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:

$$\begin{gathered} P_{11}=1000\text{ W } \\ {P}_{21}=1000\text{ W } \\ {P}_1=P_{11}+P_{21}=1000+1000=2000\text{ W} \end{gathered}$$

$$\begin{gathered} f_1=\cos {\phi }_1 \\ {t}_1=\sqrt{3}\cdot {\displaystyle \frac{P_{11}-P_{21}}{P_{11}+P_{21}}}=\sqrt{3}\cdot {\displaystyle \frac{1000-1000}{1000+1000}}=0 \\ {f}_1=\cos (\arctan (t_1))=\cos (\arctan (0))=1 \end{gathered}$$

$$\begin{gathered} P_{12}=1000\text{ W } \\ {P}_{22}=-1000\text{ W } \\ {P}_2=P_{12}+P_{22}=1000+(-1000)=0\text{ W} \end{gathered}$$

$$\begin{gathered} f_2=\cos {\phi }_2 \\ {t}_2=\sqrt{3}\cdot {\displaystyle \frac{P_{12}-P_{22}}{P_{12}+P_{22}}}=\sqrt{3}\cdot {\displaystyle \frac{1000-(-1000)}{1000+(-1000)}}=\mathrm{\infty } \\ {f}_2=\cos (\arctan (t_2))=\cos (\arctan (\mathrm{\infty }))=0 \end{gathered}$$

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