Watts Calculation Method for 240V Circuits

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Cybatrex

Cybatrex

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Florida
I need assistance with determining my Watts correctly with Ohm's law using the amperage reading from my Fluke clamp meter. I welcome anyone who can correct the errors of my ways and offer a brief explanation so I can learn correctly.

Amperage Readings
L1: 17.1 Amps
L2: 17.0 Amps

Ohms Law
P(w) = V x I

Method 1: Calculated ohms law for each leg of the circuit, calculated them individually, then added the wattage together.
L1: 17.1A x 120V = 2,052 Watts
L2: 17.0A x 120V = 2,040 Watts

Total Watts1: 4,092 Watts
Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
L1: 17.1 Amps
L2: 17.0 Amps

Lt = 34.1 Amps x 240V = 8,184 Watts

Total Watts2: 8,184 Watts




Any help with this would be greatly appreciated.
:):):)
 
Cybatrex said:
I need assistance with determining my Watts correctly with Ohm's law using the amperage reading from my Fluke clamp meter. I welcome anyone who can correct the errors of my ways and offer a brief explanation so I can learn correctly.

Amperage Readings
L1: 17.1 Amps
L2: 17.0 Amps

Ohms Law
P(w) = V x I

Method 1: Calculated ohms law for each leg of the circuit, calculated them individually, then added the wattage together.
L1: 17.1A x 120V = 2,052 Watts
L2: 17.0A x 120V = 2,040 Watts

Total Watts1: 4,092 Watts
Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
L1: 17.1 Amps
L2: 17.0 Amps

Lt = 34.1 Amps x 240V = 8,184 Watts

Total Watts2: 8,184 Watts




Any help with this would be greatly appreciated.
:):):)
Click to expand...
The problem with your second method is that the same current (approximately) is flowing through the same series circuit between L1 and L2. You only count that current once! That is where your extra factor of two came from.

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GoldDigger said:
The problem with your second method is that the same current (approximately) is flowing through the same series circuit between L1 and L2. You only count that current once! That is where your extra factor of two came from.

Sent from my XT1585 using Tapatalk
Click to expand...

Thank you so much. To clarify,

I would take the amperage to either L1 or L2 and then multiply by 120V or 240V?

L1 X 240V = CORRECT 1? :D

(L1+L2) X 120V = CORRECT 2? :D
 
Cybatrex said:
Thank you so much. To clarify,

I would take the amperage to either L1 or L2 and then multiply by 120V or 240V?

L1 X 240V = CORRECT 1? :D

(L1+L2) X 120V = CORRECT 2? :D
Click to expand...
1. Yes.
2. Yes.
For 240 you can use one L current value if the two are identical
If they are not identical, you can take the average and multiply it by 240. Algebraically this is 100% identical to the formula in 2. :)

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180526-0821 EDT

Cybatrex:

Change the word watts to volt-amperes. It may be watts, but only if the load is pure resistance like an incandescent bulb, or resistive heater,

.
 
A common rookie mistake is to think that a 240V 100A 2 pole breaker delivers 200A @240V. 100A on L1 + 100A on L2.
 
Cybatrex said:
I need assistance with determining my Watts correctly with Ohm's law using the amperage reading from my Fluke clamp meter. I welcome anyone who can correct the errors of my ways and offer a brief explanation so I can learn correctly.

Amperage Readings
L1: 17.1 Amps
L2: 17.0 Amps

Ohms Law
P(w) = V x I

Method 1: Calculated ohms law for each leg of the circuit, calculated them individually, then added the wattage together.
L1: 17.1A x 120V = 2,052 Watts
L2: 17.0A x 120V = 2,040 Watts

Total Watts1: 4,092 Watts
Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
L1: 17.1 Amps
L2: 17.0 Amps

Lt = 34.1 Amps x 240V = 8,184 Watts

Total Watts2: 8,184 Watts




Any help with this would be greatly appreciated.
:):):)
Click to expand...

If it is 120-0-120V then I suppose L1*120 + L2*120 would be the more accurate. But the 0.1A would be in the neutral but probably not significant. If is is actually there.
 
Besoeker said:
If it is 120-0-120V then I suppose L1*120 + L2*120 would be the more accurate. But the 0.1A would be in the neutral but probably not significant. If is is actually there.
Click to expand...

that is more accurate
in this case the currents happened to be balanced
not always (or even usually) the case
 
Cybatrex said:
I need assistance with determining my Watts correctly with Ohm's law using the amperage reading from my Fluke clamp meter. I welcome anyone who can correct the errors of my ways and offer a brief explanation so I can learn correctly.

