Mathcad - 02-11 - kVAR kW Problem 2-11 Solution Repeat 2-10...

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When connected in WYE, the impedance use one-third of the power they use when connected in Δ . The reason of course is they only have the line to neutral voltage imposed on them in WYE versus the line voltage when connected in Δ . Q 3.84 kVAR = P y 7.68 kW = Q3 V I L sin θ () := P y 3 V I L F p := Calculating the real and reactive power, yields: θ 26.565 deg = θ arg I L := F p 0.894 = F p cos atan X R := The power factor can be found from the impedance: arg I L 26.565 deg = I L 10.328amp = I L 9.238 4.619j amp = I L V br Z := Thus the line current is: V br 277.128volt = The branch voltage lags the line voltage by 30 degrees

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Mathcad - 02-11 - kVAR kW Problem 2-11 Solution Repeat 2-10...

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