooting at a moving target as the battery voltage decreases, is much more likely.
    $endgroup$
    – TimWescott
    9 hours ago














1












1








1





$begingroup$


In the circuit below, the Ohm's law calculations of the parts do not agree with the whole:




  1. the voltage drops of the parts do not add up to the total voltage, and

  2. the current does not calculate to be the same throughout the entire circuit.


Why? Actual VOM readings are in red and done with a digital VOM.



Circuit anomaly










share|improve this question









New contributor




Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




In the circuit below, the Ohm's law calculations of the parts do not agree with the whole:




  1. the voltage drops of the parts do not add up to the total voltage, and

  2. the current does not calculate to be the same throughout the entire circuit.


Why? Actual VOM readings are in red and done with a digital VOM.



Circuit anomaly







voltage voltage-measurement






share|improve this question









New contributor




Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 9 hours ago









Dave Tweed

120k9149257




120k9149257






New contributor




Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 9 hours ago









Sanity CheckSanity Check

93




93




New contributor




Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sanity Check is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    What instrument was used to measure the voltages?
    $endgroup$
    – TimWescott
    9 hours ago










  • $begingroup$
    I assume you realize that I should have asked "Why?" before adding the last sentence, but I can see how that might be confusing. Maybe I can still edit my original comment and put "why?" in the correct place. The device I used is a E SUN DT830 Digital Multimeter.
    $endgroup$
    – Sanity Check
    9 hours ago








  • 1




    $begingroup$
    Is there a resistance or loss between the components? Meaning if you check from the lead of 1 bulb to the other lead of the other is there a loss there? Is there a voltage drop between the last bulb and the battery? Voltage doesn't just disappear there is something that is measured when you measure both that is not being measured when measuring individually. If not then you have a bad meter.
    $endgroup$
    – Robert Fay
    9 hours ago












  • $begingroup$
    Mine was a poorly-phrased request for instrument type and model number -- because a meter with an offset could give you results such as you see. Voltage loss in the wires, or shooting at a moving target as the battery voltage decreases, is much more likely.
    $endgroup$
    – TimWescott
    9 hours ago














  • 1




    $begingroup$
    What instrument was used to measure the voltages?
    $endgroup$
    – TimWescott
    9 hours ago










  • $begingroup$
    I assume you realize that I should have asked "Why?" before adding the last sentence, but I can see how that might be confusing. Maybe I can still edit my original comment and put "why?" in the correct place. The device I used is a E SUN DT830 Digital Multimeter.
    $endgroup$
    – Sanity Check
    9 hours ago








  • 1




    $begingroup$
    Is there a resistance or loss between the components? Meaning if you check from the lead of 1 bulb to the other lead of the other is there a loss there? Is there a voltage drop between the last bulb and the battery? Voltage doesn't just disappear there is something that is measured when you measure both that is not being measured when measuring individually. If not then you have a bad meter.
    $endgroup$
    – Robert Fay
    9 hours ago












  • $begingroup$
    Mine was a poorly-phrased request for instrument type and model number -- because a meter with an offset could give you results such as you see. Voltage loss in the wires, or shooting at a moving target as the battery voltage decreases, is much more likely.
    $endgroup$
    – TimWescott
    9 hours ago








1




1




$begingroup$
What instrument was used to measure the voltages?
$endgroup$
– TimWescott
9 hours ago




$begingroup$
What instrument was used to measure the voltages?
$endgroup$
– TimWescott
9 hours ago












$begingroup$
I assume you realize that I should have asked "Why?" before adding the last sentence, but I can see how that might be confusing. Maybe I can still edit my original comment and put "why?" in the correct place. The device I used is a E SUN DT830 Digital Multimeter.
$endgroup$
– Sanity Check
9 hours ago






$begingroup$
I assume you realize that I should have asked "Why?" before adding the last sentence, but I can see how that might be confusing. Maybe I can still edit my original comment and put "why?" in the correct place. The device I used is a E SUN DT830 Digital Multimeter.
$endgroup$
– Sanity Check
9 hours ago






1




1




$begingroup$
Is there a resistance or loss between the components? Meaning if you check from the lead of 1 bulb to the other lead of the other is there a loss there? Is there a voltage drop between the last bulb and the battery? Voltage doesn't just disappear there is something that is measured when you measure both that is not being measured when measuring individually. If not then you have a bad meter.
$endgroup$
– Robert Fay
9 hours ago






