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=2000  W P_{ 11 } = 1000 \ W \ \\ P_{ 21 } = 1000 \ W \ \\ \ \\ P_{ 1 } = P_{ 11 }+P_{ 21 } = 1000+1000 = 2000 = 2000 \ \text { W }
f1=cos⁡φ1  t1=3⋅ P11−P21P11+P21=3⋅ 1000−10001000+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=0  W P_{ 12 } = 1000 \ W \ \\ P_{ 22 } = -1000 \ W \ \\ \ \\ P_{ 2 } = P_{ 12 }+P_{ 22 } = 1000+(-1000) = 0 = 0 \ \text { W }
f2=cos⁡φ2  t2=3⋅ P12−P22P12+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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Following knowledge from mathematics are needed to solve this word math problem:

Most natural application of trigonometry and trigonometric functions is a calculation of the triangles. Common and less common calculations of different types of triangles offers our triangle calculator. Word trigonometry comes from Greek and literally means triangle calculation.

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