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

Ohm's Law for voltage dividerWhy doesn't it matter if a resistor is before or behind an LED wrt voltage drop?brief explanation of Ohm's lawOhm's Law confusion — can there be voltage without current?Voltage Drop Between Battery and LoadPlacing just an LED in a circuit?Voltage drops are not adding up to voltage of the sourceHow is open-circuit voltage calculated?What is the problem with this circuit? ( MOSFET + OPAMP) Battery protection circuit (draw current from battery)Resistor for buzzer current consumption limiting calculation

0

$begingroup$

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.

voltage voltage-measurement

share|improve this question

edited 59 mins ago

Dave Tweed♦

120k9149257

asked 1 hour ago

Sanity CheckSanity Check

43

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
  1 hour 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
  1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

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.

voltage voltage-measurement

share|improve this question

edited 59 mins ago

Dave Tweed♦

120k9149257

asked 1 hour ago

Sanity CheckSanity Check

43

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
  1 hour 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
  1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

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.

voltage voltage-measurement

share|improve this question

edited 59 mins ago

Dave Tweed♦

120k9149257

asked 1 hour ago

Sanity CheckSanity Check

43

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.

$endgroup$

share|improve this question

edited 59 mins ago

Dave Tweed♦

120k9149257

asked 1 hour ago

Sanity CheckSanity Check

43

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

edited 59 mins ago

Dave Tweed♦

120k9149257

asked 1 hour ago

Sanity CheckSanity Check

43

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.

edited 59 mins ago

Dave Tweed♦

120k9149257

edited 59 mins ago

Dave Tweed♦

120k9149257

asked 1 hour ago

Sanity CheckSanity Check

43

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 1 hour ago

Sanity CheckSanity Check

43

2 Answers
2

active

oldest

votes

4

$begingroup$

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

edited 1 hour ago

Dave Tweed♦

120k9149257

answered 1 hour ago

BarryBarry

9,95211516

$endgroup$

add a comment | 

1

$begingroup$

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

edited 28 mins ago

answered 1 hour ago

Sunnyskyguy EE75Sunnyskyguy EE75

68k22398

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Close approximation tinyurl.com/yxvdmqyk
  $endgroup$
  – Sunnyskyguy EE75
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here using 1.05W@3V bulb and 1.1W bulb @ 3V tinyurl.com/y6etgsxk
  $endgroup$
  – Sunnyskyguy EE75
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, give me a chance to digest all that. Very sharp action diagram, but with my limited knowledge I don't see the + end distinguished on the circuits. And right now the wife says dinner is ready...later!
  $endgroup$
  – Sanity Check
  11 mins ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

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

edited 1 hour ago

Dave Tweed♦

120k9149257

answered 1 hour ago

BarryBarry

9,95211516

$endgroup$

add a comment | 

4

$begingroup$

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

edited 1 hour ago

Dave Tweed♦

120k9149257

answered 1 hour ago

BarryBarry

9,95211516

$endgroup$

add a comment | 

$begingroup$

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

edited 1 hour ago

Dave Tweed♦

120k9149257

answered 1 hour ago

BarryBarry

9,95211516

$endgroup$

share|improve this answer

edited 1 hour ago

Dave Tweed♦

120k9149257

answered 1 hour ago

BarryBarry

9,95211516

edited 1 hour ago

Dave Tweed♦

120k9149257

edited 1 hour ago

Dave Tweed♦

120k9149257

answered 1 hour ago

BarryBarry

9,95211516

answered 1 hour ago

BarryBarry

9,95211516

1

$begingroup$

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

edited 28 mins ago

answered 1 hour ago

Sunnyskyguy EE75Sunnyskyguy EE75

68k22398

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Close approximation tinyurl.com/yxvdmqyk
  $endgroup$
  – Sunnyskyguy EE75
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here using 1.05W@3V bulb and 1.1W bulb @ 3V tinyurl.com/y6etgsxk
  $endgroup$
  – Sunnyskyguy EE75
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, give me a chance to digest all that. Very sharp action diagram, but with my limited knowledge I don't see the + end distinguished on the circuits. And right now the wife says dinner is ready...later!
  $endgroup$
  – Sanity Check
  11 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

edited 28 mins ago

answered 1 hour ago

Sunnyskyguy EE75Sunnyskyguy EE75

68k22398

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Close approximation tinyurl.com/yxvdmqyk
  $endgroup$
  – Sunnyskyguy EE75
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here using 1.05W@3V bulb and 1.1W bulb @ 3V tinyurl.com/y6etgsxk
  $endgroup$
  – Sunnyskyguy EE75
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, give me a chance to digest all that. Very sharp action diagram, but with my limited knowledge I don't see the + end distinguished on the circuits. And right now the wife says dinner is ready...later!
  $endgroup$
  – Sanity Check
  11 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 28 mins ago

answered 1 hour ago

Sunnyskyguy EE75Sunnyskyguy EE75

68k22398

$endgroup$

share|improve this answer

edited 28 mins ago

answered 1 hour ago

Sunnyskyguy EE75Sunnyskyguy EE75

68k22398

answered 1 hour ago

Sunnyskyguy EE75Sunnyskyguy EE75

68k22398

answered 1 hour ago

Sunnyskyguy EE75Sunnyskyguy EE75

68k22398