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14.5.5 Example 1 Part 2

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14.5.5 Example 1 Part 2

  1. 1. Apparent power (S) Reactive power (Q) 2.308 kVA 49.46° True power (P) 1.5 kW
  2. 2. S2 = P2 + Q2 Q2 = S2 - P2 Q= √(S2 - P2)
  3. 3. Reactive Power (Q) = √ (S² - P²) Q = √ (2308² - 1500²) = √(5326864 -2250000) = √3076864 =1754 VAR Reactive Power (Q) = 1.754 kVAR

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