Unformatted text preview: real, reactive, and apparent power. R 20 Ω ⋅ = X C 13 Ω ⋅ = V AB 240 V ⋅ = Solution: Since this is a wyeload 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 9.139jΩ = Z 10.9 Ω = arg Z ( ) 56.976deg = V ph V AB 3 e j30 ⋅ deg ⋅ ⋅ = V ph 138.564V = arg V ph ( 29 30deg = I V ph Z =...
View
Full Document
 Spring '12
 Piller
 Alternating Current, 1.477j, 2.88 kW, 5.767j, 9.139j, 26.976 deg

Click to edit the document details