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Unformatted text preview: real, reactive, and apparent power. R 20 Ω ⋅ = X C 13 Ω ⋅ = V AB 240 V ⋅ = Solution: Since this is a wye-load with a series impedance, we need to find the total impedance and then solve for the line current. Recall that the phase voltage lags the line voltage by 30 deg Z 1 1 R 1 j X C ⋅ ( 29- = define kVA kW ≡ kVAR kW ≡ Z 5.94 9.139j-Ω = Z 10.9 Ω = arg Z ( ) 56.976-deg = V ph V AB 3 e j-30 ⋅ deg ⋅ ⋅ = V ph 138.564V = arg V ph ( 29 30-deg = I V ph Z =...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University.
- Spring '12