$begingroup$
Is there a resistance or loss between the components? Meaning if you check from the lead of 1 bulb to the other lead of the other is there a loss there? Is there a voltage drop between the last bulb and the battery? Voltage doesn't just disappear there is something that is measured when you measure both that is not being measured when measuring individually. If not then you have a bad meter.
$endgroup$
– Robert Fay
9 hours ago














$begingroup$
Mine was a poorly-phrased request for instrument type and model number -- because a meter with an offset could give you results such as you see. Voltage loss in the wires, or shooting at a moving target as the battery voltage decreases, is much more likely.
$endgroup$
– TimWescott
9 hours ago




$begingroup$
Mine was a poorly-phrased request for instrument type and model number -- because a meter with an offset could give you results such as you see. Voltage loss in the wires, or shooting at a moving target as the battery voltage decreases, is much more likely.
$endgroup$
– TimWescott
9 hours ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

the most probable causses for the discrepancies are the following:




  1. The DT830 is not known as a quality meter. Inaccuracies in the readings are one cause of the problem.

  2. You are drawing almost 300 ma from a D battery. At that level the voltage of the battery will tend to drop fairly quickly. Depending on how fast you take the measurements, the readings can vary due to that factor. Note that your own readings show a 0.2 drop in voltage from no load to full load.
    As a test of your meter try measuring the voltage of each of your D cells and compare it to the total voltage.






share|improve this answer











$endgroup$













  • $begingroup$
    Barry Meter error would be gain or offset but even 5% would not account for these results. But I agree the battery could drop 2x (.4 to .2V) this much on start up as the load resistance starts at 11 Ohms (10.1+1) and rises to > 21 ohms but even this wont explain the errors
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago





















2












$begingroup$

Ohm's Law and KVL always applies but the voltage drop is nonlinear with time due to thermal effects so bulbs never match perfectly and the voltages are never equal.



Bulbs are nonlinear PTC ( positive temp coefficient ) conductors that rise in R by 10x at rated power with temperatures of say 2500'K.



So R (hot/cold) ratio ~ 10:1 for bright warm white light (hot) and room temp (cold).



This means if you put two 6V bulbs in series with a 6V battery, you will never ever get 3V each.



The bulb with even the slightest higher cold resistance heats up faster in temperature will rise in resistance faster and thus drop more voltage than the other bulb, resulting in a runaway condition where that bulb will have full power and the other about 10 %. Putting a resistor in series as you have done reduces the balance difference between R cold= est. 0.5 Ohms and R hot est= 5 Ohms x 0.3A 1.5V



Although this is the extreme case where the bulbs are most sensitive to resistance changes at half voltage, it means your measurements probably changed while you were taking the readings.



Is it repeatable? Is it stable? Measure again.



If the load voltage is 5.9V @ 0.3A and the no-load Vbat=6.1V then the 4 battery cells have an internal resistance ESR = 0.2V/0.3A= 0.67 Ohms total.






share|improve this answer











$endgroup$













  • $begingroup$
    let me run a quick simulation to prove that
    $endgroup$
    – Sunnyskyguy EE75
    9 hours ago






  • 1




    $begingroup$
    a nitpick: saying that "Ohm's [l]aw always applies" is a heavy overstatement. No physical "law" known to human always applies. Every "law" is just a model - and every model is false, because every model is just an approximation - albeit some, as OL or KVL, are indeed extremely useful.
    $endgroup$
    – vaxquis
    8 hours ago






  • 1




    $begingroup$
    @vaxquis I see no reason to doubt my conclusions or my statement. The readings can change from test to test reading as heat transfers between bulbs and changes the results. But Ohm's Law using simultaneous readings will not fail since this is DC and there is no interference from the DMM or VOM. Do you have an experience to contradict this?
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago








  • 1




    $begingroup$
    Since it only becomes a Law when experimental data supports it after Peer review in the Physics community, do you have any data to support your doubt? Or just a hypothetical one. Just remember the time constants involved resemble a reactor but not. Ohm's Law still applies to nonlinear resistive parts like LEDs within the constraints of linear regression. over a limited range. I need data, to accept your doubt.
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago








  • 1




    $begingroup$
    Question for the mods - is it possible to down-vote a comment?
    $endgroup$
    – AnalogKid
    2 hours ago



















2












$begingroup$


The Ohm's law calculations of the parts do not agree with the whole




Actually, there's a very plausible explanation which would mean that Ohm's law is alive and well, and does explain what you report.



