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Question 24

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Solve this proportion. k W load =3(E)(I)(PF)1,000 where k W load =3 -phase power in kilowatts I= current in amperes X= voltage involts P F= power factor\begin{array}{llcc}\text {k W load =} \frac { \sqrt { 3 } ( E ) ( I ) ( P F ) } { 1,000 }& \text { where k W load =3 -phase power in kilowatts}\\& \text { I= current in amperes}\\& \text { X= voltage involts}\\& \text { P F= power factor}\\\end{array}

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