3 phase load

Two wattmeters are connected to measuring power in a 3 phase balanced load. Determine the total power and power factor if the two wattmeters read 1000 watts each (1) both positive and (2) second reading is negative

Correct result:

P1 =  2000 W
f1 =  1
P2 =  0 W
f2 =  0

Solution:

P11=1000 W P21=1000 W  P1=P11+P21=1000+1000=2000 W
f1=cosφ1  t1=3 P11P21P11+P21=3 100010001000+1000=0  f1=cos(arctan(t1))=cos(arctan(0))=1
P12=1000 W P22=1000 W  P2=P12+P22=1000+(1000)=0 W
f2=cosφ2  t2=3 P12P22P12+P22=3 1000(1000)1000+(1000)=  f2=cos(arctan(t2))=cos(arctan())=0



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