1.
Three Phase Circuits Problem 1: For the system given in figure 1, compute the each line-line
voltage at load side, phase voltage at load side. line current, and current on each phase loads
(both magnitude and angles). Problem 2: For the system given in figure 2 draw the one-line
diagram and determine 1. The source line current (Is), load line current for both delta (Idelta) and
wye (Iy) loads 2. The load line (VLPhi) and phase (VLL) voltages 3. Real and reactive power for
both the loads and source
Solution
problem 1) For the given circuit
Van=120/_0, Vbn=120/_-120 V
Van=207.8/_30 V
now to find current iab in delta connected load each impedance is 9/_20 ohms
then Iab=Vab/( Ztotal)= 207/_30/(4+8i+9/_20)=12.446/_-11.6 A
now I_line(delta) =1.732XIab=21.5/_-11.6 A
now to find Vab in delta connected load we have the following equation
Vab,l=Vab-i_lineXZtotal=207.8/_30-21.5/_-11.6XZtotal=150.6/_-149.8 V
where Vab,l load side voltage for delta connected load
problem 2)
now Vab=207.8/_30 V
for star connected load: to find the current Ia= Vab/(Ztotal)=207.8/_30/(4+8j +60+80j)=1.9/_-23
A
for star connected load line currents = phase currents =1.9/_-23 A

2.
Vph, Y= Vab/1.732=120/_30 V
for delta connected load: to find the cuurent Iab= Vab/(Ztotal)= 207.8/_30/(4+8j+60-
9j)=3.24/_30 A
now line current I_line= 1.732X3.24/_30=5.62/_30A
1) Is=I_line (of delta)=5.62/_30A, Is=I_line(of star)=1.9/_-23 A
2) for delta = V_loadphase= V_loadline
V_loadphase=207.8/_30-(1.9/_-23)(4+8j+60-9j)=167.8/_65.8
for star =V_loadlilne=207.8/_30-(21.5/_-11.6)(4+8j+60+80j)=