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 =...
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- Spring '12
- Alternating Current, Electrical impedance, AC power, Power factor, Volt-ampere, Iph