Originally posted by AtTheKeyboard
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Watts Calculation Method for 240V Circuits
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Originally posted by Besoeker View PostA 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.
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Originally posted by kwired View PostI 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.
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Originally posted by Besoeker View PostIf it is 1200120V 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.
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.
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I might be overanswering 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.
Originally posted by Cybatrex View PostMethod 1:
L^{1}: 17.1A x 120V = 2,052 Watts
L^{2}: 17.0A x 120V = 2,040 Watts
Total Watts^{1}: 4,092 Watts
Method 2:
L^{1}: 17.1 Amps
L^{2}: 17.0 Amps
L^{t }= 34.1 Amps x 240V = 8,184 Watts
Total Watts^{2}: 8,184 Watts
With a 120/240 splitstyle transformer, each LinetoNeutral 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 linetoline voltage, 240V. 17.05A * 240V = 4,092VA
Revised Method 2  34.1A * 120V = 4,092VA
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Originally posted by Ingenieur View Postthat is more accurate
in this case the currents happened to be balanced
not always (or even usually) the case
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.
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Originally posted by Cybatrex View PostI 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
L^{1}: 17.1 Amps
L^{2}: 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.
L^{1}: 17.1A x 120V = 2,052 Watts
L^{2}: 17.0A x 120V = 2,040 Watts
Total Watts^{1}: 4,092 Watts
Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
L^{1}: 17.1 Amps
L^{2}: 17.0 Amps
L^{t }= 34.1 Amps x 240V = 8,184 Watts
Total Watts^{2}: 8,184 Watts
Any help with this would be greatly appreciated.
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.
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Originally posted by Besoeker View PostIf it is 1200120V 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.
in this case the currents happened to be balanced
not always (or even usually) the case
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Originally posted by Cybatrex View PostI 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
L^{1}: 17.1 Amps
L^{2}: 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.
L^{1}: 17.1A x 120V = 2,052 Watts
L^{2}: 17.0A x 120V = 2,040 Watts
Total Watts^{1}: 4,092 Watts
Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
L^{1}: 17.1 Amps
L^{2}: 17.0 Amps
L^{t }= 34.1 Amps x 240V = 8,184 Watts
Total Watts^{2}: 8,184 Watts
Any help with this would be greatly appreciated.
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A common rookie mistake is to think that a 240V 100A 2 pole breaker delivers 200A @240V. 100A on L1 + 100A on L2.
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1805260821 EDT
[COLOR=#000000]Cybatrex:
[/COLOR]Change the word watts to voltamperes. It may be watts, but only if the load is pure resistance like an incandescent bulb, or resistive heater,
.
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Originally posted by Cybatrex View PostThank 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?
(L1+L2) X 120V = CORRECT 2?
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.
Sent from my XT1585 using Tapatalk
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Originally posted by GoldDigger View PostThe 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
I would take the amperage to either L1 or L2 and then multiply by 120V or 240V?
L1 X 240V = CORRECT 1?
(L1+L2) X 120V = CORRECT 2?
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Originally posted by Cybatrex View PostI 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
L^{1}: 17.1 Amps
L^{2}: 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.
L^{1}: 17.1A x 120V = 2,052 Watts
L^{2}: 17.0A x 120V = 2,040 Watts
Total Watts^{1}: 4,092 Watts
Method 2: Combined amperage for both legs and used 240V by combined average to calculate watts which is double Method 1.
L^{1}: 17.1 Amps
L^{2}: 17.0 Amps
L^{t }= 34.1 Amps x 240V = 8,184 Watts
Total Watts^{2}: 8,184 Watts
Any help with this would be greatly appreciated.
Sent from my XT1585 using Tapatalk
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