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θ
21.801deg
=
F
p
cos
θ
( 29
=
F
p
0.928
=
b.
Since this is a deltaload with a series impedance, we can use the total impedance to
solve for the phase current.
Recall that the phase current leads the line current by 30 deg
V
ph
V
AB
=
V
ph
480V
=
arg V
ph
( 29
0deg
=
I
ph
V
ph
Z
=
I
ph
4.138
1.655j

A
=
arg I
ph
( 29
21.801

deg
=
magnitude:
I
ph
4.457A
=
c. The magnitude of the line current is the square root of three times the magnitude of the
phase current
I
L
I
ph
3
⋅
e
j

30
⋅
deg
⋅
⋅
=
arg I
L
( 29
51.801

deg
=
magnitude:
I
L
7.719A
=
P216.
A threephase load consists of three, identical impedances connected in Delta.
The impedances have a value of (100 + j 40)
Ù
.
Line voltage = 480 V
a.
Find the power factor of the circuit.
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 Spring '12
 Piller

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