Amperage Readings
L1: 17.1 Amps
L2: 17.0 Amps

Ohms Law
P(w) = V x I

Method 1: Calculated ohms law for each leg of the circuit, calculated them individually, then added the wattage together.
L1: 17.1A x 120V = 2,052 Watts
L2: 17.0A x 120V = 2,040 Watts

Total Watts1: 4,092 Watts
Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
L1: 17.1 Amps
L2: 17.0 Amps

Lt = 34.1 Amps x 240V = 8,184 Watts

Total Watts2: 8,184 Watts




Any help with this would be greatly appreciated.
:):):)
Click to expand...
If you only have a two wire circuit you can't have a different current on one conductor then you have on the other, there is only one current path and current is same everywhere in the circuit. I'm sure accuracy of the measuring method is what resulted in the minor difference you did measure. That said and as others have been pointing out - you have ~17 amps at 240 volts and not ~17 amps at 120 volts - times two.

Add a third point in the system it gets more complicated, especially if you have 208/120 system. That situation you could possibly have 17 amps at 120 volts - times two, or even three. But that never can happen when there is only a two wire circuit.
 
Ingenieur said:
that is more accurate
in this case the currents happened to be balanced
not always (or even usually) the case
Click to expand...
Yes. And Gar's point about it being VA rather than W should be noted.

A point in passing. I usually use the letter or symbol for the unit - W rather that Watts.
This stems from losing a mark on an exam question for spelling out the unit in the answer.
I was told that spelling it out was the incorrect thing to do. IDK whether that's true or not. But it stuck with me.

Mods, forgive the digression.
 
I might be over-answering on this one, please forgive me if so.

Besoeker is correct - this is in units of VA, not watts, based on the inputs. Watts would actually be equal or less, but isn't relevant to this problem.

Cybatrex said:
Method 1:
L1: 17.1A x 120V = 2,052 Watts
L2: 17.0A x 120V = 2,040 Watts

Total Watts1: 4,092 Watts

Method 2:
L1: 17.1 Amps
L2: 17.0 Amps

Lt = 34.1 Amps x 240V = 8,184 Watts

Total Watts2: 8,184 Watts
Click to expand...

Method 1 is good, but Method 2 is toast.

With a 120/240 split-style transformer, each Line-to-Neutral leg is 120V, so it's 34.1 Amps (added across 120V legs), times 120V, not 240V, which is why the result doubled. Another way to look at it is - The average current on each side of the transformer - at (17.1+17.0)/2 - times the line-to-line voltage, 240V. 17.05A * 240V = 4,092VA

Revised Method 2 - 34.1A * 120V = 4,092VA
 
Besoeker said:
If it is 120-0-120V then I suppose L1*120 + L2*120 would be the more accurate. But the 0.1A would be in the neutral but probably not significant. If is is actually there.
Click to expand...
I would say that is "more accurate" if there is a neutral conductor involved.

If it is just a two wire circuit - then it is simply V x A. The fact he measured .1 amps difference on one line vs the other is an accuracy issue with measuring method - the current has to be the same on both lines as there is no where for any differential current to go to, outside of potential fault current issues, but if those are present we have an unintended multiwire circuit and current can be flowing in more then just a single loop.
 
kwired said:
I would say that is "more accurate" if there is a neutral conductor involved.

If it is just a two wire circuit - then it is simply V x A. The fact he measured .1 amps difference on one line vs the other is an accuracy issue with measuring method - the current has to be the same on both lines as there is no where for any differential current to go to, outside of potential fault current issues, but if those are present we have an unintended multiwire circuit and current can be flowing in more then just a single loop.
Click to expand...
Kirchoff lives! :D
 
Besoeker said:
A point in passing. I usually use the letter or symbol for the unit - W rather that Watts.
This stems from losing a mark on an exam question for spelling out the unit in the answer.
I was told that spelling it out was the incorrect thing to do. IDK whether that's true or not. But it stuck with me.
Click to expand...

IMO you had a jerk for an instructor who warped you for life. :D
 
AtTheKeyboard said:
I might be over-answering on this one, please forgive me if so.

Besoeker is correct - this is in units of VA, not watts, based on the inputs. Watts would actually be equal or less, but isn't relevant to this problem.



Method 1 is good, but Method 2 is toast.

With a 120/240 split-style transformer, each Line-to-Neutral leg is 120V, so it's 34.1 Amps (added across 120V legs), times 120V, not 240V, which is why the result doubled. Another way to look at it is - The average current on each side of the transformer - at (17.1+17.0)/2 - times the line-to-line voltage, 240V. 17.05A * 240V = 4,092VA

Revised Method 2 - 34.1A * 120V = 4,092VA
Click to expand...

Thats exactly what I needed to see. Perfect, thank you so much.
 
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