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With a clamp on amp meter, I have 8.2 amps at 123.6 volts on one leg and 7.2 amps at 123.9 volts on the 2nd leg. When I touch the volt meter to both legs I get 247.8 volts.

What is the total wattage? I get 3816 watts total if I add the amps together and divide by 247.8 or 1905.5 total if I calculate the watts (v x a) for each leg and add together.

What is the correct way to do this.

The measurement was taken from the two main wires that feed the house panel box from the transformer, so everything that was on in the house was showing in the measurment, there can be an amp difference since single pole breakers only draw from one of the wires.

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    \$\begingroup\$ You just measure the current on one leg. You don't measure on two legs. However, the two legs should have the same current. If they don't, then there must be a third leg somewhere. If you have 8.2-7.2=1A flowing through a protective earth ground conductor, that is not a good thing. It indicates a serious problem. But maybe the current just changed a little from the time you measured one leg to the time you measured the other. \$\endgroup\$ – mkeith Sep 16 '17 at 3:30
  • \$\begingroup\$ Can you please edit your question and add information about what is consuming the power? Is it a welder, a battery charger, a ballast for lighting, a motor, an appliance (and if so what type) or a heater? This information will actually help make a better answer. That is why I am requesting it. \$\endgroup\$ – mkeith Sep 16 '17 at 3:36
  • \$\begingroup\$ Which legs is the load connected to? If it is one leg and neutral then measure the voltage and current on that leg. If it is connected between two legs then measure current on either leg and voltage between the legs. If it is connected to (and drawing power from) both legs and neutral then it is not single phase. \$\endgroup\$ – Bruce Abbott Sep 16 '17 at 5:06
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What is the correct way to do this?

The correct way to measure AC power is using a wattmeter. This is because you cannot know (or reasonably estimate) the phase angle between voltage and current and measuring amps and volts and simply multiplying them together gives a falsely high reading in all but simple cases of a purely resistive load.

So, if you had two wattmeters you can measure the wattage entering one circuit to neutral and also measure the wattage entering the 2nd circuit with respect to neutral.

If you loads are resistive the powers will be: -

  • 8.2 x 123.6 = 1013 watts
  • 7.3 x 123.9 = 892 watts

Total power (if resistive load) = 1905 watts.

However, if the load is resistive AND non-linear there will be harmonics in the current waveform that don't contribute to power loss but will register on an RMS measuring current meter so here's another error source.

Best way - use a proper wattmeter.

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  • \$\begingroup\$ The loads were not resistive as all the lights are led in my home. I had lights, fans and ac units and the fridge running. Is there a way to get a close estimate using the measurements I have? \$\endgroup\$ – grantr Sep 16 '17 at 13:46
  • \$\begingroup\$ Not really possible, sorry. \$\endgroup\$ – Andy aka Sep 16 '17 at 14:02
  • \$\begingroup\$ So would the real total watts be more or less that what I measured? \$\endgroup\$ – grantr Sep 16 '17 at 16:09
  • \$\begingroup\$ It will never be more and it could be substantially less. \$\endgroup\$ – Andy aka Sep 16 '17 at 16:17
  • \$\begingroup\$ To the OP, there is a device called a "kill a watt" meter that can accurately measure power one device at a time. You should buy one. They are not expensive. \$\endgroup\$ – mkeith Sep 16 '17 at 23:51
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Residential Power Meters only the real component power a product of only the RMS V*I that are in phase.

This method cannot measure phase or power factor and thus only measures the apparent power not the "real power."

The phase of the voltages in your case is assumed to be split phase or 180 deg. The neutral current may account for the difference in L1 and L2 or not if each current has a different phase shift.

There are other sources of errors with out of phase harmonics.

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  • \$\begingroup\$ So I am being billed for apparent power which is volts x amps on each leg Summed? When sizing a generator, do you use apparent power? \$\endgroup\$ – grantr Sep 16 '17 at 16:59
  • \$\begingroup\$ Sizing a generator uses apparent power, but billing is for real power. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 16 '17 at 17:40
  • \$\begingroup\$ @grantr No, you are being billed real power. You can get real power by multiplying the V and I waveforms. (without any phase shift) \$\endgroup\$ – Jeroen3 Sep 16 '17 at 17:41
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utilizing a one phase transformer with a central common--neutral-- terminal in order to get 110/220 VAC the aproximate total power is: P =( V max) I (prom). I(prom)=(I(line A)+ I(line B))/2. If you are using standard meters -calibrated for sinusoidal signals- the difference between line A and line B is normal because of the electrical connection of neutral and ground but is really harmful if major to 10%.(this disrupts and distorts the sinusoidal low voltage signal) We are talking about VA -reactive power-. Watt or active power depends on power factor: W= VA (power factor), and oscillates between 0.7 and 0.87 for AC equipment-HVAC- and one phase motors. You can add a capacitive bank to raise power factor. Check your electrical bill and note if a charge exists by this--low power factor-.

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