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Average power can be measured by wattmeter, so why in the attached example it doesn't function properly- doesn't show 0 W?enter image description here

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  • \$\begingroup\$ See the last paragraph of the passage you quoted. The wattmeter in question isn't measuring the total power of the load. \$\endgroup\$ – Dave Tweed Aug 4 '13 at 13:25
  • \$\begingroup\$ @Dave Tweed But doesn't it suppose to be 0 because that part is also purely inductive? \$\endgroup\$ – user7777777 Aug 4 '13 at 13:33
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To corectly measure power on a three phase system (without neutral) you need two wattmeters. For a purely reactive load, one wattmeter will read positive X watts and the other will read negative X watts meaning the power consumed is X-X watts or 0 watts (as one would expect).

You have to remember that the line voltage and line/phase current on a purely reactive load will not be 90º apart because it is three phase. Neither will the phase angle be 0º for a purely resistive load. This is because phase voltage is not in-phase with line voltage.

On a resistive circuit the phase voltage and phase current will be exactly in phase BUT, phase voltage leads line voltage by 30º and in the OPs question the wattmeter is connected across two lines (i.e. it is measuring line voltage) hence the purely reactive load has a phase angle of 60º instead of 90º as you would get on a single phase circuit: -

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If you added a 2nd wattmeter in the conventional way that two wattmeters are connected for three-wire, 3-phase power measurements you'd find that the 2nd wattmeter reads a negative power balancing completely the 1st wattmeter reading: -

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Shown is a single wattmeter method that works if the load is balanced. The 2 wattmeter method works with balanced or unbalanced loads whether thy be delta or star connected loads.

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