(Note: I'm assuming that the bulbs have reached their steady-state "hot" resistances, so that those resistances don't change during the measurements. I'm also assuming that the battery voltage doesn't drop significantly during the measurements. It will drop, but I'm assuming that doesn't happen to a significant amount while measurements are being made.)



Hypothesis: Additional wiring resistances



Your diagram doesn't show additional resistances which must be present in your setup.



Let's look at just one area in your diagram as an example - the voltages across the two bulbs together and individually.



Here is your diagram again (so that readers don't have to keep scrolling up to the question):



wiring diagram from the question



Voltage drop across bulb 1 = 1.13V

Voltage drop across bulb 2 = 1.24V

Voltage drop across both bulbs and the wiring between bulbs = 2.6V

Therefore "unexpected" additional voltage drop = 2.6V - (1.13V + 1.24V) = 0.23V



Although the 10.2 Ω resistor value might not be very accurate (e.g. multimeter lead resistance is probably included in that, so the resistor's true value is likely lower) for now, let's use your approximate current value of (2.85V / 10.2 Ω =) 0.28A (to 2 decimal places).



Therefore the additional voltage drop between the two bulbs is only a resistance of (0.23V / 0.28A =) 0.82 Ω which is easy to believe, due to cable type / length / connection resistance etc.





The same reasoning explains why 2.85V (across the resistor) + 2.6V (across the bulbs) doesn't equal 5.9V - there are additional resistances (and therefore voltage drops) in the wiring / connections, which aren't included.





Summary: Go through the actual circuit hardware whose diagram you gave, measuring the voltage drops across not only the obvious parts, but also measure the voltage drop across every wire and connection of any kind.



You will find additional voltage drops which, when added to the voltage drops across the main components already given, I expect will show that the total voltage drops do all sum to the expected value.






share|improve this answer











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    the most probable causses for the discrepancies are the following:




    1. The DT830 is not known as a quality meter. Inaccuracies in the readings are one cause of the problem.

    2. You are drawing almost 300 ma from a D battery. At that level the voltage of the battery will tend to drop fairly quickly. Depending on how fast you take the measurements, the readings can vary due to that factor. Note that your own readings show a 0.2 drop in voltage from no load to full load.
      As a test of your meter try measuring the voltage of each of your D cells and compare it to the total voltage.






    share|improve this answer











    $endgroup$













    • $begingroup$
      Barry Meter error would be gain or offset but even 5% would not account for these results. But I agree the battery could drop 2x (.4 to .2V) this much on start up as the load resistance starts at 11 Ohms (10.1+1) and rises to > 21 ohms but even this wont explain the errors
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago


















    5












    $begingroup$

    the most probable causses for the discrepancies are the following:




    1. The DT830 is not known as a quality meter. Inaccuracies in the readings are one cause of the problem.

    2. You are drawing almost 300 ma from a D battery. At that level the voltage of the battery will tend to drop fairly quickly. Depending on how fast you take the measurements, the readings can vary due to that factor. Note that your own readings show a 0.2 drop in voltage from no load to full load.
      As a test of your meter try measuring the voltage of each of your D cells and compare it to the total voltage.






    share|improve this answer











    $endgroup$













    • $begingroup$
      Barry Meter error would be gain or offset but even 5% would not account for these results. But I agree the battery could drop 2x (.4 to .2V) this much on start up as the load resistance starts at 11 Ohms (10.1+1) and rises to > 21 ohms but even this wont explain the errors
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago
















    5












    5








    5





    $begingroup$

    the most probable causses for the discrepancies are the following:




    1. The DT830 is not known as a quality meter. Inaccuracies in the readings are one cause of the problem.

    2. You are drawing almost 300 ma from a D battery. At that level the voltage of the battery will tend to drop fairly quickly. Depending on how fast you take the measurements, the readings can vary due to that factor. Note that your own readings show a 0.2 drop in voltage from no load to full load.
      As a test of your meter try measuring the voltage of each of your D cells and compare it to the total voltage.






    share|improve this answer











    $endgroup$



    the most probable causses for the discrepancies are the following:




    1. The DT830 is not known as a quality meter. Inaccuracies in the readings are one cause of the problem.

    2. You are drawing almost 300 ma from a D battery. At that level the voltage of the battery will tend to drop fairly quickly. Depending on how fast you take the measurements, the readings can vary due to that factor. Note that your own readings show a 0.2 drop in voltage from no load to full load.
      As a test of your meter try measuring the voltage of each of your D cells and compare it to the total voltage.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 9 hours ago









