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

Result

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

Solution:

P11=1000 W P21=1000 W  P1=P11+P21=1000+1000=2000 WP_{11}=1000 \ \text{W} \ \\ P_{21}=1000 \ \text{W} \ \\ \ \\ P_{1}=P_{11}+P_{21}=1000+1000=2000 \ \text{W}
f1=cosφ1  t1=3 P11P21P11+P21=3 100010001000+1000=0  f1=cos(arctan(t1))=cos(arctan(0))=1f_{1}=\cos φ_{ 1 } \ \\ \ \\ t_{1}=\sqrt{ 3 } \cdot \ \dfrac{ P_{11}-P_{21} }{ P_{11}+P_{21} }=\sqrt{ 3 } \cdot \ \dfrac{ 1000-1000 }{ 1000+1000 }=0 \ \\ \ \\ f_{1}=\cos ( \arctan(t_{1}))=\cos ( \arctan(0))=1
P12=1000 W P22=1000 W  P2=P12+P22=1000+(1000)=0 WP_{12}=1000 \ \text{W} \ \\ P_{22}=-1000 \ \text{W} \ \\ \ \\ P_{2}=P_{12}+P_{22}=1000+(-1000)=0 \ \text{W}
f2=cosφ2  t2=3 P12P22P12+P22=3 1000(1000)1000+(1000)=INF  f2=cos(arctan(t2))=cos(arctan(INF))=0f_{2}=\cos φ_{ 2 } \ \\ \ \\ t_{2}=\sqrt{ 3 } \cdot \ \dfrac{ P_{12}-P_{22} }{ P_{12}+P_{22} }=\sqrt{ 3 } \cdot \ \dfrac{ 1000-(-1000) }{ 1000+(-1000) }=INF \ \\ \ \\ f_{2}=\cos ( \arctan(t_{2}))=\cos ( \arctan(INF))=0



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