PART II

•
For a Y-connected load with Z
Y
=
Z
∠ θ
, the phase voltages and currents are: v
AN
=
2 V p v
CN i
AN
=
= cos
ω t , v
BN
=
2
2 V p
I p cos(
ω t
+
120 o
) cos(
ω t
− θ
), i
BN i
CN
=
2 I p cos(
ω t
− θ +
120 o
)
2 V p
=
2 cos(
ω t
I p cos(
−
120 o
),
ω t
− θ −
120 o
),
where V p
and I p
are the rms voltage and current, respectively.
•
The total instantaneous power absorbed by the load is p ( t )
=
=
=
2 v p
V p
A
AN
I
( t i p
)
+
AN
[cos p
+
B v
ω t
( t
BN
) i
+
BN p
C
+ cos(
ω t
( t ) v
CN
− θ
) i
CN
+ cos(
ω t
+ cos(
ω t
+
120 o
) cos(
ω t
− θ
−
120 o
) cos(
ω t
+
120 o
)]
− θ −
120 o
)

•
Applying cos p ( t )
=
V p
I p
[ 3
A cos cos
θ +
B
= cos
1
2
α
[cos(
+
A cos(
+
α
B )
+ cos( A
−
B )] gives
−
240 o
)
+ cos(
α +
240 o
)]
where
α =
2
ω t
− θ
•
Applying cos( A
+
B )
= cos A cos B
− sin S sin B yields p ( t )
=
V p
I p
[ 3 cos
θ
+ cos
α
+ cos
α + cos 240 o cos
α
− sin
α cos 240 o + sin 240 o
] sin
α sin 240 o
=
V p
I p
[ 3 cos
θ + cos
α +
2 (
−
1
2
) cos
α
]
=
3 V I p cos
θ
→
The total instantaneous power in a balanced three-phase system
is regardless of whether the load is Y- or
Δ
-connected while that of each phase is still time varying.

→
The average power consumed by each phase of the load is one-
third p ( t ).
P p
=
V p
I p cos
θ
→
The reactive power consumed by each phase of the load is
Q p
=
V p
I p sin
θ
→
The apparent power consumed by each phase of the load is
S p
=
V p
I p
→
The complex power consumed by each phase of the load is
S p
=
P p
+ jQ p
=
V p
I
∗ p
Note that V p
is the phasor rms voltage whose magnitude is V p
I p
is the phasor rms current whose magnitude is I p

•
The total complex power is
S
=
3 S p
=
3 V p
I
∗ p
=
3 I
2 p
Z p
=
3 V p
2
Z
* p where Z p
=
Z
Y
∠ θ
or Z
Δ
∠ θ
(the load impedance per phase)
•
Alternatively, since I
L or I
L
=
3 I p
, V
L
=
V p
=
I p
, V
L
=
3 V p
for a Y-connected load
for a
Δ
-connected load,
P
Q
=
=
3 V p
I
3 V p
I p p cos
θ sin
θ
=
=
3 V
L
I
L
3 V
L
I
L cos
θ sin
θ
S
=
P
+ jQ
=
3 V p
I p
∠ θ =
3 V
L
I
L
∠ θ

Ex. Practice problem 12.6
For the Y-Y circuit of Practice Prob.12.2, calculate the complex power at the source and at the load

Ex. Practice problem 12.7
Calculate the line current (magnitude only) required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 440 V.

•
An unbalanced system is due to unbalanced voltage sources or
an unbalanced load.
•
Solved by direct application of mesh and nodal analysis or
ohm’s law
•
For a four-wire Y-Y system, the neutral line current is not zero
I n
= −
( I a
+
I b
+
I c
)
•
Each phase of the load consumes unequal power. Thus, the total power is the sum of the powers in the three phases.

Ex. Practice problem 12.9
The unbalanced
Δ
-load shown is supplied by balanced line-to-line voltages of 240 V in the positive sequence. Find the line currents using V ab
as reference.

Ex. Practice problem 12.9
For the unbalanced circuit shown, find:
(a) the line currents,
(b) the total complex power absorbed by the load, and
(c) the total complex power absorbed by the source.

RESIDENTIAL WIRING

SINGLE-PHASE THREE-WIRE RESIDENTIAL WIRING

GROUNDING & SAFETY

GROUND FAULT CIRCUIT INTERRUPTER (GFCI)

GENERATION & DISTRIBUTION OF AC POWER