    Dave Tweed

    120k9149257




    120k9149257










    answered 9 hours ago









    BarryBarry

    9,96211516




    9,96211516












    • $begingroup$
      Barry Meter error would be gain or offset but even 5% would not account for these results. But I agree the battery could drop 2x (.4 to .2V) this much on start up as the load resistance starts at 11 Ohms (10.1+1) and rises to > 21 ohms but even this wont explain the errors
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago




















    • $begingroup$
      Barry Meter error would be gain or offset but even 5% would not account for these results. But I agree the battery could drop 2x (.4 to .2V) this much on start up as the load resistance starts at 11 Ohms (10.1+1) and rises to > 21 ohms but even this wont explain the errors
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago


















    $begingroup$
    Barry Meter error would be gain or offset but even 5% would not account for these results. But I agree the battery could drop 2x (.4 to .2V) this much on start up as the load resistance starts at 11 Ohms (10.1+1) and rises to > 21 ohms but even this wont explain the errors
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






    $begingroup$
    Barry Meter error would be gain or offset but even 5% would not account for these results. But I agree the battery could drop 2x (.4 to .2V) this much on start up as the load resistance starts at 11 Ohms (10.1+1) and rises to > 21 ohms but even this wont explain the errors
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago















    2












    $begingroup$

    Ohm's Law and KVL always applies but the voltage drop is nonlinear with time due to thermal effects so bulbs never match perfectly and the voltages are never equal.



    Bulbs are nonlinear PTC ( positive temp coefficient ) conductors that rise in R by 10x at rated power with temperatures of say 2500'K.



    So R (hot/cold) ratio ~ 10:1 for bright warm white light (hot) and room temp (cold).



    This means if you put two 6V bulbs in series with a 6V battery, you will never ever get 3V each.



    The bulb with even the slightest higher cold resistance heats up faster in temperature will rise in resistance faster and thus drop more voltage than the other bulb, resulting in a runaway condition where that bulb will have full power and the other about 10 %. Putting a resistor in series as you have done reduces the balance difference between R cold= est. 0.5 Ohms and R hot est= 5 Ohms x 0.3A 1.5V



    Although this is the extreme case where the bulbs are most sensitive to resistance changes at half voltage, it means your measurements probably changed while you were taking the readings.



    Is it repeatable? Is it stable? Measure again.



    If the load voltage is 5.9V @ 0.3A and the no-load Vbat=6.1V then the 4 battery cells have an internal resistance ESR = 0.2V/0.3A= 0.67 Ohms total.






    share|improve this answer











    $endgroup$













    • $begingroup$
      let me run a quick simulation to prove that
      $endgroup$
      – Sunnyskyguy EE75
      9 hours ago






    • 1




      $begingroup$
      a nitpick: saying that "Ohm's [l]aw always applies" is a heavy overstatement. No physical "law" known to human always applies. Every "law" is just a model - and every model is false, because every model is just an approximation - albeit some, as OL or KVL, are indeed extremely useful.
      $endgroup$
      – vaxquis
      8 hours ago






    • 1




      $begingroup$
      @vaxquis I see no reason to doubt my conclusions or my statement. The readings can change from test to test reading as heat transfers between bulbs and changes the results. But Ohm's Law using simultaneous readings will not fail since this is DC and there is no interference from the DMM or VOM. Do you have an experience to contradict this?
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Since it only becomes a Law when experimental data supports it after Peer review in the Physics community, do you have any data to support your doubt? Or just a hypothetical one. Just remember the time constants involved resemble a reactor but not. Ohm's Law still applies to nonlinear resistive parts like LEDs within the constraints of linear regression. over a limited range. I need data, to accept your doubt.
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Question for the mods - is it possible to down-vote a comment?
      $endgroup$
      – AnalogKid
      2 hours ago
















    2












    $begingroup$

    Ohm's Law and KVL always applies but the voltage drop is nonlinear with time due to thermal effects so bulbs never match perfectly and the voltages are never equal.



    Bulbs are nonlinear PTC ( positive temp coefficient ) conductors that rise in R by 10x at rated power with temperatures of say 2500'K.



    So R (hot/cold) ratio ~ 10:1 for bright warm white light (hot) and room temp (cold).



    This means if you put two 6V bulbs in series with a 6V battery, you will never ever get 3V each.



