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This answer was edited. If needed, see history

It is known that the sum of the two wattmeters is the total power of the load, but it doesn't mean that both of them will present the same value, since the sources are out of phase. As noted by Transistor, the wattmeters read the lines in sequence.

Found a document with an exercise. Refer to example 20.1 or 20.3

https://nptel.ac.in/content/storage2/courses/108105053/pdf/L-20(NKD)(ET)%20((EE)NPTEL).pdf

Also, from the same document:

PF=1 → W1=W2
PF=0.5 (cos60) → W1=Total power
0.5<PF<1 → W1>W2

For the proposed problem:

Total power (all three loads):

P=|VL|²/R*cos(<Z)
OR
P=3*|Vp|*|Ip|*cos(<Z)

Line power:

W1=|VL|*|IL|*cos(<V+(<I))
W2=|VL|*|IL|*cos(<V-(<I))

With <V and <I being the angle between Line voltage-Reference and Line current-Reference

In my case:

W1= 380*220/10*cos(30+60)= 0
W2= 380*220/10*cos(30-60)= 7240

I wish I could find a shorter formula that takes the angles into account (0, -120, 120) instead

This answer is free to be improved

This answer was edited. If needed, see history

It is known that the sum of the two wattmeters is the total power of the load, but it doesn't mean that both of them will present the same value, since the sources are out of phase. As noted by Transistor, the wattmeters read the lines in sequence.

Found a document with an exercise. Refer to example 20.1 or 20.3

https://nptel.ac.in/content/storage2/courses/108105053/pdf/L-20(NKD)(ET)%20((EE)NPTEL).pdf

Also, from the same document:

PF=1 → W1=W2
PF=0.5 (cos60) → W1=Total power
0.5<PF<1 → W1>W2

For the proposed problem:

Total power (all three loads):

P=|VL|²/R*cos(<Z)
OR
P=3*|Vp|*|Ip|*cos(<Z)

Line power:

W1=|VL|*|IL|*cos(<V+(|<I|))
W2=|VL|*|IL|*cos(<V-(|<I|))

With <V being the angle between the Line voltage and the Reference (usually, the angle is 30°) , and |<I| being the positive angle between the Line current and the Reference

In my case:

W1= 380*220/10*cos(30+60)= 0
W2= 380*220/10*cos(30-60)= 7240

I wish I could find a shorter formula that takes the angles into account (0, -120, 120) instead

This answer is free to be improved

This answer was edited. If needed, see history

It is known that the sum of the two wattmeters is the total power of the load, but it doesn't mean that both of them will present the same value, since the sources are out of phase. As noted by Transistor, the wattmeters read the lines in sequence.

Found a document with an exercise. Refer to example 20.1 or 20.3

https://nptel.ac.in/content/storage2/courses/108105053/pdf/L-20(NKD)(ET)%20((EE)NPTEL).pdf

Also, from the same document:

PF=1 → W1=W2
PF=0.5 (cos60) → W1=Total power
0.5<PF<1 → W1>W2

For the proposed problem:

Total power (all three loads):

P=|VL|²/R*cos(<Z)
OR
P=3*|Vp|*|Ip|*cos(<Z)

Line power:

W1=|VL|*|IL|*cos(30+(<Z))
W2=|VL|*|IL|*cos(30-(<Z))

In my case:

W1= 380*220/10*cos(30+60)= 0
W2= 380*220/10*cos(30-60)= 7240

I'm yet to understand where this 30+ and 30- comes from...

Also, I wish I could find a shorter formula that takes the angles into account (0, -120, 120) instead

This answer is free to be improved

This answer was edited. If needed, see history

It is known that the sum of the two wattmeters is the total power of the load, but it doesn't mean that both of them will present the same value, since the sources are out of phase. As noted by Transistor, the wattmeters read the lines in sequence.

Found a document with an exercise. Refer to example 20.1 or 20.3

https://nptel.ac.in/content/storage2/courses/108105053/pdf/L-20(NKD)(ET)%20((EE)NPTEL).pdf

Also, from the same document:

PF=1 → W1=W2
PF=0.5 (cos60) → W1=Total power
0.5<PF<1 → W1>W2

For the proposed problem:

Total power (all three loads):

P=|VL|²/R*cos(<Z)
OR
P=3*|Vp|*|Ip|*cos(<Z)

Line power:

W1=|VL|*|IL|*cos(<V+(<I))
W2=|VL|*|IL|*cos(<V-(<I))

With <V and <I being the angle between Line voltage-Reference and Line current-Reference

In my case:

W1= 380*220/10*cos(30+60)= 0
W2= 380*220/10*cos(30-60)= 7240

I wish I could find a shorter formula that takes the angles into account (0, -120, 120) instead

This answer is free to be improved