# Thread: Watts Calculation Method for 240V Circuits

1. ## Watts Calculation Method for 240V Circuits

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.     Reply With Quote

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0 Thread(s) Originally Posted by Cybatrex 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.   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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3. Originally Posted by GoldDigger 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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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? (L1+L2) X 120V = CORRECT 2?   Reply With Quote

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0 Thread(s) Originally Posted by Cybatrex 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? (L1+L2) X 120V = CORRECT 2? 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. Sent from my XT1585 using Tapatalk  Reply With Quote

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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,

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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.  Reply With Quote

7. Originally Posted by Cybatrex 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.   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.  Reply With Quote

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0 Thread(s) Originally Posted by Besoeker 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.
that is more accurate
in this case the currents happened to be balanced
not always (or even usually) the case  Reply With Quote

9. Originally Posted by Cybatrex 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.   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.  Reply With Quote

10. Originally Posted by Ingenieur that is more accurate
in this case the currents happened to be balanced
not always (or even usually) the case
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.  Reply With Quote

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