    The bulb with even the slightest higher cold resistance heats up faster in temperature will rise in resistance faster and thus drop more voltage than the other bulb, resulting in a runaway condition where that bulb will have full power and the other about 10 %. Putting a resistor in series as you have done reduces the balance difference between R cold= est. 0.5 Ohms and R hot est= 5 Ohms x 0.3A 1.5V



    Although this is the extreme case where the bulbs are most sensitive to resistance changes at half voltage, it means your measurements probably changed while you were taking the readings.



    Is it repeatable? Is it stable? Measure again.



    If the load voltage is 5.9V @ 0.3A and the no-load Vbat=6.1V then the 4 battery cells have an internal resistance ESR = 0.2V/0.3A= 0.67 Ohms total.






    share|improve this answer











    $endgroup$













    • $begingroup$
      let me run a quick simulation to prove that
      $endgroup$
      – Sunnyskyguy EE75
      9 hours ago






    • 1




      $begingroup$
      a nitpick: saying that "Ohm's [l]aw always applies" is a heavy overstatement. No physical "law" known to human always applies. Every "law" is just a model - and every model is false, because every model is just an approximation - albeit some, as OL or KVL, are indeed extremely useful.
      $endgroup$
      – vaxquis
      8 hours ago






    • 1




      $begingroup$
      @vaxquis I see no reason to doubt my conclusions or my statement. The readings can change from test to test reading as heat transfers between bulbs and changes the results. But Ohm's Law using simultaneous readings will not fail since this is DC and there is no interference from the DMM or VOM. Do you have an experience to contradict this?
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Since it only becomes a Law when experimental data supports it after Peer review in the Physics community, do you have any data to support your doubt? Or just a hypothetical one. Just remember the time constants involved resemble a reactor but not. Ohm's Law still applies to nonlinear resistive parts like LEDs within the constraints of linear regression. over a limited range. I need data, to accept your doubt.
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Question for the mods - is it possible to down-vote a comment?
      $endgroup$
      – AnalogKid
      2 hours ago














    2












    2








    2





    $begingroup$

    Ohm's Law and KVL always applies but the voltage drop is nonlinear with time due to thermal effects so bulbs never match perfectly and the voltages are never equal.



    Bulbs are nonlinear PTC ( positive temp coefficient ) conductors that rise in R by 10x at rated power with temperatures of say 2500'K.



    So R (hot/cold) ratio ~ 10:1 for bright warm white light (hot) and room temp (cold).



    This means if you put two 6V bulbs in series with a 6V battery, you will never ever get 3V each.



    The bulb with even the slightest higher cold resistance heats up faster in temperature will rise in resistance faster and thus drop more voltage than the other bulb, resulting in a runaway condition where that bulb will have full power and the other about 10 %. Putting a resistor in series as you have done reduces the balance difference between R cold= est. 0.5 Ohms and R hot est= 5 Ohms x 0.3A 1.5V



    Although this is the extreme case where the bulbs are most sensitive to resistance changes at half voltage, it means your measurements probably changed while you were taking the readings.



    Is it repeatable? Is it stable? Measure again.



    If the load voltage is 5.9V @ 0.3A and the no-load Vbat=6.1V then the 4 battery cells have an internal resistance ESR = 0.2V/0.3A= 0.67 Ohms total.






    share|improve this answer











    $endgroup$



    Ohm's Law and KVL always applies but the voltage drop is nonlinear with time due to thermal effects so bulbs never match perfectly and the voltages are never equal.



    Bulbs are nonlinear PTC ( positive temp coefficient ) conductors that rise in R by 10x at rated power with temperatures of say 2500'K.



    So R (hot/cold) ratio ~ 10:1 for bright warm white light (hot) and room temp (cold).



    This means if you put two 6V bulbs in series with a 6V battery, you will never ever get 3V each.



    The bulb with even the slightest higher cold resistance heats up faster in temperature will rise in resistance faster and thus drop more voltage than the other bulb, resulting in a runaway condition where that bulb will have full power and the other about 10 %. Putting a resistor in series as you have done reduces the balance difference between R cold= est. 0.5 Ohms and R hot est= 5 Ohms x 0.3A 1.5V



    Although this is the extreme case where the bulbs are most sensitive to resistance changes at half voltage, it means your measurements probably changed while you were taking the readings.



    Is it repeatable? Is it stable? Measure again.



