Mathcad - 02-13

Mathcad - 02-13 - = X C 9 = V AB 240 V = Solution: Since...

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arg I ph ( 29 30.964deg = I L I ph 3 e j - 30 deg = I L 23.764A = arg I L ( 29 0.964deg = We can use the complex power formula on a per-phase basis S ph V ph I ph = S ph 2.824 1.694j - kVA = S 3ph 3 S ph = the apparent power is the magnitude of S 3ph : S 3ph 9.878 kVA = real power is the real part of S 3ph : P Re S 3ph ( 29 = P 8.471kW = reactive power is the imag part of S 3ph : Q Im S 3ph ( 29 = Q 5.082 - kVAR = Since the circuit is capacitive, the reactive power is negative P2-13. Repeat P2-12 with the three loads connected in delta. R 15
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Unformatted text preview: = X C 9 = V AB 240 V = Solution: Since this is a delta-load with a series impedance, we need to find the total impedance and then solve for the phase current. Recall that the phase current leads the line current by 30 deg Z R j X C -= Z 15 9j- = define kVA kW kVAR kW V ph V AB = V ph 240V = arg V ph ( 29 0deg = I ph V ph Z = I ph 11.765 7.059j + A = I ph 13.72A =...
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