Unformatted text preview: ⋅ = 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 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 = V ph 240V = arg V ph ( 29 0deg = I ph V ph Z = I ph 12 18.462j + A =...
View Full Document
- Spring '12
- Alternating Current, Electrical impedance, AC power, Power factor, Volt-ampere, Iph