    If the load voltage is 5.9V @ 0.3A and the no-load Vbat=6.1V then the 4 battery cells have an internal resistance ESR = 0.2V/0.3A= 0.67 Ohms total.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 8 hours ago

























    answered 9 hours ago









    Sunnyskyguy EE75Sunnyskyguy EE75

    68k22398




    68k22398












    • $begingroup$
      let me run a quick simulation to prove that
      $endgroup$
      – Sunnyskyguy EE75
      9 hours ago






    • 1




      $begingroup$
      a nitpick: saying that "Ohm's [l]aw always applies" is a heavy overstatement. No physical "law" known to human always applies. Every "law" is just a model - and every model is false, because every model is just an approximation - albeit some, as OL or KVL, are indeed extremely useful.
      $endgroup$
      – vaxquis
      8 hours ago






    • 1




      $begingroup$
      @vaxquis I see no reason to doubt my conclusions or my statement. The readings can change from test to test reading as heat transfers between bulbs and changes the results. But Ohm's Law using simultaneous readings will not fail since this is DC and there is no interference from the DMM or VOM. Do you have an experience to contradict this?
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Since it only becomes a Law when experimental data supports it after Peer review in the Physics community, do you have any data to support your doubt? Or just a hypothetical one. Just remember the time constants involved resemble a reactor but not. Ohm's Law still applies to nonlinear resistive parts like LEDs within the constraints of linear regression. over a limited range. I need data, to accept your doubt.
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Question for the mods - is it possible to down-vote a comment?
      $endgroup$
      – AnalogKid
      2 hours ago


















    • $begingroup$
      let me run a quick simulation to prove that
      $endgroup$
      – Sunnyskyguy EE75
      9 hours ago






    • 1




      $begingroup$
      a nitpick: saying that "Ohm's [l]aw always applies" is a heavy overstatement. No physical "law" known to human always applies. Every "law" is just a model - and every model is false, because every model is just an approximation - albeit some, as OL or KVL, are indeed extremely useful.
      $endgroup$
      – vaxquis
      8 hours ago






    • 1




      $begingroup$
      @vaxquis I see no reason to doubt my conclusions or my statement. The readings can change from test to test reading as heat transfers between bulbs and changes the results. But Ohm's Law using simultaneous readings will not fail since this is DC and there is no interference from the DMM or VOM. Do you have an experience to contradict this?
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Since it only becomes a Law when experimental data supports it after Peer review in the Physics community, do you have any data to support your doubt? Or just a hypothetical one. Just remember the time constants involved resemble a reactor but not. Ohm's Law still applies to nonlinear resistive parts like LEDs within the constraints of linear regression. over a limited range. I need data, to accept your doubt.
      $endgroup$
      – Sunnyskyguy EE75
      6 hours ago








    • 1




      $begingroup$
      Question for the mods - is it possible to down-vote a comment?
      $endgroup$
      – AnalogKid
      2 hours ago
















    $begingroup$
    let me run a quick simulation to prove that
    $endgroup$
    – Sunnyskyguy EE75
    9 hours ago




    $begingroup$
    let me run a quick simulation to prove that
    $endgroup$
    – Sunnyskyguy EE75
    9 hours ago




    1




    1




    $begingroup$
    a nitpick: saying that "Ohm's [l]aw always applies" is a heavy overstatement. No physical "law" known to human always applies. Every "law" is just a model - and every model is false, because every model is just an approximation - albeit some, as OL or KVL, are indeed extremely useful.
    $endgroup$
    – vaxquis
    8 hours ago




    $begingroup$
    a nitpick: saying that "Ohm's [l]aw always applies" is a heavy overstatement. No physical "law" known to human always applies. Every "law" is just a model - and every model is false, because every model is just an approximation - albeit some, as OL or KVL, are indeed extremely useful.
    $endgroup$
    – vaxquis
    8 hours ago




    1




    1




    $begingroup$
    @vaxquis I see no reason to doubt my conclusions or my statement. The readings can change from test to test reading as heat transfers between bulbs and changes the results. But Ohm's Law using simultaneous readings will not fail since this is DC and there is no interference from the DMM or VOM. Do you have an experience to contradict this?
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






    $begingroup$
    @vaxquis I see no reason to doubt my conclusions or my statement. The readings can change from test to test reading as heat transfers between bulbs and changes the results. But Ohm's Law using simultaneous readings will not fail since this is DC and there is no interference from the DMM or VOM. Do you have an experience to contradict this?
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






    1




    1




    $begingroup$
    Since it only becomes a Law when experimental data supports it after Peer review in the Physics community, do you have any data to support your doubt? Or just a hypothetical one. Just remember the time constants involved resemble a reactor but not. Ohm's Law still applies to nonlinear resistive parts like LEDs within the constraints of linear regression. over a limited range. I need data, to accept your doubt.
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






