This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.
The electrical service to your home is a two-phase service. This means
that two 110 volt, 60 Hz lines, plus neutral, terminate in the panel. The lines
are π radians (180) out of phase, so we can write them as
x
1
(
t
)
=
110
c
o
s
[
2
π
(
60
)
t
+
φ
]
=
Re
{
110
e
j
[
2
π
(
60
)
t
+
φ
]
}
=
Re
{
X
1
e
j
2
π
(
60
)
t
}
X
1
=
110
e
j
φ
x
2
(
t
)
=
110
c
o
s
[
2
π
(
60
)
t
+
φ
+
π
]
=
Re
{
110
e
j
[
2
π
(
60
)
t
+
φ
+
π
]
}
=
Re
{
X
2
e
j
2
π
(
60
)
t
}
X
2
=
110
e
j
(
φ
+
π
)
.
x
1
(
t
)
=
110
c
o
s
[
2
π
(
60
)
t
+
φ
]
=
Re
{
110
e
j
[
2
π
(
60
)
t
+
φ
]
}
=
Re
{
X
1
e
j
2
π
(
60
)
t
}
X
1
=
110
e
j
φ
x
2
(
t
)
=
110
c
o
s
[
2
π
(
60
)
t
+
φ
+
π
]
=
Re
{
110
e
j
[
2
π
(
60
)
t
+
φ
+
π
]
}
=
Re
{
X
2
e
j
2
π
(
60
)
t
}
X
2
=
110
e
j
(
φ
+
π
)
.
(1)These two voltages are illustrated as the phasors X1 and X2 in Figure 1.
You may use x1(t)x1(t) to drive your clock radio or your toaster and the
difference between x1(t)x1(t) and x2(t)x2(t) to drive your range or dryer:
x
1
(
t
)
-
x
2
(
t
)
=
220
cos
[
2
π
(
60
)
t
+
φ
]
.
x
1
(
t
)
-
x
2
(
t
)
=
220
cos
[
2
π
(
60
)
t
+
φ
]
.
(2)The phasor representation of this difference is
X
1
-
X
2
=
220
e
j
φ
.
X
1
-
X
2
=
220
e
j
φ
.
(3)The breakers in a breaker box span the x1-tox1-to-neutral bus for 110 volts and
the x1x1-to-x2x2 buses for 220 volts.
Sketch the phasor X1-X2X1-X2 on Figure 1.
Most industrial installations use a three-phase service consisting of the
signals x1(t),x2(t)x1(t),x2(t), and x3(t)x3(t):
x
n
(
t
)
=
110
Re
{
e
j
[
ω
0
t
+
n
(
2
π
/
3
)
]
}
↔
X
n
=
110
e
j
n
(
2
π
/
3
)
,
n
=
1
,
2
,
3
.
x
n
(
t
)
=
110
Re
{
e
j
[
ω
0
t
+
n
(
2
π
/
3
)
]
}
↔
X
n
=
110
e
j
n
(
2
π
/
3
)
,
n
=
1
,
2
,
3
.
(4)The phasors for three-phase power are illustrated in Figure 2.
Sketch the phasor X2-X1X2-X1 corresponding to x2(t)-x1(t)x2(t)-x1(t) on
Exercise 3. Compute the voltage you can get with x2(t)-x1(t)x2(t)-x1(t). This answer
explains why you do not get 220 volts in three-phase circuits. What do you
get?
Constant Power. Two- and three-phase power generalizes in an obvious way to N-phase power. In such a scheme, the NN signals xn(n=xn(n=0,1,...,N-10,1,...,N-1) are
x
n
(
t
)
=
A
cos
(
ω
t
+
2
π
N
n
)
=
Re
[
A
e
j
2
π
n
/
N
e
j
ω
t
]
↔
X
n
=
A
e
j
2
π
n
/
N
.
x
n
(
t
)
=
A
cos
(
ω
t
+
2
π
N
n
)
=
Re
[
A
e
j
2
π
n
/
N
e
j
ω
t
]
↔
X
n
=
A
e
j
2
π
n
/
N
.
(5)The phasors
X
n
X
n
are Aej2π(n/N)Aej2π(n/N). The sum of all
N
N signals is zero:
∑
n
=
0
N
-
1
x
n
(
t
)
=
Re
{
A
∑
n
=
0
N
-
1
e
j
2
π
n
/
N
e
j
ω
t
}
=
Re
{
A
1
-
e
j
2
π
1
-
e
j
2
π
/
N
e
j
ω
t
}
=
0
.
∑
n
=
0
N
-
1
x
n
(
t
)
=
Re
{
A
∑
n
=
0
N
-
1
e
j
2
π
n
/
N
e
j
ω
t
}
=
Re
{
A
1
-
e
j
2
π
1
-
e
j
2
π
/
N
e
j
ω
t
}
=
0
.
(6)But what about the sum of the instantaneous powers? Define the instantaneous power of the nthnth signal to be
p
n
(
t
)
=
x
n
2
(
t
)
=
A
2
cos
2
(
ω
t
+
2
π
N
n
)
=
A
2
2
+
A
2
2
cos
(
2
ω
t
+
2
2
π
N
n
)
=
A
2
2
+
Re
{
A
2
2
e
j
(
2
π
/
N
)
2
n
e
j
2
ω
t
}
.
p
n
(
t
)
=
x
n
2
(
t
)
=
A
2
cos
2
(
ω
t
+
2
π
N
n
)
=
A
2
2
+
A
2
2
cos
(
2
ω
t
+
2
2
π
N
n
)
=
A
2
2
+
Re
{
A
2
2
e
j
(
2
π
/
N
)
2
n
e
j
2
ω
t
}
.
(7)The sum of all instantaneous powers is (see Exercise 3)
P
=
∑
n
=
0
N
-
1
p
n
(
t
)
=
N
A
2
2
,
P
=
∑
n
=
0
N
-
1
p
n
(
t
)
=
N
A
2
2
,
(8)and this is independent of time!
Carry out the computations of Equation 3.33 to prove that
instantaneous power
P
P is constant in the N-phase power scheme.
(What does "Endorsed by" mean?)
"Reviewer's Comments: 'I recommend this book as a "required primary textbook." This text attempts to lower the barriers for students that take courses such as Principles of Electrical Engineering, […]"