When connected in WYE, the impedance use onethird of the power they use
when connected in
Δ
.
The reason of course is they only have the line to neutral
voltage imposed on them in WYE versus the line voltage when connected in
Δ
.
Q
3.84 kVAR
=
P
y
7.68 kW
=
Q3
V
⋅
I
L
⋅
sin
θ
−
()
⋅
:=
P
y
3
V
⋅
I
L
⋅
F
p
⋅
:=
Calculating the real and reactive power, yields:
θ
26.565
−
deg
=
θ
arg I
L
:=
F
p
0.894
=
F
p
cos atan
X
R
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
:=
The power factor can be found from the impedance:
arg I
L
26.565
−
deg
=
I
L
10.328amp
=
I
L
9.238
4.619j
−
amp
=
I
L
V
br
Z
:=
Thus the line current is:
V
br
277.128volt
=
The branch voltage lags the line voltage by 30 degrees
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 Spring '12
 Piller
 Alternating Current, Volt, Complex number, Electrical impedance, line voltage

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