    $begingroup$
    Since it only becomes a Law when experimental data supports it after Peer review in the Physics community, do you have any data to support your doubt? Or just a hypothetical one. Just remember the time constants involved resemble a reactor but not. Ohm's Law still applies to nonlinear resistive parts like LEDs within the constraints of linear regression. over a limited range. I need data, to accept your doubt.
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago






    1




    1




    $begingroup$
    Question for the mods - is it possible to down-vote a comment?
    $endgroup$
    – AnalogKid
    2 hours ago




    $begingroup$
    Question for the mods - is it possible to down-vote a comment?
    $endgroup$
    – AnalogKid
    2 hours ago











    2












    $begingroup$


    The Ohm's law calculations of the parts do not agree with the whole




    Actually, there's a very plausible explanation which would mean that Ohm's law is alive and well, and does explain what you report.



    (Note: I'm assuming that the bulbs have reached their steady-state "hot" resistances, so that those resistances don't change during the measurements. I'm also assuming that the battery voltage doesn't drop significantly during the measurements. It will drop, but I'm assuming that doesn't happen to a significant amount while measurements are being made.)



    Hypothesis: Additional wiring resistances



    Your diagram doesn't show additional resistances which must be present in your setup.



    Let's look at just one area in your diagram as an example - the voltages across the two bulbs together and individually.



    Here is your diagram again (so that readers don't have to keep scrolling up to the question):



    wiring diagram from the question



    Voltage drop across bulb 1 = 1.13V

    Voltage drop across bulb 2 = 1.24V

    Voltage drop across both bulbs and the wiring between bulbs = 2.6V

    Therefore "unexpected" additional voltage drop = 2.6V - (1.13V + 1.24V) = 0.23V



    Although the 10.2 Ω resistor value might not be very accurate (e.g. multimeter lead resistance is probably included in that, so the resistor's true value is likely lower) for now, let's use your approximate current value of (2.85V / 10.2 Ω =) 0.28A (to 2 decimal places).



    Therefore the additional voltage drop between the two bulbs is only a resistance of (0.23V / 0.28A =) 0.82 Ω which is easy to believe, due to cable type / length / connection resistance etc.





    The same reasoning explains why 2.85V (across the resistor) + 2.6V (across the bulbs) doesn't equal 5.9V - there are additional resistances (and therefore voltage drops) in the wiring / connections, which aren't included.





    Summary: Go through the actual circuit hardware whose diagram you gave, measuring the voltage drops across not only the obvious parts, but also measure the voltage drop across every wire and connection of any kind.



    You will find additional voltage drops which, when added to the voltage drops across the main components already given, I expect will show that the total voltage drops do all sum to the expected value.






    share|improve this answer











    $endgroup$


















      2












      $begingroup$


      The Ohm's law calculations of the parts do not agree with the whole




      Actually, there's a very plausible explanation which would mean that Ohm's law is alive and well, and does explain what you report.



      (Note: I'm assuming that the bulbs have reached their steady-state "hot" resistances, so that those resistances don't change during the measurements. I'm also assuming that the battery voltage doesn't drop significantly during the measurements. It will drop, but I'm assuming that doesn't happen to a significant amount while measurements are being made.)



      Hypothesis: Additional wiring resistances



      Your diagram doesn't show additional resistances which must be present in your setup.



      Let's look at just one area in your diagram as an example - the voltages across the two bulbs together and individually.



      Here is your diagram again (so that readers don't have to keep scrolling up to the question):



      wiring diagram from the question



      Voltage drop across bulb 1 = 1.13V

      Voltage drop across bulb 2 = 1.24V

      Voltage drop across both bulbs and the wiring between bulbs = 2.6V

      Therefore "unexpected" additional voltage drop = 2.6V - (1.13V + 1.24V) = 0.23V



      Although the 10.2 Ω resistor value might not be very accurate (e.g. multimeter lead resistance is probably included in that, so the resistor's true value is likely lower) for now, let's use your approximate current value of (2.85V / 10.2 Ω =) 0.28A (to 2 decimal places).



      Therefore the additional voltage drop between the two bulbs is only a resistance of (0.23V / 0.28A =) 0.82 Ω which is easy to believe, due to cable type / length / connection resistance etc.





      The same reasoning explains why 2.85V (across the resistor) + 2.6V (across the bulbs) doesn't equal 5.9V - there are additional resistances (and therefore voltage drops) in the wiring / connections, which aren't included.





