{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Mathcad - 02-13 - Ω ⋅ = X C 9 Ω ⋅ = V AB 240 V ⋅...

This preview shows page 1. Sign up to view the full content.

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 9.878kVA = 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ω ⋅ = X C 9 Ω ⋅ = 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-Ω = 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 =...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online