Mathcad - 02-14 - real, reactive, and apparent power. R 20...

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Since the circuit is capacitive, the reactive power is negative Q 4.431 - kVAR = Q Im S 3ph ( 29 = reactive power is the imag part of S 3ph : P 2.88kW = P Re S 3ph ( 29 = real power is the real part of S 3ph : S 3ph 5.285 kVA = the apparent power is the magnitude of S 3ph : S 3ph 3 S ph = S ph 0.96 1.477j - kVA = S ph V ph I = We can use the complex power formula on a per-phase basis arg I ( ) 26.976deg = I 12.713A = I 11.329 5.767j + A = P2-14. Three of the parallel-connected impedances of P2-4 are connected in a three-phase, wye load. A balanced, three-phase set of voltages is applied to the load, and VAB = 240 p 0o V. Calculate the total, three-phase
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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.

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