      Summary: Go through the actual circuit hardware whose diagram you gave, measuring the voltage drops across not only the obvious parts, but also measure the voltage drop across every wire and connection of any kind.



      You will find additional voltage drops which, when added to the voltage drops across the main components already given, I expect will show that the total voltage drops do all sum to the expected value.






      share|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$


        The Ohm's law calculations of the parts do not agree with the whole




        Actually, there's a very plausible explanation which would mean that Ohm's law is alive and well, and does explain what you report.



        (Note: I'm assuming that the bulbs have reached their steady-state "hot" resistances, so that those resistances don't change during the measurements. I'm also assuming that the battery voltage doesn't drop significantly during the measurements. It will drop, but I'm assuming that doesn't happen to a significant amount while measurements are being made.)



        Hypothesis: Additional wiring resistances



        Your diagram doesn't show additional resistances which must be present in your setup.



        Let's look at just one area in your diagram as an example - the voltages across the two bulbs together and individually.



        Here is your diagram again (so that readers don't have to keep scrolling up to the question):



        wiring diagram from the question



        Voltage drop across bulb 1 = 1.13V

        Voltage drop across bulb 2 = 1.24V

        Voltage drop across both bulbs and the wiring between bulbs = 2.6V

        Therefore "unexpected" additional voltage drop = 2.6V - (1.13V + 1.24V) = 0.23V



        Although the 10.2 Ω resistor value might not be very accurate (e.g. multimeter lead resistance is probably included in that, so the resistor's true value is likely lower) for now, let's use your approximate current value of (2.85V / 10.2 Ω =) 0.28A (to 2 decimal places).



        Therefore the additional voltage drop between the two bulbs is only a resistance of (0.23V / 0.28A =) 0.82 Ω which is easy to believe, due to cable type / length / connection resistance etc.





        The same reasoning explains why 2.85V (across the resistor) + 2.6V (across the bulbs) doesn't equal 5.9V - there are additional resistances (and therefore voltage drops) in the wiring / connections, which aren't included.





        Summary: Go through the actual circuit hardware whose diagram you gave, measuring the voltage drops across not only the obvious parts, but also measure the voltage drop across every wire and connection of any kind.



        You will find additional voltage drops which, when added to the voltage drops across the main components already given, I expect will show that the total voltage drops do all sum to the expected value.






        share|improve this answer











        $endgroup$




        The Ohm's law calculations of the parts do not agree with the whole




        Actually, there's a very plausible explanation which would mean that Ohm's law is alive and well, and does explain what you report.



        (Note: I'm assuming that the bulbs have reached their steady-state "hot" resistances, so that those resistances don't change during the measurements. I'm also assuming that the battery voltage doesn't drop significantly during the measurements. It will drop, but I'm assuming that doesn't happen to a significant amount while measurements are being made.)



        Hypothesis: Additional wiring resistances



        Your diagram doesn't show additional resistances which must be present in your setup.



        Let's look at just one area in your diagram as an example - the voltages across the two bulbs together and individually.



        Here is your diagram again (so that readers don't have to keep scrolling up to the question):



        wiring diagram from the question



        Voltage drop across bulb 1 = 1.13V

        Voltage drop across bulb 2 = 1.24V

        Voltage drop across both bulbs and the wiring between bulbs = 2.6V

        Therefore "unexpected" additional voltage drop = 2.6V - (1.13V + 1.24V) = 0.23V



        Although the 10.2 Ω resistor value might not be very accurate (e.g. multimeter lead resistance is probably included in that, so the resistor's true value is likely lower) for now, let's use your approximate current value of (2.85V / 10.2 Ω =) 0.28A (to 2 decimal places).



        Therefore the additional voltage drop between the two bulbs is only a resistance of (0.23V / 0.28A =) 0.82 Ω which is easy to believe, due to cable type / length / connection resistance etc.





        The same reasoning explains why 2.85V (across the resistor) + 2.6V (across the bulbs) doesn't equal 5.9V - there are additional resistances (and therefore voltage drops) in the wiring / connections, which aren't included.





        Summary: Go through the actual circuit hardware whose diagram you gave, measuring the voltage drops across not only the obvious parts, but also measure the voltage drop across every wire and connection of any kind.



        You will find additional voltage drops which, when added to the voltage drops across the main components already given, I expect will show that the total voltage drops do all sum to the expected value.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 3 hours ago

























        answered 3 hours ago









        SamGibsonSamGibson

        11.2k41737




        11.2k41737






















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