# ee212_5b - Click to edit Master subtitle style EE212...

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Unformatted text preview: Click to edit Master subtitle style 9/26/10 EE212 Passive AC Circuits Lecture Notes 5 Three Phase Systems 11 9/26/10 Balanced 3-Phase Systems Example: A balanced 3-phase, line-to-line voltage of 220 V is applied to a balanced 3-phase load. The 3-phase load consists of 3 identical loads (each with an impedance of 6+j8 B ) (a) connected in Y (b) connected in ∆ Calculate the line current and the total power consumed for both cases. Draw the complete phasor diagrams for each case. 22 9/26/10 A 3-phase load in ∆-connection consumes 3 times more power than in Y- connection . Balanced 3-Phase Systems A 3-phase load in ∆-connection draws 3 times more current than in Y- connection . Y- ∆ motor starting method to reduce starting current 33 9/26/10 A C B Z a Z b Z c B A C Z a b Z b c Z c a ZY = ( product of adjace nt Z ∆ ) sum of all 3 Z ∆ De lt a -W ye ( ∆-Y) Tra nsform a t ion Za = (Zab. Zca) / (Zab + Zbc + Zca) Zb = (Zab. Zbc) / (Zab + Zbc + Zca) Zc = (Zca. Zbc) / (Zab + Zbc + Zca) For ba lance d load, Zab = Zbc = Zca = Z ∆ The re fore , ZY = Z ∆ / 3 44 9/26/10 A C B Z a Z b Z c B A C Z a b Z b c Z c a Z ∆ = (product of ZY taken in pairs) the opposite ZY W ye -De lt a ( Y- ∆ ) Tra nsform a t ion Zab = (Za.Zb + Zb.Zc + Zc. Za) / Zc Zbc = (Za.Zb + Zb.Zc + Zc. Za) / Za Zca = (Za.Zb + Zb.Zc + Zc. Za) / Zb For balance d load, Za = Zb = Zc = ZY The re fore , Z ∆ = 3 .ZY 55 9/26/10...
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ee212_5b - Click to edit Master subtitle style EE212...

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